Chem 212 Final
Face Centered Cubic
(FCC) Volume efficiency is 72% 4 atoms per unit cell (8 corners all contain 1/8 of an atom) (6 half spheres) 8(1/8) + 6(1/2) =4 atoms C=12
Body centered cubic
- (every corner atom is 1/8 of an atom)(there are 8 corners)+(1 atom in middle) (1/8)8 +1 = 2 atoms per unit cell -x=4/√3 R
Simple Cubic
-1/8(8)-1 atom per unit cell - C=6 - x=2R v= x^3
How does a catalyst speed up a reaction?
-a catalyst lowers the activation energy (Ea) - this in turn raises K value - thus, this speeds up the rate (increasing K, increases the rate)
Dispersion forces
-found in every molecule. -most significant in non-polar molecules -a temporary distortion of the electron cloud (momentary dipole, where one side is negative and the other is positive) -induced dipole, a dipole creates a temporary dipole in another molecule ex. CH4, C6H6, C2H6
What is true concerning this reaction? H20(l) <--> H20 (s)
-water is freezing -intuitively, we know temperature is decreasing due to the freezing of water, and thus is spontaneous. But why? -is exothermic, since heat is leaving the system -entropy is negative, since from liquid to solid we are losing entropy, -when both H and S are negative, the reaction is only spontaneous at low temperature - this is why the reaction is spontaneous
H2 (g) ---> 2H(g) spontaneous? nonspontaneous?
-whenever you break a bond, energy is required thus is endothermic (the reverse is true, making a bond releases energy) -entropy is also increasing, due to going from 1 mol to 2 mol, low entropy to high entropy, entropy is positive -when both entropy and enthalpy are positive, reaction is only spontaneous at high temperatures
Colligative properties: Boiling Point Elevation
1. A colligiative property depends on the quantity of the solute dissolved 2. Adding salt, increases the BP of water. The ions make the water more difficult to go into the gas phase, thus more energy via heat is required to boil the water. 3. ▲T= Tsolution-Tsolvent kb- boiling point elevation constant m= molality, moles solute/kg solvent i- van h factor ex. NaCl--> Na+ + Cl- (i=2) ex. MgCl--> Mg2+ + 2Cl- (i=3) ex. Al2(SO4)3-->(i=5)
Electrode potentials
1. Cell potentials are additive -add 2 reactions, add their cell potentials 2. Cell potentials are intensive properties -if a reaction is multiplied by a number, the potential remains the same 3. Strengths of oxidising and reducing agents -the E values are in red potential in appendix (ox+ne- --> red) -the reduction potentials are the potentials for the reaction to go forward -higher value, the higher the tendence to undergo reduction, lower tendency for oxidation .4. higher Ecell in appendix, stronger the OX agent
Colligative Property: Freezing Point Depression
1. Freezing point decreases when add salt to it. ex. adding salt to roads causes ice to melt easier.
Gibbs free energy and K
1. G*<0, -spontaneous in forward direction -shifts to the right--> -K>>1 (product favored, think k=p/r) 2. G*=o -equilibrium - k=1 3. G*>0 -nonspontaneous in forward direction -shift to left <--- -0<k<1 (k is between 1 and zero, K can't be negative)
Will it be spontaneous at low or high temp? 1. If ▲H and ▲S are + 2. if ▲H and S are - 3. If ▲H is - and ▲S is + 4. if ▲H is + and ▲S is -
1. If ▲H and ▲S are + = will be spontaneous at high Temperatures 2. if ▲H and S are - =Will be spontaneous at low temperatures 3. If ▲H is - and ▲S is + = spontaneous at all temperatures 4. if ▲H is + and ▲S is - = nonspontaneous at all temperatures ▲G= ▲H -T▲S
Unit cells
1. Simple cubic -1 atom per unit cell - coordination number is 6 (# of atoms attached to a single atom, every atom is adjacent to 6 atoms) -52% volume (52% is atoms, 48% is empty space) -Edge length (x) - (the side length of a cube) x=2R (edge length is twice the value of the radius of atom) 2. Body Centered Cubic -2 atoms per unit cell -C=8 -68% volume - x= 4/√3 R 3. Face Centered Cubic - 4 atoms per unit cell -C=12 -74% (most efficient structure) - x=√8 R
A reaction mixture initially contains 0.75M of [HI]. At equilibrium, the concentration of [I2] was found to be 0.300M. Calculate the equilibrium concentration of HI in the mixture 2HI(g)--> H2(g) + I2 (g)
1. Since given initial concentration have to make an ice table 2HI(g)--> H2(g) + I2 (g) i:.75M 0 0 c: -0.6 +0.3 +0.3 e: 0.15 0.3 0.3 answer is 0.15M I is .75M since that was given. at equilibrium, I2 is .3 Since I2 and H2 are in a 1:1 ratio H2 also has a 0.3 the reasion its -0.6 for 2HI, is beccause -2(.3) is -.6 due to there be a 2 infront of it and since its .3 on the right side we have to minus it on the left and when yu add .75-.6 you get .15
Which of the following is a weak acid A. HNO3 B. HNO2 C. H2SO4 D. HI
1. The 6 six common strong acids are HCl HBr HI H2SO4 HNO3 HClO4 If not these probably not a strong acid so B is correct answer
Every redox reaction is the sum of two half-reactions
1. one oxidation and one reduction Red1--> Ox1 + ne- Ox2 +ne- --> Red2 Red1 + Ox2--> Ox1 + Red2 (stronger red1) + (stronger ox2) --> (weaker ox1) + (weaker red2)
Which oxidising and reducing agents are stronger?
1. the half-cell with the higher reduction potential is the cathode 2. The higher the reduction potential, the stronger the oxidizing agent 3. the lower the reduction potential, the stronger the reducing agent (Figure out which is the cathode and which is the anode) The cathode will have the stronger Ox agent, thus the other half reaction has the weaker Ox agent. The anode will have the stronger Red agent, then the other half reaction will have the weaker red agent.
The mass of the zinc anode decreased by 1.43g in 56 minutes. Calculate the average current that passed through the solution during this time period
1.43 g Zn (1 mol zn/ 65.38g) (2 mol e-/ 1 mol zn) (96485 AS/ 1 mol e-) (1/3360seconds) =1.26 amps
In balancing half redox reactions, how do you know which side to add electrons too and how many to add
6H+ + BrO3- --> Br- 3H20 There is a +5 charge on left, and a -1 charge on the left. Subtract the charges (+5) -(-1)= +6 Add +6 electrons to side with the higher charge (less negative ex. -1 and -9, -1 has the higher charge) 6e- +6H+ + BrO3- --> Br- 3H20
What is the pH of a strong acid and base at the eq point?
7
Which of the following salts will decrease the pH of an aqueous solution
A. NaNO2. Na+ is neutral, NO2- can form HNO2 which is a weak acid, thus a basic salt B. Correct answer. NH4+ and Br-. Br- is a neutral ion. NH4+ in water generates H3O+ thus making it more acidic. C. KCL --> K+ and Cl- will not change pH significantly D. Na+ is neutral, F- is a can forms HF with H+ which makes it a basic ion
Identify the true and false statements concerning a system undergoing an exothermic spontaneous process A. the entropy of the system is increasing B. The entropy of the system is decreasing C. The entropy of the surroundings increases D. The entropy of the surroundings increases E. The entropy of the surroundings decreases
A. The entropy of the system is increasing a. In an exothermic reaction, the system is releasing energy to the surroundings. This would imply that the system has a higher temperature then the surroundings, and since heat is going to the surroundings, the entropy of the surroundings is going to be positive, while the entropy of the system is decreasing. Temperature and entropy are related. As T increases, KE increases, due to the increased number of random motions. This statement is false B. The entropy of the system is decreasing a. True C. The entropy of the surroundings increases a. True D. The entropy of the surroundings increases a. True E. The entropy of the surroundings decrease a. False
Which of the following statements about electrochemical cells is *correct*? a) in the salt bridge cations travel toward the anode and the anions towards the cathode b) the cathode is labeled as negative (-) in a voltaic cell but positive (+) in an electrolytic cell c)Reduction occurs at the cathode of both galvanic and electrolytic cells d) The free energy change is negative for an electrolytic cell e) none are true
C
CH3OH vs CH4 which has higher BP?
CH3OH -polar and has H bond -dipole-dipole -and LDF CH4 -non-polar -only LDF IMF↑ BP↑ CH3OH has higher BP What about solubility? Polar substances dissolve well in water (like dissolves like) CH3OH has higher solubility CH3CH2CH2 relatively lower solubility due to large non-polar region
Calculate the change in the entropy of the universe when 126.1g of liquid water is vaporized into steam at 1 atm. The enthalpy of vaporization is 40.7 KJ/mol. The standard entropy value for water and steam are 70 J/mol K and 189J/mol K, respectively
Calculate the change in the entropy of the universe when 126.1g of liquid water is vaporized into steam at 1 atm. The enthalpy of vaporization is 40.7 KJ/mol. The standard entropy value for water and steam are 70 J/mol K and 189J/mol K, respectively ▲Suniv= ▲Ssys + ▲Ssurr H20(l) H20(g) ▲Sys= ▲Srxn= [H20(g)] - [H20(l) = (189)-(70) ▲Sys= 119 J/mol * K ▲Surr= -▲H/T =-(40.7KJ/373 K) ▲Ssurr= -109.1 J/mol K ▲Suniv= (119) + (-109) ▲Suniv= 9.9 J/mol K 126.1g H20 * (1mol/18.016g H20) * ( 9.9 J/ mol K) = 69.3 J/ K The entropy of universe is 69.3 J/K
Which of the following is a lewis acid? A. BH3 B. AlCl3 C. FeCl3 D. All of the above
D. lewis acids are electron pair acceptors. BH3 can accept electrons. Same for others. Look and see if have lone e- pairs.
balancing half-reactions in basic solution
Do everything the same as you do acidic solution. To make it simple, at the very last step when you have combined both half reactions. Replace all the H+ with OH-. So, if on the left side you had 6H+, add 6 OH- to that and then you would have 6 H20 on left side. but don't forget to also add 6OH- to the right side as well. Also, you may be able to cancel some H2O from both sides depending.
Difference between standard cell potential and nonstandard cell potential
E= nonstandard cell potential Eo= standard cell potential Eo= is standard when concentration is 1M and temperature is either 25C or 298 K. If the values are anything but those, then that is the nonstandard cell potential E= Eo - 0.0591/ n *log(Q) n=# of electrons Q= ratio of products/reactants (have to be in aq phase) note that Q is initial products/reactants, while K is products/reactants when in equilibrium
How to calculate the electrode potentials
Ecell= Ecathode- Eanode REMEMBER this equation only works for the reduction potentials so don't flip the equations
T/F In general, water is a good solvent for both polar and non-polar compounds
False only for polar
A current of 1.25 amps pass through a solution of CuSO4 for 39 minutes. Calculate the mass of copper that was deposited on the cathode
First, convert 39minutes seconds. Then realize that Faradays constant, 98485 C/1mole- can also be written as amps multiplied by seconds F= 96485 1mol e-/ AS Takes 2 electrons mm of Cu is 63.55 39 min (60sec/ 1 min) (1.25 Amp) (1 mol e-/ 96485 A S)(1 mol Cu-/ 2 mol e-) (63.55g/ 1 mol Cu) =.963 g
If the system is at equilibrium what is the value for the free energy change?
G=0
Depsoition
Gas---> Solid
Condensation
Gas---> liquid
pH of weak base
Given Kb and concentration, same as weak acid initially Kb= x^2/ Wb (canceled the x in demoniator if kb is really small) x= concentration of OH- pOH= -log(OH-) pH= 14- pOH
Calculating pH of a weak acid given kb
Given concentration of weak acid and Kb, how do we calculate pH? auto-ionization of water: Kw= ka * kb kw= 1x10^-14 solve for ka with ka, use ka=[x]^2 / HA x= H30+ pH=-log(H30+) Can you same logic if have concentration of weak base and given Ka. Difference is X is concentration of OH-, so will have to calculate pOH then calcuate the pH
Acid Base Titration, finding missing concentration
Going to use an example to better explain. ex. 26.4mL of H2SO4 was completely titrated with 32.5mL of a 0.15M NaOH solution. What is the concentration of H2SO4? There are two ways to solve for this. First way is stoichiometry, and the second is using an equation. First way: Recognize that H2SO4 is a strong acid, and NaOH is a strong base. So will write out the equation H2SO4 +NaOH --> H2O + Na2SO4 (whenever you have a strong acid and base, will produce a salt plus water) Now we have to balance the equation H2SO4 + 2NaOH --> 2H20 + Na2SO4 So we are solving for the concentration of H2SO4, start with the other concentration. (..0325L NaOH / 1) x (.15mol NaOH/ 1L) x (1 mol H2SO4/ 2 mol NaOH) x (1/ .0264 L H2SO4) = .0923M H2SO4 That was first way, second way is using M1V1=M2V2 Recognize that H2SO4 is diprotic, has 2 Hydrogens. NaOH is monoprotic, just one hydroxide. so new formula is : 2M1V1= M2V2 2(M1)(26.4mL) = (.15M)(32.5mL) M1= .0923
Which has a higher BP I2 or Br2?
I2 and Br2 both are nonpolar so the only IMF they have is LDF I2 has a higher MW, more electrons, e-↑ polarazability↑ LDF↑ IMF ↑BP↑ I2 has higher boiling point
When does pKa=pH?
In a weak acid strong base titration, when your at 1/2 equivalence point, the pH=pKa You can find the equivalence point volume by doing M1V1=M2V2 That will be Volume at eq point. Take half of that and thats the volume at 1/2 eq point. At that volume, the pKa=pH It works because the mols of conjugate acid equal the mols of conjugate base, and when you enter that into the henderson-hasselbach equation, log of 1 is equal to 0, thus the pH= pka
Dipole-Dipole forces
Intermolecular forces between 2 polar molecules with a S+ and S- attraction. ex. H2O and H2O ex. CO and CO (opposite atoms attract) ex. HBr and HBr ex. SO2 -is polar, thus has dipole interactions -count valence electrions: 18 18-16= 2 lone pairs -bent shape thus polar
Ion-Dipole forces
Intermolecular forces between an ion and a nearby polar molecule Dipole- one size is positive, and the other is negative. ex. Na+ and H2O H2O is a polar molecule, the oxygen is attracted to the Na+ ion. Water is a permanent dipole, always polar. ex. NaCl and H2O. The oxygen is attracted to the sodium, the hydrogens will be attracted to the chlorine.
Strength of Intermolecular forces
Ion-Ion > Ion-Dipole > H-bond > dipole-dipole > Dispersion(LDF)
Hydrogen bonding
Is a special type of dipole-dipole interaction Intermolecular forces between H and F, O, or N ex. H3O, HN2, HF Remember, when H2O binds with another H2O that is an intermolecular force (Hydrogen bonding) But, the bonds between the two hydrogens and the oxygen of a H2O molecule, intramolecular bond via covalent bond.
What is the Arrhenius equation?
K= Ae^(-Ea/RT) R= 8.3145 J/mol*K Slope intercept form: lnk= (-Ea/R)(1/T) + lnA m= -Ea/R (remember that the slope is negative, keep in mind if it shows a picture, the slope will be decreasing)
pH of Weak acid
Ka= [x]^2 / HA- X if Ka is really small can cancel an x so ka= x^2/ HA x= will be the concentration of H30+ HA= given concentration of weak acid Ka= given then to find the pH take the -log(H30+) pH= -log(x)
Calculating molar solubility from ksp
Ksp of Ca3(PO4)2 is 7.1E-33 Ca3(PO4)2--> 3Ca + 2(PO4) ksp= (3x)^3(2x)2 7.1E-33= 108x^5 x= 1.46E-7 Molar solubility of Ca3(PO4)2 is 1.46E-7 Molar solubility of Ca+2 = (3x)= 4.38E-7 Molar solubility of (PO4)3- = (2x)= 2.92E-7
Vaporization
Liquid---> gas
Which, if any, of the following two elements can be isolated by electrolysis of the aqueous salt shown? a) Ni from Ni(SO4) (aq) b) I2 from NaI (aq)
Look up cell potentials of Ni and I2 Ni= -.25 I2= .053 If the numbers are between -1.00 and 1.4 then they will be isolated both numbers are, so both can be isolated
Colligative property: Osmotic pressure
M= Molarity, mols solute/ L Sol i- van h factor T- temperature in kelvin R- gas constant 0.08206 L *atm/ mol * k
Calculate the pH of the solution made by dissolving 30.5 g of NaF in enoughwater to make a 650mL solution. The KA of HF is 6.8E-4
NaF is a weak base not an acid. So can't use the KA value that was given. Na+ is a spectator ion, the equation is: F- + H2O <--> HF + OH- (30.5gNaF)(1mol/42gNaF)(1/.650L)= 1.117M I 1.117 0 0 C -X +x +x E 1.117-x +x +x kw= kw x kb kb=1.471E-11 kb= x^2/ [wb] 1.47E-11= x^2/ 1.117 x= 4.05E-6 pOH= -logx PH= 14-pOh pH= 8.61
Which , if any, of the following three metals could be used as a sacrificial anode for protection of a metal with E=-0.21v? (All E values must refer to M3+/M half-cell reactions) 1) gold, Au, E=1.50v 2)chromium, Cr, E=-0.74 3)aluminum, Al, E=-1.66 a. All three could be used b. Only (3) could be used c. only (1) and (2) can be used d. None could be used e. Only (2) and (3) could be used
One type of cathodic protection system is the sacrificial anode. The anode is made from a metal alloy with a more "active" voltage (more negative electrochemical potential) than the metal of the structure it is protecting (the cathode). The difference in potential between the two metals means the sacrificial anode material corrodes in preference to the structure. This effectively stops the oxidation reactions on the metal of the structure being protection. so answer e
Whats important to remember when calculating Q?
Only add aq, not solids Zn(s) + Cu2+ (aq)--> Zn2+(aq) + Cu (s) Q= [Zn2+] / [Cu2+]
calculating ksp from molar solubility
PbCl2 has a molar solubility of 0.0162 what is the ksp? PbCl2--> Pb + 2Cl ksp=4(x)^3 ksp=4(0.0162)^3 ksp=1.7x10^-5
Which of the following statements below is correct given the following rate law expression Rate=k[A]^2[B][C]^0 A. The rate of the reaction doubles as the concentration of A doubles B. the rate of the reaction decreases by half as the concentration of C doubles C. The rate increases by a factor of 3 as the concentration of B triples in value D. The rate of the reaction increases by a factor of 4 as the concentrations of B and C increases by a factor of 2 simultaneously
Plug one into everything when solving C. Rate=[1][3][1] =3
Q and K
Q: for initial concentrations K: equilibirum concentrations Q>K will shift <--- Q<K will shift ---> Q=K equilibrium
What formula to calculate standard free energy change(standard gibbs free energy) given K? (or vice versa)
R= 8.4145 J/mol*K (G* is in KJ traditionally, so have to convert J to KJ for R) T in kelvin Can also use this formula lnk= e^-G*/RT
1st order
Rate= k[A] ln[Af] = -kt + ln[Ai] (slope intercept form) can also use this equation Af= Ai e^-kt t/12= ln2/k graph is y=lnAf, x= time slope is negative
2nd order
Rate= k[A]^2 1/[Af] = kt + 1/[Ai] t1/2= 1/k[Ai] graph: y=1/[Af], x=time slope is positive for this, so will be increasing
Zero order reaction
Rate=k slope= -k [Af]= -kt + [Ai] t1/2= [Ai]/2k Graph is y=[Af], x= time slope is negative so will be decreasing
Which of the following statements is correct A. The reaction shifts to the right if Q> K B. The reaction is product favored if K<<1 C. increasing the temperature for an endothermic reaction causes the equilibrium constant K to decrease D. The presence of an inert gas has no effect on the equilibrium constant K
Reasons why A. shifts to left B. If K is very small its reactants favored not product favored C. K=P/R, P↑ when T↑ thus K will increase not decrease D. Correct. Inert gases means they are non-reactant thus will have no effect on the reaction. Also K doesn't change if you change reactants or products concentrations, only changes with Temperature.
Melting
Solid---> liquid
Sublimation
Solid--> gas
The partial pressures of N2O, O2, and N2O4 are initially 0.134 atm, 0.265 atm, and 0.483 am respectively. Determine if the reaction is at equilibrium or if it will shift to the right or to the left Kp= 56.8 2N2O (g) + O2 (g) ---> N2O4 (g)
Solve for Q like your solving for K 2N2O (g) + O2 (g) ---> N2O4 (g) Q= [N2O4] / [N2O]^2 [O2] Q= [.483] / [.134]^2 [.265] Q= 101.5 101.5 > 56.8 Thus, will shift to the Left
Spontaneous processes tend to flow from
Spontaneous processes tend to flow from low entropy to high entropy (more disorder)
IMF, BP and VP
Substances with lots of intermolecular forces, are harder to boil, thus will have an increased boiling point IMF↑ BP↑ VP↓ If you compare a substance with a high boiling point versus a substance with a low boiling point. -the substance with high BP will have strong IMF interactions, but a low vapor pressure - the substance with a low BP will have weaker IMF, but will have a higher vapor pressure
Use the thermodynamics data which refers to 298 K to calculate the normal melting point of the substance A. (the melting point is the temperature at which the process A(s)--> A(l) reaches equilibrium at 1.0 atm A(s) ▲H (kj/mol)= -192.8 ▲S (J/K mol)= 96.4 A(l) ▲H (kj/mol)= -147.8 ▲s (J/K mol) = 201.5
Tf= ▲Hf/▲Sf do products minus reactants for ▲H and ▲S ▲H= (-147.8)-(-192.8)=45 kj ▲S= (201.5)-(96.4)=105.1 J Convert J to KJ for entropy T= (45)/.1051 =428.2 K
How to find BP of a substance
The BP is the temperature at which the atmospheric pressure equals the Pvap pressure Pvap of water is 760 torr at 100C, (760 torr is 1atm at sea level)
Vapor pressure
The vapor pressure of a substance is the partial pressure when the rate of evaporation equals the rate of condensation -partial pressure has many values -the vapor pressure is a specific value at a given temperature. -vapor pressure for water is 23.8 torr at 25C -remember vapor pressure is an equilibrium pressure(rate of condensation is equal to rate of evaporation)
H2O, H2S, H2Se Rank by increasing BP
Tricky question because, H20 has a H-bond, so will have higher BP due to higher IMF. H2S and H2Se both have dipole-dipole interactions, so will look at number of e-, which H2Se has higher MW H2S< H2Se < H2O
T/F An elementary reaction is one occurring in a single step on a molecular level
True
T/F The rate law cannot be predict from the stoichiometry of a reaction
True
TRUE/FALSE All atoms exhibit dispersion forces
True
Colligative propery: Vapor pressure
VP is inversely related to BP Vp↓ Bp↑ adding salt to water, causes water harder to escape to gas phase Raoult's law: the vapor pressure of the solution is equal to the mole fraction of the solvent multipled by the vapor pressent of the solvent Psoln= xsolvent * Psolvent Psoln<Psolvent X (mole fraction) = moles of solute/ mols of solute+mols of solvent (or total moles or solution) (note that if you have an ionic compound dissolved in water, you have take the total moles of the ions not the moles of the compound. ex. CaCl2, becomes 1 mol CaCl2/ 111 g/mol * 3 mole ions/1 mol CaCl2
Which of the following processes is nonspontaneous? A. A ball rolls down a hill B. iron metal turns into rust in the presence of air and water C. Cleaning your bedroom D. Heat flowing from a hot object to a cold object E. Gasoline reacts with air to produce oxygen and carbon dioxide
Which of the following processes is nonspontaneous? A process is spontaneous if it can happen on its own without any outside influence. A. A Ball rolls down a hill a. Yes, spontaneous. B. Iron metal turns into rust in the presence of air and water a. Yes, spontaneous C. Cleaning your bedroom a. No, non-spontaneous D. Heat flowing from a hot object to a cold object a. Yes, is spontaneous, because it will happen on its own E. Gasoline reacts with air to produce oxygen and carbon dioxide a. Yes, spontaneous Spontaneous processes tend to go to more probable state
What will happen If you raise the temperature of an exothermic reaction?
Will shift to the left <——
According to the Arrhenius equation, if you increase the concentration of [A], as in Rate=k[A], does the rate of the reaction increase?
Yes. Increasing the concentration of reactants increases rate. Also increasing the temperature causes K to increase and thus the rate to increase.
Determining the order based of units of k
Zero order = L^-1 mol s^-1 First order= S^-1 Second order= L mol^-1 s^-1 Third order= L^2 mol^-2 s^-1
Which of the following relationships best describes ▲S for the following reaction K2SO4(s) --> 2K+(aq) + SO4(2-) (aq) a. ▲S<0 b. ▲S>0 c. ▲S=T▲H d. ▲S=0
b
Which of the following should have the greatest molar entropy at 298K? a. NO(aq) b. NO2 (aq) c. NO2 (g) d. NO (g)
c gasses more entropy than solid and liquids ▲S↑ Molar mass ↑ More atoms ↑
equation for finding out reaction order
c= concentration R= rate at the concentration (c2/ c1) ^x = (R2/R1) x= log(answer from R division) / log(answer from c division) x is order of reaction overall order is when you add all the orders of the reactants
if E cell of a metal is < o then the metal can
displace H2 from acid
pH in a weak acid and strong base titration
ex. 70mL of 1.25M HCN is titrated with 0.25 M NaOH. Calculate the pH after the addition of 100mL of NaOH. The Ka of HCN is 6x10^-10 1) recognize that a weak acid and strong base will produce a pH>7 2) write the balanced equation. 3) set up a BCA table. Need mols of acid and base 4) (.070)(1.25)= .0875 mol of HCN 5) (.25)(.100)= .025 mol of NaOH ....HCN....+.....NaOH...-->.....H2O...+....NaCN........................ B...0.0875.....+...0.025..............ignore.......0...............................C..-0.025...........-0.025..............................+0.025.......................A...0.0625..........0........................................0.025 HCN and NaCN are conjugate acid and base, thus this is a buffer solution. So we can use the henderson-hasselbach equation. For the equation, it doesn't matter if we use mols or Molarity of the acid and base. pH= pka + log[B/A] Need the pKA pka= -log(Ka) pka= -log(6x10^-10) pka= 9.22 pH= 9.22 + log(.025/.0625) pH= 8.82 (slightly basic)
how to find the pH of a weak acid and strong base titration if your at the equivalence point
ex. 70mL of 1.25M HCN is titrated with 0.25 M NaOH. Calculate the pH after the addition of 350mL of NaOH. The Ka of HCN is 6x10^-10 1) in this problem, 350mL is eq. point (we can calulate it if we didn't know via M1V1=M2V2) 2) First set up a BCA table, since at eq. the mols of the WA and SB will be the same and so will have just NaCN left over. 3) Find the molarity of NaCN, then make a ICE table. From the ICE table (or use the equation Kb= x^2/wb, find the pOH, then find the pH ....HCN....+.....NaOH...-->.....H2O...+....NaCN........................ B...0.0875.....+...0.0875..............ignore.......0.............................C..-0.0875...........-0.0875..............................+0.0875..............A...0......................0.......................................+0.0875..................... Take 0.0875 mol of NaCN and convert to Molarity. M= (.0875mol)/(.0070+.350) M= .2083 NaCN ....CN-......+.............H20...........-->...........HCN-...+.....OH-....... I...0.2083.................ignore..................0.........................0............C..-x.........................................................+x.......................+x........... E...0.2083-x............................................+x.....................+x..... kb= [x^2]/(.2083) Given Ka, have to calculate kb kb= 1.67E-5 x= 0.00186 pOH= -log(0.00186) pH= 14-2.73 pH= 11.27
Ion-Ion interaction
ex. Na+ and Cl- -the higher the charge, the greater the ionic interaction -Charge ↑ electrostatic force↑ - Size ↑ electrostatic force ↓ So Ca2+ and O-2 have a stronger ioni-ion interaction than Na and Cl Also, those with the higher electrostatic force will have a higher melting point
T/F All particles exhibity dipole-dipole forces
false
Freezing
liquid--> solid
Calculating heat of vaporization
ln(p2/p1) = (-Hvap/R) (1/T2- 1/T1) This is known as the Clausius Clapeyron equation Hvap has to be in J/mol R= 8.3145 J/mol *K Also can write as P2= P1 e^ -Hvap/R*(1/T2-1/T1) *enter* in calculator like this: P2= P1e^-Hvap(1/T2-1/T1)/R OR Hvap= -R ln(p2/p2)/[1/T2 - 1/T1) or T2= [ 1/T1 - Rln(P2/P2)/HVap]^-1 T has to be in kelvin
Metals that can displace H2 from water
metals that have Emetal < -0.42 can displace H2 from water at pH of 7
If Ecell of a metal is > o. the metal can
not displace H2 from acid
For a spontaneous reaction in the forward direction, the cell potential will be
positve
phase diagram
solids= keeps shape and volume liquids- conforms to shape and keeps volume gas- conforms to shape and volume
oxidation occurs at
the anode
reduction occurs at
the cathode
How to calculate the cell potential for an electrolytic cell
the half reactions CAN be -, 0, or +. If given a reaction, split into two half reactions. Then have to look up the reduction potentials for the half reactions. may have to flip one of the half reactions, based on however the original reaction was given to you.
How to calculate the cell potential for an galvanic cell
the half reactions have to add to either 0 or a + number. (can't be negative)
The volume to neutralize the solution is the same as
volume at the equivalence point
Finding pH wit h acid base titration (this is for a strong acid and strong base)
will use example to illustrate ex. 50mL of 0.50 M of a HCl solution is titrated with a 0.20M NaOH solution. What is the pH after the addition of 200mL of NaOH? 1) we are going to write balanced equation. note that a SA and SB make salt and water 2) going to set a BCA table, table has to be in moles, so if given M need to convert to moles via n=M*L 3) also need to know the total volume of solution. 50mL + 200ml= 250mL= .250L 4) (.50)(.050)= 0.025 mol of HCl 5) (.20)(.200) = .04 mol of NaoH 6) in BCA table, will subtract limiting reagant .................HCl...+....NaOH......-->........H20.....NaCl................ B..............0.025.......0.04........................ignore....0..................... C............-0.025.....-0.025.....................................+0.025............A.............0..................0.015...................................+0.025 so we have 0.015 mol of OH- solution left. So are pH will be basic. Now, we need to convert to molarity of OH-. Take mols and divide by the total volume which we calculated to be .250L. M= (.015/.250)= 0.06M of OH-. pOH= -log(OH-) pOH= -log(0.06) pOH= 1.22 pH= 14-pOH pH= 14-1.22 pH=12.78
Entropy of the universe
▲Suniverse= ▲Ssys + ▲Ssurr -▲Suniverse is always positive in spontaneous process
