Chapter 18 Chemical Equilibrium 18.1 The Nature of Chemical Equilibrium.

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Chapter 18Chemical Equilibrium

18.1 The Nature of Chemical Equilibrium

Reversible Reactions

• A chemical reaction in which the products re-form the reactants

Reactants ↔ Products

• Le Chatelier’s PrincipleWhen system at equilibrium is disturbed by application of a stress, it attains a new equilibrium position that minimizes the stress.

Chemical Equilibrium

• The state where the concentrations of all reactants and products remain constant with time

• All reactions carried out in a closed vessel will reach equilibrium

Chemical Equilibrium

• Sometimes, if little product is formed, equilibrium lies far to the left (towards reactants)

Reactants Products

• Sometimes, if little reactant remains, equilibrium lies far to the right (towards products)

Reactants Products

Dynamic Equilibrium• Reactions continue to take place

• Reactant molecules continue to be converted to product

• Product continues to be converted to reactant (reverse reaction)

• Forward and reverse reactions take place at the same rate at equilibrium

Causes of Equilibrium

• Beginning of Reaction

Only reactant molecules exist, so only reactant molecules may collide

• Middle

As product concentration increases, collisions may take place that lead to the reverse reaction

• At Equilibrium

Rates of forward and reverse reactions are identical

Suppose that compound A reacts with B to form C and D.

The symbol is used for system at equilibrium.

If we add more A what would happen to the amount of C at equilibrium?

Adding A increases the amount of C and D at equilibrium.

If we add more C what would happen to the amount of A at equilibrium?

adding more C increases the amount of A and B at equilibrium.

If we add A what would happen to the amount of B at equilibrium?

A reacts with B to form C and D. Therefore, adding A decreases B.

Example - The Haber Process

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

• Hydrogen is consumed at 3x the rate of nitrogen

• Ammonia is formed at 2x the rate at which nitrogen is consumed

Points to Remember

1. Rxns in a closed system always reach equilibrium

2. When a rxn has reached equilibrium, that means the rate of the forward rxn equals the rate of the reverse rxn

3. A system will remain at equilibrium until the system is disturbed in some manner

The Equilibrium Constant• For the balanced forward equation:

jA + kB → lC + mD (j, k, l, m) are coefficients

• K is a constant called the equilibrium constant• K varies depending on temperature and upon the

coefficients of the balanced equation• K is determined experimentally• [X] represents concentration of chemical species at

equilibrium

K = [C]l [D]m

[A]j [B]k

Equilibrium Constant

4NH3(g) + 7O2(g) 4NO2(g) +6H2O(g)

K = [NO2]4[H2O]6

[NH3]4[O2]7

The value of K will tell us the

position of equilibrium

PracticeN2(g) + 3H2(g) 2NH3(g)

K = [NH3]2

[N2] [H2]3[NH3] = 3.1 x 10-2 M[N2] = 8.5 x 10-1 M[H2] = 3.1 x 10-3 M

Calculate K

K = [3.1x 10-2]2

[8.5 x 10-1] [3.1 x 10-3]3 K = 3.8 x 104

no units

[equilibrium]

[NH3] = 3.1 x 10-2 M[N2] = 8.5 x 10-1 M[H2] = 3.1 x 10-3 M

K = 3.8 x 104

The value of K will tell us the

position of equilibrium

If K >1, the rxn favors products

If K <1, the rxn favors reactants

If K = 1, equilibrium lies in the middle

[NH3] = 3.1 x 10-2 M[N2] = 8.5 x 10-1 M[H2] = 3.1 x 10-3 M

Calculate new KK = 3.8 x 104

2NH3(g) N2(g) + 3H2(g)

Original rxn

New rxn

Same [ ]Original K

The rxn was inversed so K is inversed

K = 1 3.8 x 104

K = 2.6 x 10-5

PracticeN2(g) + 3H2(g) 2NH3 (g)

[NH3] = 3.1 x 10-2 M[N2] = 8.5 x 10-1 M[H2] = 3.1 x 10-3 M

Calculate new KK = 3.8 x 104

Original rxn

New rxn

Same [ ]Original K

The coefficients were reduced by ½ K = (3.8 x 104)1/2

K = 2.0 x 102

½N2(g) + 3/2H2(g) NH3 (g)

Interpreting K

• If K = 1, then there are equal concentrations of reactants & products

• If K is small, the forward rxn occurs slightly before equilibrium is established, reactants are favored

• If K is large, reactants are mostly converted into products when equilibrium is established, products are favored

Heterogeneous Equilibria

• The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present

• If pure solids or liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction

• Pure liquids are not the same as solutions, whose concentration can change

More Practice!1. At 25 °C, an equilibrium mixture of gases

contains 6.4 x 10-3 M PCl3, 2.5 x 10-2 M Cl2, & 4.0 x 10-3 M PCl5. What is the equilibrium constant for the following reaction?

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

2. At equilibrium, a 2.0 L vessel contains 0.36 mol of H2, 0.11 mol of Br2, & 37 mol of HBr. What is the equilibrium constant for the reaction at this temperature?

H2 (g) + Br2 (g) ↔ HBr (g)

K = 4.0 x 10-2

K = 3.5 x 104

18.2 Shifting Equilibrium

The Effect of Change in Pressure: Ways to Change Pressure

1. Add or remove a gaseous reactant or product

2. Add an inert gas (one not involved in the reaction)

An inert gas increases the total pressure but has no effect on the concentrations or partial pressures of the reactants or products

The Effect of Change in Pressure3. Change the volume of the container

When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume.

This is done by decreasing the total number of gaseous molecules in the system

N2 (g) + 3H2 (g) → 2NH3 (g)

shifts to the right to decrease the total molecules of gas present

The Effect of Change in Pressure

• When the container volume is increased, the system will shift so as to increase its volume

N2 (g) + 3H2 (g) ← 2NH3 (g)

shifts to the left to increase the total number of molecules of gas present

The Effect of Change in Concentration

• If a reactant or product is added to a system at equilibrium, the system will shift away from the added component (it will attempt to "use up" the added component)

• If a reactant or product is removed from a system at equilibrium, the system will shift toward the removed component (it will attempt to "replace" the removed component)

The Effect of a Change in Temperature

• An increase in temperature increases the energy of the system. Le Chatelier's principle predicts that the system will shift in the direction that consumes the energy

• For an exothermic rxn, energy is a product. The rxn will shift to the left to use up the excess energy

• For an endothermic rxn, energy is a reactant. The rxn will shift to the right to use up the energy

• A decrease in temperature will cause a system shift in the direction that "replaces" the lost energy

Le Chatelier’s Principle

• When a stress is applied to a system at equilibrium, the system will shift in a direction that will relieve the stress

• [changes] do not affect the value of K• pressure changes do not affect the value of K• temperature changes will affect the value of K

Practice ([change]) As4O6(s) + 6C(s) As4(g) + 6CO(g)

Which direction will the rxn shift to re-establish equilibrium if:

Add CO? left right no shift

Add or remove C or

As4O6?left right no shift

Remove As4?

left right no shift

Practice (change in pressure)Which direction will the rxn shift to reestablish equilibrium if

the volume is reduced?

P4(s) + 6Cl2(g) 4PCl3(l) left right no shift

PCl3(g) + Cl2(g) PCl5(g) left right no shift

O2(g) + N2(g) 2NO(g) left right no shift

Practice (change in temperature)Which direction will the rxn shift to reestablish equilibrium if

the temperature is increased?

N2(g) + O2(g) 2NO(g) ΔHo = 181kJ left right no shift

2SO2(g) + O2(g) 2SO3(g) ΔHo = -198kJ left right no shift

How will the change affect K?

N2(g) + O2(g) 2NO(g) ΔHo = 181kJ

2SO2(g) + O2(g) 2SO3(g) ΔHo = -198kJ

increase decrease

Completed Reactions• You do not need to consider equilibrium if

you have a reaction that goes to completion.

Examples:

• Formation of a gas in an open container

• Formation of a precipitate

• Formation of a slightly ionized product (polar) – neutralization

The Common-Ion Effect

• When the addition of an ion common to two solutes brings about precipitation or reduced ionization.

Solubility Product and the Common Ion Effect

AgCl(s) Ag+(aq) + Cl-(aq)

Ksp = [Cl-][Ag+] = 1.8 x 10-10

solublility product

What happens if a solution of NaCl is

added to this system?

NaCl(s) Na+(aq) + Cl-

rxn shifts

AgCl(s) Ag+(aq) + Cl-

The solubility of AgCl has

actually decreased

18.3 Equilibria of Acids, Bases and Salts

Ionization Constant of a Weak Acid

• Since weak acids only partially ionize in solution, an equilibrium exists between the weak acid + water, reactants and the hydronium + conjugate base ions, products.

• Ka= acid ionization constant

• Ka, like Keq, is temperature-dependant

• For a weak acid:

HF (g) + H2O (l) ↔ H3O+ (aq) + F- (aq)

Ka = [H3O+] [F-]

Assume [H2O] is constant

Ionization Constant of Water

Flashback to chapter 15…

The self-ionization of water:

H2O (l) + H2O (l) ↔ H3O+ (aq) + OH-

KKWW = [H3O+][OH-] = 1.0 x 101.0 x 10-14-14

Assume [H2O] is constant

Hydrolysis of Salts

• Recall that the conjugates of strong acids & bases are weak, and the conjugates of weak acids & bases are strong:

• The salt produced from a strong acid-strong base would be neutralneutral.

• The salt produced from a strong acid-weak base would be acidicacidic.

• The salt produced from a weak acid-strong base would be basicbasic.

Hydrolysis

• A reaction between water molecules and ions of a dissolved salt.

• If the anions react with water, the process is called anion hydrolysis, and produces a basic pH

• If the cations react with water, the process is called cation hydrolysis, and produces an acidic pH

Anion Hydrolysis• For a weak acid:

HA (g) + H2O (l) ↔ H3O+ (aq) + A- (aq)

Ka = [H3O+] [A-]

[HA]• The A- can further react with water:

A- (aq) + H2O (l) ↔ HA (aq) + OH- (aq)

Anion Hydrolysis

A- (aq) + H2O (l) ↔ HA (aq) + OH- (aq)

• The forward reaction is dependant on the strength of the A-, so if the Ka of HA is low, the strength of A- will be high.

Cation Hydrolysis

• For a weak base:

B (aq) + H2O (l) ↔ BH+ (aq) + OH- (aq)

Kb = [BH+ ] [OH-]

[B]• The BH+ can further react with water:

BH+ (aq) + H2O (l) ↔ H3O + (aq) + B (aq)

Cation Hydrolysis

BH+ (aq) + H2O (l) ↔ H3O + (aq) + B (aq)

• The forward reaction is dependant on the strength of the BH+, so if the Kb of B is low, the strength of BH+ will be high.

Hydrolysis in Acid-Base Reactions

• Since the product of a neutralization is a salt & water, depending on the type of salt formed, the pH of the solution may be acidic, neutral or basic.

Strong Acid-Strong Base

• The dissociated ions from salts produced by these reactions do not undergo hydrolysis, so the pH of these resulting solutions will be neutral.

Weak Acid-Strong Base

• The dissociated anions produced by these reactions will undergo hydrolysis to produce hydroxide ions that will raise the pH, creating a basic solution

Strong Acid-Weak Base

• The dissociated cations produced by these reactions will undergo hydrolysis to produce hydronium ions that will lower the pH, creating an acidic solution

18.4 Solubility Equilibrium

Solubility

• Substances are considered soluble if at least 1 g will dissolve per 100 g water.

• Substances are considered insoluble if less than 0.1 g will dissolve per 100 g water.

• Substances that fall in-between 0.1-1 g are considered partially soluble.

• The solubility product constant indicates the respective molar concentrations of dissolved ions at equilibrium in a saturated solution.

• The smaller the Ksp value, the more insoluble the compound.

• See Appendix Table A-13, page 861

PracticeIf the solubility of magnesium sulfate at 20°C is 33.7g per 100.g of water, Calculate Ksp.

2) Calculate solubility in mol/L

33.7 g x 1 g H2O x 1000 mL x 1 mol MgSO4

100. g 1 mL H2O 1 L 120.4 gMgSO4

= 2.80 mol/L

4) Solve Ksp expression

1) Write the rxn MgSO4 (s) + H2O (l) Mg+2 (aq) + SO4

-2 (aq)

3) [Mg+2] = [SO4-2] [Mg+2] = [SO4

-2] = 2.80 mol/L

Ksp = [Mg+2] [SO4-2] = [2.80] [2.80] = 7.84

PracticeCalculate the solubility of CaCO3, in mol/L at 25°C given Ksp = 2.8 x 10-9

2) [Ca+2] = [CO3-2]

So, [Ca+2] = x and [CO3-2] = x

Then, Ksp = x2

Solubility of calcium carbonate = 5.3 x 10-5

1) Write the rxn & Ksp expression

CaCO3 (s) + H2O (l) Ca+2 (aq) + CO3-2 (aq)

Ksp = [Ca+2] [CO3-2]

3) Solve for xKsp = 2.8 x 10-9 = x2

X = √ 2.8 x 10-9

More Practice!1. What is the value of Ksp for tin (II) sulfide, given

that its solubility is 5.2 x 10-12 g / 100. g water?

2. What is the solubility in mol/L of manganese (II) sulfide given that its Ksp value is 2.5 x 10-13?

3. Calculate the concentration of Zn+2 in a saturated solution of zinc sulfide given that Ksp

of zinc sulfide = 1.6 x 10-24.

1) 1.2 x 10-25

2) 5.0 x 10-7 mol/L

3) 1.3 x 10-12 mol/L

Practice

500 mL of 0.010 M Ba(NO3)2 and 300 mL of 0.020 M NaF solutions are mixed. Will a precipitate of

BaF2 (Ksp = 1.7 x 10-6) form?

Ksp = [Ba2+][F-]2 = 1.7 x 10-6

BaF2(s) Ba2+(aq) + 2F-

[initial]

Qsp = (.00625)(.0075)2 = 3.52 x 10-7

Q < K, so the rxn will

shift in favor of the products =

no precipitate

More Practice!1) Will a precipitate form if 20. mL of 0.034 M

sodium chloride and 15 mL of 0.083 M copper (I) nitrate are mixed?

2) Does a precipitate form if 100. mL of 0.0014 M calcium nitrate and 200. mL of 0.00020 M sodium sulfate are mixed?

Yes, CuCl precipitates

No, CaSO4 does not precipitate

What if you are asked to find an [equilbrium]?

• Use the R.I.C.E method!

Practice4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)

[NH3]o = .0150 M[O2]o = .0150 M[N2]eq = 1.96 x 10-3 M

When 0.0150 mol of NH3 and 0.0150 mol of O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 1.96 x 10-3 M. Calculate Kc for the reaction at this temperature.

.0150 .0150 .0150 .0150 0 0 0 0

- 4x - 3x - 4x - 3x + 2x + 6x + 2x + 6x

.0150 - 4x .0150 - 3x .0150 - 4x .0150 - 3x 1.96 x 10 1.96 x 10-3-3 6x 6x

4NH4NH3(g)3(g) + 3O + 3O2(g)2(g) 2N 2N2g)2g) + 6H + 6H22OO(g)(g)

Practice4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)

[NH3]o = .0150 M[O2]o = .0150 M[N2]eq = 1.96 x 10-3M

When 0.0150 mol of NH3 and 0.0150 mol of O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 1.96 x 10-3 M. Calculate Kc for the reaction at this temperature.

K = [H2O]6[N2]2

[NH3]4 [O2]3

K = (6x)6(1.96 x 10-3 )2

(.0150 – 4x)4 (.0150 – 3x)3

.0150 - 4x .0150 - 3x .0150 - 4x .0150 - 3x 1.96 x 10 1.96 x 10-3-3 6x 6x

2x =1.96 x 10-3 x =9.80 x 10-4

K =5.95 x 10-6

Substitute x for this value

More Practice Mustard gas, used in chemical warfare in WWI, has been found to be an effective agent in the chemotherapy of Hodgkins disease. It is produced by the following reaction:

SCl2(g) + 2C2H4(g) S(CH2CH2Cl)2(g)

An empty five-liter tank is filled with 0.258 mol SCl2 and 0.592 mol C2H4. After equilibrium is established, 0.0349 mol of mustard gas is present.

a)Calculate [equilibrium] for each reactant gas.

b)Determine the value of Keq for the reaction.

b)Keq = 14.3a)[SCl2] = 0.0446 M [C4H4] = 0.104 M

Chapter 18 Chemical Equilibrium 18.1 The Nature of Chemical Equilibrium.

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