Chapter 14

54
CHAPTER 14 Chemical Equilibrium

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Chapter 14. Chemical Equilibrium. 14.1: Equilibrium Constant, K eq. Objective: 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. - PowerPoint PPT Presentation

Transcript of Chapter 14

Page 1: Chapter 14

CHAPTER 14

Chemical Equilibrium

Page 2: Chapter 14

14.1: Equilibrium Constant, Keq

Objective: (1)To write the equilibrium constant

expression for a chemical reaction.

Page 3: Chapter 14

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.

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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 BA

DCK

][][

][][

concentration

coefficient

Page 5: Chapter 14

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.

Page 6: Chapter 14

Example

Write the equilibrium constant expression for the following reaction:

CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3

-(aq)

][

]][[

2

23

2

CO

HCOCaKeq

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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)

Page 8: Chapter 14

Answers

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

(aq)

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

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

][

]][[

32

33

COH

HCOOHKeq

][

]][[

2

2

COCl

ClCOKeq

22

][

][

CO

COKeq

Page 9: Chapter 14

14.1: Equilibrium Constant, Keq

Objective: (1)To calculate the equilibrium constant.

Page 10: Chapter 14

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 = 13.) Keq = 50

Page 11: Chapter 14

Calculating Keq

Step 1: Write the balanced chemical equation.

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

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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. 7

2

44

32

33 1029.4)103.3(

)1019.1)(1019.1(

][

]][[

xx

xx

COH

HCOOHKeq

Note: Keq does not have units!

Page 13: Chapter 14

Practice

1.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?

Page 14: Chapter 14

Practice

2.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?

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Practice

3.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?

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Calculating Concentrations from Keq

4. 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

NH

OHNHKeq

][

)105.3)(105.3(108.1

3

445

NH

xxx

[NH3] = 6.8 x10-3 M

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Practice

5. 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?

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Practice

6.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?

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

constant, Ksp.

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Solubility

The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water.

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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)

basp BAK ][][

Page 22: Chapter 14

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

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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

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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.

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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)

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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!

Page 27: Chapter 14

Practice

1. 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.

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Practice

2. 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?

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Practice

3. If a saturated solution of silver chloride contains an AgCl concentration of 1.34 x 10-5 M, what is the solubility product constant?

Page 30: Chapter 14

Practice

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

Page 31: Chapter 14

Calculating Concentration from Ksp

5. 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.

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Practice

6. 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.

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Practice: Chem 331

7. 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?

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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)

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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 expression22 ]][[ IPbK sp

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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

Page 37: Chapter 14

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

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Practice

8. The Ksp of calcium sulfate is 9.1 x 10-6. What is the molar concentration of calcium sulfate in a saturated solution?

Page 39: Chapter 14

Practice

9. The Ksp of CdF2 is 6.4 x 10-3. What is the molar concentration of cadmium fluoride in a saturated solution?

Page 40: Chapter 14

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.

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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

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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

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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

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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 RIGHT

2.) Decrease I2 LEFT

3.) Increase HI LEFT4.) Decrease HI RIGHT

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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

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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

Page 47: Chapter 14

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

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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

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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

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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)

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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

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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

Page 53: Chapter 14

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

Page 54: Chapter 14

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 LEFT

2.) Increase Pressure LEFT3.) Increase Temperature LEFT