Chapter 15

Chemical Equilibrium

15.1

15.1

Cold Temp Hot Temp

15.1

15.1

15.1

15.2: Law of Mass Action

Derived from rate laws by Guldberg andWaage (1864) For a balanced chemical reaction

in equilibrium:

a A + b B ↔ c C + d D Equilibrium constant expression (Keq):

ba

dc

c [B] [A]

[D] [C] K b

Ba

A

dD

cC

p )(P)(P

)(P)(PK

Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).

Units: Keq is considered dimensionless (no units)

Cato Guldberg Peter Waage

(1836-1902) (1833-1900)

or

15.2 **Relates Kc to Kp

For Example

Write the equilibrium expression Kc for the following reactions:

15.2

Example #2

In the synthesis of ammonia from nitrogen and hydrogen,

Kc = 9.60 at 300°C:

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

Calculate Kp for this reaction at this temperature

Hint..you will need to use..

15.2

What does the Equilibrium constant tell us??

15.2

Looking at reversible reactions So far we have only written the expression forwards, but it can also

be written equally backwards! The constants are the recipricals of each other

Kc = 0.212 Kc = 4.72

15.2

15.3: Types of EquilibriaHomogeneous: all components in same phase

(usually g or aq)

N2 (g) + H2 (g) ↔ NH3 (g)

3H

1N

2NH

P )(P)(P

)(PK

22

3

bB

aA

dD

cC

P )(P)(P

)(P)(PK

3 2 1

Fritz Haber(1868 – 1934)

15.3

Heterogeneous: different phases

CaCO3 (s) ↔ CaO (s) + CO2 (g)

Definition: What we use:

][CaCO

)(P [CaO] K

3

COeq

22COp P K

Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.

Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq

15.3

For Example

Write equilibrium-constant expressions for Kc and Kp for each of the following reactions:

15.3

Clickfor answers

15.4: Calculating Equilibrium Constants

Steps to use “ICE” table:1. “I” = Tabulate known initial and equilibrium

concentrations of all species in equilibrium expression

2. “C” = Determine the concentration change for the species where initial and equilibrium are known

• Use stoichiometry to calculate concentration changes for all other species involved in equilibrium

3. “E” = Calculate the equilibrium concentrations

Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction:

NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

 

Initial

Change

Equilibrium

][NH

]][OH[NH K

3

-114

c

0.0124 M

- x

0.0119 M

0 M 0 M

+ x + x

4.64 x 10-4 M 4.64 x 10-4 M

5-2-4

101.81x 0.0119

)10(4.64

NH3 (aq)H2O

(l)NH4

1+ (aq) OH1- (aq)

XXX

x = 4.64 x 10-4 M

Ex: A 5.000-L flask is filled with 5.000 x 10-3 mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium?

H2 (g) + I2 (g) ↔ 2 HI (g)

H2 (g) + I2 (g) ↔ 2 HI (g)

 

Initial

Change

Equilibrium

]][I[H

[HI] K

22

2

c

1.000x10-3 M

- x M

(1.000x10-3 – x) M

2.000x10-3 M 0 M

- x M + 2x M

(2.000x10-3 – x) M 2x M

33.1x)-10x)(2.000-10(1.000

(2x)3-3-

2

4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6]

0 = -2.67x2 – 3.99x10-3x + 2.66x10-6

Using quadratic eq’n: x = 5.00x10-4 or –1.99x10-3; x = 5.00x10-4

Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M

H2 (g) I2 (g) HI (g)

15.6 Le Chatelier’s Principle

Changes that do not affect Keq:

1. Concentration Upon addition of a reactant or product, equilibrium shifts to re-

establish equilibrium by consuming part of the added substance.

Addition of solids/liquids do not appreciably shift the system and can be ignored. But they are still made/consumed!

Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance.

Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6

H2O (l)

•Add HCl, temporarily inc forward rate

Volume, with a gas present (T is constant)

Upon a decrease in V (thereby increasing P),equilibrium shifts to reduce the number of moles of gas.

Upon an increase in V (thereby decreasing P),equilibrium shifts to produce more moles of gas.

Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) If V of container is decreased, equilibrium shifts right.

XN2 and XH2 dec

XNH3 inc

3HN

2NH

P )(PP

)(PK

22

33

THTN

2TNH

)P)(XP(X

)P(X

22

3 4T

3HN

2T

2NH

P)(XX

P)(X

22

3 2T

3HN

2NH

P)(XX

)(X

22

3

23HN

2NH

P )(

)(K

22

3

Since PT also inc, KP remains constant.

3. Pressure, but not Volume

Usually addition of a noble gas, p. 560 Avogadro’s law: adding more non-reacting particles “fills in”

the empty space between particles. In the mixture of red and blue gas particles, below, adding

green particles does not stress the system, so there is no Le Châtelier shift.

Catalysts Lower the activation energy of both forward and

reverse rxns, therefore increases both forward and reverse rxn rates.

Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the Keq)

Energy

Rxn coordinate

Ea, uncatalyzed

Ea, catalyzed

Change that does affect Keq:

Temperature: consider “heat” as a part of the reaction Upon an increase in T, endothermic reaction is favored

(equilibrium shifts to “consume the extra heat”) Upon a decrease in T, equilibrium shifts to produce more

heat.Effect on Keq

1. Exothermic equilibria: Reactants ↔ Products + heat

• Inc T increases reverse reaction rate which decreases Keq

2. Endothermic equilibria: Reactants + heat ↔ Products

• Inc T increases forward reaction rate increases Keq

Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l);

H=+?•Inc T temporarily inc forward rate•Dec T temporarily inc reverse rate

Effect of Various Changes on Equilibrium

Disturbance Net Direction of Rxn

Effect of Value of K

Concentration

Increase (reactant) Towards formation of product

None

Decrease(reactant) Towards formation of reactant

None

Increase (product) Towards formation of reactant

None

Decrease (product) Towards formation of product

None

Effect of Pressure on Equilb.

Pressure

Increase P

(decrease V)

Towards formation of fewer moles of gas

None

Decrease P

(Increase V)

Towards formation of more moles of gas

None

Increase P

( Add inert gas, no change in V)

None, concentrations unchanged

None

Disturbance Direction of Reaction Effect of K

Effect of Temperature on Equilb

Temperature

Increase T

Towards absorption of heat

Increases if endothermic

Decreases if exothermic

Decrease T Towards release of heat

Increases if exothermic

Decreases if endothermic

Catalyst Added None, forward and reverse equilibrium attained sooner

None

Disturbance Direction of Reaction Effect of K

For ExampleFor the following reaction

5 CO(g) + I2O5(s) I2(g) + 5 CO2(g) + 1175 kJ

for each change listed, predict the equilibrium shift and the effect on the indicated quantity.

Change Directionof Shift

( ; ; or no change)

Effect onQuantity

Effect(increase, decrease,

or no change)

(a) decrease in volume Kc No Change

(b) raise temperature amount of CO(g) Increase

(c) addition of I2O5(s) No Change amount of CO(g) No Change(d) addition of CO2(g) amount of I2O5(s) Increase

(e) removal of I2(g) amount of CO2(g) Increase