CHAPTER 14Chemical Equilibrium

14.1: Equilibrium Constant, Keq Objective: (1)To write the equilibrium constant

expression for a chemical reaction.

Reversible Reactions and Equilibrium

Reversible Reaction: A chemical reactions in which products re-form the original reactants.

Arrows that point in opposite directions are used to indicate a reaction is reversible. Example: H2(g) + I2(g) 2HI(g)

Chemical Equilibrium: A state of balance in which the rate of a forward reaction equals the rate of the reverse reactions and the concentrations of products and reactants remain unchanged.

Equilibrium Constant, Keq Equilibrium Constant, Keq: a number

that relates that concentrations of starting materials and products of a reversible chemical reaction to one another at a given temperature.

aA + bB cC + dD

ba

dc

eq BADCK][][][][

concentration

coefficient

Writing an Equilibrium Constant Expression• Step 1: Balance the chemical equation.• Step 2: Set up your Keq expression with the

products on the top of a fraction and the reactants on the bottom of a fraction.

• Step 3: Raise each substance's concentration to the power equal to the substance’s coefficient in the balanced equation.

• Note: Solids (s) and pure liquids (l) are not used in the expression because their concentrations do not change.

Example Write the equilibrium constant expression for

the following reaction: CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) +

2HCO3-(aq)

][]][[

2

23

2

COHCOCaKeq

Practice Write the equilibrium constant expression for the

following chemical reactions at equilibrium (don’t forget to balance the equation):

1.) H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+

(aq)

2.) COCl2 (g) CO(g) + Cl2 (g)

3.) CO(g) C(s) + CO2 (g)

Answers1.) H2CO3(aq) + H2O(l) HCO3

-(aq) + H3O+ (aq)

2.) COCl2 (g) CO(g) + Cl2 (g)

3.) 2CO(g) C(s) + CO2 (g) ][]][[

32

33

COHHCOOHKeq

][]][[

2

2

COClClCOKeq

22

][][

COCOKeq

14.1: Equilibrium Constant, Keq Objective: (1)To calculate the equilibrium constant.

What does the Keq tell us? Keq < 1 Favors Reactants Keq = 1 Same amount of Reactants and

Products Keq > 1 Favors Products

Practice: Determine if the following Keq values favor the reactants, products, or neither.

1.) Keq = 0.02 2.) Keq = 1 3.) Keq = 50

Calculating Keq

Step 1: Write the balanced chemical equation.

Step 2: Set up your Keq expression. Step 3: Substitute concentrations. Step 4: Calculate!

Example An aqueous solution of carbonic acid reacts

to reach equilibrium as described below:H2CO3(aq) + H2O(l) HCO3

-(aq) + H3O+

(aq)The solution contains the following solute

concentrations: H2CO3 = 3.3 x 10-2 M; HCO3-

= 1.19 x 10-4 M; H3O+ = 1.19 x 10-4 M. Determine the Keq.

72

44

32

33 1029.4)103.3(

)1019.1)(1019.1(][]][[

xx

xxCOHHCOOHKeq

Note: Keq does not have units!

Practice1.a. Calculate the equilibrium constant for

the following reaction:COCl2(g) CO(g) + Cl2(g)[CO] = 0.0178 M[Cl2] = 0.0178 M[COCl2] = 0.00740 M

b. Are the reactants for products favored?

Practice2.a. For the system involving dinitrogen

tetraoxide and nitrogen dioxide at equilibrium at a temperature of 100⁰C, the product concentration of N2O4 is 4.0 x 10-2 M and the reactant concentration of NO2 is 1.4x 10-1 M. What is the Keq value for this reaction?

NO2(g) N2O4(g)

b. Are the reactants or products favored?

Practice3.a. An equilibrium mixture at 852 K is

found to contain 3.61 x 10-3 M of SO2, 6.11 x 10-4 M of O2, and 1.01 x 10-2 M of SO3. Calculate the equilibrium constant for the reaction.

SO2 (g) + O2 (g) SO3 (g)

b. Are the reactants or products favored?

Calculating Concentrations from Keq4. Keq for the equilibrium below is 1.8 x 10-5

at a temperature of 25⁰C. Calculate [NH3] when [NH4

+] and [OH-] are 3.5 x 10-4 M.

NH3(aq) + H2O(l) NH4+ (aq) + OH-

(aq)][

]][[

3

4

NHOHNH

Keq

][)105.3)(105.3(108.1

3

445

NHxxx

[NH3] = 6.8 x10-3 M

Practice5. a. If the equilibrium constant is 1.65 x

10-3 at 2027⁰C for the reaction below, what is the equilibrium concentration of NO when [N2] = 1.8 x 10-3 M and [O2] = 4.2 x 10-3 M.

N2(g) + O2(g) NO(g)

b. Are the reactants for products favored?

Practice6.a. At 600⁰C, the Keq for the reaction

below is 4.32 when [SO3] = 0.260 M and [O2] = 0.045 M. Calculate the equilibrium concentration for sulfur dioxide.

SO2(g) + O2(g) SO3(g)

b. Are the reactants or products favored?

14.2 Solubility Product Constant, Ksp Objective:(1) To calculate the solubility product

constant, Ksp.

Solubility The maximum concentration of a salt in

an aqueous solution is called the solubility of the salt in water.

Solubility Product Constant, Ksp

Solubility Product Constant, Ksp: the equilibrium constant for a solid that is in equilibrium with the solid’s dissolved ions.

How much of a partially soluble salt will dissolve?

AaBb (s) aA (aq) + bB (aq)ba

sp BAK ][][

Calculating Ksp

The lower the value of Ksp, the less soluble the substance.

Practice:Rank the following substances from least

soluble to most soluble:Salt Ksp

Ag2CO3 8.4 x 10-12

BaSO4 1.1 x 10-10

Ca3(PO4)2

2.1 x 10-33

CuS 1.3 x 10-36

Calculating Ksp

The lower the value of Ksp, the less soluble the substance.

Practice:Rank the following substances from least

soluble to most soluble:Salt Ksp

Ag2CO3 8.4 x 10-12

BaSO4 1.1 x 10-10

Ca3(PO4)2

2.1 x 10-33

CuS 1.3 x 10-36

CuS Least solubleCa3(PO4)2

Ag2CO3

BaSO4 Most soluble

Calculating Ksp

Step 1: Write and Balance the equation. Step 2: Determine the concentration of

the ions. Step 3: Write the solubility product

expression. Step 4: Substitute values and calculate.

Example Most parts of oceans are nearly

saturated with calcium fluoride. A saturated solution of CaF2 at 25⁰C has a solubility of 3.4 x 10-4 M. Calculate the solubility product constant for CaF2.

CaF2(s) Ca2+(aq) + F-(aq)

Solution

1. Balance equation: CaF2(s) Ca2+(aq) + 2F-

(aq)2. Determine Concentrations:

CaF2(s) Ca2+(aq) + 2F-(aq) 3.4 x 10-4 3.4 x 10-4 6.8 x 10-4

3. Write solubility product expression:4. Substitute values and calculate:

22 ]][[ FCaK sp

1024422 106.1)108.6)(104.3(]][[ xxxFCaK sp

Note: Ksp does not have units!

Practice1. Copper(I) bromide is dissolved in water

to saturation at 25⁰C. The concentration of Cu+ and Br- ions in solution is 7.9 x 10-5 M. Calculate the Ksp for copper(I) bromide at this temperature.

Practice2. What is the Ksp value for calcium

phosphate at 298 K if the concentrations in a solution at equilibrium with excess solid are 3.42 x 10-7 M for Ca2+ and 2.28 x 10-7 M for PO4

3- ions?

Practice3. If a saturated solution of silver chloride

contains an AgCl concentration of 1.34 x 10-5 M, what is the solubility product constant?

Practice4. A saturated solution of magnesium

fluoride contains a MgCl2 concentration of 1.19x10-3 M. What is the Ksp for magnesium fluoride?

Calculating Concentration from Ksp5. What is the concentration of Ca2+ in a

saturated solution of CaF2 if the concentration of F- is 2.20 x 10-3M and Ksp = 5.30 x 10-9.

Practice6. What is the concentration of Al3+ in a

saturated solution of Al(OH)3 if the OH- concentration is 7.90 x 10-9 M. Ksp = 1.30 x 10-33.

Practice: Chem 3317. The Ksp for lead(II) iodide is 7.08 x 10-9 at

25⁰C. What is the molar concentration of PbI2 in a saturated solution?

The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution?

Step 1: Write and Balance EquationPbI2 (s) Pb2+ (aq) + 2I- (aq)

The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution?

Step 1: Write and Balance EquationPbI2 (s) Pb2+ (aq) + 2I- (aq)

Step 2: Write the Ksp expression 22 ]][[ IPbK sp

The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution?

Step 1: Write and Balance EquationPbI2 (s) Pb2+ (aq) + 2I- (aq)

Step 2: Write the Ksp expression

Step 3: Assign x values to concentrationsPbI2 (s) Pb2+ (aq) + 2I- (aq)

x x 2x

22 ]][[ IPbK sp

The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution?

Step 1: Write and Balance EquationPbI2 (s) Pb2+ (aq) + 2I- (aq)

Step 2: Write the Ksp expression

Step 3: Assign x values to concentrationsPbI2 (s) Pb2+ (aq) + 2I- (aq)

x x 2xStep 4: Solve

22 ]][[ IPbK sp

932222 1008.74)4)((]2][[]][[ xxxxxxIPbK sp

x = [PbI2] = 1.21 x 10-3 M

Practice8. The Ksp of calcium sulfate is 9.1 x 10-6.

What is the molar concentration of calcium sulfate in a saturated solution?

Practice9. The Ksp of CdF2 is 6.4 x 10-3. What is the

molar concentration of cadmium fluoride in a saturated solution?

14.3 LeChatelier’s Principle Objective:(1) To use LeChatelier’s Principle to

determine how a system at equilibrium will respond to an external stress.

LeChatelier’s Principle LeChatelier’s Principle: When a

system at equilibrium is disturbed, the system adjusts in a way to reduce the change.

There are 3 possible disturbances:Change in (1) concentration, (2)

temperature, or (3) pressure

1. Change in Concentration Increase concentration of reactant

Equilibrium shifts toward products Decrease concentration of reactant

Equilibrium shifts toward reactants

Increase concentration of product Equilibrium shifts toward reactants

Decrease concentration of product Equilibrium shifts toward products

Use the following reaction to answer the questions below:

H2 (g) + I2 (g) 2HI (g)In which direction (left or right) does the equilibrium

shift in each of the following situations:1.) Increase H2

2.) Decrease I23.) Increase HI4.) Decrease HI

Use the following reaction to answer the questions below:

H2 (g) + I2 (g) 2HI (g)In which direction (left or right) does the equilibrium

shift in each of the following situations:1.) Increase H2 RIGHT2.) Decrease I2 LEFT3.) Increase HI LEFT4.) Decrease HI RIGHT

2. Change in Temperature Think of heat as a reactant or product Exothermic: heat is a product Endothermic: heat is a reactant

For an exothermic reaction: Increasing temperature equilibrium favors reactants Decreasing temperature equilibrium favors products

For an endothermic reaction Increasing temperature equilibrium favors products Decreasing temperature equilibrium favors reactants

Use the following reaction to answer the questions below:

2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift in each of the following situations:

1.) Increase the temperature2.) Decrease the temperature

Use the following reaction to answer the questions below:

2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift in each of the following situations:

1.) Increase the temperature RIGHT2.) Decrease the temperature LEFT

3. Change in Pressure Only affects gases!

Increasing pressure Equilibrium shifts toward the side with fewer moles of gas

Decreasing pressure Equilibrium shifts toward the side with more moles of gas

Use the following reaction to answer the questions below:

2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift in each of the following situations:

1.) Increase the pressure2.) Decrease the pressure

Use the following reaction to answer the questions below:

2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift in each of the following situations:

1.) Increase the pressure LEFT (3 moles gas)2.) Decrease the pressure RIGHT (5 moles gas)

Use the following reaction to answer the questions below:

H2 (g) + I2 (g) 2HI (g)In which direction (left or right) does the equilibrium

shift in each of the following situations:

1.) Increase Pressure 2.) Decrease Pressure

Use the following reaction to answer the questions below:

H2 (g) + I2 (g) 2HI (g)In which direction (left or right) does the equilibrium

shift in each of the following situations:

1.) Increase Pressure NO CHANGE2.) Decrease Pressure NO CHANGE

Practice What direction will the equilibrium shift

(left or right) in the reaction:___POCl3(g) ___PCl3(g) + ___O2 (g) +

heat

1.) Add PCl32.) Increase Pressure3.) Increase Temperature

Practice What direction will the equilibrium shift

(left or right) in the reaction:_2_POCl3(g) _2_PCl3(g) + _1_O2 (g) +

heat

1.) Add PCl3 LEFT2.) Increase Pressure LEFT3.) Increase Temperature LEFT