AP Chemistry - ABSOLUTE COMPLETE FINAL!!!!!!!

Vapor Pressures of Ionic Solids

= Ionic compounds have very low vapor pressures - Strong Coulombic interactions btwn cations and anions

If given an acid and a nonmetal...

(remember nonmetals want to be reduced) the nonmetal will gain e-s from the anion of the acid. The hydrogen ion of the acid is a spectator

If given a metal and water...

(remember that metals want to be oxidized) oxidize the metal and reduce the water. Write the water as HOH and recall this works just like any other single replacement

Brass - a __________ alloy

(substitutional) • Zn atoms r substituted for some Cu atoms • Alloys remain malleable and ductile • Density normally lies btwn those of the component metals

Atomic number

(z) the number of protons (or electrons) in a neutral atom

aminopropane, CH3CHNH2CH3, and isobutane, C4H10 • Identify the types of intermolecular forces that exist in a pure sample of aminopropane • and isobutane

= LDF and H-bonds - H atoms bonded to highly electronegative N atom, which has a LP of e-s = Non-polar so only LDF

- Heat must be added in order for a solid to liquid phase change to occur, but the temp. of the system doesn't change during this process - Heat must be added in order for a liquid to gas phase change to occur, but the temp. of the system doesn't change during this process

- The added heat is used to weaken intermolecular forces - The added heat is used to completely sever intermolecular forces

Dipole-Dipole

- The attractive forces between the negative end of one polar molecule and the positive end of another polar molecule - Molecules with dipole moments experience Coulombic interactions when they r in close proximity to one another. - FoA increase as magnitude of dipoles increases

Bond Energy

- The potential energy of Ve- decreases as they approach the nucleus of another atom - Energy is released during the formation of a bond — Potential energy decreases as the atoms move close together - The same amount of energy must be added in order to break that specific bond — Potential energy increases as the atoms move away from one another

Possible arrangements of Ions - Body Center Cubic

- This unit cell contains 8 corners x 1/8 of an ion + 1 central atom = 2 ion/unit cell

Possible arrangements of Ions - Space center cubic

- This unit cell contains 8 corners x 1/8 of an ion + 6 faces x 1/2 of an ion = 4 ion/unit cell

In a 0.450 M HONH2 solution, [OH-] = 5.28x10^-6 M HONH2 + H2O <-> HONH3+ + OH- f) What conc. of NaOH required to make a solution with pH of 8.723

5.28 x 10^-6 M

238 / 92 U isotope Protons? Neutrons? Electrons? Mass?

92 protons 146 neutrons 92 electrons 238 amu

Calculating Formal Charge

= (# of Ve- assigned to neutral atom) - (# of e- assigned to atom in structure) (# of e- assigned to atom in structure) = [# of lone e- around atom + (# of bonding e-)/2]

Boiling Points

= A liquid boils when its vapor pressure = atmospheric pressure - Evaporation occurs inside the liquid when the vapor pressure = atmospheric pressure

Temperature

= A measure of the average KE of atoms and molecules in a system - Kelvin (K) temperature scale is proportional to this - When KE doubles, the K temperature doubles

exothermic vs endothermic reactions

= ^_E is energy lost or gained in reaction ^_E = q+w • If volume doesn't change (^_E = ^_H) Exo - Bonds in products contain less PE than bonds in reactants (goes down) Endo - Bonds in products contain more PE than bonds in reactants (goes up)

Beer-Lambert Law

A = Ebc A = absorbance E = molar absorptivity (1/Mcm) b = path length of sample (cm) c = concentration (M) • E describes how intensity a sample of ions or molecules absorbs light at a specific wavelength • In most experiments, path length and wavelength remain constant, so E also remains constant, and absorbance, A, is only proportional to concentration, c

Label acid, base, conjugate acid, and conjugate base a) HNO3 + H2O <—> H3O+ + NO3-

A B CA CB

Label acid, base, conjugate acid, and conjugate base f) HCN + H2O <—> H3O+ + CN-

A B CA CB

Identify the CA/CB pairs in following reactions a) HF + H2O <—> H3O+ + F-

A B CA CB HF and F- r C acid base pair H2O and H3O+ r C acid base pair

Label acid, base, conjugate acid, and conjugate base b) HCO3- + OH- <—> CO3- + H2O

A B CB CA

Label acid, base, conjugate acid, and conjugate base i) HClO4 + H2O —> ClO4- + H3O+

A B CB CA

Label acid, base, conjugate acid, and conjugate base j) HI + H2O —> I- + H3O+

A B CB CA

Identify the CA/CB pairs in following reactions b) C3HOOH + H2O <—> CH3COO- + H3O+

A B CB CA CH3COOH and CH3COO- r C acid base pair H2O and H3O+ r a C acid base pair

Solution

A homogeneous mixture of two or more substances

Acid-Base Catalysis

A reactant gains or loses a proton, which forms a new intermediate

Alloy

A solid solution composed of two or more metals or one or more metals and one or more non-metals Ex. Steel

pH

A solution with [H3O+] = 10^-4 had a pH of 4

Label acid, base, conjugate acid, and conjugate base h) 2NH3 <—> NH4+ + NH2-

A&B CA CB

Label acid, base, conjugate acid, and conjugate base g) 2H2O <—> OH- + H3O+

A&B CB CA

Elementary steps that involve 3 or more reacting species r very rare. Explain why this is.

All termolecular elementary steps r very rare, as they require the simultaneous collision of 3 particles with correct orientations and enough energy. Elementary steps that require the simultaneous collision of more than 3 reactant particles r very rare for the same reason

Trigonal Planar Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

BF3; 3; 3; 0; 120 degrees; sp^2

Explain why the standard enthalpy of vaporization values for each set of compounds below rn't the same d) BH3 and OF2

BH3 has LDF and OF2 has LDP and D^2 forces. Bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Which of the following salts is more soluble in water at 25'C? BaF2 Ksp = 2.4 x 10^-5 BaCO3 Ksp = 1.6 c 10^-9

BaF2 is more soluble, as it has a larger Ksp value

Write balanced net ionic equations for reactions Solid beryllium hydroxide is placed in a solution of acetic acid

Be(OH)2 + 2CH3COOH —> Be 2+ + 2H2O + 2CH3COO-

Which atom in each set has the larger radius? a) Be or O b) Cu or Br c) F or I d) O or As e) Kr or K f) Ba or Li

Be, Cu, I, As, K, Ba

Is ozone polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Bent so 2 O-O bonds and 1 lone pair - non-polar individual bonds bc electronegativity difference in an O-O bond is zero - Therefore, non-polar overall compound, as there r no dipoles

Identify strongest base and justify a) I- or Br-

Both I- and Br- r weak bases as they r CB of strong acids HI and HBr • Br- is stronger base, as it has a smaller ionic radius, and has slightly greater ability to attract H+ ions

Which isotope is more likely to bond with oxygen, C-12 or C-14? Explain.

Both isotopes are equally likely to bond with oxygen. Having more or less neutrons in an isotope doesn't change its chemical reactivity

C2H2 or C2H6 H-C=_C-H or 2 C surrounded by 6 H w/ single bonds b. Which structure has the shortest C-C bond length? justify

C2H2 bc has single bonds - triple bonds have greater e- densities between bonding atoms than single bonds - Greater e- density between bonding atoms increases force of attraction between protons of each nucleus and centrally located bonding e-s which shortens bond length

According to ____________ the forces of attraction between electrons and the nucleus in n=2 are less than the forces of attraction between electrons and the nucleus in n=1

Coulomb's Law

Beer-Lambert Law

A=Ebc A = absorbance E = molar absorptivity (M^-1 cm^-1) b= path length of sample (cm) c = concentration (M) • E describes how intensely a sample of ions or molecules absorbs light at a specific wavelength • In most experiments, path length and wavelength remain constant, so E also remain constant, and absorbance, A, is only proportional to concentration, c

Free Energy of Dissolution (^_G)

AB <-> A+ + B- • if ionic compound is soluble, ^_G<0 and equilibrium lies to right • if ionic compound is not soluble, ^_G>0 and equilibrium lies to left

Solubility Rules

All salts containing Na+, K+, Nh4+, or NO3- r soluble in water • Ksp > 1 for all salts containing the ions above which is why they r considered to be soluble • Equilibrium lies to the right

What is the strongest base in each following reactions? Justify based on solution equilibrium or FoA btwn particles b) HNO3 + H2O —> NO3- + H3O+

H2O is strongest base • Strong acids (HNO3) experience ~100% dissociation and have weak CB • Thus, NO3- is a weak base • H2O and NO3- compete for H+ ions • H2O acquired H+ ions most of the time, as the reactions goes to completion

In these experiments, ____ ______ _______ remove electrons from any shell, not just the outer shell. KE of ________ electrons is determined. Frequency of _________ is recorded.

High energy photons; ejected; photons

Great deviation from Ideal at _____ P and low T bc conditions _____ attractive forces btwn particles

High; max

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces NH3 and CH4

NH3, bc both LDF and NH3 has H-bonds but CH4 is non-polar. • Both species r approximately same size, and H bonds much stronger than LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of NH3 is higher

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces b. NH3 and NCl3

NH3, bc both LDF but NH3 has H-bonds and NCl3 has dipole-dipole forces • H-bonds r much stronger than normal dipole-dipole forces • As a greater amount of energy is required to break stronger FoA, boiling pt of NH3 is higher

Solid sodium fluoride is dissolved in distilled water a) chemical equation b) Will final solution be acidic or basic? Justify

NaF (s) + H2O (L) —> HF (aq) + OH- (aq) + Na+ (aq) Final solution will be basic • F- is a base, as it's the conjugate base of the weak acid HF • Thus, F- will react with water to produce OH- ions. The increased [OH-] makes the solution basic

Shell models of 2 neutral atoms of Ne and Ar. Which atom would have higher 1st ionization energy? Justify

Ne -Ar has larger atomic radius than Ne, as it has an additional shell. -As radius of Ar is larger, its Ve- r held with a smaller force of attraction and requires less energy to remove them -According to Coulomb's Law, forces of attraction increases as distance between e-s and protons decreases.

Enthalpy of Sublimation (∆Hsub)

Sublimation and deposition of water at 0'C and 1 atm • Sublimation = ∆Hsub = + kJ/mol • Deposition = ∆Hsub = - kJ/mol ∆Hsub = ∆Hfus + ∆Hvap • This is true at constant temp. and pressure

Brass — A __________ Alloy

Substitutional • Zn atoms r substituted for some Cu atoms • Substitutional alloys remain malleable and ductile • Retains a "sea of e-" so it can conduct electricity • Density normally lies btwn those if component metals

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces h. CH4 and SnH4

SnH4, bc both LDF but SnH4 is larger and has more e-s, making it more polarizable than CH4 • SnH4 experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of SnH4 is higher

Strong Bases

Soluble compounds containing the OH- Possible Cations: • All Group 1A Cations, Ca2+, Sr2+, or Ba2+

A 1.0 mole sample of HNO3 is added to water. Final volume of solution is 1.5 L and final temp. of solution is 25'C D) Is solution acidic or basic? Explain

Solution is acidic as pH is less than 7

Volumes Changes from Heating

• When gaseous system expands bc heat is added (system heated) overall process has 2 variables - heat, q, and work, w ^_E = q + w • heating is endothermic and work is exothermic, so ^_E must be calculated

system vs surroundings

System - Actual chemical reaction that is taking place Surroundings - entire universe outside of the chemical reaction

calculating ^_S^0

^_S^0 (rxn) = EnS^o (products) -EnS^o (reactants)

Combustion Analysis

a method of obtaining empirical formulas for unknown compounds, especially those containing carbon and hydrogen, by burning a sample of the compound in pure oxygen and analyzing the products of the combustion reaction

sigma bond

a single covalent bond that is formed when an electron pair is shared by the direct overlap of bonding orbitals - 1st bond between 2 atoms (hybridized orbital) - contain more energy as the overlap between the orbitals is stronger

Location highest on these trends... a) Atomic radius b) Ionic radius c) Ionization energies d) Electron affinity e) Electronegativity

a) Atomic radius - Bottom left (Fr) b) Ionic radius - Bottom left (Fr) c) Ionization energies - Top right (He) d) Electron affinity - Top right minus Noble gases, N, and Alkali Earth Metals (F) e) Electronegativity - Top right minus Noble gases (F)

Classify as a physical change, chemical change, or both. Justify by identifying the types of intermolecular or intramolecular forces that r involved in each processes and describing what happens to those forces while the processes r occurring. e) NH2F(L) —> 1/2 N2(g) + H2(g) + 1/2 F2(g)

chemical change • covalent bonds in the NH2F molecule r broken and new covalent bonds r formed

If given a metal or nonmetal (including diatomics — usually "bubbled through") and the soluble ionic compound...

cross out the always soluble ion and replace the respective metal or nonmetal ion from the compound. Single replacement reaction

A solution of the amino acid alanine, NC3O2H8+, was created and titrated with NaOH. The data from this experiment was used to plot following titration curve d) Identify species that have highest conc. in this solution at first half equivalence pt e) Identify species that have highest conc. in this solution at first equivalence pt

d) NC3O2H8+, NC3O2H7, Na+, and H3O+ e) NC3O2H7, Na+, and H3O+

Electron Affinity: Energy change that occurs when an electron is added to a ________ atom to form a negative ion. It is a measure of how much an element want to __________ another e-.

gaseous; accept

Incomplete Octets

molecules or ions with fewer than eight electrons around an atom

smaller ion ______ condensed than larger ion and attract ____ more molecules to itself

more; more

Suppose SO2 is added to system when it's at equilibrium SO2 (g) + H2O (L) <-> H2SO3 (aq) c) Will the value of the equilibrium constant, Keq, change in response to this stress? Justify

no, changes in temperature r the only stresses that alter the value of K

Hybrid orbitals: Hybridization and Ideal Bond Angle with charge clouds (2-6) around the central atom

sp (180 degrees), sp^2 (120 degrees), sp^3 (109.5 degrees), sp^3d (120 and 90 degrees), sp^3d^2 (90 degrees)

Electrons in __________ ____ spend more of their time further away from the nucleus. Electrons in outer shells are partly "_________" from forces of attraction by inner core electrons. This ________ ________ (repulsion from inner electrons) reduces electrostatic attractions between the outer electrons and the nucleus.

successive shells; shielded; shielding effect

Electron Affinity

the energy change that occurs when an electron is added to a gaseous atom to form a negative ion

Unit Cells

the smallest group of particles within a crystal that retains the geometric shape of the crystal

Weak acid - Strong base reactions

• Strong bases completely dissociate • Each OH- ripe H+ ion off a weak acid molecule • This produces water and CB of the weak acid

Collision Energy and Temperature

• At higher temperatures, more reactants collide with enough energy to cause a reaction

Boltzmann Distribution and Temperature

• Distribution of KE incr. as temp. incr. • Average KE of the particles in a system incr. as temp. incr.

Acids vs Bases

Acids - Sour taste • Acetic acid in vinegar • Citric acid in citrus fruits (Ex. Lemon) Bases - Bitter taste • Poisonous plants (people reject this flavor) • Broccoli, Turnip, baking soda - Feels slippery - React with oils in ur skin to form soap - Many cleaning products contain NH3 - Blood - Drain Cleaner contains KOH

Formula mass

(Molar Mass) Usually refers to the basic unit in a network solid. It is the mass of one of these basic units

Molecular mass

(Molar of molecular compound) Is used to refer to molecules that can exist independently. It is the mass of one molecule

Avogadro's number

(N_A) 6.022 x 10^23 mol^-1

Van der Waals equation

(P+n^2a/V^2)(V-nb)=nRT = P = actual or measured P (atm) = n = moles of gas = a and b = constants for the specific gas in ? = V = actual or measured V (L) = T = temp. (K) = R = 0.0821 L•atm/mol•K

Bond Energy - The amount of energy that is...

...required to break the bond between 2 specific atoms in a molecule = ... released when a bond forms between the same 2 atoms

Limitations of the Lewis Structure Model (4 parts)

1) Resonance 2) Octet rule fails when there r odd numbers of Ve- 3) Incomplete octets and other issues 4) Expanded octets also fail the octet rule

3 types of acid base titrants

1) Strong acid - Strong base titration 2) Titration of weak acid by strong base 3) Titration of weak base by strong acid

All 3 of the following conditions must be met in order for a reaction to occur

1) There must be a collision 2) The collision must occur with an orientation that could cause a reaction 3) E(collision) >_ E(a) • In most cases, only a small percentage of collisions result in reactions

1st Ionization Energy

A(g) --> A+(g) + e- The minimum amount of energy that is required to remove an outermost, least tightly held, electron from an atom in the gas phase

Electrons do not follow ________. A _________ describes an electron's possible positions in three-dimensional space, and is often called an _________

Orbits, Wave function; orbits

How can u tell what gets oxidized and what gets reduced??

Oxidation numbers = Positive or negative number that indicates how many electrons an atom has gained, lost, or shared to become stable

Use list of standard reduction potential values to indicate which species is most easily reduced a) Sn 2+ or Pb 2+ b) Cu + or Cu c) Ag + or Cu d) Ag + or Ag e) Fe 2+ or Cu f) Na + or Na

Pb 2+, Cu +, Ag +, Ag +, Fe 2+, Na + Solid metals not easily reduced and don't tend to want to gain extra e-s

Solubility and the common Ion effect • Predicting precipitates and the common ion effect

PbCl2 <-> Pb^2+ + 2Cl- (+ AgCl) AgCl —> Ag+ + Cl- AgCl doesn't experience experience 100% dissociation (Ksp = 1.6 x 10 @ 25'C) Equilibrium of 1st reaction shifts left • Cl- is common ion that cause equilibrium in 1st reaction to shift (Le Chatelier) • The equilibrium in the 2nd reaction will lie further to the left than it would, due to the presence of Cl- from PbCl2

Solubility and the Common Ion effect

PbCl2 <-> Pb^2+ + 2Cl- (+ NaCl) NaCl —> Na+ + Cl- NaCl soluble and experience 100% dissociation Equilibrium of 1st reaction shifts left • Cl- is common ion that cause equilibrium in 1st reaction to shift (Le Chatelier)

Conc and Nernst Equation

Q < 1 : More reactants : - : E > E^o Q > 1 : More products : + : E < E^o

The equilibrium constant, K(p), is 0.140 for CIF (g) <-> F2 (g) + CIF (g) at 427'C. partial pressures r 0.632 atm for CIF3, 0.025 atm for F2, and 0.097 atm for CIF b) Will the partial pressure of CIF3 increase, decrease, or stay the same as the system approaches equilibrium

Qp = 3.8 x 10^-3 As Qc < Kc the numerator must increase and the denominator must decrease for Qc = Kc. • Equilibrium will move to the right to increase the concentrations and partial pressures of F2 and CIF until the system reaches equilibrium and Qc = Kc • Thus, partial pressures of CIF3 will decrease, to produce more F2 and CIF, until the system reaches equilibrium

Ionization Energy: 2nd ionization potential for an atom is always greater than the 1st.

Radius reduced after 1st e- is removed, and ration of protons to e-s increases

The electrons that are furthest from the nucleus are partly "_________" by the inner core electrons. This shielding effect (__________ from the inner core electrons) reduces _____________between the outer electrons and the nucleus.

Shielded; electrostatic repulsion; electrostatic attractions

Which mixture in each set has the highest pH? Justify a) SiO2 and water or Cu(NO3)2 and water

SiO2 and water bc it's pH will be 7 • As SiO2 is non water-soluble • Cu(NO3)2 and water will have a pH that is lower than 7, as Cu 2+ is an acidic cation

What is the difference between a sigma and pi bond?

Sigma- result from the overlap of orbitals Pi- result from the side to side interaction between unmorphed p-orbitals - Pi bonds r only present in double and triple bonds, whereas sigma bonds r present in single, double, and triple bonds

^G^o found to be +32 kJ. What can be said about relative proportion of reactants and products when system is at equilibrium?

Since + ^G^o, know system contains mostly reactants at equilibrium. Since ^G^o > + 20 kJ, equilibrium system contains virtually all reactants

Why do noble gases not have electronegativity?

Since the noble gases already have eight electrons in their outer shells, they don't want to attract any more. Since electronegativity measures the amount of attraction between an atom and an electron, noble gases do not have electronegativity.

Taking the natural log of both sides of the Arrhenius Equation gives...

ln k = -E(a)/R (1/T) + ln A • A graph of ln k versus 1/T produces a straight line • The activation energy, E(a), can be determined after calculating the slope of the line • Slope = -E(a)/R

Electronegativity: An elements ability to ________ e-s in a chemical bond. E-s will spend ______ time around the more electronegative element in a chemical bond

attract; more

Saturated solution

The solvent has dissolved the maximum amount of solute that it can at a certain temperature, and some solid solute remains on the bottom • A solution is at equilibrium when it is saturated

In all shells, electron density near the nucleus ___________ with each subsequent subshell that is added. Thus, the shielding effect __________ for each subshell that is added.

decreases; increases

= As the number of bonds between two atoms increases, the bond length ______, the bond energy _______, and the PE ________. = As the electron density between the positive nuclei increases, the attractive forces between the protons and the bonding electrons _________. = Do single or triple bonds have the hardest melting/boiling point??

decreases; increases; decreases; increases; Triple

If given an ionic compound and told it sits for an extended period of time...

it is a disproportionation reaction and the metallic ion should be in an intermediate oxidation state. This metal will get both oxidized and reduced. Oxidize to highest ox state — if u don't know what that is increase it by one, and reduce it to the pure metal

If given a strong acid and a strong base...

net ionic always H+ + OH- —> H2O

Ratio of isotopes

of a certain element in any pure sample of a given compound is constant. If we could calculate the average mass of all the atoms of an element in a pure sample, we would obtain the average atomic mass of that element, which is given on the periodic table. As the ratio of the different atoms of each element that form a given compound is constant, the ratio of the masses of the constituent elements in a pure sample of that compound will also be constant

Percentage by mass

of an element in a pure sample of a compound or a component in a substance.

Atomic radius

one-half the distance between the nuclei of two atoms of the same element when the atoms are joined

^H^o = -106.7 kJ/mol and ^S^o = -142.2 J/molK C(graphite) + 2Cl2 <-> CCl4 ^G^o = -64.3 kJ/mol b) Does system contain mostly reactants or mostly products at equilibrium? Justify

products bc ^G^o less than 0

Explain why (2NiS + 3O2 —> 2SO2 + 2NiO) has - ^_S^o value whereas (2CH3OH + 3O2 —> 2CO2 + 4H2O) has + ^_S^o

products in 1st much more ordered than reactants • Reactants composed of 5 mol gas and products composed of 2 mol gas and 4 mol L • Less disorder on products side yields -^_S^o value • Products on 2nd r less ordered than reactants • Reactants r composed of 5 mol gas and products composed of 6 mol gas • More disorder on products side yields + ^_S^o value

When the rate of the reverse reaction is greater than the rate of the forward reaction, there is a net conversion of _____ into_______ The reversible reactions is proceeding to the _____

products into reactants Left

Isoelectronic species

share the same electronic configurations, but have different radii

If given a weak acid and weak base...

show H+ transfer

If given a soluble ionic compound and water...

simply show the dissociation of the compound

Ionic Radius: Cations are __________ than neutral atoms. Anions are ___________ than neutral atoms.

smaller; larger

Solvent vs solute

solvent = Substance that is more plentiful in a solution Solute = Substance that is less plentiful in a solution

Aufbau Principle

states that each electron occupies the lowest energy orbital available

Aufbau

states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels.

If given nitric acid and a metal...

the metal will lose e-s to the nitrate ion. (See stuff sheet) Non-trivial redox

quantum mechanical model

uses complex shapes of orbitals (sometimes called electron clouds), volumes of space in which there is likely to be an electron.

Humans started using elemental Cu about 6000 yrs ago and started using elemental tin about 3800 yrs ago. Explain why humans were able to use elemental Cu before they were able to use elemental tin. Cu(s) + 1/2 O2(g) —> CuO(s) H'(f) = -156 kJ/mol Sn(s) + O2(g) —> SnO2 (s) H'(f) = -581 kJ/mol

Both of these metals exist as oxides in nature, so humans had to transform these oxides into elemental metals. Since the forward reactions r exo, the reverse reactions r endo. Thus, both of these metals oxides must be heated in order to drive off the oxygen and produce the elemental metals. It requires +581 kJ/mol to decompose SnO2 and only +156 kJ/mol to decompose CuO. Thus, it requires more energy to decompose SnO2. As yrs went by, human cultures got better at concentrating heat from fires, ovens, and furnaces. As these technologies improved, they developed the ability to decompose metal oxides which more - enthalpy of formation values

Is CO2 an empirical formula, a molecular formula, or both? Explain.

Both! There is exactly 1 C atom and 2 O atoms in each CO2 molecule, so CO2 is the molecular formula. 1 C : 2 O is also the smallest whole number ratio of C atoms to O atoms, making it an empirical formula as well.

Identify strongest acid. Justify based on molecular structure and electronegativity a) I(CH2)2COOH or Br(CH2)2COOH

Br(CH2)2COOH bc Br is more electronegative than I • highly electronegative Br reduces electron density btwn H and O to a greater degree than does I • Bc electron density btwn H and O in Br(CH2)2COOH is less than it is in I(CH2)2COOH, FoA holding onto H in Br(CH2)2COOH is also less

Explain why Cl2 is a gas and Br2 is a liquid at 25'C and 1 atm

Br2, both non-polar and only LDF but Br2 is larger and has more e-s, making it more polarizable than Cl2 • Br2 experiences greater LDF • Bc of stronger FoA, Br2 exists as a liquid at 25'C and 1 atm, while Cl2 exists as a gas

Diatomic molecules

BrINClHOF = 2 diatomic molecules that contain atoms from the same groups in the same proportions: — Will have the same shape, and — will both be either polar or non-polar = This can help chemists design new materials: — Replacing an element from one group with another from the same group could lead to a new substance with similar properties — Ex. SiO2 can be used to make ceramics. SnO2 might work just as well — or better

Chlorine gas is bubbled through a solution of potassium iodide. a. What is/are the spectator ion(s) in the reaction? b. Write the balanced oxidation half reaction.

Cl2 + 2I- → I2 + 2Cl- K+ 2I- → I2 + 2e-

F2 or Cl2 - Which structure had the longest bond length? Justify

Cl2 has the longest bond lengths. Cl has a larger atomic radius which means that the nuclei will be further apart

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces M. F2 or Cl2

Cl2, bc both LDF but Cl2 is larger and has more e-s, making it more polarizable than F2 • Cl2 experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of Cl2 is higher

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces f. Cl2 and H2

Cl2, bc both LDF but Cl2 is larger and has more e-s, making it more polarizable than H2 • Cl2 experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of Cl2 is higher

Predict approx magnitude of ^S for following reaction. Justify 2CH4 —> C2H6 + H2

Close to 0 • = #s mol on both sides of chemical equation and all molecules r in g phase • Greater incr in disorder on one side of equation could only be due to different potential arrangements of particles due to differences in molecular shape

The force of attraction _____________ as the distance between the outermost electron and the protons ___________

Decreases; increases

Suppose u have 2 identical 1.0 L sealed containers. Both containers r kept at exactly 25'C. One vessel contains only Ne gas at 1.5 atm, and the other contains only Xe gas at 2.5 atm b. What variable must be changed in order to decrease the average KE of the Xe atoms??

Decreasing the T is the only method for decreasing the average KE of any system

Will decreasing the temperature of the following equilibrium system cause the ratio if [CO]/[CO2] to increase, decrease, or remain the same? Justify 2CO(g) + O2(g) <-> 2CO2 (g) ^_H(rxn) = -566.0 kJ

Decreasing the temperature will cause the reaction to shift in the exothermic direction - to the right in this case - to produce more heat • A shift to the right will cause [CO2] to increase and [CO] to decrease; therefore, this will decrease the ratio of [CO]/[CO2]

Why is the 1st ionization energy for Li less than that of Ne?

For both atoms, the e- is removed from the n=2 shell. However, Ne has more protons in its nucleus. According to Coulomb's Law, the force of attraction between charged particles increases as the charge (in this case the number of protons) increases. As the force of attraction of the outermost e- in greater in Ne, it requires more energy to remove one of those e-

Not all collisions btwn reactants result in the formation of products for any elementary reaction. Explain why this is.

For most reactions, most collisions don't cause a chemical reaction • Orientation of each collision is completely random • Reactants must collide with an orientation that is conductive to the formation of a bond • Also, collisions must posses sufficient energy (activation energy) to produce a reaction

2 different 1.2 L buffered solutions were prepared using HOBr and LiOBr. Both buffered solutions had pH of 5.2 at 25'C. After 0.17 moles of HI were added to each solution, found that pH of one solution had dropped to 4.9 and pH of other had dropped to 3.1 b) What trend must be true when comparing conc. HOBr and OBr- in 2 solutions if they shared same pH before HI was added? Justify

For the 2 buffered solutions to share the same pH, they needed to share the same [H+] • Also have same K(a) value, as they r both at same temp • Equilibrium expression for both solutions can be rearranged as follows: HOBr —> OBr- + H+ K(a) = [OBr-][H+] / [HOBr] K(a) / [H+] = [OBr-] / [HOBr] Since K(a):[H+] is same for both when share same pH, ratio of conc. of acid to CB must also be same

The value of K(eq) for a certain reaction is 5.6 at 650 K and 1.2 at 125 K. Is the forward reaction endothermic or exothermic? Justify

Forward reaction is endothermic • Reducing the temperature causes the equilibrium to shift in the exothermic direction • In this case, equilibrium must have shifted to the left when the temperature dropped, as this would increase the value of the denominator and reduce the value of the numerator, resulting in an overall reduction in the magnitude of the equilibrium constant

Square Pyramidal Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

IF5; 6; 5; 1; <90 degrees; sp^3d^2

A 65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl at 25'C and curve was plotted. Which indicators should be used to signal endpoint of titration: methyl red (pK(a) = 5.5), litimus (pK(a)=7.0), or phenolphthalein (pK(a)=8.7)? Explain

Methyl red bc pK(a)(methyl red) is very close to pH of solution at equivalence pt

Magnesium ribbon is placed into a solution of copper (II) chloride. a. What is reduced in the reaction?

Mg + Cu2+ → Cu + Mg2+ Cu 2+

Mg^2+ and F- r isoelectronic - which ion has smaller radius? - Explain why radii of these 2 ions r different sizes. Justify.

Mg^2+ -Both species have same e- config., and same # of e-s. Mg^2+ has 12 protons, while F- only has 9 protons. -Bc Mg^2+ has more protons, its nucleus applies greater force of attraction on its e-s, thereby pulling them closer to nucleus. -According to Coulomb's Law, force of attraction acting on an e- increases as more protons r added to nucleus which gives Mg^2+ a smaller radius.

1st Ionization energy

Minimum amount of energy required to remove the least tightly held e- from an atom or ion in gas phase.

1st Ionization energy: ______________ amount of energy required to remove the least tightly held e- from an atom or ion in gas phase. Energy must always be ________ to remove an e-

Minimum; added

Benzene, a cyclic hydrocarbon, C6H6, is burned in air. a. Identify the substance oxidized and the substance reduced in the reaction.

2C6H6 + 15O2 → 12CO2 + 6H2O Carbon

Solid potassium metal is added to water. a. Write the balanced reduction half reaction.

2Na + 2H2O → 2Na+ + 2OH- + H2 2H2O → 2OH- + H2

Using only the info given in the thermochemical equation below, make a comparison between the sum of the bond enthalpies of the reactants and the sum of the bond enthalpies of the products. Explain why one value is greater than the other. CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g) ^_H(rxn) = -802.3 kJ/mol

The reaction is exothermic, meaning that the overall reaction releases energy. For this to be possible, the sum of the bond energies in the reactants must be less than the sum of the bond energies in the products. A greater amount of energy is released during the formation of 2 C=O bonds and 4 H-O bonds in the products, than is required to break the 4 C-H bonds in methane and 2 O=O bonds of oxygen gas

Coulomb's Law

The relationship among electrical force, charges, and distance: The electrical force between two charges varies directly as the product of the charges and inversely as the square of the distance between them.

The mass of one 12-C atom is exactly _________. All other atoms are measured on a scale relative to 12-C.

12 amu

Oxoacid Strengths

2 types of Oxoacids 1) An OH group bonded to an element that is not bound to other Oxygens (HOY acids) H-O-Y 2) An OH group bonded to an element that is not bound to other Oxygens (HYO(n) acids) O H-O-Y-O

Concentrated hydrobromic acid is poured into a solution of potassium permanganate. a. How many electrons are exchanged according to the reaction? b. Identify the oxidizing agent.

2(5e- + 8H+ + MnO4- → Mn2+ + 4H2O) 5(2Br- → Br2 + 2e-) 10e- + 16H+ + 2MnO4- → 2Mn2+ + 8H2O 10Br- → 5Br2 + 10e- 16H+ + 2MnO4- + 10Br- → 2Mn2+ + 8H2O + 5Br2 10e- MnO4- or KMnO4

What charge do Group 2A elements acquire as ions?

2+

1. Short saturated fatty acid 2. Long saturated fatty acid highest viscosity at 25'C??

2, bc both non-polar and LDF but 2 is larger and contains more e-s, which makes it more polarizable • intermolecular FoA btwn the molecules in a pure sample of 2 would be stronger • 2 would higher viscosity, became of stronger LDF

What charge do Group 6A elements acquire as ions?

2- for non-metals and 2+ or 4+ for metals

Which mixture in each set has the highest pH? Justify a) 2.0M K2CO3 or 2.0M LiI

2.0M K2CO3 bc pH will be greater than 7 • CO3 2- ion is basic, and therefore, it will raise [OH-] in the solution through the reaction CO3 2- (aq) + H2O (L) —> HCO3 - (aq) + OH- (aq) Increasing [OH-] increases the pH of the solution • 2.0M LiI will lower the pH of the solution, as Cu+ is an acidic cation • Its pH will be less than 7 • Li and I r both neutral ions as Li+ is the conjugate acid of a strong base, and I- is the conjugate base of a strong acid

Which mixture in each set has the highest pH? Justify a) 2.0M NH4F or 2.0M NaF

2.0M NaF bc pH will be greater than 7 • F- ion is basic, as it's conjugate base of a weak acid • Na+ ion is neutral, as it's the conjugate acid of a strong acid • Basic F- ion will raise [OH-] in the solution through the reaction F- (aq) + H2O (L) —> HF (aq) + OH- (aq) Increasing [OH-] increases the pH of the solution • 2.0M NH4F will lower the pH of the solution, as Cu+ is an acidic cation • Its pH will be less than 7 • F- ion is basic, as it's the conjugate bade of a weak acid • NH4+ ion is acidic as it's conjugate acid of a weak base • The 2 ions will work against each other to make the solution more or less neutral

Solution

(Homogenous mixture) • homogenous mixture of 2< substances • Macroscopic properties don't vary within the sample • Components cannot be separated by filtration • Components can be separated by methods that alter intermolecular forces - distillation and chromatography • No components r large enough to scatter visible light

London Dispersion Forces (LDF)

(Induced Dipole - Induced Dipole) = These forces exist btwn all species: atoms, ions, non-polar molecules and polar molecules - They contribute to the overall FoA btwn all particles = LDF r only intermolecular force that keep assemblages of non-polar species together = Caused by Coloumbic interactions btwn temporarily induced dipoles of neighboring species that result from their e- distributions = Instantaneous charge distributions r polar = Averaged over time this atom is non-polar

The Law of Definite Proportions

(Joseph Proust) Different pure samples of the same compound always contain the same proportions of each element by mass. e.g. Water is always 88.8% oxygen and 11.2% hydrogen by mass. Water is always found in these proportions by mass.

Suspension

(Mechanical mixture) • Heterogeneous mixture of 2< substances • Macroscopic properties r different at different locations within the sample - sizes, shapes, and concentrations of particles can vary • In some cases, components can be separated through filtration

Physical Processes

(Phase Change) - Involve changes in intermolecular interactions - Properties change but the composition remains the same - Ex. = phase changes and formation or separation of mixtures - Ex = NO(s) —> NO (L) NH3 (L) —> NH3 (aq)

Covalent Network Solids (SiO2) (Si and SiC)

(Quartz) (SiC - empirical formula for Quartz) = covalent network of SiO4 tetrahedra = every Si atom is covalently bonded to 4 O atoms = Every O atom is covalently bonded to 2 Si atoms (Si and SiC) (Si - Forms a covalent network with itself) = A 3-D network with a geometry that is similar of that of a diamond

Oxidation Reduction Reactions

(Redox Reactions) = e- transfer reactions - oxidized • loses e-s - reduced • gains e-s

mass number

(a) equal to the number of protons plus the number of neutrons (a/z C)

intramolecular forces

(bonds) forces within molecules. Forces caused by the attraction and repulsion of charged particles

Steel - an ______ alloy

(interstitial) • Carbon fills some spaces btwn Fe atoms • Pure Fe lacks directional bonds • Steel is more rigid, less malleable, and less ductile than pure iron, as a result of the strong directional bonds that form btwn C and Fe atoms • Density of steel is greater than that of pure Fe, as interstitial atoms don't expand the lattice by much

Activity series order of increasing oxidation and which order to write elements in

(least) Au, Pt, Ag, Cu, Fe, Zn, Mg, Ca, Ba, K (most) • Elements will oxidize anything below it in this series • easier (s bottom) + harder (aq top) —> harder (s top) + easier (aq bottom)

If given an acid (except nitric) and a metal...

(remember the metals want to be oxidized) the metal will lose e-s to the hydrogen forming H2 and the metal ion. The anion of the acid is a spectator

Is sign for entropy change + or -?? Shorter vectors —> longer vectors

+ bc temp system incr as velocity vectors indicate that avg velocity of particles incr • Range in amount of KE contained by gaseous particles within a system broadens as temp incr (Boltzmann distribution) • entropy incr when energy is dispersed

What possible charge(s) can iron acquire as an ion?

+2 and +3

What possible charge(s) can lead acquire as an ion?

+2 and +4

Although we draw all possible locations of a double bond for resonance structures, we know that the idea of the double resonating between the locations is ultimately incorrect. we know this bc...

- Bond length of all bonds is the same and lies in between the length of a single bond and length of a double bond

The space filling models for 2 diatomic molecules r shown below 1. Smaller radius 2. Larger radius b. Which structure has the greatest bond energy between the central and a terminal atom? Justify

- 1 bc 1 has a smaller radii which means that nuclei will be closer together - According to Coulomb's Law, forces of attraction between nucleus of one atom and the Ve- of another increases as distance decreases - Also, PE decreases as Ve- from one atom approach the nucleus of another - To separate the atoms and break the bond, the same amount of energy that evolved must be added - Thus, it would require more energy to break a bond in 1 than it would to break a bond in 2

The space filling models for 2 diatomic molecules r shown below 1. Smaller radius 2. Larger radius c. Which structure has the least PE associated with its bonds? Justify

- 1 bc PE decreases as Ve- from 1 atom approaches the nucleus of another - Since the bonds in 1 r shorter, they contain less PE

Non-Bonding Orbital

- A MO that is at the same energy level as the one atomic orbital that it was derived from - E-s that occupy these orbitals don't cause stability or instability - Orbitals that contain lone pairs

Anti-Bonding Orbital

- A MO that is higher in energy than any atomic orbitals from which it was derived - E-s that occupy these orbitals cause instability

Bonding Orbital

- A MO that is lower in energy than any atomic orbitals from which it was derived - E-s that occupy these orbitals cause stability

Pure sample of iron and a sample of steel - What type of alloy is steel?

- A interstitial alloy

Buffering capacity

- Ability of a buffered solution to accept protons or hydroxide ions without a significant change in pH - Higher conc. of Ha and a- in solution result in greater buffering capacities • More weak acid to react with added OH- • More CB to react with added H+ • Best buffers have equal conc. of HA and A-

Limitations of the Lewis Structure Model Part 3) Incomplete Octets and other issues

- Accepted Lewis structure may have a center element without a full octet but the formal charges work out - Try to double bond: 2 non-zero formal charges and negative formal charge should be on most electronegative element - However, we don't have evidence that supports this structure (Double bond structure) — Experiments have shown that single bond length is less than it should be — suggests resonance and bond order greater than 1 - Electronegative difference suggests that bonds have a fairly high ionic character - None of these structures r 100% correct — Suggested that actual structure is a resonance of all 3, and incomplete octet is dominant form

Inductive effect and HOY Oxoacids

- Acid strength increases as electronegativity of Y increases - Electron density is pulled toward electronegative Y electron density —-> H-O-Y - Electron density is pulled out of the O-H bond bu the inductive effect of Y • Weakens bond btwn H and O • H+ falls off more easily, as force of attraction btwn hydrogen and oxygen is reduced - Acid strength increases as the electronegativity of Y increases HOI < HOBr < HOCl Acid strength and K(a) increases —>

Inductive effect and HYO(n) Oxoacids

- Acid strength increases as more oxygens r added to the central Y Acid strength and K(a) increase O added—-> • Increasing # of electronegative oxygen atoms increases electron density around Y • Reduces electron density btwn hydrogen and oxygen in O-H bond

Brônsted - Lowery Acids and Bases

- Acids donate protons • H+ ions fall off of the acid - Bases accept protons • Pick up H+ ions

Charge clouds of both methane (CH4) and ammonia (NH3) r arranged in a tetrahedral geometry. Explain why the actual bonds angles in methane r 109.5 degrees and actual bond angles in ammonia r 107.3 degrees

- All 4 charge clouds around C atoms in CH4 r bonds, so they r same distance apart and repel one another with the same force - NH3 has 3 bonds and 1 lone pair and 2 lone e-s spend most of their time closer to N atom than the bonding pairs - This also puts the lone pair in a position that is closer to the bonding pairs so lone pair exerts a slightly greater repulsive force on the bonding pairs, which causes them to move a little closer together

Some Anions make solutions Basic

- Anions that r conjugate bases of weak acids accept protons from water to produce OH- • CB of strong acids will not raise the pH of a solution • Ex- Cl-, ClO4-, I-, NO3-, Br-, and HSO4 - do not accept protons from water Ex - NaCH3CO2 is added to water CH3COO- + H2O <-> CH3COOH + OH- • pH rises above 7 as [OH-] increases

The equilibrium constant, K(c), is 9.8 x 10^5 for H2 (g) + S (s) <-> H2S (g) [H2] = 0.762 M and [H2S] = 0.483 M Has the process established equilibrium? If not, in which direction will it proceed? Justify

- As Q(c) < K(c) the system has not reached equilibrium • It will continue to proceed to the right until reaches equilibrium • Because Q(c) < K(c), [H2S], the numerator, is too small; and [H2], the denominator, is too large • [H2S] must increases and [H2] must decrease until the system reaches equilibrium where Q(c) = K(c)

Metallic Solids

- Bonding is not covalent — not enough e- to fill octets - Bonding results from the attractions between nuclei and delocalized Ve- moving throughout the structure - Bond strength increases as the number of bonding e- increases

Identify which ionic solid will have the higher melting temperature. Justify a. NaCl or KBr

- Both ionic solids, so they have very high melting temperatures due to strong forces of attraction between ions - According to Coulomb's Law, forces of attraction (lattice energy) increase as distance between ionic centers decrease and as charges on ions increase - 2 sets of ions carry the same charges; however, Na+ has a smaller ionic radius than K+, and Cl- had a smaller ionic radius than Br-. - Thus, distance between ionic centers of Na+ and Cl- is less than distance between ionic centers of K+ and Br- - This means that NaCl must absorb more energy to overcome these forces of attraction, and thus, it has a higher melting temperature

Identify which ionic solid will have the higher melting temperature. Justify e. MgO or Al2O3

- Both ionic solids, so they have very high melting temperatures due to strong forces of attraction between ions - According to Coulomb's Law, forces of attraction (lattice energy) increase as distance between ionic centers decrease and as charges on ions increase - An increase in ionic charge is; however, the most significant factor in determining the strength of the bonds between atoms - As the ions in Al2O3 carry charges of 3+ and 2-, while the ions in MgO carry charges of 2+ and 2-; forces of attraction between the ions in Al2O3 r greater - This means that Al2O3 must absorb more energy to overcome these forces of attraction, and thus, it has a higher melting temperature

Identify which ionic solid will have the higher melting temperature. Justify c. BeF2 or LiF

- Both ionic solids, so they have very high melting temperatures due to strong forces of attraction between ions - According to Coulomb's Law, forces of attraction (lattice energy) increase as distance between ionic centers decrease and as charges on ions increase - An increase in ionic charge is; however, the most significant factor in determining the strength of the bonds between atoms - As the ions in BeF2 carry charges of 2+ and 1-, while the ions in LiF carry charges of 1+ and 1-; forces of attraction between the ions in BeF2 r greater - This means that BeF2 must absorb more energy to overcome these forces of attraction, and thus, it has a higher melting temperature

Identify which ionic solid will have the higher melting temperature. Justify d. CaO or CaCl2

- Both ionic solids, so they have very high melting temperatures due to strong forces of attraction between ions - According to Coulomb's Law, forces of attraction (lattice energy) increase as distance between ionic centers decrease and as charges on ions increase - An increase in ionic charge is; however, the most significant factor in determining the strength of the bonds between atoms - As the ions in CaO carry charges of 2+ and 2-, while the ions in CaCl2 carry charges of 2+ and 1-; forces of attraction between the ions in CaO r greater - This means that CaO must absorb more energy to overcome these forces of attraction, and thus, it has a higher melting temperature

Identify which ionic solid will have the higher melting temperature. Justify b. NaCl or MgS

- Both ionic solids, so they have very high melting temperatures due to strong forces of attraction between ions - According to Coulomb's Law, forces of attraction (lattice energy) increase as distance between ionic centers decrease and as charges on ions increase - An increase in ionic charge is; however, the most significant factor in determining the strength of the bonds between atoms - As the ions in MgS carry charges of 2+ and 2-, while the ions in NaCl carry charges of 1+ and 1-; forces of attraction between the ions in MgS r greater - This means that MgS must absorb more energy to overcome these forces of attraction, and thus, it has a higher melting temperature

CCl4 is non-polar, but CH2Cl2 is polar? Explain why this this is by discussing shape, polarity of bonds, and whether or not dipoles cancel

- Both r tetrahedral and all bonds r polar - CCl4 cancel, making structure non-polar but dipoles in CH2Cl2 don't cancel - Cl is more electronegative than H, so 2 H atoms acquire partial + charges - Polar overall compound as it acquires + and - end

Properties of Ionic solids -Cleave along planes:

- Brittle 3D structure - Ions line up in a repetitive pattern that maximizes attractive forces and minimizes repulsive forces - Not malleable or ductile (able to be hammered or pressed permanently out of shape without breaking or cracking)

Common ion effect is used to create buffer solutions

- Buffers resist changes in pH • When small quantities of strong acids or strong bases r added to a buffered solution, changes in pH r small - Buffer • Weak acid and its salt (extra CB) • Weak base and its salt (extra CA)

C2H2 or C2H6 H-C=_C-H or 2 C surrounded by 6 H w/ single bonds e. C-C bond contains the greatest amount of PE in which structure?? Justify

- C2H6 bc OE of Ve- decreases as they approach the nucleus of another atom - Since the C-C bond length in C2H6 is longer, it has more PE

CB strengths of Binary acids

- CB strength decreases when moving down a group Base strength decreases ——-> F- > Cl- > Br- > I- Anion Radius increases ——> • Small radius has a greater ability to attract and accept H+ ions

Extended Pi Bonding Ex. Benzene C6H6

- CH is every corner - Alternating single and double bonds that can flip-flop - Each p-orbital can overlap with 2 different p-orbitals - Leads to delocalization of e-s - Can be used to explain resonance in Lewis structures

In which structure, CO3 2- or CO2 , is the bond energy between the C atom and each O atom the greatest? Explain

- CO2 bc it has 2 bonds between each C and O whereas CO3 2- has only 1.33 bonds between each C and O - Bc there is a greater force of attraction between bonding atoms in CO2, it requires more energy to break the atoms apart in this structure

In which structure, CO3 2- or CO2 , is the bond length between the C atom and each O atom the shortest? Explain

- CO2 bc it has 2 bonds between each C and O whereas CO3 2- has only 1.33 bonds between each C and O - Greater number of bonds between atoms means the e- density between the nuclei of the atoms is greater - As the e- density increases, the attractive forces between protons of the bonding atoms and the e- involved in bond increase - Bond length decreases as the force of attraction increases

In which structure, CO3 2- or CO2 , has the most PE associated with its individual bonds? Explain

- CO3 2- bc PE decreases as Ve- from one atom approaches nucleus of another - Since bond lengths in CO3 2- r longer, they contain more PE

Pure sample of iron and a sample of steel - Explain how carbon changes the properties such as more rigid, less malleable, less ductile, and density of steel is also greater than the density of pure iron

- Carbon atoms from strong directional bonds with neighboring iron atoms, and pure iron lacks directional bonds - This is why steel is much more rigid, less malleable, and less ductile than pure iron - Density of steel is also greater than the density of pure iron bc interstitial carbon atoms do not increase overall volume or expand the lattice of iron by very much

Explain why CH4 has a tetrahedral geometry, while SF4 has a seesaw shape

- Central atom in methane is surrounded by 4 charge clouds - All of these charge clouds r bonding pairs and 4 bonding pairs repel one another with the same force, causing them to move as far away from one another as possible and results in a tetrahedral with bond angles of 109.5 degrees - central atom in SF4 is surrounded by 5 charge clouds. 4 of these charge clouds r bonding pairs and 1 is a lone pair. A seesaw shale results from this configuration

Polyatomic Ions Defined - Polyatomic Ions:

- Combination of non-metals or metals and non-metals bonded together to form a polyatomic ion - Bonds between atoms r non-polar covalent or polar covalent

Grey Area (Both Chemical and Physical Process)

- Dissolving a salt in water involves breaking ionic bonds and creating ion-dipole interactions - Ex. = NaCl(s) —> Na+(aq) + Cl-(aq)

C=C=C - How is the double bond between the 1st and 2nd C formed?? Reference types of orbitals involved in bonds and the methods of bonding

- Double bond contains a sigma and double bond - Sigma bonds results from overlap of the sp^2 hybrid orbital on 1st C and the sp hybrid orbital on the 2nd C - Pi bond results from side to side interaction between unmorphed p-orbital on 1st C and one of the unmorphed p-orbitals on the 2nd C

Energy Changes and Chemical Reactions

- During chemical reactions, internal energy of system can: • Decrease (exo) • Incr. (endo.) • Remain same

Pi or sigma bonds - Extended networks of sigma bonds can cause a delocalization of e-s. Explain

- E-s shared in each pi bond - If a pi bond flip-flop back and forth between adjacent p-orbitals on 2 sides of an atom, the shared e-s in p-orbitals can be considered delocalized

VSEPR Theory Step 1. Count # of charge clouds, bonds, and lone pairs around the central atom

- Each item in this list is a single charge bond — Single bond = has 2 e-s — Double bond = has 4 e-s — Triple bond = has 6 e-s — Lone pair = had 2 e-s — Single unpaired e- = 1 lone e- - Each item on this list is considered 1 bond: — single bond = has 2 e-s — double bond = has 4 e-s — triple bond = had 6 e-s - Each item on list is considered a lone pair: — Lone pair = has 2 e-s — Single unpaired e- = had 1 e-

Enthalpy of Dissolution (^_H)

- Endothermic processes • Ionic bonds must be broken • H-bonds btwn some water molecules must be broken - Exothermic processes • Ion-dipole FoA r established btwn ions & H2O • If overall process is endothermic, more energy was required to break ionic bonds and H-bonds than was released when ion-dipole attractions were formed — Dissolution of most ionic compounds is endothermic, bc ionic bonds r very strong • If overall process is exothermic, more energy was released when ion-dipole attractions were established than was required to break the ionic bonds and H-bonds

Bond Enthalpy

- Energy is always released during the formation of a bond • PE decreases as the atoms move closer together - The same amount of energy must be added in order to break the specific bond • PE increases as the atoms move away from one another

maintaining equilibrium

- Equilibrium is maintained in a closed system • Reactants and products r not able to leave the system, so concentrations and partial pressures remain constant • New species that could influence the equilibrium cannot enter the system - Volume of the system remains constant • If volume changes, partial pressures of gases would change - Temperature of system doesn't change • Temperature changes affect reaction rates and the equilibrium position

Using external energy when ^_G>0

- External energy can drive reactions when ^_G>0 • Electricity can be used to recharge battery • Photons can supply energy required to remove electrons from atoms Ex. - conversion of CO2 and H2O to glucose through photosynthesis 6CO2+6H2O—>C6H12O6+6O2 ^_G = + 2880 kJ/mol • occurs through multiple steps that r initiated by absorption of several photons - Coupling reactions when ^_G>0 • Initial step in metabolic breakdown of glucose • Coupling a non-TFP with conversion of ATP to ADP makes overall process TFP

F2 or Cl2 - Which structure has the least PE associated with its bond? Justify

- F2 bc PE decreases Ve- from one atom approaches nucleus of another - Since bond length in F2, is shorter, it contains less PE

F2 or Cl2 - Which bond has the greatest bond energy? Justify

- F2 bc it has a smaller radius which means that the nuclei will be closer together - According to Coulomb's Law, forces of attraction between nucleus of one atom and the Ve- of another increases as distance decreases - Also, PE decreases as Ve- from one atom approach the nucleus of another - To separate the atoms and break the bond, the same amount of energy that evolved must be added - Thus, it would require more energy to break a bond in F2 than it would to break a bond in Cl2

Resonance Structures

- For many molecules, double or triple bonds r located between different atoms - This can result in two or more possible Lewis structures that r equally valid — If u r asked to draw a Lewis structure for a molecule of polyatomic ion that exhibits resonance, must draw all possible structures with 2 way arrows

AB (g) <-> A (g) + B (g) Suppose a chemist adds 1 mole of pure AB (g) to a sealed vessel d) What can be said about the rates of the forward and reverse reactions when the system is in a state of equilibrium?

- Forward and reverse reactions continue to occur after equilibrium is established • Concentrations stay the dame bc the rate of the forward reaction is equal to the rate of the reverse reaction, when the system is at equilibrium • When the system reaches equilibrium, rate of the forward reaction is equal to the rate of the reverse reaction • As a result, each species maintains a set concentration, although the concentrations of the different species involved in the reaction will almost never be equal to one another

Polar Bonds

- Greater electronegativity differences lead to greater partial charges and greater bond dipoles

Types of solutions - Gas - Solid Solutions

- H2 gas can occupy the spaces btwn some metal atoms such ad iron, and palladium

^_S(universe) and ^_H

- Heat flow links system to surroundings • Exothermic Reactions = S(surr) increases (^_S(surr)>0) • Endothermic Reactions = S(surr) decreases (^_S(surr)<0)

Inductive Effect

- Highly electronegative elements tend to "pull" or "induce" electrons toward themselves - Inductive effect of electronegative elements can.... • Reduce size of electron clouds • Reduce electron density in certain bonds • Reduce electron density at certain locations on a molecule

Polar and non-polar molecules

- If a molecule is polar, it must have a dipole moment. μ μ = Q x r Q = absolute value of the net partial charge at each end of a molecule r = distance between positive and negative poles of a molecule

Hess's Law

- If a reaction is carried out in a series of steps, the overall change in enthalpy will be = to the sum of the enthalpy changes for the individual steps • The overall enthalpy change will be the same if a reaction is carried out in one steps or in several steps

Heat and Endothermic Processes

- In an isolated system, energy is conserved • CH3OH (pe + KE) • CO + 2H2 (PE + ke) - As bonds in the products contain more PE, product molecules must contain less KE • This is why the products of an endothermic reaction r at a lower temperature - Bc products r at a lower temp. than surroundings heat must flow from surroundings into system until thermal equilibrium is established

Heat and Exothermic Processes

- In an isolated system, energy is conserved • CH4 + 2O2 (ke + PE) • CO2 + 2H2O (KE + pe) - As bonds in products contain less PE, product molecules must contain more KE • This is why products of exothermic reaction r at a higher temperature - Bc products r at a higher temperature than surroundings heat must flow from system into the surroundings until thermal equilibrium is established

- Energy must be added to a system in order for a solid to liquid or a liquid to gas phase change to occur. - Energy must be released from a system in order for a gas to liquid or a liquid to solid phase change to occur.

- In doing so, the energy of the system increases (endo) - In doing so, the energy of the system decreases (exo)

Extended Pi Bonding

- In organic chemistry terms, it is used to describe the situation that occurs when π systems (e.g. double bonds) are "linked together" - exists over a longer series of atoms (e.g. C=C-C=C or C=C-C=O. etc.). - results in a extension of the chemical reactitvity.

Forming a NaCl(aq) Solution

- Ionic bonds must be broken - H-bonds btwn some water molecules must be broken • Both endothermic - Ion-dipole FoA r established btwn Na+ and H2O and btwn Cl- and H2O • exothermic = More energy required to break ionic bonds and H-bonds than is releases when ion-dipole forces established, so overall process endothermic

Strength of Ionic Bonds

- Ionic bonds r very strong so it requires a lot of energy to melt or vaporize these solids — Very endothermic

Some Cations Acidify Solutions • Ionic compounds

- Ionic compounds containing NH4+ can acidify solutions • NH4+ is the conjugate acid of NH3 Ex.- NH4NO3 is added to water NH4+ + H2O <-> NH3 + H3O+ pH drops below 7 bc [H3O+] increases

Crystalline Solids

- Ionic compounds do not exist as individual units containing one cation and one anion — NaCl doesn't exist as 1 Na cation bonded to 1 Cl anion - Ions r arranged in an orderly fashion that follows a pattern of repetition in 3D — Segments that repeat in 3D r called unit cells - Macroscopic structures usually have flat surfaces that make definite angles to one another — Due to that they cleave along planes

Arrangement of Ions

- Ions in an ionic solid r arranged in order to maximize Coulombic forces of attraction between cations and anions and to minimize repulsive forces between ions with like charges - Way in which ions r arranged depends on — Relative size of cations and anions — Ratio of cations to anions

Weak bases containing Nitrogen

- Lone pair on nitrogen accepts protons in solution • Any compound of nitrogen, that has 1 lone lair on nitrogen atom, can act as a weak base by accepting protons

Limitations of the Lewis Structure Model Part 1) Resonance

- Many of the double bonds r shown in accepted Lewis structures r not really double bonds - They r 1.5, 1.33, or 1.25 bonds - In some cases, resonance must be used to refine a Lewis structure in order to obtain qualitatively accurate predictions about the effective # of bonds, bond energy, and bond length

Metal Oxides are Basic

- Metal oxides can form solid OH- MgO <-> Mg 2+ + O 2- O 2- + H2O —> 2 OH- ________________ MgO + H2O —> Mg 2+ + 2 OH- MgO + H2O —> Mg(OH)2 • Slight solubility of solid hydroxides increases [OH-] in the solution Mg(OH)2 <-> Mg2+ + 2 OH-

Spectroscopy

- Method of analysis which is based upon the absorbance of electromagnetic radiation by matter - Used 2 acquire data pertaining to the concentration of colored species I(o) = I(T) + I(A) I(o) = intensity of electromagnetic radiation striking sample I(T) = intensity of electromagnetic radiation exiting sample I(A) = intensity of electromagnetic radiation absorbed by the sample

Some Cations Acidify Solutions • Monoatomic cations

- Monoatomic cations can release protons from H2O • Cations attract - poles of H2O • + charged cation decreases electron density in the O-H bonds • Some H+ ions can break free • Ex) FeCl2 is added to water • Fe 3+ can be written as Fe(H2O)6 3+ Fe(H2O)6 3+ <-> Fe(H2O)5(OH) 2+ + H+ • pH drops below 7 as [H+] increases

Limitations of the Lewis Structure Model Part 4) Expanded octets also fail octet rule

- More than 8 Ve- in octet

Properties of Covalent Compounds

- Most covalent compounds have much lower melting and boiling points than ionic compounds - Covalent solids r usually soft and flexible - Most covalent compounds do not conduct electricity when dissolved in water. Acids r an exception to this rule

Properties of Ionic solids -Solubility and Conductivity

- Most r soluble in polar solvents - They conduct electricity only when molten or dissolved in a polar solvent as the charged particles r free to move — The higher the concentration of ions in a solution, the higher the electrical conductivity - Smaller ion is, more condensed magnetic field and more attractive it is to water — smaller = stronger ability to attract water

Dynamic Equilibrium

- Most reactions don't go to completion • All reactants don't get used up • System reaches a dynamic state where reactants r continually turning into products, and products r continually turning back into reactants • 2-way arrow

Once equilibrium is reached,

- No observable changes occur ever though... • Forward and reverse reactions continue to take place • Rate of the forward reaction = rate of reverse reaction - Concentrations and/or partial pressures of all species remain constant, yet they will rarely be = to one another

Are ionic solids malleable?

- No, bc ions r lined up in a repetitive 3D pattern. - When an ionic solid is hit with a hammer, one plane will slide along an adjacent plane. - When this happens, ions with like charges move closer to one another, which will result in forces of repulsion between the planes. - These repulsive forces can cause solid to cleave along that plane. - For this reason, ionic solids r considered to be brittle, and not malleable

EMF under non-standard conditions

- Nonstandard conditions • Conc of reactants and products r non 1 M • PP of reactants and products rn't 1 atm • temp not 25'C - Conc of reactants decrease and conc. of products increase, as reaction proceeds - Cell potential (voltage) decreases as conc. of reactants decrease and conc. of products increase - cell potential decrease until E(cell) = 0.00 V

Metallic Bonding - Electron Sea Model

- Nuclei and inner core e- r localized which Ve- r free to move throughout the solid — Conduct Electricity — Conduct Heat — Malleable and Ductile —— Lack directional bonds

Bond energy / Bond length

- PE decreases as attractions between nuclei and Ve- pull the atoms closer together - Moving further to the left, bond (spring) is compressed and the PE rises - As atomic radii of bonding atoms increase, the bond length increases and the bond energy decreases — Closer proximity to the shared e- allows for a stronger force of attraction between protons and bonding e-s — When force of attraction is greater, it requires more energy to break the bond

Reaction Mechanisms

- the series of steps that sum to the overall reaction • most reactions do NOT happen in 1 step

polar covalent bonds

- Partial charges r = in magnitude and opposite in sign - atom w/ higher electronegativity develops partial - charge, as shared Ve- spends more time around it - atom w/ lower electronegativity develops partial + charge, which is = in magnitude to partial - charge of more electronegative atom - Partial charges increase as electronegativity difference increase - Sum of partial charges in any molecule or ion is always = to overall charge on species

Endergonic Reactions (^_G^o >0)

- Process favors reactants at equilibrium K(eq) < 1 - Reaction proceed to equilibrium in either direction • Process that is non-TF (^_G > 0) will produce products if system initially contains only reactants • If products r removed, equilibrium will shift to produce more products (Le Chatelier)

The macroscopic properties of a pure sample of an unknown solid were examined in order to determine the type of bonding between particles. When hit with a hammer, it dents. When cleaned with steel wool it becomes very shiny. The solid conducts electricity but doesn't dissolve in H2O. Make a prediction about the type of bonding in this compound, Justify

- Properties of this solid indicate that it is a metal and the atoms r held together by metallic bonds - Metallic bonds r not directional, so they tend to be malleable - Malleable materials will dent when they r hit with a hammer - The outer surfaces of metals will oxidize in the presence of oxygen gas - If oxidized surface is cleaned off, metals r shiny - Metals conduct electricity in solid state

The macroscopic properties of a pure sample of an unknown solid were examined in order to determine the type of bonding between particles. The solid is soft and can be cut with a knife. It starts melting at 35 degrees Celsius. It doesn't dissolve in H2O - instead it floats on the surface. Make a prediction about the type of bonding in this compound. Justify

- Properties of this solid indicate that it is composed of molecules - Individual molecules r held together with covalent bonds - Molecular solids tend to be relatively soft, have low melting points and low boiling points - Some molecular solids dissolve in H2O and some don't

The macroscopic properties of a pure sample of an unknown solid were examined in order to determine the type of bonding between particles. The solid is very hard. When it is broken, the fragments form similar 3D shapes. When dissolved in water the resulting solution conducts electricity. Make a prediction about the type of bonding in this compound. Justify

- Properties of this solid indicate that it is held together by ionic bonds - Unit cells in ionic solids repeat in a regular 3D patterns and these solids tend to cleave along planes, which would give the fragments similar 3D shapes - Angles that the planes cleave along will be the same - When dissolved in H2O, ionic compounds dissociate into individual ions which r charged particles that r free to move - For this reason, ionic compounds r able to conduct electricity when dissolved in polar solvents

Properties of Steel

- Pure iron lacks directional bonds - Steel is more rigid, less malleable, and less ductile than pure iron, as a result of the strong directional bonds that form between carbon and iron atoms - The density of steel is greater than that of pure iron, as interstitial atoms do not expand the lattice by much (D = M/V)

Pure sample of iron and a sample of steel - In what way have some of the bonds changed when this alloy was formed?

- Pure iron only has metallic bonds, which lack direction - In steel carbon forms strong directional bonds with the neighboring iron atoms

AB (g) <-> A (g) + B (g) Suppose a chemist adds 1 mole of pure AB (g) to a sealed vessel c) Describe the process of achieving equilibrium in terms of the rates of the forward and reverse reactions

- Reaction rates r directly proportional to concentration • In the beg., when [A] and [B] were very low, rate of the reverse reaction was low • As more A and B were produced by the decomposition of AB, the concentration of A and B increased • An increases in concentrations of A and B caused more collisions btwn A and B molecules, resulting in a higher rate of production of AB - In the beg., when [AB] was very high, rate of the forward reaction was high. • As [AB] decr., rate of decomposition of AB also decreased • Rate of the forward reaction decreases until the system reaches equilibrium, and the rate of the reverse reaction increases until the system reaches equilibrium • When system reached equilibrium, rate of forward reaction is equal to the rate of the reverse reaction • As a result, each species maintains a set concentration, although the concentrations of the different species involved in the reaction will almost never be equal to one another

Equal moles of F2 and ClO2 r drawn into a vacuum where the following process takes place F2 + 2ClO2 <-> 2FClO2 \_____________ \ \____________ ____________ / / d) Describe some additional experiments that would need to be conducted in order to find the rate law for the overall reaction. Justify

- Repeat the experiment with the same initial [F2], but increases the initial [ClO2] by a factor of 2 • Compare the rate of this reaction to the rate of the first reaction • If the rate didn't change, reaction is 0 order with respect to ClO2 • If the rate doubled, reaction is 1st order with respect to ClO2 • If rate increased by a factor of four, reaction is 2nd order with respect to ClO2 • We identify the rate with respect to ClO2 with the symbol (m) - Repeat the experiment once again with the same initial [ClO2], but increases the initial [F2] by a factor of 2 • Compare the rate of this reaction to the rate of the first reaction • If the rate didn't change, reaction is 0 order with respect to F2 • If the rate doubled, reaction is 1st order with respect to F2 • If rate increased by a factor of four, reaction is 2nd order with respect to F2 • We identify the rate with respect to F2 with the symbol (n) = The rate law will be: Rate = [ClO2]^m [F2]^n

Bond Polarity

- Shared e-s spend more time around the most electronegative element in the chemical bond - This gives the more electronegative element a slightly negative charge and the less electronegative element a slightly positive charge

Molecular Models - Space-filling model

- Shows difference in the atomic radii of the bonded atoms - Shows relative bond length - Doesn't show the 3D positions of atoms very well - Doesn't show the number of bonds between atoms

Molecular Models - Ball-and-Stick Model

- Shows the 3D positions of atoms well - Shows single, double, and triple bonds - Balls r not proportional to the size of the atoms - Sticks r not proportional to and greatly exaggerate the bond length

Changes of state : Cooling curve

- Some liquids supercool before they solidify • At the bottom of the supercooling curve, solid crystals form • This causes the temp. of the system to increase dramatically, as heat is released by the solvent particles during the process of solidification

Completion Reactions

- Some reactions go to completion • All reactants are used up, as all turn into products • One way arrow

Salts with weak acids and weak bases

- Some salts have an acidic cation and a basic anion • Solutions containing these salts tend to have little or no overall affect on pH

Acid strength

- Strong acids completely ionize in water HA + H2O —> H3O+ + A- • 1 way arrow as equilibrium lies right - Weak acids partially ionize in water HA + H2O <—> H3O+ + A- • 2 way arrow as equilibrium lies left or middle

Strengths of Bases

- Stronger base always accepts the most protons in an acid/base reaction - There r always 2 bases 1. Base on the reactants side of the equation, and... 2. Conjugate base on the products side on the equation

Electronegativity Difference

- The difference between these classification is not distinct — it is a continuum. - All polar covalent bonds have some ionic character and ionic bonds can have some covalent character

Bond energy/bond length graph

- The higher the bond energy, the 'stronger' we say the bond is between the two atoms, and the distance between them (bond length) is smaller. For instance, the HO-H bond in a water molecule requires 493 kJ/mol to break and generate the hydroxide ion (OH-). - At 0.074 nm, bond length of H2, energy is at its lowest point. This is how H2 exists - on average. Bonds can be thought of as springs, so 0.074 nm is the midpoint of the oscillation

Substitutional Alloys

- The radius of the solute and solvent atoms r similar Brass - A substitutional alloy — zinc atoms r substituted for some copper atoms — Alloys remain malleable and ductile - The density normally lies between those of the component metals

Q > K(eq) - Q < K(eq) - Q = K(eq) -

- The reaction will proceed to left • Continue to increase the conc. of reactants until equilibrium is established - The reaction will proceed to right • Continue to increase the conc. of products until equilibrium is established - The system is at equilibrium

Solubility PbCl2 <-> Pb^2+ + 2Cl-

- The solubility is the max molar concentration of fu that will dissolve at a given temperature • For every 1 mole of PbCl2 that dissolves, 1 mole of Pb^2+ ions and 2 moles of Cl- ions enter the solution • Solubility of PbCl2 = [Pb^2+] = 1/2 [Cl-] • When solution is saturated

Some Anions make solutions Basic • Sulfate ion

- The sulfate ion is a very weak base • HSO4- is a fairly strong acid, Ka = 1.2 x 10^-2, so SO4 2- is a very weak base Na2SO4 —> 2Na+ + SO4 2- (will accept few H+ and small change in pH)

Stress from changing temperatures

- This is the only stress that changes the equilibrium constant (K(eq)) - A + B <-> C + Heat • Exo (—>) Cooling (taking heat away) shifts equilibrium in direction that produces heat • Endo (<—) adding heat (increases temp) shifts equilibrium in direction that absorbs heat

Possible arrangements of Ions - Simple Cubic

- This unit cell contains 8 corners x 1/8 of an ion = 1 ion/unit cell

Hydrogen Bonds

- Type of Dipole-dipole - Occurs btwn H that is covalently bonded to F, O, or N and another F, O, or N with at least one lone pair - 5-10 x stronger than other d^2 attractions

Pure sample of iron and a sample of steel - Identify 4 properties that change when Carbon is added to pure iron in order to make steel?

- When carbon is added to pure iron, the metal becomes more rigid, less malleable, and less ductile - Density of steel is also greater than the density of pure iron

Limitations of the Lewis Structure Model Part 2) The Octet rule fails when there r odd #s of Ve-

- When sum of Ve- of all atoms in a molecule is not an even #, the octet rule fails

Electronegativity

- an elements ability to attract bonding electrons toward itself in a chemical bond. - Increases as atomic radius decreases. - When 2 atoms are involved in a chemical bond, bonding electrons experience a greater attraction for the positive nucleus that is closest to them in accordance with Coulomb's law

Determine the enthalpy change, ^_H(rxn), that occurs when 2 gaseous oxygen atoms combine to form a gaseous oxygen molecule (2 O(g) —> O2 (g)). + or -??? exo or endo?

- and exo bc energy is always released when a bond is formed. The magnitude of this, bc a double bond forms btwn the 2 oxygen atoms

Precipitation Reactions

- attractive forces btwn oppositely charged ions is greater than FoA btwn H2O molecules and the ions

Explain the electron sea model of metallic bonding. Explain the factors that make metals good conductors of heat and electricity

- e- sea model accounts for properties of metals. - Free movement of Ve- allows metals to conduct electricity. - e- movement also enables metals to conduct heat rapidly - If u heat 1 end of metal bar, u will transfer heat to the e-s in that area - These energized e-s will move to the other end of the bar, carrying heat with them

Formal Charge

- in some cases, atoms in a molecule can be assembled in different ways when drawing a Lewis diagram - Formal charge r calculated to identify the most stable or likely structure - Sum of the formal charges on each atom will be zero for a neutral molecule - sum of the formula charges on each atom will be = to the overall charge of the structure for a polyatomic ion

Chemical Processes

- involve the breaking and/or formation of chemical bonds - Substances r transformed into new substances that may have different properties - Temperature changes, production of light, formation of a gas, formation of a precipitate, and/or changes in color r indications that a chemical changes has occurred - Ex. = 2Na(s) + Cl2(g) —> 2NaCl(s)

Acid-Base Titrations

- often used to determine the conc. of an acid or base of an acid or base in a solution • Strong acid of known conc. (titrant) is added to a base of unknown conc. (analyte), which is mixed with an indicator • Strong base of known conc. (titrant) is added to an acid of unknown conc. (analyte), which is mixed with an indicator — Indicator changes color to signal the arrival at the endpoint

Sigma or Pi Bonds - Which type of bond has a greater amount of bond energy? Justify

- sigma bc they result from the direct overlap of orbitals, and pi bonds result from the side to side interaction between unmorphed p orbitals. - Bc of this, sigma binds have a shorter bond length, so greater amount of bond energy

What charge do Group 7A elements acquire as ions?

1-

Titration curves for polyprotic acids

- titration curves can be used to determine the following • Number of acidic protons • pK(a) value for each acidic proton of a weak polyprotic acid • Major species present at any pt along curve

To know if a molecule is polar...

- u must know if the bonds r polar - u must know the overall shape

Explain each of the following occurrences by referencing the structure of the atoms in question (energy levels, orbitals, protons, etc.). -The first ionization energy of Mg is 738.1 kJ/mol, while the first ionization energy of Al is only 577.9 kJ/mol.

-(exception to the rule) Normally ionization energy increases as move from left to right across period in periodic table, bc shielding effect is similar and effective nuclear charge increases. -Reason for exception has to do w/ the e- config. of these elements. Mg has e- config. of 1s^2 2s^2 2p^6 3s^2, and Al has e- config. of 1s^2 2s^2 2p^6 3s^2 3p^1. -Requires less energy to pull only e- from p-orbital in Al, which is higher energy subshell, than it does to pull e- from full s-orbital in Mg, which is lower energy subshell -Lone e- in p-orbital of Al is at much higher energy than those in full s-orbital of Mg and being "shielded" by other e-s in lower energy subshells.

Explain each of the following occurrences by referencing the structure of the atoms in question (energy levels, orbitals, protons, etc.). -2nd ionization energy of Na is 4560 kJ/mol, while 2nd ionization energy of Mg is only 1450 kJ/mol.

-2nd e- removed from Na is taken from n=2 shell, whereas 2nd e- removed from Mg is taken from n=3 shell. -distance between nucleus and n=2 shell e-s is much less than distance between nucleus and n=3 e-s. Reduced distance means attractive forces between protons and n=2 e-s is greater -According to Coulomb's Law, forces of attraction increase as distance between e-s and protons decreases. -Greater attractive force, more energy required to remove an e-

Interstitial Alloys

-Atoms with a small radius occupies the spaces between atoms with a larger radius Steel - An interstitial alloy — Carbon fills some spaces between iron atoms

Explain each of the following occurrences by referencing the structure of the atoms in question (energy levels, orbitals, protons, etc.). -1st ionization energy of Mg is greater than 1st ionization energy of Ca.

-Ca has larger atomic radius than Mg, due to fact that Ve- of Ca r in 4s orbital and those of Mg r in 3s orbital -As radius of Ca is larger, its Ve- r held w/ smaller force of attraction, and requires less energy to remove them. -According to Coulomb's Law, forces of attraction decrease as distance between e-s and protons increases

Why are some processes endothermic?

-Consider how much energy is required to break bonds to how much energy is released as new bonds are formed. -Energy is released as bonds are formed (even if it an overall endothermic process

Covalent or ionic??

-Electronegativity difference is not the only factor that determines if the bond is covalent or ionic - Generally, bonds between a metal and non-metal r ionic and bonds between 2 non-metal r covalent - Examining the properties of compound is the best way to characterize the bond type

Why do halogens have negative e- affinity value, while noble gases have positive e- affinity value?

-Highest energy subshell (p-orbital) is full all of these elements and added e- would be only e- in higher energy subshell. -Shielding effect from inner core e-s is no large that energy must be added for them to accept additional e-. -Adding energy is endothermic, which is + energy value -Halogens need 1 more e- to fill octets. Bc of this, they will accept e- to fill highest energy subshells (p-orbitals) -In doing so, they give off energy. Giving of energy is exothermic process, which has a - energy value.

Electron Affinity: Positive Values (Group 8A)

-Highest energy subshell (p-orbital) is full for all of these elements -Added e- would be only e- in higher energy -Shielding effect from inner core e-s is so large that energy must be added for them to accept additional e-

Electron Affinity: Positive Values (Group 2A)

-Highest energy subshell (s-orbital) is full for all of these elements -Added e- would be only e- in highest energy p-orbital -Shielding effect from inner core e-s is so large that energy must be added for them to accept additional e-

Why do the Group 1A elements have negative e- affinity values, while Group 2A elements have positive e- affinity values?

-Highest energy subshell (s-orbital) is full for all these elements and added e- would be only e- in higher energy p-orbital -Shielding effect from inner core e-s is so large that energy must be added in order for them to accept additional e-. -Adding energy is endothermic, which is + energy value. Group 1A elements need 1 more e- to fill s-orbital and bc of this, they will readily accept an e- and give off energy. -Giving energy is exothermic process, which has - energy value.

Explain each of the following occurrences by referencing the structure of the atoms in question (energy levels, orbitals, protons, etc.). -1st ionization energy of Na is less than 1st ionization energy of Cl.

-Na has larger atomic radius than Cl. -Ve- of both elements r in same shell (n=3), so shielding effect experienced by Ve- of both elements similar -However, Ve- in Cl experience greater effective nuclear charge, as it has more protons in nucleus. -According to Coulomb's Law, forces of attraction acting on e- increases as more protons r added to nucleus. -Bc force of attraction on Ve- in Na is less than that in Cl, it requires less energy to remove e- from Na.

Ionic Radius: (Down a Group from Top to Bottom)

-New shells added (Principal quantum # increases) w/ larger orbitals -Ve- have more energy, less stable, and r further from nucleus -Outer e-s experience greater shielding effect and decreased effective nuclear charge -According to Coulomb's Law, force of attraction decreases as distance between e-s and protons increases

Atomic radius (Down a Group from Top to Bottom)

-New shells added (principal quantum # increases) with larger orbitals -Ve- have more energy, r less stable, and r further from nucleus -Outer e-s experience greater shielding effect and a decreased effective nuclear charge -According to Coulomb's Law, force of attraction decreases as distance between e-s and protons increases

Electron Affinity: (Down a Group from Top to Bottom)

-New shells are added (principal quantum # increases) with larger orbitals -Outer e-s experience greatest shielding effect and decreased effective nuclear charge -According to Coulomb's Law, force of attraction decreases as distance between added e- and nucleus increases. -e- affinity values generally become less - (less energy released) bc force of attraction between nucleus and added e- is weaker

Is energy released or absorbed when neutral sodium atoms react with chlorine gas to form solid sodium chloride, NaCl?

-Released and the process of bond formation is exothermic -Energy is always released during bond formation.

Describe the modifications to the shell model that r required due to the experimental PES data that was provided in the lecture.

-Shell model stated that e- follow orbits and orbits were given a quantum # (n=1, n=2, n=3...). All e- follow given orbit shared same quantized energy level. -In other words, all e- in n=2 of given atom possesses same energy and all e-s in n=3 of given atom possesses same energy. -PES data indicated that shell model must be modified. Shows that there r 2 possible ionization energies in n=2, which indicated e-s can be found in 2 sublevels in n=2 at 2 separate energy levels. PES data also shows that there r 3 possible ionization energies in n=3, which indicated that e-s can be found in 3 sublevels in n=3 at 3 separate energy levels. -Furthermore, it is evident that n=1 has one level that can hold up to 2 e-s, and 1st sublevel in n=2 can hold 2 e-s and 2nd can hold 6 e-s.

Hund's Rule

-When you have more than one orbital in a subshell (includes p, d, and f subshells) a single spin up electron is added to each orbital before you start adding spin down electrons. -every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Hybrid Orbitals

-atomic orbitals around the central atom in a molecule must hybridize in order for bonding to occur

Properties of Ionic solids -Strong bonds

-very strong Coulombic forces of attraction between cations and anions — High melting points — Very hard — Low volatility (doesn't vaporize easily)

What compound is least soluble in water at 298K? 1) ksp = 8.7 x 10^-9 at 25 'C 2) ksp = 1.6 x 10^-5 at 25'C

1 bc smaller ksp value

H+ from autoionization water is not important when finding pH 2 Reasons

1) 1 x 10^-7 M is a small number 2) Le Chatelier's Principle • Increasing [H+] shifts the reaction fro the autoionization of water towards H2O, therby reducing the conc of H+ from this process H2O <-> H+ + OH- <————————- shifts left

Determining Rate Law Steps

1) Chemists determine the rate law through experimentation 2) They use the rate law to figure out what the elementary steps are • The rate law tells them what the slowest step is • Then they try to figure out what the fast steps are • Through experimental detection of reaction intermediates, evidence is built to support proposed mechanisms

Manipulating K and Q

1) Coefficient Rule • When coefficients r changed by a factor of n, K(eq) is raised to the power of n. {K(2c) = (K(1c))^1/2} 2) Reciprocal Rule • When a reaction is reversed, the new K(eq) value is the inverse of the old K(eq) value {K(3c) = 1/(K(1c))} 3) Multi Equilibria Rule • When two or more reactions r combined, the new K(eq) is the product of the K(eq) values from the individual reactions {(K(1c)) x (K(4c)) = (K(5c))

VSEPR Theory (2 Steps)

1) Count # of charge clouds, bonds, and lone pairs around center atom 2) Predict the shape

Why are H-bonds so strong?

1) F-H, O-H, and N-H bonds r very polar 2) These atoms r very small, so partial charges caused by the differences in electronegativity r highly concentrated 3) LP(s) on F, O, or N increases the already partially - charge on these atoms, thereby creating a stronger attraction for the slightly + H

Limiting Reactant, X(A) and P(A)

1) Find mole fraction of all gases present after the reaction is complete 2) Find total pressure in vessel after reaction 3) Find partial pressure of H2 after reaction

6 Rules for Determining Oxidation Numbers

1) For an atom in its elemental form, its oxidation # is 0 (Mg, H2, O3) 2) For a monoatomic ion, its oxidation # is = to its charge (Mg2+ = +2; Cl- = -1) 3) Oxidation # for O in molecular compound is -2 (H2O) (Exception: Peroxides: ON = -1 (Ex. H2O2) 4) Hydrogen (+1 when bonded to a nonmetal (Ex. HCl)) (-1 when bonded to a metal (Ex. MgH2)) 5) For other covalent compounds that don't contain H or O, most electronegative element has an oxidation # = to its charge as an ion (BF3: F oxidation # = -1) (PCl5: Cl oxidation # = -1) 6) Sum of oxidation #s for all atoms in a compound must = overall charge of that compound (H2O)

Ksp and Qsp 1) If Qsp = Ksp 2) If Qsp > Ksp 3) If Qsp < Ksp

1) If Qsp = Ksp • System is at equilibrium • a saturated solution with solid and aqueous species 2) If Qsp > Ksp; precipitate forms • Reaction will proceed to the left until system reaches equilibrium 3) If Qsp < Ksp; no precipitate forms • Solution unsaturated • All ions remain in solution

Melting Point of Ionic Solids - Which compound from each set has the greatest melting point? Why? 1) LiF and LiI 2) MgCl2 and MgO 3) NaF and MgI2

1) LiF, easier to melting/boiling (F smaller than I) 2) MgO, O is smaller than Cl2 3) MgI2, I2 is smaller than F

How to tell which structure is more likely from formal charges??

1) More likely Lewis Structure will have formal charges that r closer to or = to zero 2) If there r negative formal charges, they should reside on the more electronegative elements in the structure

Problems with Arrhenius Acids and Bases

1) Only true for bases that contain OH- • There r bases that don't have OH-, but still make solutions basic 2) Only true for reactions that take place in aqueous solutions

Factors Affecting Reaction Rate

1) Rate of collisions btwn reactants 2) Percentage of collisions with an orientation that could produce a reaction 3) Percentage of collisions possessing sufficient energy (activation energy) to produce a reaction

Spectroscopy - 2 methods used for spectroscopic analysis

1) Ultraviolet/Visual (UV/Vis) spectroscopy • Examines transitions in electronic energy levels - Is used to probe the electronic structure of certain compounds • Is used to determine concentration of solutions that contain certain compounds 2) Infrared (IR) Spectroscopy • Examine transitions in molecular vibrations - Is used to detect the presence of different types of bonds and to identify molecules

Balancing Redox Reactions in Acidic Solutions

1) Write 2 unbalanced half-reactions • Oxidation/Reduction half-reactions For each of the half reactions: 2) Balance all atoms except for O and H 3) Balance for O by adding H2O molecules 4) Balance for H by adding H+ ions 5) Balance the charge by adding e-s 6) Cross multiply to cancel e-s 7) Add half reactions and cancel things that r the same

Procedure for Balancing Chemical Equations

1) Write the chemical formulas for all reactants and products 2) Balance elements that don't exist independently first by adding coefficients 3) Balance elements that exist independently last 4) Check to make sure that both sides are balanced

2 methods for calculating ^_G

1) ^_G = ^_H - T^_S • Calculate ^_H using calorimetry, Hess's Law, or enthalpy of formation values • Calculate ^_S using entropy values • Calculate ^_G using ^_G = ^_H - T^_S 2) Calculate ^_G using ^_G^o (f) values

Compounds with Multiple Central Atoms (4 steps)

1) look at each central atom on its own (Everything it is bonded to is considered to be a terminal atom 2) Count the charge clouds and bonds around it 3) Predict shape around it 4) Isolate next central atom and repeat steps 1-3

What charge do Group 1A elements acquire as ions?

1+

Lewis structures (steps)

1. Count total Ve- 2. Put least electronegative atom in center and connect terminal atoms to it with single bonds 3. Complete octets for all terminal atoms except H atoms 4. Add up e- used and subtract them from total Ve-. Attach leftover e-s to central atom as lone pairs 5. Make multiple bonds to complete the octet of central atom 6. DON'T FORGET BRACKETS AND CHARGE!

Laws of Thermodynamics

1. Energy contained within the universe is constant 2. entropy of the universe is constantly increasing 1) energy can not be created or destroyed, only converted from one form to another 2) each time you convert one form of energy to another, some energy is converted to a non-usable form (more energy efficient to consume plants because they exist very close to the initial source of energy)

Intermolecular FoA order

1. Ion-Ion (Ionic - Intra) (strongest) 2. Ion-dipole 3. H-bonds (small molecules can have very strong intermolecular attractions whereas H-bonds r present) 4. Dipole-dipole 5. Ion-induced dipole 6. dipole-induced dipole 7. LDF (Often strongest FoA btwn large species) (weakest)

Study of Reaction Rates

1. Rate of disappearance of a reactant 2. Rate of appearance of a product 3. Rate at which the overall reaction proceeds

3 factors affecting solubility

1. Structure = "Like dissolves Like" (polar dissolves polar; non-polar dissolves non-polar) 2. temperature = Different rules for different types of solutions 3. pressure = applies to Gas-liquid solutions

In a 0.450 M HONH2 solution, [OH-] = 5.28x10^-6 M HONH2 + H2O <-> HONH3+ + OH- g) Find percent ionization of NaOH in above sol.

100% as NaOH is a strong base

Which mixture in each set has the lowest pH? Justify 1.0M Mg(NO3)2 or 1.0M KCl

1.0M Mg(NO3)2 have lowest pH and lesd than 7, as Mg 2+ is an acidic cation • 1.0M KCl will have larger pH • KCl and water produces a neutral solution with a pH of 7, as both ions r neutral • K+ is the conjugate acid of a strong base, and Cl- is the conjugate base of a strong acid

Which mixture in each set has the lowest pH? Justify 1.0M MgSO4 or 1.0M Li2CO3

1.0M MgSO4 have lowest bc pH close to 7 • Mg 2+ is an acidic anion, which will lower the pH and SO4 2- is a slightly basic anion that will raise the pH slightly • Ka for HSO4- is 1.2 x 10^-2 at 25'C, and therefore, it's fairly strong acid • Conjugate bases of strong acids have a very limited ability to attract H+ ions • 1.0M Li2CO3 will have highest bc pH will be greater than 7 • Li+ is conjugate acid of a strong base, and thus, it's not acidic • CO3 2- is a basic anion, and therefore, it will raise [OH-] in the solution through the reaction CO3 2- (aq) + H2O (L) —> HCO3- (aq) + OH- (aq) Increasing [OH-] increases the pH of the solution

Which mixture in each set has the highest pH? Justify a) 1.0M NaC2H3O2 or 1.0M CuNO3

1.0M NaC2H3O2, bc pH will be greater than 7 • CH3COO- is the conjugate base of a weak acid, making it a relatively strong base • Following reaction will increase [OH-] in the solution CH3COO- (aq) + H2O (L) —> CH3COOH (aq) + OH- (aq) Increasing [OH-] increases the pH of the solution • 1.0M CuNO3 will lower the pH of the solution, as Cu+ is an acidic cation • Its pH will be less than 7

Which mixture in each set has the lowest pH? Justify 1.0M NaCl or 1.0M NaF

1.0M NaCl have lowest and pH will be less than 7 • NaCl and water produces a neutral solution and Na+ is a conjugate acid of a strong base and Cl- is the conjugate base of a strong acid • 1.0M NaF have highest and Na+ ion is neutral as it's conjugate acid of a strong base. • F- ion is basic as it's conjugate base of a weak acid and it will raise [OH-] in the solution through following reaction F- (aq) + H2O (L) —> HF (aq) + OH- (aq) • Increasing [OH-] increases the pH of the solution

In a 0.032 M NH3 solution, [OH-] = 1.27x10^-3M NH3 + H2O <-> NH4+ + OH- What conc. of KOH would be required to make a solution with pH of 11.104

1.27 x 10^-3 M

Which mixture in each set has the lowest pH? Justify 1.5M NH4Cl or 1.5M NaCl

1.5M NH4Cl have lowest and pH will be less than 7 • NH4+ is an acidic cation, which will lower pH and Cl- is neutral anion, which will not affect the pH • 1.5M NaCl have highest and NaCl and water produces a neutral solution with a pH of 7, as both ions r neutral • Na+ is the conjugate acid of a strong base, and Cl- is the conjugate base of a strong acid

2nd Order Integrated Rate Law

1/[A]t = kt + 1/[A]0 • A plot of 1/[A]t vs t will give a straight line for 2nd order reactions • This is used to determine if a reaction is 2nd order, as a straight line will only occur if the reaction is 2nd order slope = k

In a 0.032 M NH3 solution, [OH-] = 1.27x10^-3M NH3 + H2O <-> NH4+ + OH- Find percent ionization of KOH

100% as KOH is a strong base

Which subshell contains the lowest energy electron(s) in Na?

1s subshell -aufbau principle, electrons enter orbitals of lowest energy first -e-s in that subshell r closest to nucleus and experience greatest force of attraction according to Coulomb's Law. -Bc of this, those e-s r most stable

1. Unsaturated fatty acid 2. Saturated Fatty acid Highest boiling pt??

2 bc LDF r greater and molecules r linear, so they stack up better (can approach each other more closely) • means that 2 can produce more induced temporary dipoles than 1

The space filling models for 2 diatomic molecules r shown below 1. Smaller radius 2. Larger radius a. Which structure has the longest bond lengths? Justify

2 bc atoms in 2 have larger radii which means that nuclei will be further apart

Nickle (I) nitrate is allowed to stand for an extended period of time. a. Write the balanced reduction half reaction for the above.

2Ni+ → Ni + Ni2+ e- + Ni+ → Ni

Sulphur is burned in air. a. Is the above reaction a redox reaction? Justify your claim in terms of electrons.

2S + 3O2 → 2SO3 (Could also go to SO2) Yes, it is redox. Each sulfur loses 6 electrons and each of the three oxygens gains two electrons. IF you wen to SO2, the S loses 4 and each of the two O's gains 2.

An electron in ___ spends more of its time further from the nucleus than one in ______, but the electron density near the nucleus is greater for 2s electrons due to the first maximum. 2p electrons experience a _________________ from the 1s electrons than do the 2s electrons

2s; 2p; greater shielding effect

which Following reactants is TFP at all temps? Justify 1) 2H2O(L) —> 2H2 + O2 ^_H^o = + 572 kJ 2) 8Fe + 6O2 —> 4Fe2O3 ^_H^o = -3296.8 kJ 3) 2CH3OH + 3O2 —> 2CO2 + 4H2O ^_H^o = - 1352 kJ

3 is TFP at all temps 2CH3OH + 3O2 —> 2CO2 + 4H2O - ^_H and + ^_S r both favorable and will produce TFP at any temp

Copper wire is placed into a solution of nitric acid. a. What is reduced in the reaction? b. Write the balanced reduction half reaction. c. What is the oxidation number of nitrogen before and after the reaction occurs?

3(Cu → Cu2+ + 2e-) 2(3e- + 4H+ + NO3- → NO + 2H2O) 3Cu → 3Cu2+ + 6e- 6e- + 8H+ + 2NO3- → 2NO + 4H2O 3Cu + 8H+ + 2NO3- → 3Cu2+ + 2NO + 4H2O NO3- 3e- + 4H+ + NO3- → NO + 2H2O Before +5, after +2

Rate Law

3) rate of overall reaction A + B —> C + D Rate = k[A]^m • [B]^n Rate = rate of disappearance of reactants (M/time) • k = rate constant, or proportionality constant, depends on the specific reaction and temperature m = reaction order in terms of A n = reaction order in terms of B

What charge do Group 3A elements acquire as ions?

3+ (Tl can be 3+ or 1+)

What charge do Group 5A elements acquire as ions?

3- for non-metals and 3+ or 5+ for metals

Solutions of lead (II) nitrate and lithium phosphate are mixed. a. Is the above reaction a redox reaction? Justify your claim in terms of electrons. b. What are the spectator ions in the reaction? c. What is the oxidation number of lead before and after the reaction occurs?

3Pb2+ + 2PO43- → Pb3(PO4)2 No, it is not redox. The oxidation numbers on all substances remains the same indicating that no electrons are exchanged. NO3- and Li+ +2 before and after

Which subshell contains the electron with the lowest ionization energy in Al?

3p. -According to Coulomb's Law, force of attraction between two charged particles decreases as the distance between them increases, as distance is the denominator of the equation. -e- in n=3 shell are furthest from nucleus. When force of attraction decreases, energy required to remove e- also decreases. -Average distance between nucleus and e- decreases as subshells within a shell are added. Leads us to believe that ionization energy for a 3s e- would be less than 3p e- in same element (Coulomb's Law), but that is not the case. -e- density close to nucleus decreases for each subsequent subshell that is added to shell. Decreases in e- density close to nucleus increases shielding effect from e- in n=1 and n=2. -Increase in shielding effect has greater impact on force of attraction than Coulomb's Law when looking at subshells within a shell.

Which subshell contains the highest energy electron(s) in Na?

3s subshell. -aufbau principle, electrons enter orbitals of lowest energy first -e- in that subshell furthest from nucleus and experiences smallest force of attraction according to Coulomb's Law. -Attraction further reduced due to shielding effect from e-s in n=1 and n=2. -Bc of this, that e- is least stable.

Covalent Network Solids

= Always composed of 1 or 2 nonmetals held together by networks of covalent bonds = C group elements often form covalent networks solids, as they can form four covalent bonds = Very high melting points and normally very hard, as atoms r covalently bonded with fixed bond angles

triangleHfus for Ionic Compounds

= As ionic bonds r much stronger than intermolecular forces, the triangleHfus values for ionic compounds r very large = melting and boiling temperatures for ionic compounds r very high

PV/RT vs P for 1.0 mole of N2(g) at Different Temperatures

= As the T increases, behavior N gas approaches that of an ideal gas • Gas only behaves ideally at P below ~ 5 atm

Crystalline Solids

= Atoms, ions, or molecules r arranged in an orderly fashion that follows a pattern of repetition in 3D - Segments that repeat in 3D r called unit cells = Macroscopic structures usually have flat surfaces that make definite angles to one another = Quartz and ionic solids r crystalline

Maxwell-Boltzmann distribution, Temperature and Pressure

= Average KE of the particles in a system increases as the temperature increases - At any temperature, there is a large range of KE = At any given temperature, the particles with less KE exert a lower pressure and the particles with more KE exert a higher pressure - Total pressure exerted by the gas particles in a system in an average

Boiling Points of Water

= Boiling points decrease as elevation increases (higher altitude)

Metallic Solids

= Bonding is not covalent - Not enough e-s to fill octet = Bonding results from the attractions btwn nuclei and delocalized Ve- moving throughout the structure = Bond strength increases as # of bonding e-s increases

Properties of Ionic Solids - Cleave along planes

= Brittle 3D structure = Ions line up in a repetitive pattern that maximizes attractive forces and minimizes repulsive forces = Not malleable or ductile

Reactions Rates and Temperature

= Collision rates increase as temperature increases - When temp. increases, the average kinetic energy of the particles in that system increases • This increases the average velocity of the particles (KE = 1/2 mv^2) - The collision rate increases as velocity increases • This rule is amplified when the reactants r gases, but it's also true for other states - Increases in temp. increase the value of the rate constant, k, which the rate for any rate law

Reactions Rates and Surface Area

= Collision rates increase as the surface area of reactants in the solid phase increases - When u break something in half, the total surface area increases - If the object is ground into a powder, the total surface area of the particles is much larger than the surface area of the original object - When the surface area increases, a larger # of reactants in the solid phase r able to collide with the other reactants

Reactions Rates and Concentration

= Collision rates increases as the concentration of reactants in the liquid or gas phase increases - If there r more reactants in a liquid or gaseous system of the same volume and temperature, there will be more collisions btwn reactants per unit of time - Reaction rates increase when collision rates increase for all processes that r higher than 0th order rate = k rate = k[A] rate = k[A]^2

Kinetic Molecular Theory 4

= Collisions experienced by gas particles r elastic • KE is conserved

Explain how absorption of electromagnetic radiation occurs in infrared (IR) spectroscopy experiments

= Different types of covalent bonds in molecular vibrate at different frequencies, which r within the IR spectrum, and IR radiation of the exact same frequency is absorbed by those covalent bonds

Avogadro's Principle

= Equal volumes of different gases at the same temperature and pressure contain equal numbers of particles • Each balloon contains different types of gaseous particles at 0*C and 1 atm • However, they all contain 6.7x10^-2 moles of gaseous particles

Explain how Einstein was able to use experimental evidence related to the photoelectric effect to conclude that quantized energy must be contained by individual particles

= E-s don't eject until the threshold frequency is reached, even after prolonged exposure • This would lead us ti believe that there is a one to one relationship btwn the e- and another particle • If it was not one to one, e- would slowly absorb enough energy to break free from the metal's surface = Once the threshold frequency is reached, ejections take place immediately • This further supports the idea that an e- absorbs a single particle containing a set quantity of energy • If they slowly absorbed continuous energy, ejections wouldn't take place immediately = At a set frequency, velocity of all ejections is the same; however, increasing the frequency increases the velocity of the ejections • Further supports the one to one relationship of e- to photon = For there to be a one to one relationship, photons must be particles

Covalent Network Solids (Graphite)

= Each C forms 3 sp^2 hybrid orbitals that bind with 3 other C atoms = These sheets sit on top of one another = Delocalized pi-bonds btwn sheets = Weak Pi-bonds and LDF allow sheets to slide over one another = If hooked up to a potential difference, e-s will flow = High melting point, as covalent bonds btwn C in each layer r relatively strong

Conservation of Energy

= Energy can be neither created not destroyed, but it can be transformed from one form to another - Energy of a system changes through: • chemical reactions, phase changes, or heating and cooling - Results in energy being transferred into or out of the system in the form of heat or work

What factors r necessary in order to cause an e- transition during an ultraviolet/visual (UV/Vis) spectroscopy experiment?

= Energy contained by the incident photons must be exactly the same as the difference in energy btwn the 2 MOs in order for the photons to be absorbed by e-s and transmit the e-s to the higher energy MO • Molecule must have Pi-orbitals (which means that it must contain double or triple bonds within its structure) or it must have non-bonding MOs (which means it must contain MOs with LP) • Light btwn the wavelength of 200 nm and 800 nm can only transmit e-s from pi (bonding orbitals) to Pi+ (anti-bonding orbitals), or from non-bonding orbitals to a pi+ (anti-bonding orbitals) or sigma+ (anti-bonding orbitals)

Species with more electrons and larger electron clouds are more polarizable

= Have a weaker hold on their outer electrons = When moving down a group, or constructing molecules with more atoms, the resulting species has more e-s - More e-s a species has, the more polarizable it is - This is not always true when moving from left to right across a period, as atomic radius and therefore the size of the e- cloud decreases

Limiting Reactant vs Excess Reactant

= If u have set quantities of 2 different reactants, one will get used up and some amount of the other will be leftover - Limiting • reactant that used up limits how far the reaction will proceed - Excess • reactant that is leftover when reaction is complete

Explain how solutes can be separated through paper chromatography based in intermolecular forces

= In paper chromatography, the bottom of a strip of chromatography paper is placed in the solution in question • Solution is drawn up the paper due to capillary action • stationary phase is the chromatography paper, and the moving phase is the solvent used. • As solvent moves up the piece of paper it carries solute particles with it • However, distances that the different solute particles travel up the paper depends on their relative attractions for the moving solvent and the stationary paper = Paper is made of cellulose fibers, which have polar and non-polar components • Cellulose fibers have non-polar C chains with polar -OH groups that stick out at various locations along the structure • Solutes that r considered to be "most polar" can form H-bonds at several locations on their structures • These structures will have a greater attraction for the stationary paper than they do for the mobile mostly non-polar solvent = If there r strong intermolecular FoA btwn the solvent and solute, it will travel a fairly far distance up the paper • If there t weak intermolecular FoA btwn the solvent and solute, solute will fall out of solution and cling to the paper fairly close to the solution's surface • All identical solute particles will fall out if the solution and be deposited on the paper at the same height, but this height will be different for different types of solute particles • As the different solute particles r often different colors, experiment produces different colored stripes at different heights above the surface of the solution • Pure solvent will always travel further up the paper than any of the solutes it originally contained

Molecular Shape and Dispersion Forces

= Molecular shape plays a role in the strength of LDF and physical state - Saturated fat, straight, solid at room temp - Unsaturated fat, bent, liquid at room temp = Dispersion forces increase as contact area btwn molecules increases

Molecular Solids

= Molecular shape plays a role in their physical state - Saturated and unsaturated fat = In a solid, molecules r held close together in a regular pattern by intermolecular forces that attempt to maximize attractions and minimize repulsions

Molecular Liquids

= Intermolecular forces also attempt to maximize attractions and minimize repulsions = These forces r strong, but not as strong as they r in a solid, so the molecules have more freedom to move

Vapor Pressures of Ionic Solids

= Ionic compounds have very low vapor pressures and very high boiling points - Strong Coulombic interactions btwn cations and anions

Would it be possible to conduct an UV/Vis spectroscopy experiment on a solution containing a colored compound? Justify

= It is possible to conduct UV/Vis spectroscopy experiments on solutions containing colored compounds, as all colored compounds absorb light in the visual spectrum

Would it be possible to conduct an UV/Vis spectroscopy experiment on a solution containing a colorless compound? Justify

= It is possible to conduct UV/Vis spectroscopy experiments on solutions containing colorless compounds, as some colorless compounds absorb light in the ultraviolet spectrum

Indications of Energy Changes

= Macroscopic indications that an energy change has occurred include: • Changes in temp. - Ex. After mixing 2 solutions that r same temp., temp. of final solution is either higher or lower than original solutions - Ex. Heating or cooling a substance changes its internal energy • Changes in volume - Ex. Volume of gaseous system incr. or decr.

Covalent Network Solids (Diamond)

= Many C atoms bound together with sp^3 hybrid orbitals = Each C makes a single covalent bind with 4 other C atoms = Very hard and very high melting point (3550 Degrees C)

Properties of Molecular Solids

= Most don't conduct electricity when molten or dissolved in water - Individual molecules have no net charge, as Ve- r tightly held within covalent bonds and LP - Acids r molecules that can ionize and conduct electricity = Most molecular solids r held together by intermolecular forces, which r much weaker than ionic or covalent bonds - Have much higher vapor pressures than ionic solids - Have much lower melting and boiling points than ionic solids

Properties of Ionic Solids - Solubility and Conductivity

= Most r soluble in polar solvents = Conduct electricity only when molten or dissolved in a polar solvent, as the charged particles r free to move - Higher the concentration of ions in a solution, higher the electrical conductivity

Particulate Characteristics of Solids

= Motion of individual particles is limited - Particles don't undergo translation with respect to one another = Particles r very close together = Structures r held together by intermolecular forces and/or chemical bonds = Structure is influenced by the ability of the particles to pack together

Net Ionic Equation vs Complete Ionic Equations

= Net - only shows the reaction that actually took place - Normally used when dealing with a problem that only involves the precipitation reaction = Complete - Shows all of the species that r present - Normally used when it is necessary to identify all of the species in a system (reacting species and spectator ions)

Metallic Bonding - Electron Sea Model

= Nuclei and inner core e-s r localized, while Ve- r free to move throughout the solid 1. Conduct electricity 2. Conduct Heat 3. Malleable and Ductile - Lack directional bonds

Redox Titrations

= Often used to determine the concentration of a species that can be oxidized in a solution • A solution with a known concentration of a species that is easily reduced is added to a solution with an unknown concentration of a species that will be oxidized in the reaction • A color change signals the arrival at the endpoint, which is extremely close to the equivalence point = Equivalence Point • Occurs when equal # of moles of the species being oxidized and the species being reduced have reacted = Titrant • + to solution

Protein Function

= Overall surface shape determines function = Groove creates the opportunity for a protein to interact with other molecules = Enzymes break down other molecule through this type of interaction = Water soluble proteins have polar 'R' groups that face out and non-polar 'R' groups that face in - H-bonds form with water - LDF between side chains on the inside

Explain why bubbles of CO2 (g) form when u open a bottle of soda

= Partial P of CO2(g) in the neck of the bottle is 4 atm. • If the system is at 25'C, it will maintain a constant concentration of 0.1 M CO2 in the solution • When the lid is removed, the equilibrium shifts • The system then contains more aqueous CO2 than it will allow with the reduced partial P of CO2 (g) on the surface of the liquid • In response to this, CO2 (g) bubbles form and the excess CO2 starts to leave the solution = If u leave the lid off the bottle for long enough, system will re-establish its equilibrium with a smaller concentration of aqueous CO2 (9.9x10^-6 atm)

Properties of Liquids

= Particles r constantly moving and colliding with one another - Particles undergo translation with respect to one another - Movement is influenced by the strength of the intermolecular forces that r present btwn the particles and the Temp = Particles r very close together

Explain what happens within the structure of a molecule during an ultraviolet/visual (UV/Vis) spectroscopy experiment

= Photon absorbs an e- for a pi (bonding orbital) and transmits it to a pi+ (anti-bonding orbital); or a photon absorbs an e- from a non-bonding MO and transmits it to a pi+ (anti-bonding orbital or sigma+ (anti-bonding orbital) • Photon must contain the same amount of energy as the difference in energy btwn the 2 MOs

Synthetic Polymers - Polyethylene

= Plastic r non-polar, so they r held together by LDF = LDF hold chains together

Properties of Synthetic Polymers

= Plastics r generally flexible solids or viscous liquids = Heating plastics increases flexibility and allows them to be molded - Molecular vibrations increase - LDF weaken = Properties of synthetic polymers can be modified by manipulating their structures

Pi-bonds and Polarizability

= Presence of pi-bonds increases polarizability = e-s in pi-bonds r more delocalized, and thus, have more freedom to move and assist with the polarization of the species

ethanol, C2H4O, and acetic acid, CH3COOH • Identify types of intermolecular forces that exist in a pure sample of acetic acid • Highest boiling point??

= Pure acetic acid has LDF and H-bonds • also have some ion-dipole = Acetic acid, LDF in both structures r similar, but acetic acid forms H-bonds which r much stronger than the dipole-dipole forces btwn ethanol molecules • as greater amount of energy is required to break stronger FoA, the boiling pt of acetic acid is higher

Amorphous Solids

= Random arrangement of particles = Particles have no orderly structure = Macroscopic structures lack well defined faces and shapes = Many r mixtures of molecules that don't stack up well together = Glass and rubber r examples of amorphous solids

Reaction Orders (m and n)

= Reaction orders cannot be predicted from the balanced equation = There r only found through experimental data = Reaction orders may be: • Zero Order • First Order • Second Order meaning that m and n may be 0, 1, or 2

Reaction Rates r Influenced by Collision Rate and Catalysts

= Reaction rates increase in the presence of a catalyst = Reaction rates increase when collision rates increase = Collision rates increase 1) as the concentration of reactants in the liquid or gas phase increase 2) as the surface area of reactants in the solid phase increases 3) as temperature increases = Not every collision triggers a chemical reaction

Electrolysis of Aqueous CaI2

= Reduction could be + ion or water Ca 2+ + 2e- —> Ca E^o (red) —> -2.87 V 2H2O + 2e- —> H2 + 2OH- E^o (red) —> -0.83 V • species with most +, least -, reduction potential reduced = Oxidation could be - ion or water 2I- —> I2 + 2e- E^o (red) —> +0.53 V 2H2O —> O2 + 4H+ + 4e- E^o (red) —> +1.23 V • species with most +, least -, oxidation potential is oxidized

Charles' Law

= Relationship btwn Temperature and Volume of Gases V1/T1 = V2/T2 - V is proportional to Temp - must use absolute Temp (K)

Photoelectric effect

= Science world knew that certain clean metal surfaces would shed e-s when certain frequencies of light were shined on them • 1st Fact: Highly intense low frequency light doesn't eject any e-s, even if it shines on the surface for several days • 2nd fact: When threshold frequency is reached, e-s r ejected immediately • 3rd fact: Increasing the intensity of the light at a frequency that will cause e-s to eject results in a higher ejection rate. However, all rejected e-s share the same velocity • 4th fact: Increasing the frequency of light increases the velocity of the ejected e-s. However, all ejected e-s share the same velocity

Sublimation

= Solids can evaporate and have a vapor pressure = As intermolecular forces r stronger in solids, the vapor pressures of solids r normally low = Solids with high vapor pressures, have relatively weak intermolecular forces

Gas solubility and temperature

= Solubility of most gases decreases as temperature increases • Gases tend to have weak intermolecular forces - N2 and O2 form weak dipole-induces dipole forces and weak dispersion forces with water • As the kinetic energy of particles within a solution increases, aqueous particles break free from these weak attractions and re-enter the gas phase

One of the factors that influences the rate of any chemical reaction is the collision rate of the reactants a) Identify the variable that influence the rate at which reactants collide and explain how each of these variable can increase the rate of a chemical reaction

= Temperature - is a measure of the average KE of the particles in a system. When temp. incr., average KE of reactants incr.. KE = 1/2 mv^2, so as average KE incr. average velocity of the reactants also incr.. When velocity of reactants incr., # of collisions per unit time incr. = Concentration of reactants - When more reactants per unit vol., and temp. remains same, # of collisions per unit time incr. = Surface Area of reactants - As particle size incr., SA decr.. When break something in half, total SA incr.. If object is ground into powder, total SA of particle is much larger than the SA of the original object. When SA incr., larger # of molecules in solid phase r able to collide w/ the other reactants per unit of time

PE, Polarity, and Strength of Intermolecular Forces

= The PE of e-s associated w/ - pole of a molecule decreases as approach + pole of another molecule = Molecules w/ stronger dipoles have stronger attractions for one another, which pull them closer together - PE decreases as molecules move closer together = Energy must be added to weaken or break intermolecular forces - PE increases as molecules move away from one another

Kinetic Energy of Gas Molecules

= Transitional Energy - Gas molecules move through space in straight lines = Rotational Energy = Vibrational Energy = Most of a gas particle's KE is related to its translational velocity

If u were to design an experiment to find the concentration of a compound in solution, would u choose to use an infrared (IR) spectrometer or an ultraviolet/visual (UV/Vis) spectrometer? Justify

= Ultraviolet/visual (UV/Vis) spectrometer • If molar absorptivity, a, of the compound at the wavelength being used in the experiment is known, one can use an UV/Vis spectrometer to measure the absorbance, A, of a sample of the solution • Concentration of the solutions can then be calculated using the Beer-Lambert law equation: A = Ebc • In this equation, 'b' is the oath length of the curvette, which would be known, and 'c' is the concentration of the solution

If u were to design an experiment to determine the differences btwn some of the energy levels in a pure sample of a molecular compound, would u choose to use an infrared (IR) spectrometer or an ultraviolet/visual (UV/Vis) spectroscopy? Justify

= Ultraviolet/visual (UV/Vis) spectrometer • In this, photons cause e-s to more from lower energy MOs to higher energy MOs • For this to happen, the difference btwn the energy levels of the 2 MOs must be exactly the same as the energy contained by the photon • After one had determined the frequencies of electromagnetic radiation that r being absorbed by the sample, the energy contained by the photons in that electromagnetic radiation can be calculated using E=hv

Boiling Points of Different Liquids

= Vapor Pressure on the surface of a liquid depends on the strength of its intermolecular forces - A molecule restrained by strong intermolecular forces requires more energy to break free from the liquid state - When a system requires more energy to cause its molecules to enter the gas phase, it will also require more energy to cause its vapor pressure to = the atmospheric pressure

Biomolecules - Protein Structure

= Very long chains can be constructed through the reaction from the previous slide = Chains may contains hundreds of amino acids = H-bonds form btwn Os, and Hs that r bonded to Ns, within the same chain = Following secondary structures can form - a-helix structure is held together by H-bonds - B-pleated sheet structure is also held together by H-bonds = Tertiary structures r caused by intermolecular interactions btwn R groups = Quaternary structures r caused by intermolecular interactions btwn different chains

Properties of Ionic Solids - Strong Bonds

= Very strong Coulombic FoA btwn cations and anions - High melting points - Very hard - Low volatility

Beverages and Pollution

= Warm carbonated drinks go flat faster than cold carbonated drinks = Thermal Pollution • During many industrial processes, water is pumped out of lakes or rivers and used to cool equipment, gases, or other liquids • Heat flows into this water • Then returned to the lake or river that it came from • Temperature of that body of water increases, (O2 (aq)) drops, and the fish die

Orientation and Dipole-Dipole

= When + and - dipoles line up well, attractive forces r stronger and repulsive r weaker = When + and - dipoles do not line up well, attractive forces r weaker and repulsive r stronger

Vapor Pressure

= When molecules leave the surface of a liquid to enter the gas phase, they exert a pressure = Vapor pressure exerted depends on the rate of evaporation per unit area of the liquid's surface = Rate of evaporation and vapor pressure increase as temperature increases = When 2 substances r at the same temperature, rate if evaporation and vapor pressure will be higher in the substance that has weaker intermolecular forces

Mass % Element Mass % Component

= [(#of atoms of element)(element's atomic mass)] / [(Formula weight of compound)] x100 = (mass of component in question) / (total mass of substance) x100

Explain why every element has a distinct atomic emission spectrum

= amount of energy possessed by the e-s of an element is fixed according to its principle energy levels and sublevels • Change in energy that an e- experiences when it drops from a higher to a lower energy level is also fixed • This means that the energy possessed by a photon that is emitted when an e- drops from a higher to a lower energy level is also an exact quantity • As the energy levels possessed by different types of elements r different (due to different nuclear charges and different #s of energy levels and sublevels), the changes in energy that the e-s experience as they drop from higher to lower energy levels r also different • This means that the energies contained in the photons emitted by different elements r different, so the wavelength (color) the light is also different

dipole-induced dipole

= attractions resulting from these forces r stronger when magnitude of dipole in polar molecule is larger - Molecules that have larger dipoles have greater ability of induce a larger dipole in a non-polar molecule = Strength of these forces increases when non-polar molecule has larger e- cloud and is more polarizable

What is the difference btwn a continuous spectrum and an atomic emission spectrum?

= continuous spectrum contains every possible wavelength • Rainbow is an ex. of a continuous spectrum of visible and each color blends into the next • Color gradually changes as the wavelength of light changes • If rainbows were missing certain wavelengths of light, they would have transparent bands that r devoid of color • Every element has a distinct atomic spectrum and in an atomic spectrum, only certain wavelengths of light r present, so atomic spectrums don't possess every possible wavelength of light

Would it be possible to conduct an UV/Vis spectroscopy experiment on a sample of the compound? Justify H-C=C=C-OH II I O H

= would be possible to conduct a UV/Vis spectroscopy experiment on a sample of this compound, as it has pi-orbitals associated with its double bonds and non-bonding MOs (LP) • Light btwn the wavelength of 200 nm and 800 nm can only transmit e-s from pi (bonding orbitals) to pi+ (anti-bonding orbitals), or from non-bonding orbitals to a pi+ (anti-bonding orbitals) or sigma+ (anti-bonding orbitals)

Atomic mass =

=(a1)(%1) + (a2)(%2) +...

Electronegativity: Reasoning

=Electronegativity increases as atomic radius decreases -When 2 atoms r involved in a chemical bond, bonding e-s will experience a greater attraction for the + nucleus that is closest to them.

First Ionization Energy: Reasoning

=Generally increases as atomic radii decrease -According to Coulomb's Law, force of attraction decreases as distance between e-s and protons increases

Ionic Bonding Defined

=Metals transfer e-s to non-metals, and 2 form bonds due to electrostatic attractions between them (Coulomb's Law) -or... =Cations and anions form electrostatic bonds based on opposite charges. Cations and anions may be polyatomic

Ionic Radius: Cations (Across Period from Left to Right)

=Radii of cations decrease -# of e-s remains same and protons added -Shielding effect experienced by outer e-s of cations in period similar, so addition of protons increases effective nuclear charge on those e-s. -According to Coulomb's Law, increase in + charge increases force of attraction experienced by each e-, thereby decreasing radius

First Ionization Energy: Exceptions

=Requires more energy to pull up e- from full d-subshell than it does to pull the only e- in a p-subshell (when in same period). =Requires more energy to pull up e- from full s-subshell than it does to pull the only e- in a p-subshell (when in same period). -That single e- is easier to remove, bc only e- in higher energy subshell, and being "shielded" by electrons in lower energy shells

Electron Affinity: (Across Period from Left to Right)

=Ve- r in same shell -Shielding effect experienced by outer e-s of elements in a period similar so addition of protons increases effective nuclear charge of those e-s =More protons r added and atomic radii decrease. -According to Coulomb's Law, increase in + charge and decrease in atomic radii increase force of attraction experienced by added e-

Atomic radius (Across a Period from Left to Right)

=Ve- r in same shell -Shielding effect experienced by outer e-s of elements in period r similar so addition of protons increases effective nuclear charge on those electrons =Moving from left to right, more protons are added. -According to Coulomb's Law, increase in + charge increases force of attraction experienced by each e-, thereby, decreasing radius.

Ionic Radius: Anions (Across Period from Left to Right)

=When u encounter 1st anion, radius increases dramatically -Repulsions from additional e-s cause e- cloud to expand =Radii then continue to decrease -Shielding effect experienced by outer e-s of anions in period similar so addition of protons increases effective nuclear charge on those e-s -According to Coulomb's Law, increase + charge increases force of attraction experienced by each e-, thereby decreasing radius

Multistep Reaction Energy Profiles

A + B —> AB (Step 1) AB + CD —> ABC + D (step 2) === A + B + CD —> ABC + D (overall) Step 1 A + B —> AB Step 2 AB + CD —> ABC + D

Kinetics

A + B —> D + E 1) Rate of disappearance of a reactant Rate(A) = -delta[A] / delta t (- sign needed, as rates r always + values) 2. Rate of appearance of a product Rate(D) = delta[D] / delta t = rate at which 1 species appears or disappears can be used to find other rates aA + bB —> dD + eE

A 0.57 M solution of propanoic acid, HOC6H5, 0.0684% of acid has ionized. What conc. of HBr would produce a solution with the same pH as 0.57 M solution of propanoic acid, HOC6H5? Justify

A 3.9x10^-4 M solution of HBr would have the same pH as a 0.57 M solution of propanoic acid. Bc HBr experiences 100% ionization pH = -log[H3O+] = -log(3.9x10^-4 M) = 3.41

A 0.45 M solution of propanoic acid, HC3H5O2, experiences 1.58% ionization. What conc. of HCl would produce a solution with the same pH as 0.45 M solution of propanoic acid, HC3H5O2? Justify

A 7.1x10^-3 M solution of HCl would have the same pH as a 0.45 M solution of propanoic acid. Bc HCl experiences 100% ionization pH = -log[H+] = -log(7.1x10^-3 M) = 2.15

Label acid, base, conjugate acid, and conjugate base d) HF + H2O <—> H3O+ + F-

A B CA CB

Concentration Cells

A voltaic cell can function with same species at anode and cathode • E^o (cell) = O, but • E(cell) will be +, if Q < 1, • When Q < 1, [anode] < [cathode] • As reaction proceeds, con. at anode increases and conc. at cathode decreases • when [anode] = [Cathode], E=O • deviations from standard conditions where Q < 1 will increase the cell potential • When Q < 1, [anode] < [cathode] • As reaction proceeds, conc. at anode increases and conc. at cathode decreases • When [anode] < [cathode], system at equilibrium and E = O • e-s always flow from anode to cathode

Use Coulomb's law to explain why this electron has the lowest ionization energy.

According to Coulomb's Law, F=k (q1q2)/d^2, the force of attraction between 2 charges particles decreases as the distance between them increases, as distance is the denominator of the equation. As you can see from the shell model above, the e- that was identifies as having the lowest ionization energy is the furthest from the nucleus. When the force of attraction decreases, the energy required to remove the e- also decreases.

K(a) for acetic acid is 1.8x10^-5, and K(a) for hypochlorous acid is 3.5x10^-8 at 25'C. If 500.0 mL of 1.0 M acetic acid was mixed with 500.0 mL of 1.0 M hypochlorous acid, which CB would have highest conc.? Justify

Acetic acid is stronger, as it has a larger K(a) value • Acid strength increases as K(a) increases • Larger K(a) value, further to the right the equilibrium position (more products) • For this reason, [CH3COO-] > [ClO-]

Suppose a system operating in accordance with the chemical equation below is in a state of equilibrium at 25'C. In which direction will the reaction shift when the temperature is increased to 75'C? CO(g) + NO(g) <-> CO2 (g) + 1/2 N2 (g) ^_H(rxn) = -373 kJ

Adding heat will cause the reaction to shift in the endothermic direction, to the left in this case, as this will consume some of the energy

NO2 (g) is a reddish-brown color and N2O4 (g) is colorless. Suppose the 2 gases establish the following equilibrium NO2 (g) <-> N2O4 (g) ^_H' = -57.2 kJ If temperature increased from 25'C to 45'C, and the volume remained the same, what would happen to the overall color of the gaseous system? Justify

Adding heat will cause the reaction to shift to the left (endothermic direction) as that will consume some of the energy • This will increase the concentration of NO2, which will cause the system to become a darker reddish-brown color

Suppose a system operating in accordance with the chemical equation below is in a state of equilibrium. In which direction will the reaction shift when the heat is added to the system? 2N2O5 (s) <-> 4NO (g) + 3O2 (g) ^_H(rxn) = +247.4 kJ

Adding heat will cause the reaction to shift to the right • Adding heat will cause a reaction to shift in the endothermic direction, as this will consume done of the added energy

2 different 1.2 L buffered solutions were prepared using HOBr and LiOBr. Both buffered solutions had pH of 5.2 at 25'C. After 0.17 moles of HI were added to each solution, found that pH of one solution had dropped to 4.9 and pH of other had dropped to 3.1 c) Explain was pH of 2 solution ended up being different after same amount of HI was added to each

After = amounts HI added to both solutions, ratios of K(a):[H+] changed by a greater degree in the solution that ended up with pH = 3.1 than did in solution ended up with pH=4.9 • Solution ended up with pH=3.1 obtained a higher conc. of H+ • For this to happen, must have lower conc. of OBr- than did solution that ended up with higher pH • Since ratios were originally the same, solution ended up with a pH=3.1 also had lower conc. of HOBr when shared same pH

Two separate pure samples of CO2 were analyzed. Both samples were found to contain 27.29% C by mass. Justify these findings on the basis of atomic molecular theory.

All CO2 molecules contain exactly 1 C and 2 O atoms. Although different isotopes of C and different isotopes of O have different masses, the ratio of isotopes of C and the ratio of isotopes of O in any pure sample of CO2 is constant. If we could calculate the average mass of all the atoms of an element in a pure sample, we would obtain the average atomic mass of that element, which is given on the periodic table. As the ratio of C atoms to O atoms in both pure samples is the same, the ratio of the masses of C atoms to O atoms in that pure sample will also be the same.

Solubility Rules

All salts containing: Group IA, NH4+ (Ammonium), NO3- (Nitrate), CH3COOH- (Acetate)

On average, e- from the 2s subshell of a element are further from the nucleus than e- from the 2p subshell of an element; however, the energy required to remove an e- of an element from 2s is greater than that of 2p. Explain!

Although the average distance between the nuclues and an e- in 2s is greater than that of an e- in 2p, electrons in 2s spend some of their time much closer to the nucleus than those in 2p. As a result, e- in 2p experience a greater shielding effect. The greater repulsive forces that the 2p e- experience from the 1s e- result in a lower activation energy for the 2p e-.

What is the only polyatomic cation you need to know??

Ammonium (NH4 +)

Electronegativity

An elements ability to attract e-s in a chemical bond

Pure sample

Contains particles of one type of atom, molecule, or formula unit (e.g. an ionic solid such as NaCl)

Which atoms has the smallest radius? Na vs Ar. Justify

Ar, bc Ve- of both elements r in same shell (n=3). Thus, shielding effect experienced by outer e-s of both elements similar -Na has 11 protons in nucleus, as it is neutral and has 11 e-s. Ar has 18 protons in nucleus, as it is neutral and has 18 e-s. -Addition of protons increases effective nuclear charge on Ve-s -According to Coulomb's Law, increase in + charge increases force of attraction experienced by each e-, thereby decreasing radius. This gives Ar a smaller atomic radius.

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces e. He and Ar

Ar, bc both LDF but Ar is larger and has more e-s, making it more polarizable than He • Ar experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of Ar is higher

Summary of EMF under non-standard conditions and equilibrium

As cell approaches equilibrium, magnitude of cell potential decreases until reaches 0 at equilibrium where Q = K • Deviations from standard conditions where Q < 1 will increase cell potential • Deviations from standard conditions where Q > 1 will decrease cell potential • Deviations from standard conditions where Q = 1 will not change cell potential

Thermodynamically Unfavored Process

Assistance from outside system is necessary to induce desired change - H2O doesn't bool at 75^o and 1 atm - H2O doesn't freeze at 15^oC

AB (g) <-> A (g) + B (g) Suppose a chemist adds 1 mole of pure AB (g) to a sealed vessel b) What can be said about the concentrations of AB(g), A(g), and B(g) when the system is at equilibrium?

At equilibrium, each species ends up maintaining a specific concentration • Equilibrium doesn't mean that all species have equal concentrations

A 55 mL solution of 0.15 M HF is titrated with 0.30 M NaOH c) Is pH at equivalence pt for situation (28mL) greater than, equal to, or less than that for the situation (14 mL)?? Justify

At equivalence pt, only species that will have an effect on pH is basic F- • F- will react with water to produce OH- according to equation F- + H2O —> HF + OH- # moles OH- produced will be about the same in both situations • Conc. OH- ions in 2 situations will be different, as overall vol. of final solutions r different • Situation (a) has a larger vol. [55 mL + 28 mL = 83 mL] than situation (b) [55 mL + 14 mL = 69 mL] • (b) has higher pH as it's a higher conc. of OH- ions

equivalence point

At the equivalence point, the number of moles of titrant added is equal to the number of moles of analyte that were originally present

The __________ given to you on the periodic table is an average of all the isotopes of that element. The __________ of any large number of atoms of a given element is always the same for that given element. The ratio of isotopes for each given element is constant.

Atomic mass Average mass

Expanded Octets

Atoms in period 3-7 can bond with other atoms in such a way that they end up with more than 8 e-s in their octets -this is bc the d-orbitals r empty which means they can be used during bonding

Label acid, base, conjugate acid, and conjugate base c) H2PO4- + H2S <—> H3PO4 + HS-

B A CA CB

Label acid, base, conjugate acid, and conjugate base e) CN- + NH4+ <—> HCN + NH3

B A CA CB

Write balanced net ionic equations for reactions Solid Beryllium carbonate is placed in a solution of 0.5 M hydrofluoric acid

BeCO3 + 2HF —> Be 2+ + H2O + CO2 + 2F-

Is H2O polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Bent so 2 H-O bonds and 2 lone pairs - Polar individual bonds bc O is more electronegative than H - Polar overall compound bc dipoles don't cancel due to asymmetrical arrangement of 2 O-H bonds - O end is slightly - and H ends r slightly +

Enthalpy Change (^_H) Hydrogen Fuel for ur Car

Burning Hydrogen Fuel ^_H = -# kJ Making Hydrogen Gas ^_H = # kJ

Bonds btwn the carbon atoms in C2H6 and C2H4 b) Which C-C bond contains the least amount of PE? Justify

C-C bond in C2H4 contains the least amount of PE. PE of Ve-s decreases as they approach the nucleus of another atom. Since the bond length btwn the C atoms in C2H4 is shorter, that bond contains less PE than the bond btwn the C atoms in C2H6

Bonds btwn the carbon atoms in C2H6 and C2H4 c) Which C-C bond requires the greatest input of energy in order to be broken? Justify

C-C bond in C2H4 requires the greatest input of energy in order to be broken. PE of Ve-s decreases as they approach the nucleus of another atom. The same amount of energy that is evolved when 2 atoms approach one another to form a bond must be added in order to separate those atoms and break that bond. Since the bond length btwn the C atoms in C2H4 is shorter, a greater amount of energy was released when the bond was formed, and a greater amount of energy must be added in order to break that bond

Which is most bond in each set? Explain a. C-F or C-O b. P-O or P-F c. As-S or As-F

C-F; P-F; As-F; greater electronegativity difference, the more polar the bond

explain why propane, C3H8, is a gas and decane, C10H22, is a liquid at 25'C and 1 atm

C10H22, both non-polar and only LDF but C10H22 is larger and has more e-s, making it more polarizable than C3H8 • C10H22 experiences greater LDF • Bc of stronger FoA, C10H22 exists as a liquid at 25'C and 1 atm, while C3H8 exists as a gas

C2H2 or C2H6 H-C=_C-H or 2 C surrounded by 6 H w/ single bonds a. What is the bond order for the C-C bond in each structure

C2H2 = 3 C2H6 = 1

C2H2 or C2H6 H-C=_C-H or 2 C surrounded by 6 H w/ single bonds c. Which structure has the greatest C-C bond energy? Justify

C2H2 bc has single bonds - triple bonds have greater e- densities between bonding atoms than single bonds - Greater e- density between bonding atoms increases force of attraction between protons of each nucleus and centrally located bonding e-s which shortens bond length - Force of attraction is greater, energy required to break bond is also greater

which species has the highest boiling point? C2H6 or C3H8

C3H8 bc although both non-polar (LDF), larger molecule has higher boiling point bc more polarizable

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces n. C2H6 or C4H10

C4H10, bc both LDF but C4H10is larger and has more e-s, making it more polarizable than C2H6 • C4H10 experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of C4H10 is higher

Conc. of C5H5NH+ as 1.8x10^-5 M in 0.25 M C5H5N solution at 25'C and conc. of C6H5NH+ is 1.2x10^-4 M in a 0.25 M C6H5NH2 solution at 25'C. Identify stronger base. Justify

C6H5NH2 is stronger base, as accepts a greater # of protons from water

explain why ethane, C2H6, melts at -183'C and nonane, C9H20, melts at -54'C

C9H20, both non-polar and only LDF but C9H20 is larger and has more e-s, making it more polarizable than C2H6 • C9H20 experiences greater LDF • Bc of stronger FoA, C9H20 exists as a liquid at 25'C and 1 atm, while C2H6 exists as a gas

CB stability through resonance

CB of 3 of 6 strong acids - H2SO4, HNO3, HClO4 - r stabilized by resonance due to presence of 1, 2, or 3 double bonds btwn oxygen and central atom

Identify strongest acid. Justify based on molecular structure and electronegativity a) CH3COOH or CCl3COOH

CCl3COOH bc Cl is more electronegative than H • Conc. of highly electronegative elements at opposite end of structure from H that is donated in CCl3COOH indicated that electron density btwn H and O in CCl3COOH is less than that in CH3COOH • Bc electron density btwn H and O in CCl3COOH is less than it is in CH3COOH, FoA holding in CCl3COOH is also less

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces d. CH4 and CCl4

CCl4, bc both LDF but I2 is larger and has more e-s, making it more polarizable than CH4 • CCl4 experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of CCl4 is higher

Identify strongest acid. Justify based on molecular structure and electronegativity a) CH2ClCOOH or CHCl2COOH

CH2ClCOOH bc Cl is more electronegative than H • Conc. of highly electronegative Cl atoms at opposite end of structure from H that is donated in Cl is more electronegative than H • Conc. of highly electronegative elements at opposite end of structure from H that is donated in CH2ClCOOH indicates that electron density btwn H and O in CH2ClCOOH is less than that in CH2ClCOOH • Bc electron density btwn H and O in CHCl2COOH is less than it is in CH2ClCOOH, FoA holding onto H in CHCl2COOH is also less

Identify strongest acid. Justify based on molecular structure and electronegativity a) CH3CHCl2COOH or CH3CH2ClCOOH

CH3CHCl2COOH bc Cl is more electronegative than H • Conc. of highly electronegative Cl atoms at centre of structure in CH3CHCl2COOH indicates that electron density btwn H and O in CH3CHCl2COOH is less than that in CH3CHCl2COOH • Bc electron density btwn H and O in CH3CHCl2COOH is less than it is in CH3CH2ClCOOH, FoA holding onto H in CH3CHCl2COOH is also less

Equal vol. of 1.0 M NaCH3CO2 and 1.0 M HCl r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

CH3CO2- + H3O+ —> CH3COOH + H2O • System contains equal # moles NaCH3CO2 and HCl. React in 1:1 mole ratio and reaction goes to completion • aqueous species that have the highest conc. at equilibrium r Na+ (spectator ion), Cl- (spectator ion), and acetic acid • Conc. of spectator ion will be slightly higher than that of the acetate acid

Solutions of acetic acid and sodium fluoride r poured into a beaker

CH3COOH + F- <-> CH3COO- + HF

Equal vol. of 0.25 M CH3COOH and 0.25 M KOH r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

CH3COOH + OH- —> CH3COO- + H2O • System contains equal # moles CH3COOH and KOH. React in 1:1 mole ratio and reaction goes to completion • No an excess reactant and aqueous species that have the highest conc. at equilibrium r the acetate ion (CB of acetic acid) and K+ (spectator ion) • Conc. of spectator ion will be slightly higher than that if the acetate ion • As acetate is a base, it will accept some protons from water to form acetic acid

Required to create a buffer solution where acid and its salt have very similar conc.. select weak acid and its salt use to create buffered solution with pH of 4.70 HCN = 6.2 x 10^-10 HOCl = 3.5 x 10^-8 CH3COOH = 1.8 x 10^-5 HF = 7.2 x 10^-4

CH3COOH bc pK(a) = 4.74 and desired pH is 4.7 • Since fairly equal conc. of a weak acid and its salt, it's necessary to choose acid where pK(a) = pH • Equation demonstrated how pK(a) = pH when conc. r equal, as log(1) = 0 pH = pK(a) + log [A-]/[HA] NaCH3CO2 or another soluble salt containing CH3CO2- ion would be used to prepare this buffer

Explain why standard enthalpy of vaporization, DeltaH(vap), values for each set of compounds below rn't the same a. CH4 and H2O

CH4 has LDF and H2O has LDF and H-bonds • bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Explain why the standard enthalpy of vaporization values for each set of compounds below rn't the same a) CH4 and H2O

CH4 has LDF and H2O has LDP and H-bonds. Bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Tetrahedral Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

CH4; 4; 4; 0; 109.5 degrees; sp^3

2.2 M solution of hydrocyanic acid, HCN, at 25'C. pK(a) = 9.21 at 25'C c) Identify strongest base in this system

CN- is strongest base in this system. H2O and CN- compete for protons and CN- wins most of time. Bc of equilibrium lies far to left

CO2 or CS2 O=C=O or S=C=S a. Which structure has the shortest bond length between the central and each terminal atom? Justify b. Which structure has the greatest bond energy in its individual bonds? justify

CO2 and CO2 - Both held together by double bonds but S has larger atomic radius bc of another shell - S has larger radius than O, the distance between nuclear centers in the C-S bond is greater than the distance between nuclear centers in the C-O bond - bc of this, O's nucleus is closer to shared e-s than is S's nucleus - A closer proximity to shared e-s allows for a stronger force of attraction between protons and bonding e-s - When force of attraction is greater, more energy is required to break the bond

Which mixture in each set has the lowest pH? Justify CO2 and water or O2 and water

CO2 and water will form a solution with lowest pH and it's less than 7 • CO2 dissolves in water to form H2CO3 • O2 doesn't change the pH of water

Linear Geometry Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

CO2; 2; 2; 0; 180 degrees; sp

Equal vol. of 0.2 M Na2CO3 and 0.4 M HCl r combined. Write net ionic equation for reaction

CO3 2- + 2H+ —> H2CO3 —> H2O + CO2 CO3 2- + 2H+ —> H2O + CO2

Equal vol. of 0.1 M Na2CO3 and 0.1 M H2SO4 r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

CO3 2- + H3O+ —> HCO3- + H2O • System contains equal # moles of Na2CO3 and H2SO4. React in a 1:1 mole ratio and reaction goes to completion • aqueous species that have the highest conc. at equilibrium r Na+, HCO3 -, and HSO4 - • Conc. of Na+ will be about twice that of HCO3 - or HSO4 -

Which mixture in each set has the highest pH? Justify a) CaO and water or NaCl and water

Ca and water bc metal oxides r basic and pH will be greater than 7 • CaO reacts with water to form Ca(OH)2 • Ca(OH)2 is slightly soluble, and it increases the [OH-] in the solution • NaCl and water produces a neutral solution with a pH of 7, as both ions are neutral • Na+ is the conjugate acid of a strong base, and Cl- is the conjugate base of a strong acid

Write balanced net ionic equations for reactions Solid calcium hydroxide is placed in a solution of nitric acid

Ca(OH)2 + 2H+ —> Ca 2+ + 2H2O

Write balanced net ionic equations for reactions Solid calcium oxide is placed in a solution of hydrbromic acid

CaO + 2H+ —> Ca 2+ + H2O

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces k. CaOH or H2O

CaOH, bc has LDF and H-bonds but CaOh had ionic bonds • Ionic bonds r much stronger than H-bonds • As a greater amount of energy is required to break stronger FoA, boiling pt of CaOH is higher

After 20.0g of Na2SO4 r added to a 0.5 L saturated solution of CaSO4, does conc. of Ca 2+ increase, decrease, or stay the same? Justify. Assume overall volume of solution does not change

CaSO4 <—> Ca 2+ + SO4 2- Na2SO4 —> Na+ + SO4 2- NaSO4 dissolve completely and addition of excess SO4 2- shift equilibrium of 1st reaction to left, according to Le Chatelier's principle • Excess SO4 2- combine with Ca 2+ to form precipitate CaSO4 until product of [Ca 2+][SO4 2-] equals K(sp) once again • Will decrease the conc. of Ca 2+ in solution

Which species has a smaller radius: Ca or Ca^2+? Provide an explanation and justify ur claims using your knowledge of atomic structure.

Ca^2+ -Both have same # of protons, but Ca^2+ has 2 less e-s which causes protons to pull on e-s in Ca^2+ with greater force of attraction. -Ca^2+ has one less shell and Ca has e-s in n=4 shell and Ca^2+ only has e-s in n=3 shell (lost its 2 e-s in n=4 shell) -By losing n=4 shell, radius of Ca^2+ decreased dramatically -As there r fewer core e-s in Ca^2+ its Ve-s experience less of a shielding effect.

Calorimetry

Calorimetry - Measurement of heat transfer Calorimeter - Device used to determine amount of heat transferred • Heat flow cannot by measured directly • Measure temp. before and after a reaction, and use that data to calculate heat transfer

Ion Formation Metals lost e- to form ______ K --> ___ + ____ Non-metals gain e- to form _______ Cl + ___ --> _____

Cations; K+ + e- Anions; e- --> Cl-

Galvanic cell operates through reaction 3 Pb 2+ + 2 Al —> 3 Pb + 2 Al 3+ What happen to voltage of conc Pb 2+ decr. Justify

Cell potential decr according to equation. When [Pb 2+] < [Ni 2+], +LnQ which decr value of E(cell) E(cell) = E^o - RT/nF x Ln [Ni 2+] / [Pb 2+]

1) strong acid and strong base titration curve 2) Weak acid and strong base titration curve

Change is pH less than ~ 1.5 over region where most base required to reach equivalence pt is added. Change is pH is very large in the vicinity of equivalence pt 1) Initial pH for a strong acid is close to 1. Equivalence pt pH=7 2) Initial pH for a weak acid is greater than 1. Equivalence pt in basic range. pK(a) = pH at half equivalence pt.

Classify as a physical change, chemical change, or both. Justify by identifying the types of intermolecular or intramolecular forces that r involved in each processes and describing what happens to those forces while the processes r occurring. b) CO2(g) —> C(s) + O2(g)

Chemical change • Covalent bonds in the CO2 molecule r broken

Molecular Formulas

Chemical formulas that provide the actual number of each type of atom in molecule

Empirical Formulas

Chemical formulas that provide the relative number of each type of atom in molecule. (a ratio in simplest form)

Cis vs trans isomers

Cis-2-butune = Carbon chain follows the same side Trans-2-butune = Carbon chain follows opposite sides

Identify strongest acid. Justify based on molecular structure and electronegativity a) I(CH2)2COOH or Cl(CH2)2COOH

Cl(CH2)2COOH bc Cl is more electronegative than I • highly electronegative Cl reduces electron density btwn H and O to a greater degree than does I • Bc electron density btwn H and O in Cl(CH2)2COOH is less than it is in I(CH2)2COOH, FoA holding onto H in Cl(CH2)2COOH is also less

Which bonds from each set is most covalent? Explain a. Cl-Cl or Al-F b. N-O or C-O c. Ca-O or N-O

Cl-Cl (electronegativity difference between like elements is 0); N-O; N-O; smaller electronegativity difference between two atoms, the more covalent the bond. This is bc e-s r more evenly shared

Standard conditions for all cells

Conc : Reactants = 1.0 M & Products = 1.0 M Pressure : Reactants = 1 atm & Products = 1 atm Temp. : Reactants = 25^oC & Products = 25^oC A + B 2+ —> A 2+ + B Q(c) = [A 2+]/[B 2+] = [1 M]/[1 M] = 1.0 • Value for Q(c) or Q(p) always be 1.0 under standard conditions

Mayan and Aztec limestone (CaCO3) pyramids r being eroded away by acid rain. Use chemical principles and a balanced net ionic equation to explain why acid rain is a problem for these ancient monuments

Concentration of H+ is high in acid rain • H+ ions react with solid calcium carbonate according to following process CaCO3 + 2H+ —> Ca 2+ + H2O + CO2 • Solid material, with which the pyramids were built, is converted into aqueous, liquid, and gaseous components

Conjugate Acid-Base Pairs

Conjugate acid-base pair are: • Acid and its conjugate base • bade and its conjugate acid

Mixture

Contains particles of more than one type atom, formula unit or molecule.

Photoelectron Spectroscopy (PES) data indicated that ionization energy for an e- in 2s orbital of Ca is 42.7 MJ/mol and ionization energy for e- in 3s orbital of Ca is only 4.65 MJ/mol. Explain.

Coulomb's Law and shielding effect -According to Coulomb's Law, force of attraction between 2 charged particles decreases as distance between them increases, as distance is the denominator of equation (F=k(q1q2/d^2) -e- in 3s orbital r, average, furthest from nucleus than e- in 2s orbital. When force of attraction decreases, energy required to remove e- also decreases. -e-s in 2s orbital r shielded by e- in n=1, and e- in 3s orbital r shieded by e- in n=1 and n=2 shells. Shielding effect, caused by repulsion by inner core e-s, reduces force of attraction between e- and nucleus. As e- in 3s orbital r being repelled away from nucleus by more inner core e-s, their ionization energies r lower.

Kinetic Molecular Theory 3

Coulombic FoA or repulsion don't exist btwn gas particles in a system

Copper turnings are added to a solution of copper (I) nitrate. a. Write the balanced oxidation half reaction.

Cu + Cu2+ → 2Cu+ Cu → Cu+ + e-

What are the 3 polyatomic ions that end in -ide??

Cyanide (CN-), Hydroxide (OH-), and Peroxide (O2 2-)

1) Ionization energy _______ when moving down group in PT. 2) Ionization energy __________ moving left to right across period in PT

Decreases Increases

Ionic Radius

Distance from the center of an ion's nucleus to its outermost electron

Double and triple bonds - charge clouds = - Hybridization = - Sigma bonds = - Pi bonds=

Double: 3; sp^2; 1; 1 Triple: 2; sp; 1; 2

Nernst equation

E = E° - (RT/nF)lnQ E = emf under non-standard conditions (V) E° = emf under standard conditions (V) R = 8.314 J/molK T = temp (K) n = mol e-s transferred in balanced equation (mol e-) F = Faraday's constant (96,500 C/mol e-) Q = reaction quotient

What is the difference btwn E(a) and Delta E?

E(a) - (activation energy) is the minimum amount of energy that is required for a reaction to occur Delta E - (change in energy) is the energy that is lost or gained in the chemical reaction. • It's the difference btwn PE locked up in the bonds of the reactants and the PE locked up in the bonds of the products • Most of the energy is lost or gained in the form of heat, delta H • Remaining portion of this energy change comes from the work done on or by the system, which is related to changes in pressure and volume that occur during the chemical reaction

Planck's equation

E=hv Energy per Quantum (J) = Planck's constant (6.63x10^-34 Js) x frequency (s^-1)

Planck and Einstein's formula

E=hv Energy per photon (J) = Planck's Constant (6.63x10^-34 Js) x frequency (s^-1)

HOY Oxoacids and CB stability through Induction

EN of Y increases —> HOI < HOBr < HOCl Acid Strength of K(a) increases —> OI- < OBr- < OCl- CB stability increases • When LP r pulled away from oxygen and toward electronegative Y, r less available to bond with an H+ ion

TF Redox reactions

E^o(cell) value : + > 0; TF = yes E^o(cell) value : + < 0; TF = no E^o(cell) = E^o(red) (cathode) - E^o(red) (anode) • formula and table of standard reduction potentials used instead of activity series to determine if any redox reaction is TF E^o = E^o(red) (reduction) - E^o(red) (oxidation)

2 separate 250.0 g pure samples of NaCl were analyzed. Both samples were found to contain 39.3% Na and 60.7% Cl by mass. Use what you have been learning about atomic molecular theory to explain why the same results would be obtained from both samples.

Each formula unit of NaCl contains exactly one Na and one Cl atom, so both samples have a 1:1 ratio of Na to Cl. The average atomic mass of Na would be the same in both samples, as the ratio of the different isotopes of Na would be the same. The average atomic mass of Cl would be the same in both samples, as the ratio of the different isotopes of Cl would be the same. Because all the above mentioned ratios are the same, the mass ratio between Na and Cl will also be the same in both samples.

Pauli Exclusion Principle

Each orbital can hold a maximum of two electrons that will spin in opposite directions

induction, acid strength and stability of CB

Electronegative elements tend to stabilize CB

Determine Bond Type H (2.1) - H (2.1)

Electronegativity Difference = 0.0 Non-polar covalent

Determine Bond Type H (2.1) - Br (2.8)

Electronegativity Difference = 0.7 polar covalent

Determine Bond Type Li (1.0) - F (4.0)

Electronegativity Difference = 3.0 Ionic

S is more electronegative than Ca. Explain why this is and justify your claims using your knowledge of atomic structure

Electronegativity increases as atomic radius decreases. -Ve- in larger elements experience greater amount of shielding from inner core e-s than Ve- in smaller elements. -When 2 atoms r involved in chemical bond, Ve- will experience greater attraction for + nucleus that is closest to them -According to Coulomb's Law, forces of attraction increases as the distance between e-s and protons decreases.

Use 2 analogies to describe continuous change

Elevation - It can increase or decrease by infinitely small quantities Velocity - It can increase or decrease by infinitely small quantities

2 Solutions at 25'C were mixed in a beaker. A precipitate formed and temperature of new solution dropped to 19'C b) Was chemical reaction that took place endothermic or exothermic?

Endothermic

The reaction-energy profile diagram below outlines an uncatalyzed process Potential Energy ______________ / \ / \ l l / \ _________________/ \__________________ Reaction progress d) Is this process endothermic or exothermic? Explain.

Endothermic • bc PE contained within the bonds of the products is greater than that of the reactants • energy was absorbed from the surroundings during the reaction

A chemical reaction takes place in a beaker that u r holding. While the reaction is taking place, ur hand feels cold. b) Is the reaction endothermic or exothermic?

Endothermic • products contain more PE

Energy must always be added to remove electrons means are _______________. Values obtained by shining different frequencies light on pure sample of neutral atoms in gas phase, recording frequency that removes electrons and using E=hv to find 1st ionization energy

Endothermic process

System 1 (479K) vs. System 2 (298K) a) If insulation of outside of 2 systems prevents any energy from flowing into surroundings, will energy lost by System 1 be greater, less, or = to energy gained by system 2?

Energy lost by system 1 will = to energy gained by system 2

Activation Energy (E(a))

Energy must be absorbed to contort or weaken bonds in reactant molecules so that they can break and new bonds form. • If E(collision) < E(a) the reaction will not happen • Reaction rates depend on the magnitude of E(a) — Generally a slower rate — Generally a faster rate, when at the same temperature • Generally, if less energy is required to get over the hump, more reactants will collide with enough energy

Experiment conducted in order to determine the enthalpy change that occurs when 1.0 mole of ice at 0'C melts and becomes 1.0 mole of water at 0'C. Enthalpy change associated with this process is referred to as heat of fusion, ^_H_(fus) of ice. In experiment, a 9.68 g sample of ice at 0'C was added to a coffee cup calorimeter containing 278.25 mL of distilled water. Temperature of the water was 22.485'C before the ice was added. The lowest temp. that was recorded after ice had melted was 19.050'C. g) Was energy conserved in the process? Justify

Energy was conserved in this process. All of the heat lost by the water was gained by ice as it melted and by the melted ice as its temperature increased from 0'C to 19.050'C

A 85.2 g copper bar was heated to 221.32'C and placed in a coffee cup calorimeter containing 425.0 mL of water at 22.55'C. Final temperature of water was recorded to be 26.15'C d) Was energy conserved in the process? Justify

Energy was conserved in this process. All the heat that was lost by the Cu was gained by the water

Enzymes

Enzyme binds to the reactant(s) to form a new reaction intermediate

Equal volumes of 0.2 M CH3COOH (K(a) = 1.8x10^-5) and 0.2 M C6H5NH2 (K(b) = 3.8x10^-10) r mixed at 25'C a) Which aqueous compound will have the highest conc. when equilibrium is established in final solution? Justify

Equal # moles CH3COOH and C6H5NH2 r present in solution • K values for both reactions r less than 1, so both equilibrium lie to the left • [C6H5NH2] > [C6H5NH3+] and [CH3COOH] > [CH3COO-] • Bc K(a) value for CH3COOH is larger than K(b) value for C6H5NH2, the equilibrium for following reaction lies further to the left: C6H5NH2 + H+ <-> C6H5NH3 + • [C6H5NH2] > [CH3COOH], and thus, C6H5NH2 is aqueous compound that has highest conc.

Equal volumes of 0.2 M CH3COOH (K(a) = 1.8x10^-5) and 0.2 M C6H5NH2 (K(b) = 3.8x10^-10) r mixed at 25'C b) Is final solution acidic or basic? Justify

Equal #s moles of CH3COOH and C6H5NH2 r present in solution • K(a) for CH3COOH is 1.8x10^-5 and K(b) for C6H5NH2, CH3COOH is a stronger acid than C6H5NH2, is a base • Equilibrium for reaction lies further to right: CH3COOH <-> CH3COO- + H+ • [H+] > [OH-] so final solution would be acidic

Which species has a smaller radius: F or F-? Provide an explanation and justify ur claims using ur knowledge of atomic structure.

F -Both have same # of protons, but F has 1 less e- which causes protons to pull on e-s in F with greater force of attraction. -Addition of extra e- in F- causes greater electron-electron repulsion which increases distance between Ve- and increases its radius.

Ion-Dipole and Coulomb's Law

F = k (Q1Q2)/d^2 — Ion-Dipole FoA increase as: - Radius of the ion decreases - Charge of the ion increases - Magnitude of the dipole on the polar molecule increases

Ion-Dipole and Coulomb's Law

F = k(Q1Q2)/d^2 • Some ionic compounds don't dissolve in water • Solubility of ionic compounds can be explained through Coulomb's Law • If cation-anion attractions r stronger than ion-dipole attraction, compound won't be soluble • Ionic compounds don't dissolve in non-polar solvents as non-polar solvent don't carry permanent dipoles • However, ion-dipole FoA btwn Mg 2+ and water r stronger than those btwn Ba 2+ and water for same reason

What 2 elements don't double bond??

F and Cl

Which element from each set is most electronegative? a) F or C b) Al or Cl c) Po or S d) Cs or S e) Ca or Cl f) O or Se g) Zn or K h) C or Pb i) Ga or O

F, Cl, S, I, Cl, O, Zn, C, O

Identify strongest base and justify a) F- or Cl-

F- bc it's the CB of the weak acid HF • Relatively strong base, as it has a great ability to attract H+ ion • Cl- is CB of strong acid, HCl • Makes it a weak base, as it does not have ability to attract H+ ions in an aqueous solution • HCl experience ~100% ionization, so equilibrium lies far to the right

Identify strongest base and justify a) I- or F-

F- bc it's the CB of the weak acid HF • Relatively strong base, as it has a great ability to attract H+ ion • I- is CB of strong acid, HI • Makes it a weak base, as it does not have ability to attract H+ ions in an aqueous solution • HI experience ~100% ionization, so equilibrium lies far to the right

What is strongest base in reaction? Provide justification based on solution equilibrium or FoA btwn particles HF + H2O <—> F- + H3O+

F- bc stronger base • HF is a weak acid and weak acids have strong CB • F- is a relatively strong base, bc it's very electronegative • Water only moderate strength base, as it can act as an acid or base

What is the strongest base in each following reactions? Justify based on solution equilibrium or FoA btwn particles d) HF + H2O <—> H3O+ + F-

F- is strongest base • HF is a weak acid, and weak acids have strong CB • Also know that F- is a relatively strong base, bc it's very electronegative • H2O has only moderate strength as a base, as it can act as an acids or a base

Solid iron (III) nitrate is added to water.

Fe(NO3)3 → Fe3+ + 3NO3-

Activation Energy E(a) AB + C <-> A + BC Forward Reaction vs Reverse Reaction

For - AB + C <-> A + BC Reverse - A + BC <-> AB + C The E(a) value changes for the reverse reaction

In a lab experiment, some sodium fluoride was dissolved in distilled water at 25'C. pH if the final solution was measured to be 8.6 at 25'C. After heating the solution on a hot plate, and pH was measured to be 8.9. Explain why pH increased after the solution was heated F- + H2O <-> FH + OH- ^_H' = + 118.4 kJ/mol

Forward reaction is endothermic • According to Le Chatelier's principle, stress caused by adding heat to a system at equilibrium causes the reaction to shift in the endothermic direction. This shift increased [OH-] which caused the pH to increase

Enthalpy Change (^_H) Freezing and Melting Water at 0'C

Freezing Water ^_H = -# kJ/mol Melting Water ^_H = # kJ/mol

pH of HOCl solution is 3.9 at 25'C. pK(a) = 7.46 at 25'C. Is conc. HOCl greater than, less than, or equal to the conc. of OCl-? Justify

From Henderson-Hasselbalch equation, when pH is lees than pK(a), log[OCl-]/[HOCl] must be negative #. Only occurs when [HOCl] > [OCl-] pH = pK(a) + log[OCl-]/[HOCl]

Which depicts most - charge in entropy? Justify L —> S G —> L

G —> L bc G have very high entropy values cuz gas particles have greatest freedom to move and ideally have no attraction to one another • Results in an extremely large # of possible arrangements • When G condenses, FoA btwn particles formed and randomness decr • Change in entropy associated with solidification of L is less -

The electromagnetic spectrum is a continuous spectrum of light

Gamma Rays = V X-rays = I UV = B Visible light = G IR = Y Microwaves = O Radio = R

What charge do Group 4A elements acquire as ions?

Ge is 4+. Pb and Sn can be 2+ or 4+

IE_Li is less than ________ as large as IE_H+ If the most loosely held electron in Li was about the same distance from the nucleus as those in H and He, we would expect IE_Li to be about _______ times greater than IE_H+

Half; 3

Hydrogen gas is blown over hot lead (II) oxide. a. Explain your thought process about how to write the equation above.

H2 + PbO → Pb + H2O Hydrogen is already reduced as far as it can go, as such, it must oxidize. Now we need to think about what would get reduced. In this case, as oxygen is already reduced to O2- and cannot be reduced further, the lead will reduce to lead metal.

Write balanced chemical equation that outlines the reaction used to determine the enthalpy of formation for one mole of H2CO(g)

H2(g) + 1/2 O2(g) + C(s) —> H2CO (g)

Write balanced chemical equation that outlines the reaction used to determine the enthalpy of formation for one mole of water

H2(g) + 1/2 O2(g) —> H2O (g)

What is strongest base in reaction? Provide justification based on solution equilibrium or FoA btwn particles HClO4 + H2O —> ClO4- + H3O+

H2O bc strong acids (HClO4) experience ~100% ionization and have weak CB • ClO4- is a weak base • H2O and ClO4- compete for H+ ions • H2O acquires H+ ions most of the time, as reaction goes to completion

What is strongest base in reaction? Provide justification based on solution equilibrium or FoA btwn particles HNO3 + H2O —> NO3- + H3O+

H2O bc strong acids (HNO3) experience ~100% ionization and have weak CB • NO3- is a weak base • H2O and NO3- compete for H+ ions • H2O acquires H+ ions most of the time, as reaction goes to completion

What is the strongest base in each following reactions? Justify based on solution equilibrium or FoA btwn particles a) HClO4 + H2O —> ClO4- + H3O+

H2O is strongest base • Strong acids (HClO4) experience ~100% dissociation and have weak CB • Thus, ClO4- is a weak base • H2O and ClO4- compete for H+ ions • H2O acquired H+ ions most of the time, as the reactions goes to completion

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces g. H2O and H2S

H2O, bc both LDF but H2O has H-bonds and H2S has dipole-dipole forces • H-bonds r much stronger than normal dipole-dipole forces • As a greater amount of energy is required to break stronger FoA, boiling pt of NH3 is higher

Bent (2 lone pairs) Ex. - Charge Clouds - Bonds - Bond Angle - Hybridization -

H2O; 4; 2; <109.5 degrees; sp^3

Equal vol. of 0.15 M H2SO3 and 0.30 M KOH r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

H2SO3 + 2 OH- —> SO3 2- + 2 H2O • Mole ratio of H2SO3 to KOH is 1:2 and H2SO3 is a polyprotic acid with 2 labile protons • React in a 1:2 mole ratio and reaction goes to completion • Aqueous species that have highest conc. at equilibrium r SO3 2- (CB of H2SO3) and K+ (spectator ion) • Conc. of K+ will be 0.15 M. • Conc. slightly less than 0.075 M, as weak baseand will accept some protons from water

Equal vol. of 0.15 M H2SO3 and 0.15 M LiOH r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

H2SO3 + OH- —> HSO3- + H2O • System contains equal # moles H2SO3 and LiOH. React in 1:1 mole ratio and reaction goes to completion • aqueous species that have the highest conc. at equilibrium r the HSO3- (CB of H2SO3) and Li+ (spectator ion)

H+ and H3O+ equations for acids

HA + H2O —> H3O+ + A- H+ and H3O+ r used interchangeable for aqueous ion of hydrogen

Strong acids have very weak CB

HBr + H2O —> H3O+ + Br- • H2O and Br- compete for protons • H2O is stronger base, so wins most of time, and reaction goes to completion

Identify strongest acid in following sets. Justify a) HF or HBr

HBr bc ionic radius of Br- is larger than that of F- • FoA btwn Br- and H+ is less than that btwn H+ and F- • Br- will lose H+ ions easier than F-

Big 6 strong acids

HBr, HI, HNO3, HCl, HClO4, H2SO4

Draw Lewis structures of acids and identify strongest acid in each sets. Justify based on electronegativity and molecular structure a) HOBr or HBrO3

HBrO3 bc pattern of bonding for 2 structures is as follows: HOBrO2 and HOBr • HBrO3 has 2 more highly electronegative terminal O atoms attached to Br atom than does HOBr • Higher conc. of electronegative elements around Br in HBrO3 results in a greater reduction in electron density btwn O and H over that in HOBr • Bc electron density btwn H and O in HBrO3 is less than it is in HOBr, FoA holding onto H in HBrO3 is also less

Required to create a buffer solution where acid and its salt have very similar conc.. select weak acid and its salt use to create buffered solution with pH of 9.3 HCN = 6.2 x 10^-10 HOCl = 3.5 x 10^-8 CH3COOH = 1.8 x 10^-5 HF = 7.2 x 10^-4

HCN bc pK(a) = 9.21 and desired pH is 9.3 • Since fairly equal conc. of a weak acid and its salt, it's necessary to choose acid where pK(a) = pH • Equation demonstrated how pK(a) = pH when conc. r equal, as log(1) = 0 pH = pK(a) + log [A-]/[HA] NaCN or another soluble salt containing CN- ion would be used to prepare this buffer

"Big six" strong acids that completely ionize in water

HClO, HI, HBr, HCl, HNO3, H2SO4

Draw Lewis structures of acids and identify strongest acid in each sets. Justify based on electronegativity and molecular structure a) HOBr or HClO3

HClO3 bc pattern of bonding for 2 structures is as follows: HOClO2 and HOBr • HClO3 has 2 more highly electronegative terminal O atoms at one end of its structure. Cl is also more electronegative than Br • Higher conc. of electronegative elements at one end of HClO3 results in a greater reduction in electron density btwn O and H over that in HOBr • Bc electron density btwn H and O in HClO3 is less than it is in HOBr, FoA holding onto H in HClO3 is also less

Draw Lewis structures of acids and identify strongest acid in each sets. Justify based on electronegativity and molecular structure a) HBrO3 or HClO3

HClO3 bc pattern of bonding for 2 structures is as follows: HOClO2 and HOBrO2 • Cl more electronegative than Br, resulting in a greater reduction in electron density btwn O and H over that in HClO3 over that in HOBr • Bc electron density btwn H and O in HClO3 is less than it is in HOBr, FoA holding onto H in HClO3 is also less

Draw Lewis structures of acids and identify strongest acid in each sets. Justify based on electronegativity and molecular structure a) HClO4 or HOCl

HClO4 bc pattern of bonding for 2 structures is as follows: HOClO3 and HOCl • HOClO3 has 3 more highly electronegative terminal O atoms attached to Cl atom than does HOCl • Higher conc. of electronegative elements around Cl in HClO4 results in a greater reduction in electron density btwn O and H over that in HOCl • Bc electron density btwn H and O in HClO4 is less than it is in HOCl, FoA holding onto H in HClO4 is also less

Common Ion Effect

HF + H2O <-> H3O+ + F- • HF is a very weak acid. Equilibrium lies to left (+ KF) KF —> K+ + F- KF is soluble and experience 100% dissociation • F- is common ion that causes the first reaction to shift to the left (Le Chatelier) and pH of system to rise

A 100.0 mL sample of 0.40 M HF is mixed with 100.0 mL of 0.40 M LiOH a) write balanced net ionic equation b) Will pH of final solution be less than 7, equal to 7, or greater than 7. Justify

HF + OH- —> H2O + F- pH of final solution will be greater than 7. F- ion will act as base and remove H+ ions from water to form OH- ions. Increase in [OH-] will cause pH to rise F- + H2O —> HF + OH-

Strong bases have very weak CA

HF + OH- —> H2O + F- • CA of strong base is H2O • Strong base could be any soluble Group 1A or Group 2A hydroxide

Required to create a buffer solution where acid and its salt have very similar conc.. select weak acid and its salt use to create buffered solution with pH of 3.25 HCN = 6.2 x 10^-10 HOCl = 3.5 x 10^-8 CH3COOH = 1.8 x 10^-5 HF = 7.2 x 10^-4

HF bc pK(a) = 3.14 and desired pH is 3.25 • Since fairly equal conc. of a weak acid and its salt, it's necessary to choose acid where pK(a) = pH • Equation demonstrated how pK(a) = pH when conc. r equal, as log(1) = 0 pH = pK(a) + log [A-]/[HA] NaF or another soluble salt containing F- ion would be used to prepare this buffer

Identify strongest acid in following sets. Justify a) HCl or HI

HI bc ionic radius of I- is larger than that of Cl- • FoA btwn I- and H+ is less than that btwn H+ and Cl- • I- will lose H+ ions easier than Cl-, although both r strong acids

Identify strongest acid in following sets. Justify a) HF or HI

HI bc ionic radius of I- is larger than that of F- • FoA btwn I- and H+ is less than that btwn H+ and F- • I- will lose H+ ions easier than F-

What is the strongest acid in each of the following reactions? Provide justification based on solution equilibrium a) HI + H2O —> I- + H3O+ b) H2SO4 + H2O —> HSO4- + H3O+ c) HNO3 + H2O —> NO3- + H3O+

HI, H2SO4, and HNO3 r all very string acids • all 3 of the "Big six" strong acids • As they experience ~100% dissociation, equilibrium for these reactions lies very far to the right

Strongest acid in each reaction? Justify a) HI + H2O —> I- + H3O+ b) H2SO4 + H2O —> HSO4- + H3O+ C) HNO3 + H2O —> NO3- + H3O+

HI, H2SO4, and HNO3 r all very strong acids • 3 of the 'Big Six' strong acids • As they experience ~ 100% ionization, the equilibrium for these reactions lies very far to the right

Draw Lewis structures of acids and identify strongest acid in each sets. Justify based on electronegativity and molecular structure a) HIO3 or HIO2

HIO3 bc pattern of bonding for 2 structures is as follows: HOIO2 and HOIO • HIO3 has 1 more highly electronegative terminal O atoms attached to I atom than does HOI2 • Higher conc. of electronegative elements around I in HClO3 results in a greater reduction in electron density btwn O and H over that in HIO2 • Bc electron density btwn H and O in HIO3is less than it is in HIO2, FoA holding onto H in HIO3 is also less

Identify strongest acid a) HOI or HOBr

HOBr bc Br is more electronegative than I, Br reduces electron density in H-O bond to a greater degree than I • Bc electron density btwn H and O in HOBr is less than it is in HOI, FoA btwn H and O in HOBr is less than they r in HOI • HOBr will lose H+ ion more easily

Identify strongest acid a) HOCl or HOBr

HOCl bc Cl is more electronegative than Cl, Br reduces electron density in H-O bond to a greater degree than Br • Bc electron density btwn H and O in HOCl is less than it is in HOBr, FoA btwn H and O in HOCl is less than they r in HOBr • HOCl will lose H+ ion more easily

Identify strongest acid a) HOCl or HOI

HOCl bc Cl is more electronegative than I, Cl reduces electron density in H-O bond to a greater degree than I • Bc electron density btwn H and O in HOBr is less than it is in HOI, FoA btwn H and O in HOBr is less than they r in HOI • HOBr will lose H+ ion more easily

A beaker containing 125 mL of 0.120 M HOCl is titrated using 0.250 M NaOH K(a) for HOCl is 3.5x10^-8 b) Is solution acidic or basic at equivalence pt? Jusitfy

HOCl is a weak acid, so its CB, OCl- is a weak base. • At equivalence pt, equal #s moles HOCl and OH- have reacted • Only species remaining in solution that can affect pH is OCl-, which is basic • OCl- accepts protons from water molecules, thereby increasing conc. OH- in solution according to the reaction below OCl- + H2O —> HOCl + OH- pH rises into basic range as [OH-] increases

The acid dissociation constants (Ka values) for hypoiodous acid, HOI, and lactic acid, HC3H5O3 at 298 K r 2 x 10^-11 and 1.38 x 10^-4 respectively. Which solution is more basic: 1.0M NaOI or 1.0M NaC3H5O3? Justify

HOI is the weaker acid, as 2 x 10^-11 < 1.38 x 10^-4 • its conjugate base, OI-, is stronger • Weaker acids have stronger conjugate bases • equilibrium for OI- + H2O <-> HOI + OH- lies further to the right than it does in C3H5O3- + H2O <-> HC3H5O3 + OH- Thus, 1.0M NaOI produces a higher concentration of OH-, making a more basic solution

Acid dissociation constants for HOI and HC3H5O3 at 298 K r 2x10^-11 and 1.38x10^-4 respectively. Which solution is more basic: 1.0 M NaOI or 1.0 M NaC3H5O3? Justify

HOI is weaker acid, as 2x10^-11 < 1.38x10^-4 • Its CB, OI-, is stronger • Weaker acids have stronger CB • Equilibrium for OI- + H2O <-> HOI + OH- lies further to right than it does in... C3H5O3- + H2O <-> HC3H5O3 + OH- • Thus, 1.0 M NaOI produces a higher conc. of OH-, making a more basic solution

Half Life for 1st Order Reactions

Half Life - The time it takes for the conc of a reactant to decrease by half • the half life for a first order reaction is a CONSTANT. plug in the half amount for the concentration to find the half life. = Deriving the half-life equation for 1st order processes • These equations can also be used for any radioactive decay problems, as those processes r always 1st order

John Dalton

He presumed that if elements were made of tiny indivisible particles, then molecules of a particular compound would always be composed of equal numbers of each type of element. Thus, each element in a given compound would account for a consistent percentage of that compound's mass. Ex. Water is always composed of the same number of H atoms and the same number of O atoms

Heat vs. Temperature

Heat - Form of energy (J) Temp. - A measure of average KE of atoms and molecules in a system • K temp. scale proportional to this • When KE doubles, K temp. doubles

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces a. Br2 and I2

I2, bc both LDF but I2 is larger and has more e-s, making it more polarizable than Br2 • I2 experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of I2 is higher

Explain why H2 is a gas and I2 is a solid at 25'C and 1 atm

I2, both non-polar and only LDF but I2 is larger and has more e-s, making it more polarizable than H2 • I2 experiences greater LDF • Bc of stronger FoA, I2 exists as a liquid at 25'C and 1 atm, while H2 exists as a gas

T-Shaped Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

ICl3; 5; 3; 2; <90 degrees; sp^3d

Ionization energy for any electron is calculated using:

IE = hv - KE

If u were to design an experiment to identify an unknown compound, would u choose to use an infrared (IR) spectrometer or an ultraviolet/visual (UV/Vis) spectrometer? Justify

IR spectrometer • in this, different frequencies of electromagnetic radiation in the IR spectrum r absorbed by different types of bonds in the molecule that vibrate at the same frequencies • spectrum that is obtained from these experiments can be used to identify the different types of bonds in the structure and the elements that r involved in those bonds • spectrum for a compound also provides a 'fingerprint' that can be used to identify it

If u were to design an experiment to determine the different types of atoms and bonds that make up a pure sample of an unknown compound, would u choose to use an infrared (IR) spectrometer or an ultraviolet/visual (UV/Vis) spectrometer? Justify

IR spectrometer • in this, different frequencies of electromagnetic radiation in the IR spectrum r absorbed by different types of bonds in the molecule that vibrate at the same frequencies as the radiation • spectrum that is obtained from these experiments can be used to identify the different types of bonds in the structure and the elements that r involved in those bonds

FoA in Real Gases

Ideal Gas Law assumes there r no FoA btwn gaseous particles, but force is proportional to 1/d^6

Isotopes of an element exhibit ____________ behavior. This is why 1/1 H and 2/1 H will both bond with oxygen to form water. C-14 and C-12 bond with oxygen to form CO2.

Identical chemical

The energy profile diagram below was plotted for the following second order elementary reaction: A + BC <-> AB + C Potential Energy ______________ / \ / \ l l / \ _________________/ \__________________ a) At low temperature a relatively small fraction of collisions btwn reactants result in the formation of products. Describe the interactions btwn reactants that collided with an orientation that would produce products, but failed to do so

If 2 species collide with the correct orientation, but with insufficient energy, 2 molecules will bond together temporarily (A + BC —> A-B-C); however, they will not produce products • Instead, the temporary bonds will lose PE, and they'll once again separate reactants (A-B-C —> A + BC)

Autoionization of Water

In pure water at 25'C, [H+] = 1.0 x 10^-7 and [OH-] = 1.0 x 10^-7 As [H3O+] = [OH-] the solution is normal K(w) = [H3O+] [OH-] K(w) = (1.0 x 10^-7) (1.0 x 10^-7) K(w) = 1.0 x 10^-14 (at 25'C) Constant and if know conc of [H3O+] u can calculate the conc of [OH-] or vise versa

The table below outlines the results from 3 experiments, which were conducted at the same temperature, involving the following reaction Ex: [O2]initial [NO]initial Initial Reaction Rate 1 0.022 0.026 0.0256 M/s 2 0.022 0.013 0.0064 M/s 3 0.044 0.013 0.0128 M/s b) Which environmental factor could be altered in order to increase the initial reaction rate in experiment one if the initial conc. of the reactants remained the same? Justify

Increasing the temp. would incr. the initial reaction rate • By increasing temp., u r increasing the average KE of the molecules in that system • In doing so, u r increasing the average velocity of these particles • Collision rates incr. as molecular speed increases • This rule is amplified when when the reactants r gases, but it's also true for other states

What is the only thing that one can do to reduce the value of the equilibrium constant, K(eq), for the following system? 2SO2 (g) + O2 (g) <-> 2SO3 (g) ^_H(rxn) = -198.4 kJ

Increasing the temperature is the only thing that can be done to reduce the value of the equilibrium constant for an exothermic reaction • Adding heat causes the equilibrium to shift in then endothermic direction - to the left in this case • This will reduce the value of the numerator, [SO3], and increase the value of the denominator, [SO2]^2 [O2], in the equilibrium expression, resulting in an equilibrium constant with a smaller magnitude

ICE Chart

Initial • Given #s for reactants and 0 for products Change • Identical coefficients as equations and Reactants negative while products positive • Ex: Reactants: -2x; Products: +2x and +x Equilibrium • Total

Bohr called the shell __________, as each orbit was associated with quantized energy level

Integers quantum numbers

Steel — an ________ Alloy

Interstitial • C fills some spaces btwn Fe atoms • Interstitial C atoms make the lattice more rigid, less malleable, and less ductile • Retains a "sea of e-" so it can conduct electricity

Conducts electricity in molten state, dissolves water and is brittle. Molecule??

Ionic

Atoms from a pure sample of an element are ___________ and __________ through a magnetic field. Isotopes with smaller masses experience a greater degree of ____________

Ionized and accelerated; deflection

Citric Acid H3C6H5O7 is a polyprotic acid. K(a1) = 8.4x10^-4, and K(a2) = 1.8x10^-5 at 25'C c) What us equilibrium constant for reaction below? H3C6H5O7 + H2O <-> H3O+ + H2C6H5O7 -

K(a1) = 8.4 x 10^-4

The Equilibrium Constant (K(eq))

K(c)=[E]^e[D]^d/[A]^a[B]^b Units for Conc - mol/L = M

Write the equation for the 1st ionization of Potassium

K(g) --> K+(g) + e-

K(p)

K(p)=(P(E))^e(P(D))^d/(P(A))^a(P(B))^b

pK(a) = 14 at 25'C

K(w) = [H3O+] [OH-] -log K(w) = (-log[H3O+]) + (-log[OH-]) pK(w) = pH + pOH = 14 at 25'C In pure water, pH = pOH = 7.0 at 25'C

pH of distilled water at 25'C is 7.0. When temp. is increased to 37'C the pH drops to 6.8. At both temps. the water is considered to be neutral as [H3O+]=[OH-]. Explain why pH drops when temp. increases.

K(w) is temperature dependent and autodissociation of water is endothermic process. H2O + heat <—> H+ + OH- As heat is added, equilibrium shifts to the right to use up that heat. Causes [H3O+] and [OH-] to increase at the same rate. pH is lower bc [H3O+] is higher

At -92'C, a pure sample of HBr has a higher vapor pressure than a pure sample of KBr b. Explain why the vapor pressure of HBr is higher than the vapor pressure if KBr at -92'C

KBr held together by ionic bonds, which r very strong, and HBr held together by dipole-dipole interactions, which r weaker • Bc FoA btwn particles in HBr r weaker, molecules or HBr can enter the gas phase more easily

At -92'C, a pure sample of HBr has a higher vapor pressure than a pure sample of KBr. Explain why the vapor pressure of HBr is higher than the vapor pressure of KBr at -92'C

KBr id held together by ionic bonds, which r very strong; and HBr is held together by D^2 interactions, which r much weaker. Bc the FoA btwn particles in HBr r weaker, molecules or HBr can enter the gas phase more easily

Potassium bromide is least soluble in which of the 2 liquids from each set below. Justify c. NH3 or Br2

KBr is least soluble in liquid bromine, bc it's non-polar. Ionic compounds r soluble in polar solvents, and NH3 is a polar solvent

Potassium bromide is least soluble in which of the 2 liquids from each set below. Justify a. H2O or CH4

KBr is least soluble in methane bc it's non-polar. Ionic compounds r soluble in polar solvents such as water

Potassium bromide is least soluble in which of the 2 liquids from each set below. Justify b. CH3OH or CH3CH2OH

KBr is soluble in both CH3OH and CH3CH2OH, as they r both polar solvents. KBr is less soluble in CH3CH2OH bc it has a longer non-polar C chains

Which substance in each set has the highest melting pt? Justify ur answer using chemical principles a. KCl or SiO2

KCl has ionic bonds and SiO2 is a network solid held together with covalent bonds • Network solids t stronger than ionic bonds, so SiO2 will have a higher melting temperature

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces L. KCl or Cl2

KCl, bc has ionic bonds and Cl2 has LDF • Ionic bonds r much stronger than LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of KCl is higher

KE per Gas Particle

KE = 1/2 mv^2 v = Velocity of a specific gas particle (m/s) m = mass of that particle (kg)

Equation for finding the number of orbitals and electrons??

Orbitals = n^2 Electrons = 2n^2

Use list of standard reduction potential values to indicate which species is most easily ox a) Li + or Li b) Au or Fe c) Au 3+ or Ba d) Ag or Cr e) Mn or H2O f) Cu or Sn

Li, Fe, Ba, Cr, Mn, Sn

A solution of KOH was titrated with HCl and curve was plotted. Which indicators should be used to signal endpoint of titration: methyl red (pK(a) = 5.5), litimus (pK(a)=7.0), or phenolphthalein (pK(a)=8.7)? Explain

Litimus bc pK(a)(litimus) is same as pH of solution at equivalence pt

Molar Mass (w/ density) formula

MM = (D/P)•RT

Metal Refinement and Electrolysis

Metals can be refined through electrolysis • solid cooper is plated out on cathode of electrolytic cell through following process Cu 2+ + 2Br- —> Cu + Br2 • Calculations can be made to determine how long will take to produce certain quantities of pure metals

The table below outlines the results from 3 experiments, which were conducted at the same temperature, involving the following reaction Ex: [O2]initial [NO]initial Initial Reaction Rate 1 0.022 0.026 0.0256 M/s 2 0.022 0.013 0.0064 M/s 3 0.044 0.013 0.0128 M/s h) If the same 3 experiments were conducted at a lower temperature, would the magnitude of the rate constant be greater than, less than, or equal to the value calculated above? Justify

Magnitude of rate constant would be less than the value calculated above • At lower temps., frequency of collisions btwn reactants is reduced • Also, average KE of of the reactants is less at lower temps., so fewer collisions would produce enough energy (activation energy) to cause a reaction • Thus, rate at which the reaction proceeds is reduced • As initial conc. would be the same, rate constant must have a smaller magnitude, as Rate=k[NO]^2[O2]

Expressing Concentration - 2 methods for expressing concentration

Molarity (M) = moles solute / liters solution • Molarity can change with temperature Mole fraction = X(A) = (moles A)/(moles A + moles B + ... + moles Z) • Mole fractions don't change w/ Temp

Calculating Mole Fractions

Mole Fraction - % composition by mikes of a single component in a mixture, represented in its decimal form X(A) = (Moles if 1 component (n(A)) in a mixture) / (Sum of moles of all components in mixture) X(A) = n(A) / [n(A) + n(B) + n(C) + n(D) + ... + n(Z)]

Rate Laws of Elementary Steps

Molecularity Elementary Reaction Rate Law Unimolecular A—>products Rate=k[A] Bimolecular A+A—>products Rate=k[A]^2 Bimolecular A+B—>products Rate=k[A][B] Termolecular A+A+A->products Rate=k[A]^3 Termolecular A+A+B->products Rate=k[A]^2[B] Termolecular A+B+C->products R=k[A][B][C] • Termolecular elementary reactions r very rareC as they require the simultaneous collision of 3 particles with sufficient energy and the correct orientations • Value of a rate constant, k, increases as temperature increases

Which bond from each set has the greatest bond energy? a. N-O or N-F b. B-F or B-Cl c. B-C or O-F d. C-O or C=O e. P-Br or P-Cl

N-F; B-F; O-F; C=O (Double bonds r shorter than single bonds and have greater bond energy); P-Cl Smaller atomic radius = smaller bond length = greater bond energy

Which bond from each set is the longest? a. N-O or N=O b. B-N or B-F c. Si-O or Si-I d. C-Cl or C-F

N-O (Single bonds r longer than double bonds); B-N; Si-I; C-Cl Bond lengths increase as the atomic radii increase

Equal vol. of 0.2 M NH3 and 0.2 M HNO3 r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

NH3 + H3O+ —> NH4+ + H2O • System contains equal # moles ammonia and nitric acid. React in 1:1 mole ratio and reaction goes to completion • aqueous species that have the highest conc. at equilibrium r NH4+ and NO3- (spectator ion) • Conc. of spectator ion will be slightly higher than that of the NH4+

Explain why NH3 has a higher solubility in water than SbH3

NH3 is able to form H-bonds with water, whereas SbH3 is only able to form weak dipole-dipole attractions with water • Bc the intermolecular FoA btwn NH3 and water r stronger than those of SbH3 and water, water is able to hold a higher concentration of NH3

65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl at 25'C e) Identify species that have highest conc. in this solution at half equivalence pt

NH3, NH4 +, Cl-, and OH-

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces i. NH3 and PH3

NH3, bc both LDF but NH3 has H-bonds and PH3 has dipole-dipole forces • H-bonds r much stronger than normal dipole-dipole forces • As a greater amount of energy is required to break stronger FoA, boiling pt of NH3 is higher

Trigonal Pyramidal Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

NH3; 4; 3; 1; <109.5 degrees; sp^3

65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl at 25'C f) Identify species that have highest conc. in this solution at equivalence pt

NH4 +, Cl-, and H3O +

Solutions Ammonium nitrate and NaCN r poured into a beaker. Write net ionic equation for reaction

NH4+ + CN- <-> NH3 + HCN

Odor is detected when solutions of Ammonium fluoride and KOH r combined. What is the odor?

NH4+ + OH- —> NH3 + H2O Odor is ammonia

Weak acids have strong CB

NH4- + H2O <-> H3O+ + NH3 • H2O and NH3 compete for protons • NH3 is stronger base, so wins most of the time, and equilibrium lies to the left

Bent (1 Lone Pair) Ex. - Charge Clouds - Bonds - Bond Angle - Hybridization -

NO2 -; 3; 2; <120 degrees; sp^2

5.0 M solution of HNO3 is titrated with 0.30 M NaOH. Identify species that have highest conc. in the solution being titrated halfway to the equivalence pt

NO3-, Na+, and H3O+

28.0 mL sample of 0.500 M NaHSO4 is titrated with 0.250 M NaOH at 25'C d) identify species that has highest conc. in this solution at half equivalence pt

Na+, HSO4-, SO4 2-, and H3O+

which bond from each set is most ionic? Explain a. Al-O or Na-O b. K-Cl or Zn-Cl c. Fr-F or B-F

Na-O; K-Cl; Fr-F; Greater electronegativity difference between two atoms, the more ionic the bond

Ionic compounds in polar solvents

NaCl dissolves in water NaCl(s) —> Na+(aq) + Cl-(aq) • Ions drag a certain # of water molecules around with them in solution • smaller ions have stronger electric fields so they drag more water molecules around with them

which compound from each set is most ionic? Explain a. Al2O3 or NaCl b. KCl or FeO

NaCl; KCl; Greater the electronegativity difference between 2 atoms, the more ionic the bonds. the more ionic the bonds, the more ionic the compound

Common Reductions in Acidic Solutions

Nitrate: NO3- —> NO Permanganate: MnO4- —> Mn 2+ Dichromate: Cr2O7 2- —> Cr 3+ Sulfite: SO3 2- —> SO4 2- Hydrogen Peroxide: H2O2 —> O2 • All the above oxidize p and d-block metals, sulfite ions, peroxides, and substances that have a lower (less positive) oxidation state than usual in acidic solutions

Will Mg oxidize Zn? Will Sn reduce Fe? Will Ag oxidize Fe? Will Ag oxidize Cu?

No, No, Yes, Yes

Argon, Ar, and Calcium cation, Ca^2+, r isoelectronic. Will they exhibit same photoelectron spectra? Justify

No, bc Ca^2+ has 20 protons in nucleus and Ar only has 18 protons in its nucleus. -According to Coulomb's Law, force of attraction on each of 18 e-s in Ca^2+ will be greater than force of attraction on each of e-s in Ar. For this reason, ionization energies for each ionization will be greater in Ca^2+. -Since both isoelectronic, 2 species will have same # of peaks, and peaks will show that they share same e- config.; however, peaks for Ca^2+ will have different energies than those for Ar due to having a lower effective nuclear charge

Does the photoelectron spectra (H, He, Li, Be) suggest a need to refine shell model of atom? Justify

No, bc these spectra support shell model. -The show that elements in 1st period of periodic table have one shell (n=1), and elements in 2nd period of period table have two shells (n=1 and n=2). -Shell model predicts that n=1 can hold up to two e-s and n=2 can hold up to 8 e-s. -Spectrum for Be shows that n=1 shell is holding 2 e-s and n=2 shell is holding 2 e-s.

Suppose a system operating in accordance with the chemical equation below is in a state of equation. Will the reaction shift when the pressure acting on the system is increased? If so, in which direction will it shift? Br2 (g) + H2 (g) <-> 2HBr (g)

No, it will not shift, as there r equal numbers of moles of gas on each side of the balanced chemical equation

Can prolonged exposure to highly intense infrared light cause e-s to be ejected from a clean metal surface? Explain

No, the frequency of infrared light is too low to eject e-s • To eject e-s from any metal surface a certain threshold frequency must be applied • Anything below that threshold frequency will not cause a single e- to eject, even if the metal is exposed to the light for several yrs

According to kinetic molecular theory, does a gas molecular move slower after it bounces off of the wall of a container? Explain

No, will continue to move at the same speed • KMT states that all collisions btwn gas particles and the walls of a container r elastic • No KE is lost in elastic collisions and if no KE is lost, then the velocity of the gas molecule must remain the same, as KE = 1/2 mv^2. • Molecule doesn't lose any mass during the collision, so it cannot lose any velocity if it retains the same amount of KE

Electronic Difference Ranges between Ionic, Polar Covalent, and Non-polar Covalent bonds

Non-polar Covalent Bond = 0-0.5 Polar Covalent Bond = 0.5-1.9 Ionic Bond = 1.9-3.5

Rank the following single bonds in order of increasing bond length: O-C, O-N, O-O, and O-F

O-F < O-O < O-N < O-C Bond length increase as the atomic radius of one or both of the bonding atoms increases

Explain why O2 is able to dissolve in water

O2 is non-polar and water is polar • When O2 dissolves in water, the polar molecules induce a dipole in the non-polar O molecules • Dipole-induced dipole FoA r able to hold a small concentration of O molecules in water

2 different 1.2 L buffered solutions were prepared using HOBr and LiOBr. Both buffered solutions had pH of 5.2 at 25'C. After 0.17 moles of HI were added to each solution, found that pH of one solution had dropped to 4.9 and pH of other had dropped to 3.1 a) Balanced net ionic equation for reaction that occurred when HI was added to these buffered solutions?

OBr- + H+ —> HOBr OBr- + H3O+ —> HOBr + H2O

0.50 M solution of HOCl at 25'C. pK(a) = 7.46 at 25'C Identify strongest base in this system

OCl- is strongest base in this system. H2O and OCl- compete for protons and OCl- wins most of time. Bc equilibrium lies to left

Solutions of sodium hydroxide and sulfuric acid are mixed. a. What is the strongest base in the reaction?

OH- + H+ → H2O OH- or NaOH

Ionization Energy: Every element has 1 extremely large increase in ionization energy.

Occurs when e- config. drops a principal quantum #, which causes radius to shrink much more than other ionizations

An ____________ is a graphic representation of the space that an electron will occupy 90% of the time.

Orbital

Is SF6 polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Octahedral so 6 S-F bonds and no lone pairs - polar individual bonds bc F is more electronegative than S - Non-polar overall compound bc dipoles cancel due to the symmetrical arrangement of the 6 S-F bonds

Forces of attraction between the electron and the nucleus result from ________________. The nucleus contains positively charged protons and neutral neutrons

Opposite charges

Collecting Gases over water

P(atm) = P(total) = P(gas) + P(water) • When measuring V of gas collected, must line up the water levels inside and outside the graduated cylinder • ensures that P inside cylinder is = to atmospheric pressure

Pressure Adjustment for Gases Under High Pressures (Low Volumes) (equation)

P(ideal) = P(measured) + (n^2 a)/(V^2) (Correction increases P(measured) to P(ideal) = P(measured) = P measured in the lab = P(ideal) = P expected when using the ideal gas equation = n = # of moles of gas = a = a constant for the specific gas in the system = V = V of the system (V(measured))

Combined Gas Law

P1V1/T1=P2V2/T2 • If P, V, or T changes, but number of moles remains the same, combined gas law equation can be derived from the ideal gas law equation in order to solve the problem • Must use absolute temperature (K)

After 31.0g of NaCl r added to a 1.0 L saturated solution of PbCl2, does conc. of Pb 2+ increase, decrease, or stay the same? Justify. Assume overall volume of solution does not change

PBCl2 <—> Pb 2+ + 2Cl - NaCl —> Na+ + Cl- NaCl dissolve completely and addition of excess Cl- shift equilibrium of 1st reaction to left, according to Le Chatelier's principle • Excess Cl- combine with Pb 2+ to form precipitate PbCl2 until product of [Pb 2+][Cl-]^2 equals K(sp) once again • Will decrease the conc. of Pb 2+ in solution

Trigonal Bipyramidal Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

PCl5; 5; 5; 0; 90 degrees and 120 degrees; sp^3d

Explain why the standard enthalpy of vaporization values for each set of compounds below rn't the same b) PH3 and NH3

PH4 has LDF and D^2. H2O has LDP and H-bonds. Bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Explain why standard enthalpy of vaporization, DeltaH(vap), values for each set of compounds below rn't the same b. PH3 and NH3

PH4 has LDF and dipole-dipole forces and NB3 has LDF and H-bonds • bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Steps for calculating Equilibrium Partial Pressures versus concentrations

PP 1) Make an ICE chart 2) Equilibrium expression with algebraic values Conc. 1) Find conc in mol/L 2) Make ICE chart 3) Equilibrium expression with algebraic values = 1. write/check balanced equation 2. under equation make an ICE table 3. substitute equilibrium concentrations into expression and solve for x, may require quadratic equation 4. determine equilibrium concentrations based on the value of x 5. check answer by substituting back into expression. (this is when given intial concentrations, how to get the equilibrium concentrations)

O2 + 2H2S —> 2S + 2H2O d) what must PP of reactants be to produce volt of 1.09 V

PP of O2 and H2S must be 1 atm. Under standard conditions, PP of all gases must be 1 atm

Ideal Gas Equation

PV=nRT P = pressure (atm) V= volume (L) n = number of moles R = (Molar Gas constant) 0.0821 L atm/K mol T = temperature (K)

Classify as a physical change, chemical change, or both. Justify by identifying the types of intermolecular or intramolecular forces that r involved in each processes and describing what happens to those forces while the processes r occurring. c) NH2F(L) —> NH2F(g)

Physical change • LDF and H-bonds in liquid NH2F r overcome, generating individual gaseous NH2F molecules

Classify as a physical change, chemical change, or both. Justify by identifying the types of intermolecular or intramolecular forces that r involved in each processes and describing what happens to those forces while the processes r occurring. f) H2O(s) —> H2O(L)

Physical change • LDF and H-bonds in solid water r stretched to produce liquid water

Classify as a physical change, chemical change, or both. Justify by identifying the types of intermolecular or intramolecular forces that r involved in each processes and describing what happens to those forces while the processes r occurring. a) CO2(s) —> CO2(g)

Physical change • LDF in solid CO2 r overcome, generating individual CO2 molecules

Pi or sigma bonds Which type of bond causes cis and trans isomers? Explain

Pi bonds cause the isomers bc they r rigid and prevent sigma bonds from spinning

Draw a V versus T graph, and plot the expected results of heating a gas from O K to 300 K at a constant P

Plot should consist of a straight line with a + slope that starts at the origin

Periodicity

Predictable physical and chemical trends that occur as one moves across a period or down a group in the periodic table. Ex. Atomic radii, Ionic radii, ionization energies, electron affinity, and electronegativity

Pressure (formula also)

Pressure (Pa) = Force (N) / Area (m^2) = Gasses exert P by bouncing off surfaces - Gas particles r evenly distributed in a container - The same number smash off every cm^2 per unit of time • Each collision exerts a force - The pressure is constant at constant temperatures (1 atm = 760 mm Hg = 760 torr)

Thermodynamically Favored Process (TFP)

Proceeds without any assistance from outside system - Water evaporates at 25^C - Fe rusts in presence of O2 and H2O - NaCl dissolves in H2O • A process that is TFP in 1 direction is non-TFP in other direction

Which if the following processes has the longer half-life? Justify Process Rate Law Rate Constant 1 Rate=k[A][B] k=6.79 M^-1 min^-1 2 Rate=k[A][B] k=14.55 M^-1 min^-1

Process 1 bc both 2nd order. has a slower rate when both processes share the same initial conc., as its rate constant has a smaller magnitude. Rate constant is inversely proportional to half-life

Explain why reaction (I) has -^_S^o whereas (II) has + ^_S^o 1. 2CH3OH + 3O2 —> 2CO2 + 4H2O (L) ^_S^o = -387 J/molK 2. 2CH3OH + 3O2 —> 2CO2 + 4H2O (g) ^_S^o = 89 J/molK

Products in 1 r much more ordered than reactants • Reactants r composed of 5 mol gas and products composed of 2 mol g and 5 mol L • Less disorder on products side yields -^_S^o • Products in 2 less ordered than reactants • Reactants r composed of 5 mol g and products composed 6 mol g • More disorder on products side yields + ^_S^o

Overall Order for a Reaction

Rate = k [A]^m • [B]^n Overall Order of the Reaction = m + n

When calculating Equilibrium PP or conc... • Equilibrium expression with algebraic values

Reactants (bottom): When K(eq) < 1 x 10^-4, you can really ignore this (x) to avoid the difficult math

What is the difference btwn a reaction intermediate and a catalyst?

Reaction intermediate - substance that is produced in one elementary step and then consumed in another. It's not present before the reaction starts, nor is it present when the overall reaction is complete Catalyst - substance that lowers the activation energy of a reaction by providing an alternate mechanism through which the reaction can proceed. It's not produced or consumed by the chemical reaction. It's present before the reaction starts and regains its original form when the overall reaction is complete

Suppose a system is in a state if equilibrium in accordance with the chemical equation below. Will the reaction shift if distilled water is added to the system? If so, in which direction will it shift? CH3COOH (aq) <-> CH3COO- (aq) + H+ (aq)

Reaction will shift to the right in order to increase the concentration of particles in the system

Redox Reactions in Acidic Solutions (II)

Reactions btwn metals and Nitric Acid • NO3- will oxidize metals that H+ cannot

Suppose the system is in the state of equilibrium. If pressure is then reduced, in which direction will the reaction shift? Br2 (g) + 3F2 (g) <-> 2BrF3 (g)

Reduction is pressure will cause the reaction to shift to the left. A reduction in pressure favors more moles of gas

Boyle's Law

Relationship btwn P and V of gases V1P1 = V2P2 • V is inversely proportional to P

As a catalyzed reaction proceeds, does the net conc. of the catalyst increase, decrease, or remain the same? Explain

Remains the same • catalyst is added to the reaction system • It's consumed in one elementary step and regenerated in another

_______ ____ caused by shielding effect reduce the effective nuclear charge experienced by outer electrons

Repulsive forces

Seesaw Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

SF4; 5; 4; 1; <90 and <120 degrees; sp^3d

Octahedral Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

SF6; 6; 6; 0; 90 degrees; sp^3d^2

Equal vol. of 0.2 M NaSO3 and 0.4 M HI r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

SO3 2- + 2H3O+ —> H2SO3 + 2H2O • System contains twice as many moles HI as it does NaSO3 • Allows each SO3 2- to accept 2 H+ and reaction goes to completion with no limiting reagent • aqueous species that have the highest conc. at equilibrium r H2SO4 and I- (spectator ion)

Isotope

Same atomic number, but different atomic masses. Same number of protons but not neutrons.

For each pair of compounds listed below, identify the compound that has the highest boiling point. Justify ur choice in terms of intermolecular forces j. AsH3 and SbH3

SbH3, bc both LDF and dipole-dipole but SbH3 is larger and more polarizable than AsH3 • SbH3 experiences greater LDF • As a greater amount of energy is required to break stronger FoA, boiling pt of SbH3 is higher

A chemical company was producing Mo(CO)5P(CH3)3 through the following process Mo(CO)5 + P(CH3)3 <-> Mo(CO)5P(CH3)3 One of the chemists suggested that they should add Mo(CO)6 to the system, as it would create the following reactions Mo(CO)6 <-> Mo(CO)5 + CO Why would chemist make this suggestion?

Second process produces Mo(CO)5, which is one of the reactants in the first process • Addition of Mo(CO)6 effectively adde mote Mo(CO)5 to the system, which increases the rate of forward reaction in the first process and, in turn, increases the concentration of the desired product, Mo(CO)5P(CH3)3

Is XeO2F2 polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Seesaw so 2 F-O bonds and 2 lone pairs - Polar individual bonds bc F and O is more electronegative than Xe - Polar overall compound bc dipoles don't cancel

Spinning Single bond - Double bond -

Single - Sigma bonds r able to spin on an axis Double - Pi bonds prevent sigma bonds from spinning on this axis

Is XeF4 polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Square planar so 4 Xe-F bonds and 2 lone pairs - polar individual bonds bc F is more electronegative than Xe - Non-polar overall compound bc dipoles cancel - Dipoles r = and opposite to one another in the same plane

Write balanced net ionic equations for reactions Solid strontium carbonate is placed in a solution of nitric acid

SrCO3 + 2H+ —> Sr 2+ + H2O + CO2

HBr + H2O ---> H3O+ + Br- Weak or strong Acids and Conjugate Bases??

Strong acids/weak conjugate bases - H2O and Br- compete for protons - H2O stronger base, so wins most of the time and reaction goes to completion

Acid Strength

Strong acids experience ~ 100% ionization in water HA + H2O —> H3O+ + A- • 1 way arrow, as equilibrium lies far to the right Weak acids partially ionize in water HA + H2O <=> H3O+ + A- • 2 way arrow, as equilibrium lies to the left or ~ middle

A 1.0 mole pure sample of molten tin is dissolved in a 5.0 mole pure sample of molted copper. The solution is set aside to cool and solidifies. The atomic radius of tin is 140 pm and the atomic radius of copper is 128 pm a) Identify the type of alloy that is formed. Justify ur answer

Substitutional alloy is formed. Since atomic radii of the 2 elements r similar, tin will substitute for copper

Chemical company claims they r able to produce certain product through T un-FP. Explain

T un-FP do produce products. • When T un-FP at equilibrium, conc. reactants greater than conc. products • Company could create system with only reactants, wait for that system to react equilibrium, and extract desired product • According to Le Chat's principle, would create stress on system that cause reaction to shift in forward direction to produce more desired product

Is BrF3 polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

T-shaped so 3 Br-F bonds and 2 lone pairs - Polar individual bonds bc F is more electronegative than Br - Polar overall compound bc dipoles don't cancel due to asymmetrical arrangement of 3 B-F bonds

Voltaic cell operates with 0.40 M Zn^2+ at solid Zn anode, and 1.45 M Zn^2+ at solid Zn cathode at 25'C. Is reaction TFP? Explain

TFP cuz [reactants] > [products] or conc Zn 2+ at anode less than conc Zn^2+ at cathode

Explain why the temp. of a liquid remains constant while it's being boiled, although heat continues to be absorbed. Describe what happens to the heat that is absorbed.

Temp. increases as the average KE of the molecules in a system increases. As the temp. doesn't increase during the boiling process, average KE of the water molecules doesn't increase. The heat is absorbed by the liquid is used to break intermolecular FoA, allowing units it separate from one another to become independent gaseous molecules.

Is SO4 2- polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Tetrahedral so 4 S-O bonds and no lone pairs - polar individual bonds bc O is more electronegative than S - Non-polar overall compound bc dipoles cancel due to the symmetrical arrangement of the 4 S-O bonds

Surface Catalysis

The catalyst binds to or forms covalent bonds with a surface, thereby forming a new intermediate

Use the 'shielding effect' to explain why this e- has the lowest ionization energy

The e- in the n=4 shell is partially 'shielded' from the force of attraction by the inner core e- from the n=1, n=2, and n=3 shells. The e- in n=4 is repelled by the negative charges of those inner core e-, and this reduces the force of attraction between it and the positive nucleus.

Ion-Dipole

The forces of attraction between an ion and a polar molecule - smaller ion, condensed and attract more molecules to itself than larger ions

A 150.0 g pure sample of Ilmenite, FeTiO3, was analyzed and it was found to contain 36.81% Fe by mass. Would you expect this % to be the higher, lower, or the same in a 450.0 g pure sample of FeTiO3? Justify your answer.

The mass % of Fe would be the same (36.81%) in the 450.0 g pure sample of Ilmenite. Each compound in both samples contains exactly 1 Fe, 1 Ti, and 3 O. Average atomic mass of Fe would be the same in both samples as the ratio of the different isotopes of Fe would be the same in both samples, the average atomic mass of Ti would be the same in both samples as the ratio of the different isotopes of Ti would be the same in both samples, and the average atomic mass of O would be the same in both samples as the ratio of the different isotopes of O would be the same in both samples. For the reasons, the mass % of Fe would be the same in both samples.

Where is the lowest ionization energy in the shell model of Potassium located??

The outside electron of the 4th outer shell.

The reaction-energy profile diagram below outlines an uncatalyzed process Potential Energy ______________ / \ / \ l l / \ _________________/ \__________________ Reaction progress c) Explain why the presence of a catalyst increases the reaction rate.

The presence of a catalyst increases the reaction rate by providing an alternate mechanism for the reaction that has a lower activation energy. • When minimum collision energy that is required to trigger a chemical reaction is lower, a percentage of collisions btwn reactants will result in chemical reactions • If more collisions per period of time r resulting in chemical reaction, reaction rate is higher

Solute

The substance that is less plentiful in a solution

Solvent

The substance that is more plentiful in a solution

A 1.0 mole pure sample of molten tin is dissolved in a 5.0 mole pure sample of molted copper. The solution is set aside to cool and solidifies. The atomic radius of tin is 140 pm and the atomic radius of copper is 128 pm b) Identify the solvent in this solution. Justify

There r 5.0 moles of Cu and 1.0 moles of tin in the solution. Since Cu is more plentiful, it is the solvent

Do the peaks on infrared spectra point up or down? Justify

They point down, as % transmittance is in the y-axis • if the bind is absorbing most if the light, only a very small amount is transmitted through the sample

If given a metal and an ionic compound with the same metal in it...

This is also disproportionation, take the metal and the metal ion and form ions at the intermediate ox state. These often r left for an extended period of time as well

Is ammonia polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Trigonal Pyramidal so 3 N-O bonds and 1 lone pair - Polar individual bonds bc N is more electronegative than H - Polar overall compound bc dipoles don't cancel due to asymmetrical arrangement of 3 N-H bonds - N end is slightly - and H ends r slightly

Is PF5 polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Trigonal bipyramidal so 5 P-F bonds and no lone pairs - polar individual bonds bc F is more electronegative than P - Non-polar overall compound bc dipoles cancel due to the symmetrical arrangement of the 5 P-F bonds

Is BCl3 polar or non-polar?? Explain by discussing shape, polar bonds, and whether or not dipoles cancel

Trigonal planar so 3 B-Cl bonds and no lone pairs - polar individual bonds bc Cl is more electronegative than B - Non-polar overall compound bc dipoles cancel due to the symmetrical arrangement of the 3 B-Cl bonds

IE_He is almost ________ IE_H+ This increase can be explained through Coulomb's Law, as He has ______ as many protons as H

Twice

volume adjustment for gases under high pressure (equation)

V(ideal) = V(measured) - nb(Correction decreases V(measured) to V(ideal) = V(measured) = V of empty space + V of gas particles. (V measured in the lab) = V(ideal) = Corrected V, which can be used in the ideal gas equation. (V of actual empty space in the container = n = moles of gas = b = a constant for the type of gas

VSEPR Theory

Valence Shell Electron Pair Repulsion Theory - Charge clouds repel each other due to Coulombic forces - Terminal atoms move far away from one another as possible - Results in distinctive geometric shapes

Galvanic cell operates through reaction 3 Cu 2+ + 2 Cr —> 3 Cu + 2 Cr 3+ c) Would value of cell potential, E(cell), under conditions ([Cr 3+] = 1.23 M) be greater than or less than value of E^o (cell)

Value of E(cell) would be less the value for E^o (cell), as [reactants] decr from 1.0 M to 0.65 M and [products] incr from 1.0 M to 1.23 M. Would cause decr in rate of forward reaction

Explain each of the following occurrences by referring to structure of atoms in question (energy levels, orbits, protons, etc.) -The atomic radius of Mg is smaller than the atomic radius of Ca.

Ve- in Mg r contained within 3s orbital and Ve- in Ca r contained within 4s orbital. -This means that outermost e-s in Ca experience greater shielding effect, possess more energy and r further from nucleus than outermost e-s in Mg. -According to Coulomb's Law, force of attraction decrease as distance between e-s and protons increases, so Mg has smaller atomic radius.

Explain each of the following occurrences by referring to structure of atoms in question (energy levels, orbits, protons, etc.) -Atomic radius of O is smaller than atomic radius of C

Ve-s of both elements r same shell (n=2). Thus, shielding effect experienced by outer e-s of both elements similar. -O atoms has 8 protons in nucleus while C atom only has 6 protons. -Addition of protons increases effective nuclear charge on Ve-s. -Moving from left to right, more protons r added. According to Coulomb's Law, increase in + charge increases force of attraction experienced by each e-, thereby decreasing radius. This gives O a smaller atomic radius.

Titration curve for weak base that is titrated by a strong acid differs from that a strong base that is titrated by a strong acid? 2 main differences?

Weak base: Initial pH= <13 by >7 pH at equivalence pt <7 Strong base: initial pH= ~13 pH at equivalence pt = 7 pH of 0.1 M strong base is 13

Strengths of acids and bases (List)

Weak to strong Acid (CH4, H2O, NH4+, HF, HNO3, HCl) Base (Cl-, NO3-, F-, NH3, OH-, CH3-)

The table below outlines the results from 3 experiments, which were conducted at the same temperature, involving the following reaction Ex: [O2]initial [NO]initial Initial Reaction Rate 1 0.022 0.026 0.0256 M/s 2 0.022 0.013 0.0064 M/s 3 0.044 0.013 0.0128 M/s a) The results show that the initial rate increased when the initial conc. of either reactant increased. Explain why increasing the conc. or partial pressure of a reactant can increase the rate at which the reaction proceeds

When the conc. of a reactant incr., there is more of that reactant circulating within the same volume. • As a result, there will be more collisions btwn reactants per second • Reaction rate incr. when collision rate incr.

The energy profile diagram below was plotted for the following second order elementary reaction: A + BC <-> AB + C Potential Energy ______________ / \ / \ l l / \ _________________/ \__________________ b) At high temperatures a relatively large fraction of collisions btwn reactants resulted in the formation of products. Describe the interactions btwn the reactants from the moment of a collision through to the formation of separate products

When 2 molecules collide, KE from their motion is converted into PE • This PE cultivates itself in the bonds btwn all of the reacting atoms (A + BC —> A-B-C) • In this case, A forms a new bond with B, and the bond btwn B and C lengthens • At the top of the curve, PE contained within the bonds of A-B-C has reached a maximum • At this pt, we don't have reactants or products • What we have is an extremely unstable aggregation of all of the atoms contained by the reacting molecules • During the downward section of the graph, PE begins to drop • Electron density btwn A and B, causing that bond to shorten • Electron density btwn B and C decr., causing that bond to start stretching • When the graph plateaus again, there r 2 separate species: AB and C

65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl at 25'C g) Explain why the solution is acidic at the equivalence pt of the titration. Use a chemical equation

When NH3 is titrated with a strong acid the reaction produces NH4+ • At the equivalence pt, NH4 +, which has developed high conc., reacts with water to produce H3O+ • Increased conc. of H3O+ makes solution acidic NH4+ + H2O <-> NH3 + H3O+

Le Chatelier's Principle

When a system at equilibrium is subjected to a stress, equilibrium will shift in order to reduce that stress The only 3 stresses r changes in... • Pressure • Conc. • Temp.

Explain why the solubility of AgBr decreases when NaBr is added to the system

When additional Br- is added to the system, equilibrium in the following reaction [AgBr (s) <-> Ag+ (aq) + Br- (aq)] shifts to the left to relieve the stress (Le Chatelier's principle)

A student notes that the grass was damp when she arrived at school at 7am and it was dry when she went out for the break btwn morning classes. Explain this occurrence in terms of reversible processes

When it was cold at night, the following process occurred and the water condensed on the grass H2O (g) —> H2O (L) When it warmed up, the following process occurred and the water evaporated H2O (L) —> H2O (g)

Determining Rate Law • Slow initial step

When the mechanism has a slow initial step, the rate law for the overall reaction is = to the rate law for the 1st step

Linear Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

XeF2; 5; 2; 3; 180 degrees; sp^3d

Square Planar Ex. - Charge Clouds - Bonds - Lone Pairs - Bond Angle - Hybridization -

XeF4; 6; 4; 2; 90 degrees; sp^3d^2

Does the photoelectron spectra (C, N, O, F, Ne) suggest a need to refine shell model of atom? If so, is there another model that provides a better explanation of this data? Justify

Yes, bc these spectra do no support shell model. -show that elements in 2nd period of periodic table have 1 shell in n=1, and 2 subshells in n=2 that r at different energy levels. 1st subshell in n=2 can hold up to 2 e-s and 2nd subshell in n=2 can hold up to 6 e-s. -Quantum mechanical (QM) model used to explain why 2 subshells in n=2 shell. -According to QM model, n=2 contains 2s orbitals, which can hold up to 2 e-s at same quantized energy level, and 3 2p orbitals, each of which can hold 2 e-s. All e-s in 2p r at same quantized energy level.

In order to maximize the production of SO2, a chemist suggested that they increase the pressure on following system. Would this work? Justify 2PbS (g) + 3O2 (g) <-> 2PbO (s) + 2SO2 (g)

Yes, increasing the pressure on the system favors fewer moles of gas and more concentrated states, as this reduces the internal energy of the system • There r 5 moles of gas on the reactants side of the equation, and 2 moles of gas and 2 moles of solid on the products side of the equation • Thus, increasing the pressure will increase the concentration of SO2 (g) (a product) in the system

Suppose a system operating in accordance with the chemical equation below is in a state of equation. Will the reaction shift if the pressure is reduced? If so, in which direction will it shift? Cl2 (g) + 2I- (aq) <-> 2Cl- (aq) + I2 (s)

Yes, it will shift to the left • A reduction in pressure will favor less condensed states and more moles of gas

After learning about PES data, do you see a need to refine the Shell Model of an atom?

Yes, shell model states that all e- within given shell are at same energy level. PES data shows that there are 2 subshells in n=2 and 3 subshells in n=3

effective nuclear charge (Zeff)

Zeff = z - σ Zeff - charge experienced by an e- Z - actual nuclear charge (atomic number of element) σ - shielding constant (0 < σ < Z)

Reaction Orders (m and n) Zero, First, and Second Order Reactions

Zero • Doubling the concentration of that species, has no affect on the reaction rate First • Doubling the concentration of that species, doubles the reaction rate Second • Doubling the concentration of that species, quadruples the reaction rate

Zinc metal is placed into a solution of hydrochloric acid. a. What substance is reduced in the reaction?

Zn + 2H+ → Zn2+ + H2 H+

How would a straight ln[A] line change in the presence of a catalyst?

[A] and ln[A] decrease at a faster rate in the presence of a catalyst • Initial conc. of A is the same in both situations

Zeroth Order Integrated Rate Law

[A]t = -kt + [A]0 [A]0= initial conc (at t=0s) [A]t= conc after some period of time • A plot of [A]t (conc.) vs t (time) produces a straight line for 0th order reactions, as the rate does not change when conc changes • This is used to determine if a reaction is 0th order, as a straight line will only occur if reaction is 0th order Slope = -k

Citric Acid H3C6H5O7 is a polyprotic acid. K(a1) = 8.4x10^-4, and K(a2) = 1.8x10^-5 at 25'C a) which species has lowest conc. in a 1.0 M H3C6H5O7 solution: H3C6H5O7, H2C6H5O7-, or HC6H5O7 2-. Justify

[H3C6H5O7 2-] is lowest, bc K2 is smaller than K1

Citric Acid H3C6H5O7 is a polyprotic acid. K(a1) = 8.4x10^-4, and K(a2) = 1.8x10^-5 at 25'C b) which possesses highest conc. in a 1.0 M H3C6H5O7 solution: H3C6H5O7, H2C6H5O7-, or HC6H5O7 2-. Justify

[H3C6H5O7] is largest, bc it's weak acid. Equilibrium lies far to left (mostly reactants).

If pH of HBr solution is same as pH of a CH3COOH solution, is [HBr] less than, equal to, or greater than [CH3COOH]? Justify

[HBr] < [CH3COOH]. • HBr is a strong acid and experiences 100% ionization • CH3COOH is a weak acid and experiences much less than 100% ionization • Fewer moles of HBr r required to produce the same molar concentration of H3O+ in the solution

A 1.0 mole sample of HNO3 is added to water. Final volume of solution is 1.5 L and final temp. of solution is 25'C C) Molar conc. of nitrate ion in final solution?

[NO3-]=0.67 M, as HNO3 is strong acid the experiences 100% dissociation

In a 0.450 M HONH2 solution, [OH-] = 5.28x10^-6 M HONH2 + H2O <-> HONH3+ + OH- Find [HONH3+]

[OH-] = [HONH3+] = 5.28 x 10^6 M

In a 0.032 M NH3 solution, [OH-] = 1.27x10^-3M NH3 + H2O <-> NH4+ + OH- Find [NH4+]

[OH-] = [NH4+] = 1.27 x 10^-3 M

Galvanic cell operates through reaction 3 Pb 2+ + 2 Al —> 3 Pb + 2 Al 3+ What happen to standard cell potential, E^o, if conc of Al 3+ decr and conc Pb 2+ remained the same? explain

[products] < [reactants] give - value for LnQ • - value for LnQ cause cell potential to incr according to equation. (-) x (-) = + E(cell) = E^o - RT/nF x Ln [Al 3+]^2 / [Pb 2+]^3

Due to large + value for ^G^o, Zn cannot be extracted from ZnS through process ZnS —> Zn + S ^G(375K) = 198.3 kJ/mol But, combustion of S is TFP S + O2 —> SO2 ^G(375K) = - 300.1 kJ Is formation of pure Zn a TFP if 2 reactions coupled? Justify

^G(375K) = -101.8 kJ Formation of pure Zn TFP bc ^G^o < 0

At 375 K, decomposition of Cu (I) oxide is non-TFP 2Cu2O —> 4Cu + O2 ^G(375K) = 280 kJ If solid carbon added to system, following reaction occur 2C + O2 —> 2CO ^G(375K) = - 287.6 kJ Is formation of pure Cu from decomposition Cu (I) oxide in presence of solid C a TFP at 375 K? Justify

^G(375K) = -7.6 kJ Formation of pure Cu from decomposition Cu2O in presence of C is TFP bc ^G(375K)<0

Change in a System's Internal Energy (triangleE) equation

^_E = q + w ^_E = Change in PE and KE of particles in a system q = Heat transferred into (+ value) or out of (- value) the system w = Work done on the system by the surroundings (+ value as energy flows into the system) or work done on surrounding by system (- value as energy flows out of the system) • Work results from changes in the volume of a gas

Gibbs Free Energy (^_G)

^_G = ^_H - T^_S ^_G = Free energy change (kJ). Left over energy to do work. Max amount of energy that can be used to do work. ^_H = Enthalpy change (kJ). Energy transferred as heat T = temp of system (K). Energy used to create disorder ^_S = Entropy change (kJ/K) • If ^_G<0, process is TFP • Standard state: - Pure substances - Solutions with 1.0 M conc. - Gases at a pressure of 1.0 atm (1 bar)

Free Energy of Formation

^_G^o (f) = free energy change that occurs when 1 mol compound is made from its elements in their standard states ^_G^o (f) = for an element in its standard state = 0 ^_G^o (rxn) = EnS^o (products) -EnS^o (reactants) E = sum of n = stoichiometric coefficients

Estimation of K(eq)

^_G^o = -RT Ln K(eq) At 298 K, RT ~~ 2400 J/mol 8 J/molK x 300 K = 2400 J/mol If, ^_G>>2400 J/mol, then K<<1 If, 2400 J/mol > ^_G^o > -2400 J/mol, then K~~1 If, ^_G^o << -2400 J/mol, then K>>1 Can be used to estimate equilibrium position

Relating ^_G^o to K(eq)

^_G^o = -RT Ln K(eq) K(eq) = e^(-^_G^o / RT) -^_G^o = must be in J/mol R = 8.314 J/molK T = must be in K K(eq) = must be calculated using PP for gases and molar conc. for aqueous species

Gibbs Free Energy (^_G^o)

^_G^o = -nFE^o ^_G^o = Gibbs free energy (J/mol) n = # mol e-s transferred in balanced chemical equation (mol electrons) F = Faraday's constant (96,500 C/mol e-s) E^o = emf (V) or (J/C) • As ^_G^o becomes more -, E^o becomes + • As ^_G^o becomes more +, E^o becomes -

Estimation of K(eq)

^_G^o = 0; K(eq) = 1; Extent of Reaction at Equilibrium = [Reactants] = [Products] ^_G^o >>+2.4 kJ; K(eq) << 1; Extent of Reaction at Equilibrium = Mostly Reactants ^_G^o << -2.4 kJ; K(eq) >> 1; Extent of Reaction at Equilibrium = Mostly Products

Temperature and Thermodynamic favorability ^_G = ^_H - T^_S

^_H = -, ^_S=+, TFP at all temps ^_H = +, ^_S=-, non-TFP at all temps ^_H = -, ^_S=-, TFP at low temps & non-TFP at high temps ^_H = +, ^_S=+, TFP at high temps & non-TFP at low temps

An unknown ionic compound dissolves in water. During dissolving process the temp. of solution decreases. Predict signs (+ or -) for ^_H, ^_S, ^_G. Justify

^_H>0. +. • Heat pulled from surrounding H2O to system to facilitate process • Process must be thermo favored, as compound dissolved so equilibrium lies right • ^_S>0 + • ^_G= ^_H - T^_S • Since ^_H>0, ^_G<0 to produce a - ^_G

^_S(universe)

^_S(universe) = ^_S(system) + ^_S(surrounding) If ^_S(universe) > 0, reaction is TFP • If process: - Increases entropy of system (^_S(sys)>0), - and is exothermic (^_S(surr)>0), - Is must also be TFP (^_S(universe)>0)

PE (electric) & Potential Difference (^_V)

^_V (volts) = ^_PE (joules) / Charge (Coulombs) • PD (^_V) is the change in Electrical PE, ^_PE(electric), per unit charge • Takes 12 J energy to pull 6.25 x 10^18 e-s (-1C of charge) from the cathode to anode • 6.25 x 10^18 e-s now possess 12 J of PE, which will be converted into KE as spontaneously flow back to cathode

Cu2S —> 2Cu + S a) Predict sign associated with ^_S value for reaction. Justify b) does change in entropy (S) favor R or P? which condition would it prefer? c) ^_G for reactions is - at low temps and + at high temps. Predicts sign associated with ^_H for reaction Justify d) Under what conditions does TFP?? Justify

a) + ^_S bc all particles in same phase but more mol products than reactants so reactants more ordered than products b) ^_S favors products as more possible arrangements of product particles than reactant particles c) To be true, ^_H must be +. if reaction was exothermic, would be TFP under all conditions ^G = ^H - T^S d) ^G = ^H - T^S • Significance of fav + ^S incr as temp incr, bc multiplying by larger #. High temp, absolute value of -T^_S will exceed that of ^H and make ^G - • significance of fav + ^S decr as temp decr, bc multiplying by smaller #. At low temps, + ^H make ^G =+. TFP at high temps

2SO2 + O2 —> 2SO3 a) Predict sign associated with ^_S value for reaction. Justify b) does change in entropy (S) favor R or P? which condition would it prefer? c) ^_G for reactions is - at low temps and + at high temps. Predicts sign associated with ^_H for reaction Justify d) Under what conditions does TFP?? Justify

a) - ^_S as there r 3 g reactants and only 2 g products, so products r more ordered than reactants b) ^_S favors reactants as larger degree of disorder or randomness on reactants side c) To be true, ^_H must be -, as - ^_S is unfav and if reaction was endothermic, wouldn't be TFP under any conditions ^G = ^H - T^S d) ^G = ^H - T^S • Significance of unfav - ^S incr as temp incr, bc multiplying by larger #. High temp, -T^_S will make ^G + (- x - = +) • significance of unfav -^S decr as temp decr, bc multiplying by smaller #. At low temps, -^H make ^G =-. TFP at low temps

(5)__________________(7) / / __________/ (3) /(1.3) / a) During what period of time was the substance at its normal freezing point? b) Over what period of time was the substance boiling?

a) 1.3-3.1 mins b) 5-7 mins

voltaic cell: Ag on left and Pb or right a) Write balance net ionic for TFP c) What conc. Ag+ and Pb 2+ necessary for cell to produce standard cell potential d) Which electrode is anode and which cathode? e) Better to use NaCl or NaNO3 in salt bridge? f) Which direction r e-s flowing?

a) 2 Ag+ + Pb —> 2Ag + Pb 2+ c) must be 1.0 M. Standard conditions for all aqueous cells must be 1.0 M conc. of all species involved in reaction d) Pb = anode cuz ox occurs in anode Ag = cathode cuz red occurs in cathode e) Better to use NaNO3 in salt bridge. Cl- travel to anode where bond with Pb 2+ to form solid precipitate PbCl2. Will decr conc. Pb 2+ at anode f) E-s flow from anode to cathode. In cell, flowing from Pb anode to Ag cathode (R—>L)

1.0 M CuSO4; 1.0 M ZnSO4; 1.0 M Au(NO3)2 Al metal, Zn metal, Au metal; salt bridge Use this and wires and voltmeter to construct galvanic cell with largest possible value for E^o (cell) a) write balanced net equation that produces larges possible value c) which metal is used at anode? Cathode? d) What would voltage of cell if constructed without a salt bridge? Explain

a) 2 Au 2+ + 3 Zn —> 3 Zn 2+ + 2 Au c) Au = cathode and Zn = anode d) If no SB, voltage = 0. SB needed to maintain neutral solutions in 1/2 cells. Without SB, ions from SB not able to balance accumulation of anions in 1/2 cell of cathode, nor balance influx cations in 1/2 cell anode. When neutrality in 1/2 cells not maintained, e-s cease to flow through wire and voltage = 0

Direction current applied to Ni (II) F a) Write balanced equation for anode b) Write balanced equation for cathode c) Write balanced net for overall that takes place in the cell d) Predict sign for ^G

a) 2 H2O (L) —> O2 E ox = -1.23 V b) Ni 2+ + 2 e- E red = -0.25 V c) 2H2O (L) + 2 Ni 2+ —> O2 + 4H+ + 2 Ni d) + bc electrolytic reactions r always non-TPF. Non-TFP have + ^G values

Direction current applied to Mn (II) I a) Write balanced equation for ox 1/2 b) Write balanced equation for red 1/2 c) Write balanced net for overall that takes place in the cell d) Which reaction takes place at cathode? e) Predict sign for ^G

a) 2 I- —> I2 + 2 e- E ox = -.53 V b) 2 H2+ + 2 e- —> H2 + 2OH- E red = -0.83 V c) 2 I- + 2H2O —> I2 + H2 + 2 OH- d) 2 H2O + 2 e- —> H2 + 2OH- e) + bc electrolytic reactions r always non-TPF. Non-TFP have + ^G values

Direct current is applied to aqueous Ni (II) Br solution a) half reaction takes place at anode b) half reaction takes place at cathode c) Overall reaction in cell d) do E-s flow from anode to cathode or from cathode to anode? e) Predict sign for ^G

a) 2Br- —> Br2 + 2e- E ox= -1.07 V b) Ni 2+ + 2 e- —> Ni E red = -0.25 V c) 2 Br- + Ni 2+ —> Br2 + Ni d) E- always flow from anode to cathode e) + cuz electrolytic reactions r always non-TFP. Non-TFP have + value

A two step reaction is outlined below. H2O2 + I- —> H2O + IO- (slow step) H2O2 + IO- —> H2O + O2 + I- (fast step) a) overall reaction?? b) Identify intermediate in this reaction mechanism c) Rate law for each of the elementary steps?? d) Rate law for the overall reaction??

a) 2H2O —> 2H2O + O2 b) IO- bc it's produced and consumed in the reaction c) Step 1. Rate = k1 [H2O2][I-] Step 2. Rate = k2 [H2O2][IO-] d) Rate = k [H2O2][I-]

Water being electrolysed in presence of non-reactive electrolyte under standard conditions a) Write balanced equation for ox 1/2 b) Write balanced equation for red 1/2 c) Write balanced net for overall that takes place in the cell d) Label anode and cathode and write half reaction that takes place at each electrode e) In which direction do e-s flow

a) 2H2O —> O2 + 4 H+ + 4 e- E ox = -1.23 V b) 2H2O + 2e- —> H2 + 2OH- E red = -.83 V c) 2H2O —> O2 + 2H2 d) Reaction produces twice as many moles of H2, as does O2. H2 produced at cathode and O2 produced at anode. Cylinder on left contains twice as much gas as one on right. Cylinder on left is cathode and one on right is anode e) E-s flow from R —> L from anode to cathode

A two step mechanism is outlined below 2NO —> N2O2 (fast equilibrium) N2O2 + Br2 —> 2NOBr (slow) a) overall reaction?? b) Identify the intermediate in this reaction mechanism. Justify c) Rate law for the overall reaction?? d) Order of the reaction with respect to NO according to the above mechanism? Justify

a) 2NO + Br2 —> 2NOBr b) N2O2 bc it's produced and consumed in the reaction c) k1 [NO]^2 = K-1 [N2O2] [N2O2] = K1/K-1 [NO]^2 Rate = k2 [N2O2][Br2] Rate = K2 (k1/k-1 [NO]^2) [Br2] Rate = k3 [NO]^2 [Br2] d) reaction id 2nd order with respect to NO, as the [NO] is raised to the power of 2 in the rate law

A two step mechanism is outlined below. NO2Cl —> NO2 + Cl (slow step) NO2Cl + Cl —> NO2 + Cl2 (fast step) a) overall reaction?? b) Identify intermediate in this reaction mechanism. Justify c) Rate law for each of the elementary steps?? d) rate law for overall reaction??

a) 2NO2Cl —> 2NO2 + Cl2 b) Cl bc it's produced and consumed in the reaction c) Step 1. Rate = k1 [NO2Cl] Step 2. Rate = k2 [NO2Cl][Cl] d) Rate = k2 [NO2Cl]

A two step mechanism is outlined below. F2 + NO2 —> NO2F + F (slow step) NO2 + F —> NO2F (fast step) a) overall reaction?? b) Identify intermediate in this reaction mechanism. Justify c) Rate law for each of the elementary steps?? d) rate law for overall reaction??

a) F2 + 2NO2 —> 2NO2F b) F bc it's produced and consumed in the reaction c) Step 1. Rate = k1 [F2][NO2] Step 2. Rate = k2 [F][NO2] d) Rate = k [F2][NO2]

Equal moles of F2 and ClO2 r drawn into a vacuum where the following process takes place F2 + 2ClO2 <-> 2FClO2 \_____________ \ \____________ ____________ / / a) Identify which curve on the graph is associated with which molecule in reaction b) At what time does the system reach equilibrium? Justify

a) F2, ClO2, then FClO2 b) 45 minutes. The system is at equilibrium when the conc. of reactants and products remain constant. On graph, that happens when the plotted line become horizontal

Direct current applied to solution of CaBr2 and gas evolved from one of the electrodes a) Is gas evolved at anode or cathode? b) What is gas?

a) Gas evolve from cathode. E^o (red) (Ca 2+) = -2.87 V and E^o (red) (H2O) = -0.83 V • Water reduced at cathode, as more +, less - reduction potential No gas evolve from anode E^o (ox) (Br-) = -1.07 V and E^o (ox) (H2O) = -1.23 V Br- oxidized to Br2 at anode b) H2 evolved 2H2O (L) + 2 e^- —> H2 + 2 OH- E(red) = -0.83 V

a) Which color of light has the highest frequency: red or green?? b) Which color of light has the longest wavelength: green or violet?

a) Green: shorter wavelength = higher frequency b) green

What K(eq) tells us!!! a) = 0.0184 b) = 2.4 x 10^3 c) = 1

a) K(eq) >> 1 (mostly products and equilibrium lies far to the right) b) K(eq) << 1 (mostly reactants and equilibrium lies far to the left) c) ~ equal amounts of each and equilibrium lies in the middle)

Suppose NaOH is added to system when it's at equilibrium NH3 (aq) + H2O (L) <-> NH4+ (aq) + OH- (aq) a) In which direction will the reaction shift after the NaOH is added? b) Will this stress increase or decrease the value of the reaction quotient, Q? Justify

a) Left b) Increase the value of Q as [OH-] is part of the numerator in the reaction quotient expression

Ionic Compounds of Mg^2+ a) Write the ionic formula for every compound that can be formed between Mg and an element from Group 7A. b) Provide an explanation for ratio of bonding for the compounds of Mg^2+ in part (a) c) Will ratio of cation to anion be the same or different for when other Group 2A cations form ionic compounds with anions of Group 7A? Justify

a) MgF_2, MgCl_2, MgBr_2, MgI_2, MgAt_2 b) 1:2, bc Mg has 2+ and all 7A have 1- c) Same, bc all Group 2A cations have a charge of 2+ and all Group 7A anions have a charge of 1- so in order to be a neutral atom, the ratio will be the same.

The rate law for the overall reaction outlined below is: Rate = k [NO][Cl2]. NO + Cl2 —> NOCl2 NOCl2 + NO —> 2NOCl === 2NO + Cl2 —> 2NOCl a) Identify the intermediate in this reaction mechanism. Justify b) Which step occurs the fastest? c) Which step determines the overall rate of the reaction??

a) NOCl2 bc it's produced and consumed in the reaction b) Step 2. NOCl2 + NO —> 2NOCl bc it doesn't determine the rate of the overall reaction. Rate law for the overall reaction is the same as the rate law for the 1st step, which means that the first step is the slow, rate determining step c) Step 1. NO + Cl2 —> NOCl2 determines the overall reaction. The rate law of the overall reaction is: Rate = k [NO][Cl2]. This is also rate law for this step. Step 1 is the slow step, which determines the rate of the overall reaction

Reaction below was found to be second order for A and zero order for B A + B —> C + D Following mechanism was proposed for this reaction Step 1) 2A —> C + E (slow step) Step 2) B + E —> A + D (fast step) a) Rate law for this reaction?? b) Give 2 reasons why the above mechanism is possible

a) Rate = k [A]^2 b) 1. When the 2 reactions r added together, we get A + B —> C + D, which is the same as the overall reaction 2. Slow step is consistent with the rate law, as B is zero order and thus is not included in the rate law; and A is 2nd order and raised to the power of 2 in the rate law, which means that 2 A particles collide in the rate determining step

^H^o = -537.2 kJ/mol and ^S^o = 13.7 J/molK H2 + F2 <-> 2HF a) TFP?? K(p) = 6.79 x 10^94 b) Does system contain mostly reactants or mostly products at equilibrium? Justify

a) TFP bc ^H^o < 0 and ^S^o > 0 c) mostly products bc K(eq) is very large, meaning that numerator in equilibrium expression, [products], is much larger than denominator, [reactants] • Know system contains mostly products bc ^G^o = -

Galvanic cell contains 2 standard 1/2 cells. X(X^+2) (NO3-)—> Y (Y^+2) (NO2 -) Cd + Pb 2+ —> Cd 2+ + Pb a) What r Y and Y^+2 in diagram? b) Initial conc of X(NO3)2 in solution? c) which electrode is cathode and anode? d) Write balanced equation for 1/2 cell reaction that occurs at cathode

a) Y=Pb and Y^2+ = Pb 2+ b) Initial conc of X(NO3)2 must be 1.0 M, as want 1/2 cells exits under standard conditions. For aqueous solutions, standard conditions r defined as 1.0 M conc of all aqueous reactants and products and 25'C c) X = anode bc ox Y = cathode bc red d) Pb 2+ + 2 e- —> Pb

Ionic compound was dissolved in 250 mL water and temp of solution incr from 25'C to 42'C a) is ^H greater/less than 0?? b) Expect ^H has large or small magnitude? c) Is ^G greater/less than 0?? d) Determined that cation carries a +1 charge and anion carries a -1 charge. Predict sign ^S e) Expect ^S have large/small magnitude?

a) ^H < 0 and exothermic process. Heat flowed from system into surrounding water. Cause temp of solution to incr b) ^H has large - value, as temp of water incr by 17'C c) ^G<0. Since compound dissolved, must be TFP as equilibrium lies to right AB <-> A+ + B- d) ^S>0 bc ions carry small charges • Ions with small charges won't decrease entropy water that surrounds ions by much • most ionic compounds don't decrease entropy of water surrounding ions by enough to make produce an overall - ^S • Only ions with large changes and small radii, such as Al 3+ able to do this e) + ^S since arrangement ions in ionic solid is very organized and arrangement of ions in a solution is very random • Since ions carry +1 and -1 charges, they won't decrease entropy water that surrounds them by very much

Dissolving solid NaCl in water a) What sign is ^_S for dissolving NaCl b) Does entropy incr or decr during dissolving process? Justify by describing changes in entropy that occur during dissolving process

a) ^_S = + b) incr bc solid NaCl is highly organized network of Na and Cl ions • When dissolves in water, highly organized network breaks up into individual Na+ and Cl- that r free to move around within the solution • Entropy incr as start with highly organized solid and end up with a very random and constantly changing arrangement of ions in a solution

H-atoms can absorb and emit photons containing 4.8x10^-19 J of energy a. If a H atom absorbs a photon containing 4.8x10^-19 J of energy, what component of the atom experienced an increase in energy? c. If a H atom emits a photon containing 4.8x10^-19 J of energy, what component of the atom experienced a decrease in energy?

a) an e- absorbed the photon and experienced an increase in energy c) An e- emits the photon and experienced a decrease in energy

Do processes produce an incr or decr in entropy? a. H2O (L) —> H2O (g) b. N + O —> NO c. N2 + 3 H2 —> 2NH3 d. C8H18 + 25 O2 —> 16 CO2 + 18 H2O e. CaO + CO2 —> CaCO3 f. MgCl2 + H2O —> MgO + 2HCl

a) incr bc L to g b) decr bc fewer mol c) decr bc fewer mol d) incr bc more mol e) decr bc fewer mol and gas to solid f) Incr bc more mol and 2 mol gas on products side

Suppose SO2 is added to system when it's at equilibrium SO2 (g) + H2O (L) <-> H2SO3 (aq) a) In which direction will the reaction shift after the SO2 is removed? b) Will the value of the reaction quotient, Q, change in response to this stress? Justify

a) left b) Yes, will increase the value of Q as [SO2] is part of the denominator in the reaction quotient expression

Consider following processes take place at 25'C 1. A + BC <-> AB + C K(c) = 1.76 x 10^-5 2. AB + D <-> AD + B K(c) = 2.85 x 10^12 3. A + BC + D <-> C + AD + B K(c) = 5.02 x 10^7 a) Process 1 TFP?? B) 2?? d) 3?? e) can non-TFP become TFP when coupled with other TFP?

a) non-TFP bc ^G^o is + and K(c) much smaller than 1 b) TFP bc ^G^o is - and K(c) much larger than 1 d) TFP bc ^G^o is - and K(c) much larger than 1 e) yes

Suppose CO is added to system when it's at equilibrium CO (g) + PbO (s) <-> CO2 (g) + Pb (s) a) In which direction will the reaction shift after the CO is added? b) Will this stress increase or decrease the value of the reaction quotient, Q? Justify

a) right b) Decrease the value of Q as [CO] is part of the denominator in the reaction quotient expression

Percent Yield

actual yield/theoretical yield x 100

aminopropane, CH3CHNH2CH3, and isobutane, C4H10 • Explain why aminopropaneis soluble in water, whereas isobutane isn't soluble in water

aminopropane polar and capable of forming H-bonds with water • Isobutane is non-polar and can't form H-bonds • aminopropane dissolve in water bc strong H-bonds (or d^2 FoA) that form with water molecules • No strong intermolecular FoA exist btwn isobutane and water

Electric current (I)

amount of charge that passes through cross-sectional area of medium per second I = q / T I = current [amperes (A)] q = Charge [Coulombs (C)] T = Time [seconds (s)] F = Faraday's constant = 96,485 C/mol e-s

H-atoms can absorb and emit photons containing 4.8x10^-19 J of energy b. What can be said about the difference in energy btwn 2 of the sublevels in a H atom. Justify

b) Difference in energy btwn 2 of the sublevels in a H atom must be 4.8x10^-19 J • For a photon to absorb and e-, it must contain the exact amount of energy required to move that e- from a lower energy level to a higher energy level • Furthermore, when an e- drops from a higher energy level to a lower energy level, ot emits a photon that is = to the difference in energy levels • As H atoms can absorb and emit photons containing 4.8x10^-19 J of energy, this must he the difference in energy btwn 2 of its sublevels

For a reaction that occurred at 197'C, enthalpy change, ^H, was found to be +26.5 kJ/mol and free energy change, ^G, was found = -46 kJ/mol b) What is principal force driving reaction in forward direction, S or H? Explain c) If temp of system decr dramatically, could process become non-TFP? Justify

b) Large fav + change entropy, ^S, is principal force driving reaction • unfav + ^H cannot drive reaction • ^G = ^H - T^S • Large + ^S cause - ^G c) If temp reduced, ^G becomes less -, as ^S multiplied by smaller # according to equation ^G = ^H - T^S • If temp decr by large enough degree, absolute value of -T^S become less than ^H • non-TFP

Galvanic cell constructed with Cr/Cr 3+ at 1/2 cell and Cu/Cu 2+ at other. Both 1/2 cells under standard conditions. b) Which electrode does reduction occur? e) Better to use Na2SO4 or BaSO4 in salt bridge? Explain

b) Reduction takes place at solid Cu electrode - cathode e) Na2SO4 bc BaSO4 is insoluble in water. So, BaSO4 wouldn't be able to provide any ions that could travel to anode or cathode to keep solutions neutral

Galvanic cell operates through reaction Sn 2+ + Zn —> Sn + Zn 2+ b) Is value K(eq) greater than, less than, or = to 1? Explain (how does E^o affect value of ^G^o?) c) What happen to cell potential, E(cell), if conc of Sn 2+ dropped and conc Zn 2+ remained same (1.0 M)? explain

b) Since E^o (cell) +, ^G^o must be - according to equation: ^G^o = -nFE^o. When ^G^o is less than 1, K(eq) is greater than 1. Reaction yields mostly products c) When [products] > [reactants], LnQ + which cause cell potential to decr according to equation. E(cell) = E^o - RT/nF x Ln [Zn 2+] / [Sn 2+]

^H^o = -508.3 kJ/mol and ^S^o = -178 J/molK 2PCl3 + O2 <-> 2POCl3 ^G^o = -455 kJ/mol b) TFP?? e) Which direction will reaction proceed??

b) TFP cuz ^G^o less than 0 e) Bc Q<K, reaction proceed right to incr PP of products (num in equilibrium expression)

2H2 + 2NO —> N2 + 2H2O b) Is reaction TFP under standard conditions? Justify c) Predict sign associated with ^_S value for reaction. Justify d) Predict sign associated with ^_H^o value for reaction. Justify

b) TFP under standard conditions, as ^_G^o <0 c) - ^_S as there r fewer mol g on products side than reactants. Products r more ordered than reactants d) -^_H^o bc TFP under standard conditions (- ^_G^o). If +, both S and H be unfavorable, and reactions would be non-TFP at all temps

For a certain process entropy change, ^S, found to be -117.8 J/molK and free energy change, ^G, found to be -27 kJ/mol at 320K b) What is principal force driving reaction in forward direction, S or H? Explain c) If temp of system decr dramatically, could process become non-TFP? Justify

b) fav - change entropy, ^H, is principal force driving reaction • unfav + ^S cannot drive reaction • ^G = ^H - T^S • - ^H cause - ^G c) If temp incr, ^G becomes less -, as unfav -^S multiplied by larger # according to equation ^G = ^H - T^S • If temp incr by large enough degree, absolute value of -T^S become greater than ^H • when happens, ^G become + and non-TFP

^H^o = -305.3 kJ/mol and ^S^o = -155.2 J/molK Ni + Cl2 <-> NiCl2 K(p) = 2.617 x 10^45 b) Does system contain mostly reactants or mostly products at equilibrium? Justify

b) mostly products bc K(eq) is very large, meaning that numerator in equilibrium expression, [products], is much larger than denominator, [reactants] • Know system contains virtually all products bc ^G^o = -, and has a magnitude much greater than -2.4 kJ/mol

^H^o = 178.1 kJ/mol and ^S^o = -160.5 J/molK CaCO3 <-> CaO + CO2 K(p) = 1.63 x 10^-23 b) Does system contain mostly reactants or mostly products at equilibrium? Justify

b) mostly reactants bc K(eq) is very small, meaning that numerator in equilibrium expression, [products], is much smaller than denominator, [reactants] • Know system contains mostly reactants bc ^G^o = +, and has a magnitude much greater than +20 kJ/mol

Al2O3 + 3H2 —> 2Al + 3H2O b) Is reaction TFP under standard conditions? Justify c) Predict sign associated with ^_S value for reaction. Justify

b) not TFP under standard conditions, as ^_G^o >0 c) + ^_S as there r same # g particles on reactants side as products but there r 2 mol solid particles on products side and only 1 mol solid particles on reactants side. Products r less ordered than reactants

pH of 0.25 M C5H5N solution at 25'C is 9.25. C5H5N + H2O <-> C5H5NH+ + OH- Is solution acidic or basic?

basic, as pH is greater than 7

Explain why the standard enthalpy of vaporization values for each set of compounds below rn't the same c) C2H6 and C3H8

both C2H6 and C3H8 only experience LDF. Bc C3H8 is larger and has more e-s, it's polarizable and thus has larger dispersion forces. Bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Classify as a physical change, chemical change, or both. Justify by identifying the types of intermolecular or intramolecular forces that r involved in each processes and describing what happens to those forces while the processes r occurring. d) NaCl(s) —> Na+(aq) + Cl-(aq)

both bc dissolving an ionic compound in water can be classified as a physical change and a chemical change • Ionic bonds r broken (chemical change) and ion-dipole intermolecular forces btwn water and the ions r formed (physical change)

Suppose NaOH is added to system when it's at equilibrium NH3 (aq) + H2O (L) <-> NH4+ (aq) + OH- (aq) c) Will the rate of the forward reaction exceed the rate of the reverse reaction before equilibrium is re-established? Justify d) When equilibrium is re-established will the rate of the forward reaction exceed the rate of the reverse reaction? Justify

c) No, the rate of the reverse reaction will exceed the rate of forward reaction as the system works to reduce the concentrations of OH- ions d) No, when equilibrium is re-established, rate of the forward reaction will be equal to the rate of the reverse reaction

(5)__________________(7) / / __________/ (3) /(1.3) / c) What happening to the substance btwn 1-1.5 min marks? d) What happening to the substance btwn 2-3 min marks?

c) Substance remains in a solid phase; however, average KE of particles increases steadily as heat is added to the system d) Intermolecular FoA r weakening. Substance is changing from solid state to the liquid state. Average KE of particles remains the same

Suppose CO is added to system when it's at equilibrium CO (g) + PbO (s) <-> CO2 (g) + Pb (s) c) Will the rate of the forward reaction exceed the rate of the reverse reaction before equilibrium is re-established? Justify d) When equilibrium is re-established will the rate of the forward reaction exceed the rate of the reverse reaction? Justify

c) yes, rate of the forward reaction will exceed the rate of the reverse reaction as the system works to reduce the concentrations of CO gas d) Q is always equal to K when system is at equilibrium

Speed of light formula

c=λv Speed of light (3x10^8 m/s) = λ (wavelength (m)) v (Frequency (s^-1 or Hz))

Galvanic cell operates through reaction 2 Ag+ + Zn —> 2Ag + Zn 2+ d) How would cell potential, E(cell), differ from standard cell potential, E^o (cell), if cell is assembled so that [Ag+] = 1.67 M and [Zn 2+] = 0.25 M at 25'C. Justify e) Is reaction TFP under conditions specified in part d. Explain

d) Value of E(cell) greater than 1.56 V, as incr [Ag+] from 1.0 M to 1.67 M and decr [Zn 2+] from 1.0 M to 0.25 M cause incr in rate forward reaction e) React TFP under conditions specified in part d, value of E(cell) = +

Galvanic cell operates through reaction Sn 2+ + Zn —> Sn + Zn 2+ d) What happen to cell potential, E(cell), if conc of Sn 2+ incr and conc Zn 2+ incr to 1.5 M? explain e) Would non-standard cell described in part d power a device more, same, or less time than standard cell? Assume all cells contain = vol of solution at both electrodes. Justify

d) Value of E(cell) same as value for E^o(cell), as value for Q wouldn't change. Q = [Zn 2+]/[Sn 2+] E(cell) = E^o - RT/nF x Ln [Zn 2+] / [Sn 2+] e) Cell described in part d power a device for a longer period of time, bc it has a greater supply of Sn 2+ cation that r available to be reduced to solid Sn

When completing a ground state electron configuration for Zn^2+.

d-block cations lose their electrons from the highest energy s-orbitals before they start losing them from the highest energy d-orbitals.

Although the average distance between the nucleus and electrons _________ as subsequent subshells are added within a shell, the ionization energy ________ due to an __________ in the shielding effect.

decreases; decreases; increases

Electronegativity and Bond Polarity

difference in electronegativity large = POLAR (or ionic, if very large difference) electrons shared equally (electronegative values are equal) = nonpolar covalent

Ethanol _____ form H-bonds

does not - H bonded to C - H needs to bond with F, O, or N to form an H-bond

(5)__________________(7) / / __________/ (3) /(1.3) / e) What is happening to the substance btwn 3.5-4.5 min marks? f) Happening to the substance btwn 5-7 min marks?

e) Substance remains in the liquid state. Average KE of the particles increase steadily f) Intermolecular FoA btwn particles r breaking. Substance is changing from the liquid state to the gaseous state. Average KE of the particles remains the same

Representative Particles of element covalent compound ionic compound

element --> atom covalent compound --> molecule ionic compound --> Formula unit

^_H_(rxn) = + 80.3 kJ for following reaction Ba(OH)2•8H2O+2NH4Cl—>BaCl2+2NH3+10H2O a) is forward reaction endo or exo??

endo

pi bonds

forms when parallel orbitals overlap and share electrons. - unhybridized p-orbitals and occur in double or triple bonds (weaker)

Experiment conducted in order to determine the enthalpy change that occurs when 1.0 mole of ice at 0'C melts and becomes 1.0 mole of water at 0'C. Enthalpy change associated with this process is referred to as heat of fusion, ^_H_(fus) of ice. In experiment, a 9.68 g sample of ice at 0'C was added to a coffee cup calorimeter containing 278.25 mL of distilled water. Temperature of the water was 22.485'C before the ice was added. The lowest temp. that was recorded after ice had melted was 19.050'C. e) Endo or exo?? f) Energy was transferred from one system to another during this experiment. Identify the two interacting systems and outline the direction of energy flow

endo One system was the ice and water that produced by the ice as it melted, and the other system was the water that was in the liquid form for the entire experiment. The heat flowed from the system of water into the system of ice and melted ice

Coffee cup calorimeter contains 100.0 mL of 1.50 M Ba(NO3)2 at 25.0'C. A student pours 100.0 mL of 1.50 M Na2SO4 at 25.0'C into the calorimeter. A precipitate forms and temperature rises to 29.7'C. Assume that no heat was lost to the surroundings, volumes were addictive, specific heat capacity of the solution was 4.184 J/gK, and density of solution was 1.00 g/mL d) Exo or endo??

exo

Exothermic reaction vs Endothermic Reaction

exo = Bonds in products contain less PE than bonds in reactants endo = Bonds in products contain more PE than bonds in reactants • Delta E is the energy lost of gained in a reaction

A 100.0 mL sample of 0.76 M HCl at 23.0'C was mixed with 100.0 mL of 0.76 M NaOH at 23.0'C in a coffee cup calorimeter and following reaction occurred. H+ (aq) + OH- (aq) —> H2O (L) Temperature of solution increased and maximum temperature of 28.2'C was recorder. Assume that no heat was lost to surroundings, volumes were additive, specific heat capacity of solution was 4.184 J/gK, and density of solution was 1.00 g/mL b) Is the reaction endo or exo?? c) Was energy conserved in process? Justify

exo and energy was conserved in this process. All of the heat lost by the reaction was gained by the solution

A solution of the amino acid alanine, NC3O2H8+, was created and titrated with NaOH. The data from this experiment was used to plot following titration curve f) Identify species that have highest conc. in this solution at 2nd half equivalence pt g) Identify species that have highest conc. in this solution at 2nd equivalence pt

f) NC3O2H7, NC3O2H6-, Na+, and OH- g) NC3O2H6-, Na+, and OH-

intermolecular forces

forces of attraction between molecules (can be individual particles)

Do non-metals gain or lose electrons in order to acquire a full octet?

gain and in doing so, they form - ions

Although the average distance between the nucleus and an electron in 2s is ________ than that of an electron in 2p, electrons in 2s spend some of their time much closer to the nucleus than those in 2p. As a result, electrons in 2p experience a ______ shielding effect. These _____ repulsive forces that the 2p electrons experience from the 1s electrons result in a lower ionization energy for the 2p electrons.

greater

Strong FoA —> ____ boiling point —> _____ energy needed

higher; more

as Atomic radii increases, bond length ________, PE _____ and Bond energy __________

increases, increases, decreases

Endpoint

indicator, which is mixed with the analyte, changes color to signal the arrival at the endpoint • When the correct indicator is chosen, endpoint is very close to the equivalence point

Mass Spectrometry

is an analytical technique that measures the mass-to-charge ratio of ions. The results are typically presented as a mass spectrum, a plot of intensity as a function of the mass-to-charge ratio. It is used to compare the masses of isotopes.

Mole

is the fundamental unit for counting particles on the macroscopic level

If given two soluble ionic compounds...

it is a double replacement. So, cross out the always soluble ions and write the net ionic to form the precipitate

If given a hydrocarbon and O2...

it is combustion and will produce CO2 and H2O

If given a metal and told it burns in the air (read reacts with oxygen)...

it will form a metal oxide

If given a nonmetal and told it burns in air (read reacts with oxygen)...

it will form a nonmetal oxide

The Arrhenius Equation

k=Ae^(-Ea/RT) k= rate constant E(a) = Activation energy (J) R = 8.314 J/mol•K T = absolute temperature (K) A = A constant related to the frequency of collisions and the probability that the orientation could produce a reaction • The Arrhenius Equation shows that increasing the temperature increases the rate of a reaction — As the temperature increases, -E/RT becomes less negative and the rate constant, k, increases — As the rate constant, k, increases, the rate increases rate = k rate = k[A] rate = k [A]^2

Coulomb's Law F=

kq1q2/r^2 F = Force of attraction k = Constant q = Magnitude of charge associated with a particle - protons or electrons d = Distance between charged particles

If given an acid and base where one is strong and one is weak...

leave the weak together and take the acidic or basic part of the strong and react to neutralize

1st Order Integrated Rate Law

ln[A]t = -kt + ln[A]0 • A plot of ln[A]t vs t produces a straight line for 1st order reactions • Used to determine if a reaction is 1st order, as a straight line will only occur if the reaction is 1st order Slope = -k

These elements _______ electrons from their highest s-sublevel first before losing from their d-sublevel

lose

Do metals gain or lose electrons in order to acquire a full octet?

lose and in doing so, they form + ions

Electrons are added to the ________ energy orbital available. This is because nature always favors _________ energy systems. Luckily, there is a format that can be followed to determine the _______energy available

lowest; lower; lowest

pH and [A-]:[HA] Ratio

pH = pK(a) + log [A-]/[HA] When [A-]/[HA]=10, log [A-]/[HA]=1 When [A-]/[HA]=0.1, log [A-]/[HA]=-1 • Adding small amounts of acid or base to a buffered solution causes very small changes in pH

Choosing Conjugate Acid-Base pairs

pH = pK(a) + log[A-]/[HA] When [A-]/[HA]=1, log [A-]/[HA]=0 and pH=pK(a) • Helpful when choosing a CAB pair • If buffer needs to have a certain pH, 1 would choose a weak acid with a pK(a) value that is very close to the desired pH

Henderson-Hasselbalch equation

pH = pKa + log [A-]/[HA] pH = pKa + log [Base]/[Acid] pK(a) = -log10 (K(a)) [A-] = molarity of CB [HA] = molarity of weak acid (initial molarity) log [Base] / [Acid] = Conjugate Acid-base pair and weak base and its conjugate acid

A 45 mL sample of 0.175 M KOH is titrated with 0.200 M HI e) what is the pH at the equivalence pt

pH is 7.0 at equivalence pt. All of the added H+ and OH- have reacted to form water. Solution only contains K+ and I- ions, neither of which affect the pH of solution

210.0 mL of 0.10 M HI is mixed with 100.0 mL of 0.1 M NaOH and 55.0 mL 0.30 M LiOH. Would pH of final solution at 25'C be less than 7, greater than 7, or equal to 7?

pH will be greater than 7, as [OH-] > [H+]

pH and pOH Equations

pH=-log[H3O+] [H+] = 10^(-pH) pOH=-log[OH-] [OH-] = 10^(-pOH)

pK(a) and pK(b) equations

pK(a) = -log10 (K(a)) K(a) = 10^(-pK(a)) pK(b) = -log10 (K(b)) K(b) = 10^(-pK(b))

In general, the more effectively the electrons within a subshell are able to ______ ____ _________ _______, the __________ the ionization energy of those electrons

penetrate the shielding electrons; greater

Ethanol, C2H4O, and methanol, CH3OH • Identify the types of intermolecular forces that exist in a pure sample of ethanol • and Methanol

pure ethanol has LDF & dipole-dipole forces. • dipole-dipole forces occur, as the structure is slightly polar. • electronegative O will acquire a partial - charge and that each atom bonded to the same C will acquire a partial + charge = LDF and H-bonds

q_(rxn) vs. ∆H_(rxn)

q_(rxn) = Heat lost or gained in experiment that took place in the calorimeter ∆H_(rxn) = Heat lost or gained in balanced chemical equation when system is at constant pressure

When the rate of the forward reaction is greater than the rate of the reverse reaction, there is a net conversion of _____ into_______ The reversible reactions is proceeding to the _____

reactants into products Right

Ammonia is produced from a reaction btwn H2 and N2 N2 + 3H2 = 2 NH3 ^_H_(rxn) = -92.2 kJ a) Does the forward reaction release of absorb heat?

released

Electron Affinity: =If value is -, it _______ energy (___________) -The more - the more it want to ________ e- =If value is +, it _______ energy (___________) -It _______ want to accept the e-

releases; exothermic; accept requires; endothermic; does not

2nd ionization energy

same as first ionization energy, except the alkali metals are the highest; this is because once group 1A loses one electron, it has an octet, and it takes the most energy to pull an electron from a full last energy level

Acid Ionization Constant K(a) and K(b)

the equilibrium constant for a reaction in which an acid donates a proton to water K(a) = [H+][A-]/[HA] • The stronger the acid, the larger the K(a) and smaller the pK(a) • The stronger the base, the larger the K(b) and the smaller the pK(b)

solubility product constant (Ksp)

the equilibrium expression for a chemical equation representing the dissolution of a slightly to moderately soluble ionic compound • This Ksp is for ionic compounds in water • Ksp values generally range from about 1 x 10^-3 to 1 x 10^-54 for compounds that r considered to be insoluble

Bond Order (Def and equation) When the bond order increases... = Bond length _______ = PE associated with the bond _______ = bond energy ________

the number of bonds between two atoms (Bonds total / Bond sites) decreases, decreases, increases

Polyprotic Acids

they can donate more than one H+ in a solution (H2SO4 and H2CO3) they have a different Ka value for each possible dissociation (removing one H+ at a time) (Ka1 and Ka2) (K(a1) = 4.3 x 10^-7) >> (K(a2) = 5.6 x 10^-11) • Always use K(a1) to calculate [H+] and pH • Most of H+ ions come from 1st ionization • H+ from 1st ionization drive equilibrium for other ionization to the left

Heat of fusion (triangleHfus)

triangleHfus - the heat absorbed as 1 mole of a solid liquifies = Energy is required ti expand/sever the intermolecular FoA, as a molecule moves from the solid to the liquid phase = This is why molar heat of fusion, triangleHfus values r always + = Melting is always an endothermic process

Heat of Vaporization (triangleHvap)

triangleHvap - heat absorbed as 1 mole of a liquid becomes gaseous = energy is required to sever the intermolecular FoA, as a molecule moves from the liquid to the gas phase = Vaporization is always endothermic, so triangleHvap values r + = Ideally, there r no intermolecular FoA btwn gas particles

NH+ + H2O <---> H3O+ + NH3 Weak or strong Acids and Conjugate Bases??

weak acids/strong conjugate bases - H2O and NH3 compete for protons - NH3 stronger base, so wins most of the time and reaction lies to the left

Autoionization of Water

when pure water reacts with itself to for hydronium and hydroxide ions • in pure water at 25'C [H3O+] = 1.0 x 10^-7 and [OH-] = 1.0 x 10^-7 As [H3O+] = [OH-] the solution is neutral Kw = [H3O+][OH-] Kw = (1.0 x 10^-7)(1.0 x 10^-7) Kw = 1.0 x 10^-14 (at 25'C) • 1.0 x 10^-14 is a constant • If u know conc of [H3O+] u can calculate the conc of [OH-] or vise versa

Is CaSO4 more soluble in 1.0 L of 0.25 M Li2SO4 or 1.0 L of 0.25 M Al2(SO4)3? Justify

yes bc conc. SO4- ions in the Li2SO4 solution is 0.25 M whereas conc. SO4 2- ions in Al2(SO4)3 solution is 0.75 M • Increasing conc. of SO4 2- pushes equilibrium below further to left according to Le Chatelier's principle CaSO4 <-> Ca 2+ + SO4 2- More CaSO4 will dissolve in Li2SO4 bc it has lower sulfate ion conc.

Which of the following are isoelectronic? a) Ne and F- b) Ca^2+ and Se^2- c) N and F- d) I- and Ba^2+ e) K+ and Ca^2+

yes, no, no, yes, yes

Formula for Calculating the Enthalpy of a reaction under standard conditions

ΔH('rxn) = Σ n ΔH('f) (products) - Σ n ΔH('f) (reactants) Σ = Sum of n = Stoichiometric coefficients

Enthalpy of fusion (ΔHfus)

ΔHfus - the heat absorbed as 1 mole of a solid liquefies • Stretch and sever IMF • Energy is required to expand/sever the IMF of attraction, as a molecule moves from the solid to the liquid phase • This is why molar heat of fusion, ΔHfus, values r always + • Melting is always an endothermic process

enthalpy of vaporization (ΔHvap) Vaporizing and condensing Water at 0'C and 1 atm

ΔHvap - The heat absorbed as 1 mole of a liquid becomes gaseous • Sever remaining IMF • Energy is required to sever the IMF of attraction, as a molecule moves from the liquid to the gas phase • Vap is always endothermic, so ΔHvap values r + • Ideally there r no IMF of attraction btwn gas particles • Vaporizing = ΔHvap = + kJ/mol • Condensing = ΔHvap = - kJ/mol

The Structure of NaCl

— Each Na cation in surrounded by 6 Cl anions — Each Cl anion is surrounded by 6 Na cations - Arrangement of ions maximizes the Coulombic forces of attraction between cations and anions and minimizes repulsive forces between ions with like charges

^_G^o??

• "normally" indicated that the system is in a standard state • Standard state for ^_G^o tells us that all species exist as pure substances, all gases have partial pressures of 1 atm, all aqueous species have 1 M conc., and temp of system is 298K • Bc equilibrium exists at other temps, use same equations to calculate ^_G^o at other temps. Often denoted as ^_G^o (T) • ^_G^o (T) is max amount of work that can be done by a system as moves from 100% reactants in standard states to 100% products in their standard states • ^_G^o (T) is max amount of work that can be done by a system as moves from 100% reactants (1 M conc. and 1 atm) to 100% products (1 M conc. and 1 atm) at non-standard temps

C2H4 reacts with O2 to form CO2 and water b) Explain why all atoms were conserved during this reaction

• 1 C2H4 molecule reacted with 3 O2 molecules to form 2 CO2 molecules and 2 H2O molecules • Reactant side of the equation has 2 C atoms and the product side of the equation has 2 C atoms • Reactant side of the equation has 4 H atoms and the product side of the equation has 4 H atoms • Reactant side of the equation has 6 O atoms and the product side of the equation has 6 O atoms • Atoms formed new molecules, but they were all conserved

Standard Temperature and Pressure (STP) for Gases is 0*C and 1 atm

• 1 mole of any gas particles will occupy 22.4 L of space at STP

Which mixture in each set has the lowest pH? Justify 1.0M Ni(NO3)3 or 1.0M Ca(NO3)2

• 1.0M Ni(NO3)3 have lowest and pH will be less than 7, as Ni 3+ is an acidic cation • 1.0M Ca(NO3)2 have highest and pH will be a neutral solution with a pH of 7, as Ca 2+ and NO3- r neutral ions

Lowering the T of a system causes deviations from the ideal gas law. Explain

• 2 factors that cause deviations from ideal behavior r an increase in attractive forces btwn particles and a reduction in free space relative to the V of the container • At low T, gas particles move slower and r closer together • FoA exist btwn gas particles when they r in close proximity to one another and these attractive forces reduces reduce the expected pressure exerted by the gas • Ideal gas law assumes that gas particles have no V of their own, as their V is negligible when compared to the overall V of the container • When the V occupied by the particles makes up a relatively high percentage of the total V inside the container, the gad doesn't behave ideally

Ion-Dipole vs. Ionic Bonds

• 2 forces work against one another through Coulomb's law • 1 of forces will be stronger than the other • Mg(OH)2 isn't soluble in water and Ba(OH)2 is, bc, for alkaline earth metal hydroxides, FoA from ionic bonds decrease more rapidly that the ion-dipole intermolecular forces when moving down the group

Use Coulomb's law and visual representations to explain why BaSO4 is not soluble in water and why MgSO4 is soluble in water

• 2 opposing forces to consider when examining the solubility of an ionic compound in water. There r ion-dipole forces btwn the water molecules and the ions, which r trying to dissolve the compound; and there r ionic bonds btwn the anions and cations, which r trying to hold the solid structure together • Ion-dipole forces btwn water and Mg 2+ r greater than the ion-dipole forces btwn water and Ba 2+. This is bc Mg 2+ has a smaller ionic radius, so there is a shorter distance it and - ends of the water molecules • This can also be explained through Coulomb's Law, F=k (Q1Q2)/d^2 • Ion-dipole forces btwn Mg 2+ and water and SO4 2- and water r stronger than the ionic bonds btwn Mg 2+ and SO4 2-. This explains why MgSO4 is soluble • Ion-dipole forces btwn Ba 2+ and water and SO4 2- and water r stronger than the ionic bonds btwn Ba 2+ and SO4 2-. This explains why BaSO4 is soluble

Heat Transfer btwn systems

• 2 systems at different temp. that r in thermal contact with one another will exchange energy (heat) • energy transferred to system 1 = to energy transferred from system 2 • When thermal equilibrium is established, temp., and therefore average KE, of both systems will be the same • This process is referred to as heat transfer, heat exchange, or transfer of energy as heat

Standard Enthalpy of Formation

• A Hypothetical value that indicates how much heat would be lost or gained during the formation of 1 mole of a compound from the most common form of its elements in their standard states ΔH('f) —> Formation —> Standard state (25'C, 1 atm) • Heat of formation reactions r always written so that all reactants exist as they would under standard conditions, and there is 1 mole of product

Einstein's Theory (1905)

• A beam of light is a stream of particles called photons • Energy of a photon is related to its frequency according to E=hv • Quantum of Planck is a particle - a photon • If the frequency of a photon is below a certain threshold, no e-s r ejected • If the frequency of the photon is at or above a certain threshold, its energy is transferred to the e- - This causes the e- to overcome the FoA holding it to the metal - E- absorbs the photon

Catalysis

• A catalyst increases the rate of a chemical reactions by providing a new and more efficient mechanism • The new mechanism will have: — a lower activation energy, and/or — a higher frequency of collisions with an orientation that could produce a reaction • Catalysts r added to the system — They r there before the reaction starts and they return when the reaction is complete — Bc of this, the new conc. of the catalyst remains constant • A catalyst provides a mechanism with a lower activation energy, thereby increasing the reaction rate • A catalyst lowers the activation energy for the forward and the reverse reactions — K(eq) for a reaction is the same at the same temperature, with of without the catalyst

Galvanic (Voltaic) Cells (^_G<0)

• A device that transfers e-s from one reactant to another through a pathway — as opposed to a direct transfer • Process is always TFP (^_G<0) • Chemical energy is converted into electrical energy that moves e-s Zn + Cu 2+ —> Zn 2+ + Cu 2e- = 2 electrons go to cathode Zn 2+ = Solid Zn loses 2 e-s and falls off anode, as it becomes aqueous Cu 2+ = Aqueous Cu gains 2 e-s to become solid Cu atom that sticks to the cathode

Spectroscopy

• A method of analysis which is based upon the absorbance of electromagnetic radiation by matter • Used to acquire data pertaining to the structure of a molecule or the concentration of a species I(O) = I(T) + I(A) I(O) - Intensity of electromagnetic radiation striking sample I(T) - Intensity of electromagnetic radiation exiting sample I(A) - Intensity of electromagnetic radiation absorbed by the sample

Reaction Mechanisms • Reaction Intermediate

• A substance that is produced and then consumed during the overall reaction

1354 J of heat was absorbed by a balloon from surroundings, which causes gas inside balloon to expand from 2.3 L to 3.98 L under a constant external P of 1.07 atm a) Was absorption of 1354 J of heat from surrounding as exothermic or endothermic process? Justify

• Absorption of heat from surroundings is endothermic • Heat flows from surroundings into the system and internal energy of system increases

HYO(n) Oxoacids and CB stability through induction

• Acid strength and K(a) increase as oxygens r added • CB stability increases • As more oxygen atoms r added, electron density around oxygen atom that an H+ ion could form bond with its further reduced • Stabilizes the base, as H+ ion have less ability to form a bond

How acid-base indicators work

• Acid-base indicators r weak acids HIn <-> H+ + In- <——- Highly acidic sol. shift reaction left • H+ from acid in sol. being titrated cuz shift Highly basic sol. shift reaction right ——-> • Reduction of H+ in sol. being titrated causes shift, which caused by basic titrant (NaOH) • In titration, large shift in pH as equivalence pt is passed • Large shift in pH causes indicator's equilibrium to shift and its color to change • Color starts to change when [HIn]~~[In-] • Color change experienced by different indicators occurs over different pH ranges — must select an indicator that changes color at a pH that is as close as possible to pH at equivalence pt • When [HIn] = [In-] pH = pK(a) + log[In-]/[HIn] pH = pK(a) Choose indicator that has: pK(a)(indicator) ~~ pH(equivalence pt)

Arrehenius Acids and Bases

• Acids dissociate in water to produce H+ ions HA —> H+ + A- • Bases dissociate in water to produce OH- ions BOH —> OH- + B+

Is the actual V of a gaseous system less than, equal to, or greater than the V that would be predicted using the ideal gas law equation. Explain.

• Actual V of a gaseous system is less than the V that would be predicted using the ideal gas law equation • Ideal gas law assumes that the gas particles in a system don't occupy any V; however, gas particles do occupy a V and that V must be taken into consideration under conditions of high P and low T • Actual V of a gaseous system is the V of the container minus the V occupied by the gad particles

Given 10 mL of a hydrochloric acid, HCl, solution with a pH of 1.0. Required to change the pH to 2.0 be adding water. How much water do u add?

• Add 90 mL of water HCl is a string acid, thus # moles H+ will not change. To change pH we must change conc. H+ • Reducing conc. H+ by a factor of ten will cause pH to increase by 1. If vol is increased by a factor of ten, conc. is reduced by a factor of ten. • Thus, adding 90 mL of water will raise pH from 1.0 to 2.0

Given 100 mL of a potassium hydroxide with a pH of 12.0. Required to change the pH to 11.0 be adding water. How much water do u add?

• Add 900 mL of water KOH is a string base, thus # moles OH- will not change. To change pH we must change conc. OH- • Reducing conc. OH- by a factor of ten will cause pOH to increase by 1 and pH to drop by 1. If vol is increased by a factor of ten, conc. is reduced by a factor of ten. • Thus, adding 900 mL of water will reduce pH from 12.0 to 11.0

Why are liquids left out of the equilibrium expression?

• Adding or taking away a small amounts of water in a reaction that takes place in an aqueous solution does not affect the overall concentration

Combustion reactions

• All combustion reactions r redox reactions • Products for combustion of any hydrocarbon r always CO2 and H2O • O2 is required reactant in all combustion reactions • 2 C2H2 + 5 O2 —> 4 CO2 + 2 H2O

Infrared (IR) Spectroscopy

• All covalent bonds in molecules r vibrating • Bond length is the average distance btwn nuclei • Covalent bonds have a vibrational frequency that is in the IR region of the electromagnetic spectrum • IR radiation of exactly the same frequency will be absorbed by the molecule • Vibrational frequencies depend on the mass of the atoms and the strength of the bonds • Frequency is related to wavelength - c=(Wavelength) • v = IR spectra can used to identify bond types, functional groups, and compounds • Molecules the share the same functional group, such as alcohols or carboxylic acids can be identified through IR spectroscopy, as they will have absorption peaks within the same range • Every compound has a characteristic IR spectrum that it can be identified through

C2H4 reacts with O2 to form CO2 and water c) explain how the law of conservation of mass applies to this reaction

• All the atoms that were contained in the reactant molecules r present in the product molecules • No atoms were lost and no atoms were gained in this reactions • mass of these atoms is constant, so no mass was lost or gained in the chemical reaction

Breaking Bonds and/ot Intermolecular Interactions

• Always requires energy to break a bond or to sever intermolecular attractions • This can be thought of in terms of work, as a force could be exerted over a distance to pull the 2 species apart W = Fd • W = Work • F = Force • d = distance

Breaking / Establishing Bonds and/or Intermolecular Interactions

• Always requires energy to break a bond or to sever intermolecular attractions • can be thought of in terms of work, bc force could be exerted over a distance to pull 2 species apart • W (work) =F(Force) x d(distance) • Energy always evolved when bonds or intermolecular attraction r established • PE decreases as attracted species approach one another

2 Solutions at 25'C were mixed in a beaker. A precipitate formed and temperature of new solution dropped to 19'C c) Did amount of energy contained by surroundings incr., decr., or remain same? Justify

• Amount of energy contained by the surroundings decreased • Heat (energy) flowed from the water (part of surroundings) into the system

2 Solutions at 25'C were mixed in a beaker. A precipitate formed and temperature of new solution dropped to 19'C a) Did amount of energy contained by system incr., decr., or remain same? Justify

• Amount of energy contained by the system increased • Heat (energy) flowed from the water (part of surroundings) into the system

Student puts beaker w/ 150 mL of distilled water a hotplate. Temp. of water incr. from 22'C to 85'C a) Did amount of energy contained by 150 mL of water incr., decr., or remain same? Justify

• Amount of energy contained by the water increased • Energy flowed from the hotplate into the water • This caused the average KE of the water molecules to increase

Ultraviolet/Visual (UV/Vis) spectroscopy

• An absorption spectrometer is used to measure the absorbance of a sample at wavelengths btwn about 200 nm and 800 nm • The peaks represent wavelengths that correspond to the energy associated with possible electronic transitions within the molecule • The following e- transitions r possible with UV and visual light btwn 200 nm and 800 nm - transitions r only possible when the compound has at least 1 LP. LP r common on O, N, and halogens - Transition is only possible of the compound has pi-bonds, which r present in double and triple bonds - Energy contained by the incident photons must be exactly the same as the difference in energy btwn the 2 MOs in order for the photons to be absorbed - In this molecule, 3 different frequencies of light would cause e- transitions • Colorless species can only absorb UV light btwn about 200 nm and 400 nm in these experiments - The peak occurs in the UV range of the electromagnetic spectrum, as CH2=CH-CH=CH2 is colorless - CH2=CH-CH=CH2 has no LP, so the e- transitions occurred at the pi-bond portions of the double bonds • Following e- transitions r possible with UV and visual light btwn 200 nm and 800 nm - Energy contained by the incident photons must be exactly the same as the difference in energy btwn the 2 MOs in order for the photons to be absorbed - In this molecule, 3 different frequencies of light would cause e- transitions • Colorless species can only absorb UV light btwn about 200 nm and 400 nm in these experiments - Peak occurs in the UV range of the electromagnetic spectrum, as CH2=CH-CH=CH2 is colorless - CH2=CH-CH=CH2 has no LP, so the e- transitions occurred at the pi-bond portions of the double bonds • Colored species will always absorb light from visual spectrum, but could also absorb UV light in these experiments - 2 peaks occur in the visual spectrum—> Oxygenated hemoglobin is bright red - Hemoglobin's structure contains several pi-bonds and LP

A container is filled with Ar gas at -48'C and sealed. After sitting on a lab bench at room temperature for several minutes, the lid blows off and the gas escapes. Explain why this happened

• Ar gas was at a very low T when the container is sealed and when gases r at a low T, the particles/atoms r relatively close to one another • As heat from the room flows into the container, gaseous Ar atoms started moving faster • P inside the container became overpowered the strength if the seal • When the seal broke, gas expanded and when the T of a gaseous system increases, P and/or V must also increase

ΔHfus for Ionic Compounds Freezing and Melting Water at 0'C

• As ionic bonds r much stronger than IMF, the ΔHfus values for ionic compounds r very large • Melting = ΔHfus = + kJ/mol • Freezing = ΔHfus = - kJ/mol

Collision Rate increase...

• As the conc. of reactants in the liquid or gas phase increases • As the surface area of reactants in the solid phase increases • As temperature increases Not every collision triggers a chemical reaction

Why do we have different colors of light?

• As wavelength/frequency changes, color changes • But there is a duality to light - Behaves like a wave, and - Behaves like a particle (a photon) • As wavelength/frequency changes, the energy per photon changes E=hv

Molar Mass, Molecular Speed, and Temperature

• At any given temperature, average KE of all gas particles is the same • Gases with smaller molar masses will have higher average velocities • Gases with larger molar masses will have lower average velocities KE = 1/2 mv^2

Maxwell-Boltzmann distribution

• At higher temperatures, a larger percentage of particles will be able to convert their KE in into the PE needed to overcome the E(a)

Hybrid Orbital Theory

• Atomic orbitals on the same atom combine in order to form hybrids • Atomic orbitals on different atoms overlap in order to form covalent bonds • Each atom in the compound retains its associated orbitals and e-s • This theory correlates with observed bond angles in molecules

K(w) @ Temps other than 25'C 2H2O + heat <-> H3O+ + OH- K(w) = [H3O+] [OH-]

• Auto-dissociation of water is an endothermic process • When temp incr., equilibrium shifts to the right so [H3O+], [OH-], and K(w) increases • When temp decr., equilibrium shifts to the left so [H3O+], [OH-], and K(w) decreases • In any sample of pure water at any temp pH = pOH and [H3O+] = [OH-] • Pure water at temp above 25'C will have a pH that is lower than 7.0 bc [H3O+] and [OH-] r larger than 1 x 10^-7M • Pure water at temp below 25'C will have a pH that is higher than 7.0 bc [H3O+] and [OH-] r smaller than 1 x 10^-7M

A gaseous system is kept at 25.0'C. A chemist slowly increase the temperature of the system until it reaches 50.0'C. Did the average KE of the gas particles in the system double when he did this. Justify

• Average KE of the gaseous particles in the system increases, but it didn't double • Average KE of the particles in a system is proportional to the Kelvin temperature scale (absolute T) • If the T in Kelvin doubled, the average KE would have double but it didn't occur • 25.0 + 273 = 298 K 50.0 + 273 = 323K 2x298 K doesn't = 323 K

Suppose u have 2 identical 1.0 L sealed containers. Both containers r kept at exactly 25'C. One vessel contains only Ne gas at 1.5 atm, and the other contains only Xe gas at 2.5 atm a. Is the average KE possessed by the Ne atoms greater than, equal to, or less than that of the Xe atoms? Explain

• Average KE possessed by the Ne atoms is = to that of the Xe atoms • T is a measure of the average KE of the particles in a system • = T means that the average KE is also =

Explain why standard enthalpy of vaporization, DeltaH(vap), values for each set of compounds below rn't the same d. BH3 and OF2

• BH3 has LDF and OF2 has LDF and d^2 forces • bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Basic solutions Common Reduction in Basic Solutions

• Basic solutions have a higher concentration of OH- • Add same # of OH- to Eliminate however many H+ • Permanganate: MnO4- —> MnO2 • Permanganate will oxidize p and d-block metals, sulfite ions, and substances that have a lower (less +) oxidation state than usual in basic solutions

A real gas will behave more like an ideal gas when the pressure of a system decreases and the temperature remains the same. Explain

• Bc average distance btwn gaseous particles increases as the P decreases and when the gas particles further apart, attractive forces btwn particles r greatly reduced • This causes gas particles to behave as though they have no interaction with 1 another aside from elastic collisions • Thus, there is no reduction in the expected P due to FoA btwn particles

Fractional distillation was used to isolate an unknown volatile substances that had contaminated the well water at a rural property a. Is the boiling pt of the unknown substance greater than, less than, or equal to 100'C at 1.0 atm? Justify

• Boiling pt of the unknown substance is less than 100'C at 1.0 atm • Vapor pressure of the unknown substance when it's in the liquid phase must be greater than that of water at any given T in order for it to be isolated through fractional distillation • As a substance boils when its vapor P = atmospheric P, boiling pt of the unknown substance must be less than 100'C at 1.0 atm

How can process 8Fe + 6O2 —> 4Fe2O3 (^_H^o = -3296.8 kJ) bc TFP when entropy of system decreases so dramatically? Justify

• Entropy in process does decr, making - ^_S value • However, process extremely exothermic, giving ^_H large - value • With equation ^_G = ^_H - T^_S, still possible to obtain a - value for ^_G when have - value for ^_S, so long as reaction yields - ^_H value greater in magnitude than T^_S

Explain why CH3OH is miscible in water whereas CH3(CH2)6OH is not

• Both CH3OH and CH3(CH2)6OH can form H-bonds with water - CH3OH had a shorter non-polar component around the C atoms, so most significant intermolecular force possessed by each compound is H-bonds - Bc both share the same type of intermolecular forces, they r very soluble in one another • CH3(CH2)6OH has a large non-polar C chain, which is 7 C long - Water molecules have a greater attraction for one another (due to their strong H-bonds) than they do for long hydrocarbon chains, which can only offer weak dipole induced dipole and LDF. - Solubility of alcohols decreases drastically as C r added

A. Full saturated fatty acid B. double bonded at 1 part unsaturated fatty acid • most likely to be a solid at room temperatures

• Both about same size and have same # of e-s (b has 2 fewer e-s that a, but b has pi-bond which increases polarizability) • Both can form H-bonds at the end w/ the OH- group but LDF will be stronger in a due to its shape • a more likely bc it is saturated which means linear structure overall • bc linear stack up well, many pts of contact btwn neighboring molecules, which means more locations where LDF can be established • B is unsaturated as it has a double bond btwn 2 C in C chain which causes overall structure to bend at that pt • bc of kink, molecules don't stack up well, so fewer pts of contact btwn neighboring molecules which means fewer locations where LDF can be established

Which of the compounds below is most soluble in water? Justify. HOCH2CH2OH or CH3CH2OH

• Both compounds have non-polar C chains that r the same length • HOCH2CH2OH is most soluble in water, bc both ends of the molecule can form H-bonds with water, whereas only one end of CH3CH2OH can form H-bonds with water

Explain why standard enthalpy of vaporization, DeltaH(vap), values for each set of compounds below rn't the same c. C2H6 and C3H8

• Both only experience LDF but bc C3H8 is larger and has more e-s, it is more polarizable and this has larger dispersion forces • bc intermolecular FoA in these 2 substances r different, their enthalpies of vaporization will also be different

Why are solids left out of the equilibrium expression?

• Both pieces have the same density, so both contain the same number of particles per unit volume The concentration of solid is density. Density is an intensive property so it does not change with volume.

Suppose u have 2 identical 1.0 L sealed containers. Both containers r kept at exactly 25'C. One vessel contains only Ne gas at 1.5 atm, and the other contains only Xe gas at 2.5 atm C. Does the vessel with the Xe gas contains more, fewer, or the same # of gas particles as the vessel of Ne gas? Explain

• Both vessels r 1.0 L and 25'C and the P in the vessel with the Xe gas is higher, so there must be more moles of gas in that container • When there r more moles of gas occupying the same V, when both systems r at the same T, there will be more collisions per unit of time with the walls of the container, and therefore, the P will be higher

Is the dissolving of Al(NO3)3 in water a chemical process, a physical process, or both? Justify

• Both, bc chemical processes involve the breaking and/or formation of chemical bonds • Ionic bonds btwn Al 3+ and NO3 - r broken during the dissolving process • Physical process involve changes in intermolecular forces • Ion-dipole intermolecular forces r established during the dissolving of Al(NO3)3

CB stability through induction

• CB of 3 of 6 strong acids - H2SO4, HNO3, HClO4 - r stabilized by induction due to presence of multiple highly electronegative oxygen atoms

A piece of paper catches on fire and burns. Has a chemical change occurred or has a physical change occurred? Justify

• Chemical change has occurred • Production of heat and light r indications of chemical changes

Two solutions r at 25*C r mixed. The temperature of the combined solution increases to 32*C and a white precipitate forms. Has a chemical change occurred or has a physical change occurred? Justify

• Chemical change occurred • Changes in temperature and formation of precipitates r indications of chemical changes

Chromatography and Solubility

• Chromatography paper is composed of non-polar C chains with -OH groups that can form H-bonds • Max height traveled by the mainly non-polar solvent • Solution: Mainly non-polar solvent, solute A, and solute B • Stationary phase is chromatography paper, and mobile phase is solvent used • As solvent moves up the piece of paper it carries solute particles with it • Distances that the different solute travel up the paper depend on their relative attractions for the moving solvent and the stationary paper • Solute particles that can form H-bonds at several locations along their structures will not travel very far up the paper, as the molecules in the paper contain many -OH groups - These solute particles form H-bonds with paper relatively close to the solutions surface • Solute particles that r mostly non-polar will have weak attractions for the paper and relatively strong attractions for the mainly non-polar solvent - These particles will be deposited further up the paper

Buffer solutions

• Equilibrium lies far to left for weak acid CH3COOH <-> H+ + CH3COO- • Salt completely dissociate NaCH3CO2 —> Na+ + CH3COO- • System has lots of acid to react with strong bases and lots of CB to react with strong acids

The Reactions Quotient Q(p))

• Describes the relative PP of products and reactant at any point in time Q(p)=(P(E))^e(P(D))^d/(P(A))^a(P(B))^b • When system is at equilibrium Q(p) = K(p)

The Reactions Quotient Q(c))

• Describes the relative concentrations of products and reactant at any point in time Q(c)=[E]^e[D]^d/[A]^a[B]^b • When system is at equilibrium Q(c) = K(c)

What r the main factors that account for the extreme hardness of diamond?

• Diamond is a network solid, meaning that each diamond is one massive molecule held together with covalent bonds • All C r bound together with sp^3 hybrid orbitals • Means that there is a tetrahedral geometry around each C atom, which is a very strong configuration

Why do FoA btwn gas particles increase when they r closer together?

• Distance is in the denominator of the Coulomb's Law equation. • As distance btwn particles decreases the intermolecular FoA btwn the particles will increase

Disproportionation Reactions (I)

• Element in one oxidation state simultaneously undergoes both oxidation and reduction • Element must, therefore, have at least 3 oxidation states, and it must be in an intermediate state

Activity Series

• Elements that r less easily oxidized, will oxidize elements that r more easily oxidized • Elements can be listed, one on top of the other, in order of increasing or decreasing activity • Elements will oxidize anything below it in this series

Gas contracts from 4.26 L to 1.89 L in a cylinder under a constant external P of 1.10 atm. Temp. of system remained same during this process. Was process endo or exo? Justify

• Endothermic, bc P caused by the gas particles colliding w/ piston on outside cylinder was greater than P caused by the gas particles colliding w/ piston on inside of cylinder • This caused piston to move in direction that decreased the volume inside cylinder • Gas outside cylinder did work on piston

Establishing Bonds and/or Intermolecular Interactions

• Energy always evolved when bonds or intermolecular attraction r established • PE decreases as attracted species approach one another

first law of thermodynamics

• Energy is conserved in chemical and physical processes • Internal energy of a system may be transferred into or out of that system in form of heat or work, but cannot be created or destroyed

PE (electric) and ^_V Voltage : Energy per unit charge

• Energy is difference in PE btwn anode and cathode • Charges r e-s, each carrying -1.60 x 10^-19 C • Take 6.25 x 10^18 e-s to add up to -1 C of charge • In a 12 V battery, difference in PE btwn anode and cathode is 12 J per C of charge • If take 6.25 x 10^18 e-s at cathode and drag back to anode, require 12 J of work • 12 J of PE is converted into KE that moves 6.25 x 10^18 e-s from anode to cathode in 12 V battery

Enthalpy Change, ∆H

• Enthalpy change, ∆H, of a reaction is amount of heat energy that is released (exo) or absorbed (endo) by a chemical reaction at constant pressure • 4 methods for finding ∆H 1. Find it by preforming calorimetry experiments 2. Calculate it using average bond enthalpies 3. Calculate it using enthalpies of formation 4. Calculate it using Hess's Law

reversible reaction

• Evaporation and condensation of water • Dissolving and precipitating a salt • Absorption and desorption of a gas (CO2 gas is a carbonated beverage) • Biology (Hemoglobin binds with oxygen in alveoli of lungs; Oxygen later released to cells within the organism) • Recharging and discharging a lead-acid battery • Reversible acid-base reactions

Endothermic reactions can be TFP

• Evaporation is TFP • Dissolving soluble compounds is TFP

Burning or methane gas, CH4, in presence of oxygen b) is this reaction endo or exo? c) Did heat flow into or out of system?

• Exothermic • Heat flowed out of the system and into the surroundings

How to tell if process is TFP??

• Exothermic reactions r favorable ^_H<0 • Producing a greater degree of disorder is favorable ^_S > 0 • 2 methods for determining if process is TFP: ^_S(universe) and ^_G (Gibbs Free Energy)

1354 J of heat was absorbed by a balloon from surroundings, which causes gas inside balloon to expand from 2.3 L to 3.98 L under a constant external P of 1.07 atm b) Was expansion of balloon endo or exo? Justify

• Expansion of balloon was exothermic, as it expanded into the surroundings • This means that gas inside balloon did work on surorundings

Dalton's Law of Partial Pressures

• For a mixture of gases in a container, total pressure exerted is the sum of the pressures that each gas would exert if it were alone P(total) = P1 + P2 + P3 + ...

Types of solutions - Solid - Solid Solutions

• Formed by melting, mixing, and solidifying Ex - Steel, Brass

Kinetic Molecular Theory 1

• Gases consist of particles (molecules and/or individual atoms) that r in continuous random movement

Types of solutions - Gas - Gas Solutions

• Gases r always indefinitely soluble in one another • Air N2(g), O2(g), CO2(g), H2O(g), etc.

CB of H2SO4 Stability through Resonance

• H+ ions r attracted to - charges • Delocalized electrons cause -1 charge to migrate btwn 3 oxygen atoms, which makes it difficult for H+ to form a bond • Stabilizes CB of H2SO4 - making it a weak base

Identify and describe energy changes that take place when sample of sugar, C6H12O6 dissolves in a pure sample of water

• H-bonds and LDF of attraction btwn individual sugar molecules r broken, which r endothermic • Some H-bonds btwn water molecules r broken, which is endothermic • H-bonds btwn water molecules and sugar molecules r established, which is exothermic

Halogen Displacement Redox Reactions

• Halogens bonded to themselves can take e-s from (oxidize) any halide ions that is below it in the family F2 < Cl2 < Br2 < I2 • In these reactions, H atoms in H2O r reduced, as they gain e-s to form H2 2H2O + 2e- —> H2 + 2OH- • Alkali metals, Ca, Sr, and Ba will displace H from water

Which gas more soluble in water - He or CO??

• He has weak LDF but CO is polar so can form dipole-dipole attractions and dispersion attractions • CO has more e-s, and is more polarizable, it forms stronger LDF • CO id mote soluble in water as it forms D^2 attractions and stronger LDF

System 1 (479K) vs. System 2 (298K) b) Identify type of energy transferred from system 1 into system 2. Justify

• Heat is transferred from system 1 to 2 • Initially, both systems r at different temps, so heat will flow from system 1 into system 2 until both systems reach same temp. • Since systems r housed in rigid containers, changes in volume rn't possible so energy cannot be transferred btwn systems in the form of work

Specific Heat Capacity (c) Equation

• Heat transfer equation shows that transfer of = quantities of heat to 2 materials of = masses w/ different specific heat capacities will produce different temperature changes • Material w/ higher specific heat capacity will experience a smaller increase in temperature q=mc∆T q = Heat lost or gained by a substance m = mass of that substance c = Specific heat capacity of that substance ∆T = Temp. change of that substance (∆T = T_t - T_i)

Causes of Energy Changes

• Heating or cooling a substance - Adding heat to, or removing heat from, a substance changes its internal energy - If temp. changes, average KE of particles in system changes • Phase Changes - Temp. doesn't change during a phase change - Energy that is released or absorbed by the system is used to form, alter, or sever intermolecular forces • Chemical reactions - During chemical reactions, energy is released or absorbed in the form or heat and/or work

System 1 (short arrows) vs System 2 (long arrows) a) Is System #2 at higher or lower temp. than System #1? Justify

• Higher bc temp. is a measure of average KE of the particles in a system • As both systems r pure samples of same gas, all individual particles have same mass • According the equation KE = 1/2 mv^2, when all particles share same mass, vel. is only variable affects KE • As arrows in system 2 r longer, those particles r moving at higher velocities, so possess a higher average KE

Max Planck (1900)

• Hypothesized that the energy radiated fron a heated object, such as a stove element or a lightbulb filament, is emitted in discrete units, or quanta

Gases don't behave ideally at low T

• Ideal Gas Law assumes that gases experience no intermolecular FoA • At high T, KE of gas particles overcome any intermolecular FoA • At low T, gas particles move slower and r closer together. Attractions btwn molecules exist under these conditions

What assumption does the ideal gas law make about the V of gas particles in a system?

• Ideal gas law assumes that the V occupied by the gas particles in a system is negligible • V variable in the ideal gas equation is the V of empty space in the container and it assumes that the V occupied by the gas particles is 0

Work and Exothermic Processes

• If P caused gas particles colliding w/ piston on inside of cylinder is greater than outside cylinder... • Gas inside does work on piston, energy is transferred from gas to piston, and gas inside cylinder expands • When gaseous system expands at constant temperature, process is exothermic • System does work on surroundings

Work and Endothermic Processes

• If P caused gas particles colliding w/ piston on outside of cylinder is greater than that inside cylinder... • Gas outside does work on piston, energy is transferred from gas to piston, and gas inside cylinder contracts • When gaseous system contracts at constant temperature, process is endothermic • Surroundings does work on system

Stress from Decreasing Concentrations

• If conc. of one of the species in an equilibrium is decreased, Q changes and the reaction will proceed in the direction that will increase that conc. of that species • When equilibrium is re-established, Q=K(eq) once again • This decr. the equilibrium conc., but it doesn't change the Equilibrium Constant (K(eq))

Stress from increasing concentrations

• If conc. of one of the species in an equilibrium is increased, Q changes and the reaction will proceed in the direction that will reduce that conc. of that species • When equilibrium is re-established, Q=K(eq) once again • This incr. the equilibrium conc., but it doesn't change the Equilibrium Constant (K(eq))

Stress from Increasing Pressure

• If pressure of system at equilibrium is increased, Q changes and the reaction will proceed toward the side with fewer moles of gas to reduce that stress • When equilibrium is re-established, Q=K(eq) once again • This changes the equilibrium conc., but it doesn't change the Equilibrium Constant (K(eq))

Stress from Dilution

• If solvent is added to a system at equilibrium, Q changes and the reactant will proceed toward the side with more particles to reduce that stress • When equilibrium is re-established, Q=K(eq) once again • This changes the equilibrium conc., but it doesn't change the Equilibrium Constant (K(eq))

Substances with Lower (less +) Oxidation states than usual

• In C2O4 2-, C has an oxidation state of +3 • As C is in Group 4A, it would rather have an oxidation state of +4 • It will oxidize (lose e-s) to raise its oxidations state C2O4 2- —> CO2 • Some non-metals can raise their oxidation states by bonding with like elements • Happens when hydrochloric acid is combined with a species that is easily reduced

Liquids

• In a liquid the same IMF of attraction exist • These forces r strong, but not as strong as they r in a solid

Saturated solution is at equilibrium

• In a saturated solution, an equilibrium is created btwn the solid and aqueous states • Solids r constantly dissolving to form aqueous components, and aqueous components r solidifying at the same rate

Identify the types of motion that individual molecules can undergo in a molecular solid

• In molecular solid, predominant motion that molecules exhibit is vibration • In some cases, rotation is possible and molecules r not free to move through translation in molecular solids

Equal moles of F2 and ClO2 r drawn into a vacuum where the following process takes place F2 + 2ClO2 <-> 2FClO2 \_____________ \ \____________ ____________ / / c) Explain how the graph could be used to calculate the instantaneous rate of appearance of FClO2 at t=15 min..

• Instantaneous rate of appearance of FClO3 at 15 minutes is the slope of the line that is tangent to the FClO3 curve at the 15 minute mark slope = rise/run

Fractional distillation was used to isolate an unknown volatile substances that had contaminated the well water at a rural property b. What can be said about the relative strengths of intermolecular interactions among and btwn the compounds in question (contaminant and water)?

• Intermolecular FoA btwn the water molecules r greater than they r btwn the contaminant molecules • This is why the vapor P of the contaminant is higher and its boiling temperature is lower • Intermolecular FoA btwn the water and contaminant r also weaker than those btwn water molecules

Non-Ideal Behavior and Condensation

• Intermolecular FoA increase as distance btwn particles decreases • Can lead to condensation at sufficiently low T and/or exceptionally high P • This applies to all gases, even those with relatively weak intermolecular forces • Condensation can occur rapidly when T of system drops - KE of gas particles is reduced and they move closer together - Under these conditions, intermolecular forces can cause particles to stick together when they collide • Condensation can also occur at high P - Gas particles r closer together and experience a higher rate of collisions - Intermolecular forces btwn particles can cause them to condense • When a gaseous system is approaching the pt where condensation will occur, FoA btwn gas particles r at a maximum • Results in the largest possible decrease in measured P, so large deviation from ideal behavior

Some Cations acidify solutions • The ability to release protons from water increases as:

• Ionic radius decreases • Charge increases • Ex. - Fe 3+ decreases the pH more than Fe 2+ • Fe 3+ is smaller with more net + charge - Conjugate acids of strong bases will not affect pH • Group 1A cations, Ca 2+, Sr 2+, and Ba 2+ • Large radii and/or small net (+) charge

An e- is promoted from the n=2 and the n=3 energy level in a H atom. Does the e- release or absorb energy during this transitions?

• It absorbs energy, as it moves from a lower energy level to a higher energy level

The following ?s pertain to aqueous solutions of potassium bromide b. Explain why potassium bromide dissolves in water

• KBr dissolved in water bc the cations (K+) r more attracted to the - ends of the warer molecules than they r to the anions (Br-), and anions (Br-) r more attracted to the + ends of the water molecules than they r to the catjon (K+) • Furthermore, the water molecules r more attracted to the ions than they r to other water molecules • If this was not true, the solid compound would not dissolve

System 1 (short arrows) vs System 2 (long arrows) b) Will KE be transferred from one of these systems to the other? Justify in terms of molecular collisions

• KE r transferred from higher (2) to lower (1) temp. • Collisions btwn gas particles and molecules that make up the plane of thermal contact r similar to collisions btwn a series of billiard balls • One molecule collides w/ another and transfers some KE to it • Results in a decr. in KE in higher temp. and an incr. in KE in lower temp. system

∆H from Bond Enthalpies Endothermic process

• Less energy is released during the formation of bonds in the products than is required to break the bonds in the reactants • The products r at a higher PE than the reactants

Disproportionation Reactions (II)

• Like elements existing together in 2 oxidation states can undergo a redox reaction to produce a species in an intermediate oxidation state • Element must, therefore, have at least 3 oxidation states, and there must be an intermediate state that falls btwn the states of the reactants

Explain why the boiling point of water decreases as elevation increases.

• Liquid boils when its vapor P = atmospheric P. •Vapor P depends on the T of the specific liquid. • When the T of the liquid decreases, the vapor P also decreases • Atmospheric pressure decreases as elevation increases, so at higher elevations, liquid will boil at lower temperatures

Types of solutions - Solid - Liquid Solutions

• Many ionic compounds dissolve in polar solvents (ion-dipole) • Polar solids, such as glucose, dissolve in polar solvents (dipole-dipole or H-bonds) • Non-polar solids, such as mothballs, dissolve in non-polar solvents (dispersion) • Carbonated drinks H2O(L) + CO2(g) —> H3CO3(aq) • Oxygen gas dissolves in water O2(g) —> O2(aq)

Is the melting of ice to create liquid water a chemical process, physical process, or both? Justify

• Melting ice is a phase change, which is a physical process • Physical process involve changes in intermolecular forces • In this case, H-bonds r weakened

Microwave Spectroscopy

• Microwaves cause polar molecules to rotate • Each type of polar molecule has specific rotational frequencies that it can exhibit • The peaks in the microwave spectra below correlate with the different rotational frequencies for a specific polar molecule • Data from microwave spectra can be used to calculate the bond length of diatomic polar molecules and to determine the shapes of polar molecules

Collision Orientation

• Molecules must collide with an orientation that can yield a reaction

Solid

• Molecules r held close together in a regular pattern by intermolecular forces = Why + next to -?? • Maximizes attractive forces

Carboxylic acid strength and inductive effects

• More electronegative elements that make up "R", the stronger acid • Electron density in O-H bond is reduced by electronegative fluorine atoms • Decreases force of attraction on H+

∆H from Bond Enthalpies Exothermic process

• More energy is released during the formation of bonds in the products than is required to break the bonds in the reactants • The products r at a lower PE than the reactants

3 Ni 2+ + 2 Cr —> 3 Ni + 2 Cr 3+ Would voltage of a cell that operates according to chemical equation above under standard conditions be greater than, less than, or equal to that of a cell where [Ni 2+] < 1.0 M and [Cr 3+] > 1.0 M. Justify

• More products than reactants would give + lnQ value • Cause value of E to be less than E ^o according to equation E = E° - (RT/nF)lnQ

Which substance in each set has the highest melting pt? Justify ur answer using chemical principles b. NH3 or C(diamond)

• NH3 has H-bonds and LDF and Diamond is a network solid held together by covalent bonds • Covalent bonds r stronger than intermolecular forces, so C(diamond) will have a higher melting temperature

NO(g) is more soluble in water than O2 (g). Explain

• NO is larger than O2, as N has a bigger radius than O, and the bond btwn N and O is slightly polar (electronegativity different of 0.5) • Thus, NO experiences larger dispersion forces and dipole-dipole forces when it dissolves in water, making it more soluble

Exothermic reactions often TFP

• Nature tends to favor processes that causes a reduction in energy • Bonds in products contain less energy than bonds in reactants • excess energy is released as heat

A gravimetric experiment was performed to determine the # of moles of sulfate, SO4 2-, that were present in a solution of NaSO4 2-. In this experiment, an excess of barium nitrate, Ba(NO3)2, was added to the solution and a precipitate formed. The resulting mixture was filtered and the precipitate was dried. The dry mass of the precipitate was measured to be 8.642 g. c) Explain why a net ionic equation is the best way to represent this reaction??

• Net ionic equation doesn't include spectator ions. as they r not involved in the reaction • only includes the aqueous species that reacted and solid product that was formed • In this experiment, only asked to determine the moles of SO4- that was present. In order to determine this quantity, we reacted sulfate with excess Ba 2+ to form solid BaSO4

Suppose u have 2 identical 1.0 L cylinders. Both cylinders r kept at exactly 25'C. 1 cylinder contains 0.3 moles of helium gas and other contains 0.3 moles of Xe gas a. Is P in cylinder containing He less than, equal to, or greater than that of Xe? Justify ur answer without performing any calculations

• P in both cylinders is same and have same V, T, and # of moles, so must have same P • According to Avogadro's Law, = V of gas at the same T and P contain = #s of gaseous particles (n1R1T1)/P1V1 = (n2R2T2)/P2V2 P1 = (P2n1R1T1V2)/(n2R2T2V1) P1 = P2, as n, R, T, and V cancel

Suppose u have 2 identical 1.0 L cylinders. Both cylinders r kept at exactly 25'C. 1 cylinder contains 0.3 moles of helium gas and other contains 0.3 moles of Xe gas b. If V of cylinder containing He increases to 2.0 L and T remains the same, what happens to the P inside cylinder? Justify ur answer by manipulating the combined gas law equation

• P inside the cylinder would decrease to half of its original value P2V2/T2 = P1V1/T1 P2V2 = P1V1 (Temperature is the same) P2(2V1) = P1V1 (V2 = 2V1) 2P2 = P1 P2 = 1/2 P1

Suppose u have 2 identical 1.0 L cylinders. Both cylinders r kept at exactly 25'C. 1 cylinder contains 0.3 moles of helium gas and other contains 0.3 moles of Xe gas b. If V of cylinder containing Xe is reduced to 0.25 L and T remains the same, what happens to the P inside cylinder? Justify ur answer by manipulating the combined gas law equation

• P would increase by a factor or 4 P2V2/T2 = P1V1/T1 P2V2 = P1V1 (Temp is the same) P2(0.25)V1 = P1V1 (V2 = 0.25V1) (0.25)P2 = P1 P2 = P1/0.25 = 4P1

Bond Energy

• PE of Ve-s decreases as they approach the nucleus of another atom

For gases, K(p) is often used... Partial Pressure

• PP of a gas in a system is the pressure exerted by that specific gas • PP of all the gases in a system sum up to the total pressure in that system • 1 mole of one type of gas will exert the same pressure as 1 mole of a different type of gas, under the same conditions of volume and temperature • 2 moles of a gas will exert twice as much pressure

Properties of Gases

• Particles r constantly moving • Expand to fill the V of the container they occupy • Form homogeneous mixtures • Low density • Highly compressible • Exert a pressure • Gases don't have definite shape or definite volume, as they r constantly moving and, ideally, there is no FoA btwn particles • Collision frequency and density of gas particles depends on P, V, and T

Student puts beaker w/ 150 mL of distilled water a hotplate. Temp. of water incr. from 22'C to 85'C b) Was this process endo or exo for water? Explain

• Process was endothermic for the water • Energy flowed into the water from the surroundings

Carboxylic Acids Strength

• R weak organic acids • They take this form O R - C - O - H Where "R" can be just about anything Denoted as RCOOH or RCO2H

Coffee Cup Calorimetry

• Reactants and products r system • Water which they dissolved in and calorimeter make up surroundings • Heat lost or gained by solution is = to heat lost or gained by reaction Ex. If calculated q_(solution) to be +568 J, solution gained heat. Means reaction was exothermic, as heat gained by solution is = to heat lost by reaction (q(rxn) = -568 J)

Acids - Bases reactions

• Reactions btwn acids and bases r called neutralization reactions • If strong acid or base is involved, K(eq) > 1, reaction goes to completion and a 1 way arrow is used • If weak acid-weak base reactions, K(eq) < 1, a state of equilibrium is established and 2 way arrows r used • Strong Acids and strong bases experience 100% ionization • H+ and OH- ions combine to form H2O • K(eq) for reaction is 1 x 10^14 at 25'C — goes to completion — K(w) = 1 x 10^14 at 25'C • Other parts of these acid and base act as spectator ions • pH can be determined from the excess reactant

Redox Reactions in Acidic Solutions (I)

• Reactions btwn metals and all strong acids except Nitric Acid • H+ ions can oxidize many metals (X) • H+ pulls e-s out of the metal to form H2

Volume of Solid and Liquid Phases

• Solid and liquid phases for a particular substances normally have similar molar volumes • Density of particles is similar in both phases • Ice has a slightly larger molar V than liquid water • Most solids have a slightly smaller molar V than their liquids

Sublimation

• Solids r evaporate and have a vapor pressure • As IMF r stronger in solids, the vapor pressure of solids r normally low • Solids with high vapor pressures, have relatively weak IMF

Solubility of Strong Acids

• Solubility of slightly soluble salts containing basic anions increases as [H+] increases • Basic aqueous anions will accept protons from the acidic solutions. Here we add a strong acid • 2nd reactions reduces [OH-] in solution, and drives the 1st equilibrium to the right

Solubility of Weak Acids

• Solubility of slightly soluble salts containing basic anions increases as [H+] increases • Basic aqueous anions will accept protons from the acidic solutions. Here we add a weak acid • This reduces [OH-] in solution, and drives the equilibrium in 1st reaction to the right

Strong Bases

• Soluble compounds containing OH- Possible cations: • All group 1A Cations • Ca 2+, Sr 2+, or Ba 2+

Specific Heat Capacity (c)

• Specific heat capacity, c, is amount of heat requires to raise temperature of 1.0 g of a substance by 1.0 K • A large value for water (4.184 J/g•K) - Takes a lot of heat to increase the temperature of 1.0 g of water by 1.0 K • A small value for iron (0.45 J/g•K) - Takes much less heat to raise temp. of 1.0 g if iron by 1.0 K

Standard Reduction Potentials (E^o(red))

• Standard potentials for half-cell reactions r used to calculate standard cell potential (E^o (cell)) for overall reaction • By convention, all standard half-cell potentials given as reduction potentials (E^o (red)) E^o(cell) = E^o(red) (cathode) - E^o(red) (anode) E^o(cell) = standard cell potential and overall difference in PE/charge btwn anode and cathode E^o(red) (cathode) = where reduction takes place. Most +, least - E^o(red) value. Greater driving force to attract e-s E^o(red) (anode) = Where ox takes place. E^o(oxidation) = -E^o(red). If flip reactions, u can change sign

Types of solutions - Liquid - Liquid Solutions

• Strong H-bonds allow these species to form a solution • CH3OH and H2O r miscible (soluble in all proportions) • Miscible solutions never become saturated • Differences in intermolecular forces can cause solution's V to differ from the sum of the volumes before mixing • Hexanol and water r not miscible - Some hexanol will dissolve in water - Solubility of hexanol limited by the long non-polar C chain • Hexane and hexanol r miscible - Hexane completely non-polar - Hexanol is mostly non-polar - Form strong LDF of attraction for one another

Strength of acids and bases

• Strong acids always have very weak CB • Very weak acids always have very strong CB • Acids with mid-range strengths have CB with mid-range strengths • Stronger base always accepts most protons in an acid/base reaction = 2 bases • Base on reactants side of equation • CB on products side of equation

Strength of Acids

• Strong acids always have very weak conjugate bases • Very weak acids always have very strong conjugate bases • Acids with mid-range strengths have conjugate bases with mid-range strengths

3 notes with reactions

• Substances in lower than normal oxidation states tend to oxidize (they will oxidize if put with a strong oxidizer — stuff sheet) • If u need to decide which substance will oxidize, always pick the metal • If a substances is mentioned as a solid, don't break it into ions on the reaction side. Even if it is soluble

Like Dissolves Like

• Substances that share similar intermolecular interactions tend to be soluble or miscible in one another • Miscible - Soluble in all proportions • Miscible solutions never become saturated • Form strong LDF of attraction for one another

According to KMT, when 2 gas particles that share the same molar mass collide, is the sum of their velocities after the collision less than, equal to, or greater than the sun of their velocities before the collision? Justify

• Sum of their velocities would be the same before and after the collision • According to KMT, all collisions btwn gas particles don't change after they collide, the sum of their velocities doesn't change either • KE(initial(I))+KE(initial(II))=KE(final(I))+ KE(final(II)) • 1/2mv^2 (In. (1)) + 1/2mv^2 (In. (2)) = 1/2mv^2 (fin. (1)) + 1/2mv^2 (fin. (2)) • 1/2m(v^2 (In. 1) + v^2 (In. 2) = 1/2m(v^2 (Fin 1) + v^2 (fin. 2) • (v^2 (In. 1) + v^2 (In. 2)) = (v^2 (Fin. 1) + v^2 (Fin. 2)

A ridged 1.0 L cylinder contains 1.0 moles of O2 gas at 25'C and another ridged 1.0 L cylinder contains 1.0 moles of CH4 gas at 25'C. Is the average speed of the O2 gas particles greater than, less than, or the same as the average speed of the CH4 gas particles? Justify ur answer

• T is a measure of the average KE of the particles in a system and since both systems t at the same T, gas particles in both systems will possess the same average KE • Basic formula for KE is: KE = 1/2 mv^2. • Tells that molecules with smallest molar mass will have the largest average velocity • Atomic mass of O2 is 32 g/mol and the atomic mass of CH4 is 16 g/mol. • Since CH4 has the smaller molar mass, particles in that system will have the largest average velocity

Suppose u have 2 identical 2.0 L cylinders. Both cylinders r kept at exactly 25'C. 1 cylinder contains 0.250 moles of He, and the other contains 0.250 moles of krypton. V of these cylinder can change b. What variable must change in order to increase the average velocity of the molecules in either cylinder? Explain

• T must increase in order to increase the average speed of the atoms in either container

Electrolytic Cells (^_G > O)

• T un-F redox reactions (^_G > O) • Requires a source direct electrical current hooked up to 2 electrodes • Electrodes r immersed in a molten salt or an electrolytic solution • Anode takes e-s from 1 species in solution (ox) • Cathode gives e-s to another species in solution (red)

Kinetic Control

• TFP that don't occur at measurable rates r said to be under "kinetic control" • Often result of high activation energies • When a process doesn't proceed at a measurable rate, cannot be assumed that it has established equilibrium • If it is know that a certain process is TFP, but doesn't occur at a measurable rate, process is most likely under kinetic control

Use Boltzmann distribution curves to related temp. to motions of particles

• Temp. is a measure of average KE of particles in a system • Basic formula for KE is KE = 1/2 mv^2 • As the mass of atoms and molecules doesn't change, vel. incr. as KE or temp. incr. • Boltzmann distribution curves show that average vel. as well as distribution of velocities (fast to slow) incr. as temp. incr.

Suppose u have 2 identical 1.0L sealed containers. Both containers r kept at exactly 25'C. One vessel contains only Ne gas at 1.5 atm, and other contains only Xe gas at 1.9 atm. Is the value for the moles of Ne less than, equal to, or greater than that of Xe? Explain

• The # of moles of Ne must be less than # of moles of Xe • P exerted by the Ne atom is less than P exerted by the Xe atoms, when both containers r kept at the same V and T

volume adjustment for gases under high pressure

• The Ideal Gas Equation assumes that molecules don't have their own V • The V in PV = nRT is the volume of empty space in the container • Volumes of gas particles r negligible, when compared to the overall V of the container, at relatively low pressures (~ 1atm) bc they're very small and very far apart • At high P, molecules r compressed into a much smaller volume of space, and the volume occupied by the molecules becomes significant

Combustion Reactions : Combustion of Organic Compounds

• The combustion of any organic hydrocarbon in the presence of O2 produces CO2 and H2O

E(a) for Unimolecular Decomposition Reactions

• The energy required to break the bonds in the reactant is obtained during the collision with a background or solvent particle O3 + M —> O2 + O + M • For this reason, unimolecular decomposition rates increase as temperature increases

When the temperature of a solid copper pipe increase, does the motion of the free e-s increases, decrease, or remain the same?

• The free e-s would move faster when the temperature of the copper pipe increases, as the average KE of the free e-s would increase

Half-Reactions Method and Other Redox Reactions

• The half-reaction method used in the previous 2 ex. will work for any redox reactions - If there is no H2O or H+ in the balanced chemical equation, they will cancel - However H2O or H+ may appear in the half-reactions - Could be given a redox reaction and asked to write the oxidation or reduction half-reaction

The pH Scale

• The pH of a solution increases as [H3O+] decreases and [OH-] increases • The pH of a solution decreases as [H3O+] increases and [OH-] decreases

Why do electrons flow from anode to cathode??

• The potential energy for electrons is higher at the anode than it is at cathode. • All things naturally move to a state of lower potential energy

Determining a Rate Law

• The rate law for an overall reaction can only be found experimentally • But, rate laws for elementary steps r predictable • For any elementary step... — aA + bB —> dD + eE — Rate = k[A]^a [B]^b • The rate of the overall reaction = rate of the slowest elementary step

Fractional Distillation

• The separation of volatile liquids in a liquid-liquid solution on the basis of boiling points • Condensation solution has a higher concentration of the component with the higher vapor pressure • If cycle of boiling and condensing is repeated enough times, complete purification of the more volatile substance can be achieved

Explain why a glass of soda at 30'C will go flat faster than a glass of soda at 5'C

• The solubility of a gas decreases as the T if the liquid solution increases • Gases tend to have weak dipole-induced dipole forces and weak LDF, as they r not very polarizable • As the KE of the particles in the solution increases, gas particles break free from these weak intermolecular forces and re-enter the gas phase

Gas solubility and pressure - Henry's law

• The solubility of a gas is directly proportional to the partial pressure of that gas above the solution • P only affects the solubility of gases

Enthalpy of Formation

• The ΔH('f) value for the most stable form of any element in its standard state is zero - Ex. = Ca, Ag, Na, O2, Cl, H2, Fe, N2 Not Br2

Determining a Rate Law • Fast, Reversible —> Slow

• This reaction proceeds at the rate of the slow step, but the problem is that NO3 is an intermediate so its conc is unknown • To solve this problem, we must modify the rate law

Kinetic Molecular Theory 2

• Total V of all of the gas particles in a system is negligible when compared with the total V of the system • For this reason, Ideal Gas Law assumes that V of gas particles in a system is 0

Molar Heat Capacity (C_(m)) Equation

• Transfer of = quantities of heat to = #s of moles of 2 substances w/ different molar heat capacities will produce different temperature changes • Substance with higher molar heat capacity will experience a smaller increase in temperature q = nC_(m)∆T q = Heat lost or gained by a substance n = # of moles of that substance C_(m) = Molar heat capacity of that substance ∆T = Temperature change of that substance (∆T = T_t - T_i)

salt bridge

• U-shaped tube containing a solution of gel with soluble ions — Na+, SO4 2-, NO3 - r common • Ions don't play a role in redox reaction • Ions from salt bridge mustn't react with ions in electrolytic solutions to form a precipitate • Ions from salt bridge keep the solutions in half-cells neutral • If solution doesn't remain neutral, reaction will stop

All real gases do not behave ideally when ___ and ___! Under these conditions, the ___ does not make accurate predictions!

• Under high pressures (P > 5 atm) • At low temperatures • Ideal Gas Equation (PV = nRT)

According to KMT, when 2 gas particles that have different molar masses collide, is the sum of their velocities after the collision always going to be the same? Justify

• Very unlikely that the sum of their velocities would be the same • This would only happen if each particle continued to move at the exact same speed after they collided • Sum of their KE would be the same before and after the collision, but bc they have different molar masses the sum of their velocities would most likely be different • KE = 1/2 mv^2

Molecular Orbital (MO) Theory

• Views a molecule as a whole instead of a collection of individual atoms • MOs r similar to atomic orbitals - Both have specific energy levels - Both have specific sizes and shapes - Both hold a maximum of 2 e-s that spin in opposite directions • Atomic orbitals combine to form MOs - When 2 atomic orbitals combine, 2 MOs r formed — Orbitals r always 'conserved'

Relating Volume and Temperature

• Volume of a gas can be measured at different temperatures in a laboratory. • That data can then be plotted to estimate absolute zero • If u increase the number of moles of gas in the system, the slope of the line will increase • Equal numbers of moles of any gas will produce lines with the same slope

Relating Volume and Pressure

• Volume of a gas doubles when pressure is halved

Water and Acid-Base Reactions

• Water plays an important role in many acid-base reactions that take place in aqueous solutions, as its structure allows it to accept and donate protons • H+, which has no electrons, can attach itself to one of the lone pairs on a water • H+ can be pulled off a water, which leaves a OH- with 3 lone pairs around the O atoms

Quantum Theory and Planck

• We assume that energy increases in a continuous stream • But it actually increases in discrete units. zit increases by a full quantum, or not at all • The height of each step is = to 1 quantum of energy for a certain frequency electromagnetic radiation

What is the strongest base in each following reactions? Justify based on solution equilibrium or FoA btwn particles c) 2H2O <—> OH- + H3O+

• We know that equilibrium for this reaction lies far to the left, as K(W) is 1.0x10^-14 • This means that OH- is a stronger base than H2O • OH- and H2O compete for H+ ions, and OH- wins most of the time • This drives the equilibrium to the left

AB (g) <-> A (g) + B (g) Suppose a chemist adds 1 mole of pure AB (g) to a sealed vessel a) Describe the process of achieving equilibrium in terms of the changing of concentrations of AB (g), A (g), and B (g)

• We start out with only AB gas in a vessel • As time goes on, [AB] drops • At the same time, [A] and [B] r increasing • Due to the fact that every time one molecule of AB disappears one molecule (particle) of A and one molecule (particle) of B is created • Eventually, each species ends up maintaining a specific concentration • At that point, the system has reached equilibrium • Equilibrium doesn't mean that all species have equal concentrations

Strong acid - weak bade reactions

• Weak bases will accept protons from a strong acid • Weak bases may be nitrogen containing compounds such as NH3, or the CB and weak acids

Explain why the measured P of a gaseous system under conditions that r very close to those that would result in condensation will be lower than what the ideal gas law would predict. Draw a particulate representation to assist with ur explanation

• When a gaseous system is approaching the pt where condensation will occur, the FoA btwn gas particles r at a maximum. • Results in the largest possible decrease in measured P, and thus, a large deviation from ideal behavior • Attractions to neighboring gas particles will reduce impact velocities, and therefore P, by the largest possible degree under conditions that r approaching those that will result in condensation

Absorbing Photons

• When a photon is absorbed by an atom or molecule, an e- moves up one or more energy levels • Increase in energy is = to the energy of photon that was absorbed • Increase in energy is also = to the difference in energy btwn the 2 energy levels

Emitting Photons

• When a photon is emitted from an atom or molecule, an e- moves down 1 or more energy levels • Decreases in energy is = to the energy of the photon that was released • Decrease in energy is also = to the difference in energy btwn the 2 energy levels

What if u have a choice of 2 species that can be oxidized?

• When compounds contain p-block or d-block metals and non-metals with less + than usual oxidation states... • The metal will be oxidized Ex: SnCl2 Sn 2+ will be oxidized to Sn 4+

Pressure Adjustment for Gases Under High Pressures (Low Volumes)

• When gas particles r very close together, P they exert may be less than what the Ideal Gas equation would predict • Neighboring molecules exert FoA on one another when they r very close together • Such forces pull a gas molecule in the direction opposite to its motion • This reduces P resulting from impacts with the walls of the container = Low pressure system = No FoA reducing impact velocity = High pressure system = FoA reduce impact velocity

aminopropane, CH3CHNH2CH3, and isobutane, C4H10 • Which pure liquid has the highest boiling pt?

• aminopropane, bc both have similar LDF and same # of e-s so similar abilities to become polarized, but aminopropane forms H-bonds, which r much stronger than LDF • As a greater amount of energy is required to break stronger FoA, the boiling pt of aminopropane is higher

Volume Changes from Cooling

• When gaseous system contracts bc heat is removed (system cooled) overall process has 2 variables - heat, q, and work, w ^_E = q + w • cooling is exothermic and work is endothermic, so ^_E must be calculated

The following questions pertain to a system containing 122 (g) CO (g) in a 0.400 L container at -71.2'C b. Actual P exerted by the carbon monoxide gas is this system was found to be 145 atm. Explain why the actual P is less than what would be expected?

• When gases r behaving ideally, there r no FoA btwn particles • This means that there r no forces slowing them down before they strike the wall of a container • However, FoA do exist btwn gas particles when they r in close proximity to one another • Neighboring gas particles will pull other gas particles toward them as they r about to strike a wall • This reduces the force that particles exert on the walls of the container, thereby reducing the P

Heating and Cooling

• When heat flows into a system, energy of that system increases - Internal energy of system increases as heat flows into the system • When heat flows out of a system, energy of that system decreases - Internal energy of system decreases as heat flows out of system

Coffee Cup Calorimetry equation

• When measuring the heat lost or gained in a chemical reaction that takes place in a coffee cup calorimeter, we use the solution as the basis for our calculations q = mc∆T q = Heat lost or gained by the solution in reaction m = Mass of solution c = Specific heat of solution 4.184 J/g•K is often if conc. of an aqueous solution is low ∆T = Temp. change of solution (∆T = T_t - T_i)

Saturated Solution

• When the solvent has dissolved the maximum amount of solute possible at a certain temperature, and some solid particles remain undissolved • an equilibrium system where solid particles continually dissolve in the solvent and dissolved particles fall out of solution

EMF under non-standard conditions and equilibrium Cu 2+ + Ni —> Cu + Ni 2+ • If Cu 2+ is removed to a system under standard conditions

• [Cu 2+] < 1.0 M • Q > 1, so closer to K than was under standard conditions • Voltage, E(cell), decreases • Electron flow (current) decreases • Cell will power a device for a shorter time due to the additional supply of Cu 2+

EMF under non-standard conditions and equilibrium Cu 2+ + Ni —> Cu + Ni 2+ • If Cu 2+ is added to a system under standard conditions

• [Cu 2+] > 1.0 M • Q < 1, so farther away from K than was under standard conditions • Voltage, E(cell), increases • Electron flow (current) increases • Cell will power a device for a longer time due to the additional supply of Cu 2+

EMF under non-standard conditions and equilibrium Cu 2+ + Ni —> Cu + Ni 2+ • If Ni 2+ is removed to a system under standard conditions

• [Ni 2+] < 1.0 M • Q < 1, so further from K than was under standard conditions • Voltage, E(cell), increases • Electron flow (current) increases

EMF under non-standard conditions and equilibrium Cu 2+ + Ni —> Cu + Ni 2+ • If Ni 2+ is added to a system under standard conditions

• [Ni 2+] > 1.0 M • Q > 1, so closer to K than was under standard conditions • Voltage, E(cell), decreases • Electron flow (current) decreases

Entropy (S)

• a measure of the disorder of a system • This increases when matter becomes more dispersed • Greater degree of disorder is favorable • + value for ^_S is favorable ^_S = S(products) - S(reactants)

Binary Acid Strengths

• acid strength increases when moving down a group Acids strength increases ——> HF << HCl < HBr < HI Anion Radius Increases ———> • According to Coulomb's law, the greater the distance between the nucleus of the anion and its outermost electrons, the smaller the attractive force on the H+ ion

Transfer of Energy through Work

• associated w/ changes in volume of a gas w = F x ^_d w = P x A x ^_d w = -P^_V F = P x A ^_V = ^_d x A

Mole Fractions and Partial Pressure

• can find partial pressure of any component in a gas mixture by multiplying the total pressure by its mole fraction

Combustion Reactions : Combustion of Elements

• combustion of any element in the presence of oxygen produces its oxide

Acids

• donate protons in aqueous solutions • Strong acids like HCl and H2SO4 experience ~100% ionization in polar solvents

PD is also known as...

• emf = electromotive force —> force that causes e-s to move • Standard emf = electromotive force under standard conditions —> 25^o and 1 M conc. or 1 atm PP of reactants and products • E(cell) = cell potential • E^o (cell) = standard cell potential —> 25^o and 1 M conc. or 1 atm PP of reactants and products • Cell Voltage = unit for all above volt (V)

What is strongest base in reaction? Provide justification based on solution equilibrium or FoA btwn particles 2H2O <—> OH- + H3O+

• equilibrium for this reaction lies far to left, as K(w) is 1.0x10^-14 • OH- is stronger base than H2O • OH- and H2O compete for H+ ions, and OH- wins most of the time • Drives equilibrium left

EMF under non-standard conditions and equilibrium A + B 2+ —> A 2+ + B

• equilibrium lies far right so K(eq) >> 1 • Cell potential (voltage) decreases — As reactants turn into products — As the system approaches equilibrium — As Q approaches K • When Q = K the cell potential is 0.00 V • When E(cell) = 0.00 V the system is at equilibrium — Conc. or PP remain constant — cell is "dead"

Equilibrium Expressions • Include ____ and _____ • Don't include _____ or _____

• gases (g) and aqueous species (aq) • Liquids (L) or solids (s)

Weak acid - weak base reaction

• generally don't go to completion • acids and bases r mostly undissociated • We write these protons transfer reactions

If given an acidified solution of 2 ionic compounds...

• get rid of any always soluble ions and look at what is left • Decide what is getting oxidized and reduced • "Stuff sheet" includes lists of strong oxidizers — such as dichromate, permanganate, and nitrate • These reduce so then look for what could oxidize such as metals or metallic ions that could be oxidized further • The balance with half-reaction method

^_S>0 when products have more particles

• increasing # of moles increases # of potential arrangements • Changing from liquid to gas increases # of potential arrangements

^_S>0 (usually) when making solutions with solids and/or liquids

• ionic compound is very organized • Solution containing same ions is much more disordered

Entropy of Dissolution (^_S)

• ionic solid is very organized • Solution containing the same ions is much more disordered • ^_S>0 (usually) when ionic solids dissolve in water • When an ionic compound in water, entropy will usually increase greatly, giving a + value for ^_S • Crystal lattice breaks apart and ions experience a huge increase in their freedom of motion — allowing for a larger # of possible arrangements • ^_S can be - for some small ions with large charges, such as Al 3+ • Ions may attract and reduce movement of large # of water molecules, thereby reducing entropy of H2O • Entropy of H2O molecules that surround these ions may be reduced to point where entropy of overall system becomes less than that of the components before they were mixed • Entropy of ions increases, but entropy of water can decrease by a greater degree

Molar Heat Capacity (C_m)

• is the amount of heat required to raise temperature of 1.0 mole of substance by 1.0 K • A large value for water (75.3 J/mol•K) - Takes a lot of heat to increase temperature of 1.0 mole of water by 1.0 K • Small value for iron (25.1 J/mol•K) - Takes much less heat to raise temperature of 1.0 mole of iron by 1.0 K

Nylon is saturated fatty acid. Explain why Nylon is such a strong material in terms of the interactions btwn adjacent polymers

• long chains that form strong LDF of attraction • large chainlike structures with many atoms and e-s which makes them highly polarizable • Bc overall shape of individual molecules is somewhat linear, polymers stack up well and have many contact pts which increases the strength of LDF • H-bonds can also form btwn neighboring polymers

^_S>0 for following processes:

• melting/vaporization • Reactions where products r in the same phase as the reactants but contain more particles than reactants • Making most solutions • +ing heat • Increasing volume of gas

what we know about ^_G

• must be less than 0 for a reaction to proceed in a given direction • is max amount of work that can be done by a system - System is at equilibrium when it is no longer able to do work (^_G=0) - Although forward & reverse reactions continue to take place at equilibrium, net changed in conc. rn't possible • K(eq) is directly related to ^_G

Buffer Solutions and pH

• pH = pK(a) when [A-] = [HA], as log(1) = 0 • Buffers made from very weak acids and their salts have high pH values — pH = pK(a) = -log(1x10^-10) = 10.0 — basic anion from salt increases [OH-] • Buffers made from stronger weak acids and their salts have lower pH values — pH = pK(a) = -log(1x10^-4) = 4.0 — Stronger acid increases [H+]

when ^_G<0

• process favors products at equilibrium K(eq) > 1 • reactions will proceed to equilibrium in either direction • reaction is considered to be exergonic

TFO r not necessarily fast

• rate of reaction is determined by chemical kinetics, not thermodynamics • Many TFPs don't occur at a measurable rate - Ex. = Oxidizing is a TFP • can take several hundred yrs for an Fe beam to rust away

pH, pK(a), [HA], [A-]

• relative conc. of HA and A- in any solution, buffered or not, can be predicted by comparing the pH and pK(a) of acid • When pH < pK(a), [A-]<[HA] pH = pK(a) + log[A-]/[HA] pH = pK(a) + log 1/2 pH = pK(a) + (-.3) • When pH > pK(a), [A-] > [HA] pH = pK(a) + log[A-]/[HA] pH = pK(a) + log 2/1 pH = pK(a) + (0.3)

Suppose u have 2 identical 2.0 L cylinders. Both cylinders r kept at exactly 25'C. 1 cylinder contains 0.250 moles of He, and the other contains 0.250 moles of krypton. V of these cylinder can change a. Explain why these 2 gases don't share the same velocity under these conditions

• root mean square velocity of He is greater than that of krypton • This is due to the fact that the molar mass of He is less than that of krypton • Square root of the molar mass is in the denominator of the equation and a smaller denominator (molar mass), will yield a higher velocity • The temperature is the same for both gases and the molar mass is the only variable that changes

Gas solubility and pressure - The Bends

• slope of these lines increase as intermolecular forces increase • If u ascend too quickly, the reduction in pressure causes N2(aq) to form N2(g) in ur blood (can be painful/fatal) • Deep sea divers prevent this by using He(g) in place of N2(g) - He(g) exhibits low solubility under high pressures

In a laboratory, students were given 2 unlabeled beakers and told that one of the beakers contained 1.0g of solid CaCO3 and the other contained 1.0g of solid AgNO3. They were told to devise an experiment to identify which compound was which. A student did so by adding 50 mL of distilled water to each beaker. Describe the student's observation that allowed him to identify each compound.

• solid AgNO3 would dissolve, as all nitrates r soluble in water. • CaCO3 wouldn't dissolve, as it is insoluble in water

Standard Reduction (half-cell) Potentials (E^o(red)) Which species reduced??

• species with most +, least -, (E^o(red)) value reduced, as has greatest ability to attract e-s • An e- can only decrease its PE by moving to species with most +, least -, (E^o(red)) value

Kinetic Molecular Theory 5

• the average kinetic energy of the molecules is proportional to the absolute temperature • Gas particles in any system kept at the same temperature will have the same average KE

Weak bases

• weak bases react with water to produce OH- • Equilibrium for weak base reactions normally lies to the left

^_S<0 when making solutions with liquids and gases

• when dissolving a gas in liquid the entropy will decrease • Rapid and chaotic movements of gas particles r greatly reduced by molecules in the solution

Stress of Decreasing Pressure

•If pressure of system at equilibrium is decreased, Q changes and the reaction will proceed toward the side with more moles of gas to reduce that stress • When equilibrium is re-established, Q=K(eq) once again • This changes the equilibrium conc., but it doesn't change the Equilibrium Constant (K(eq))

∆H from Bond Enthalpies

∆H = Σ BE (bonds broken) - Σ BE (bonds formed) • Reactant - product BE = Bond Enthalpy - Amount of energy required to break a bond, which is = to the amount of energy released when that same bond is formed • 1 to 2: In order to break the A-A bond, a certain amount of energy must be absorbed • 2 to 1: If an A-A bond forms, same amount of energy is released