AP Chemistry - Unit 7 Equilibrium
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
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
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
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
The Equilibrium Constant (K(eq))
K(c)=[E]^e[D]^d/[A]^a[B]^b Units for Conc - mol/L = M
K(p)
K(p)=(P(E))^e(P(D))^d/(P(A))^a(P(B))^b
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
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
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)
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)
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
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
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
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
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
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
Write balanced net ionic equations for reactions Solid strontium carbonate is placed in a solution of nitric acid
SrCO3 + 2H+ —> Sr 2+ + H2O + CO2
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
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)
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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)
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
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))
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
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
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
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
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
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
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
Equilibrium Expressions • Include ____ and _____ • Don't include _____ or _____
• gases (g) and aqueous species (aq) • Liquids (L) or solids (s)
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))
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)
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
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
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
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-
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
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
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
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
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
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
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))
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
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
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
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
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
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
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
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-
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-
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
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
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
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]