Slide 1 of 33 Chapter 15: Principles of Chemical Equilibrium

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Chapter 15: Principles of Chemical Equilibrium

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Contents

15-1 Dynamic Equilibrium

15-2 The Equilibrium Constant Expression

15-3 Relationships Involving Equilibrium Constants

15-4 The Magnitude of an Equilibrium Constant

15-5 The Reaction Quotient, Q: Predicting the Direction of a Net Change

15-6 Altering Equilibrium Conditions:

Le Châtelier’s Principle

15-7 Equilibrium Calculations: Some Illustrative Examples

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15-1 Dynamic Equilibrium

Equilibrium – two opposing processes taking place at equal rates.

H2O(l) H2O(g)

I2(H2O) I2(CCl4)

NaCl(s) NaCl(aq)H2O

CO(g) + 2 H2(g) CH3OH(g)

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Dynamic Equilibrium

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15-2 The Equilibrium Constant Expression

Methanol synthesis is a reversible reaction.

CO(g) + 2 H2(g) CH3OH(g)k1

k-1

CH3OH(g) CO(g) + 2 H2(g)

CO(g) + 2 H2(g) CH3OH(g)k1

k-1

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Three Approaches to the Equilibrium

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Three Approaches to Equilibrium

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Three Approaches to Equilibrium

CO(g) + 2 H2(g) CH3OH(g)k1

k-1

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The Equilibrium Constant Expression

Forward: CO(g) + 2 H2(g) → CH3OH(g)

Reverse: CH3OH(g) → CO(g) + 2 H2(g)

At Equilibrium:

Rfwrd = k1[CO][H2]2

Rrvrs = k-1[CH3OH]

Rfwrd = Rrvrs

k1[CO][H2]2 = k-1[CH3OH]

[CH3OH]

[CO][H2]2=

k1

k-1

= Kc

CO(g) + 2 H2(g) CH3OH(g) k1

k-1

k1

k-1

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General Expressions

a A + b B …. → g G + h H ….

Equilibrium constant = Kc= [A]m[B]n ….

[G]g[H]h ….

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15-3 Relationships Involving the Equilibrium Constant

Reversing an equation causes inversion of K. Multiplying by coefficients by a common factor

raises the equilibrium constant to the corresponding power.

Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root.

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Combining Equilibrium Constant Expressions

N2O(g) + ½O2 2 NO(g) Kc= ?

Kc= [N2O][O2]½

[NO]2

=[N2][O2]½

[N2O][N2][O2]

[NO]2

Kc(2)

1Kc(3)= = 1.710-13

[N2][O2]

[NO]2

=

[N2][O2]½

[N2O]=N2(g) + ½O2 N2O(g) Kc(2)= 2.710-18

N2(g) + O2 2 NO(g) Kc(3)= 4.710-31

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KP = Kc(RT)Δn

:

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Pure Liquids and Solids

Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids).

Kc = [H2O]2

[CO][H2]

C(s) + H2O(g) CO(g) + H2(g)

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Worked Examples Follow:

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15-1 Practice Example B

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CRS Questions Follow:

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H2

CO

CH4 = H2O

time

mol

es o

f su

bsta

nce

0

1

2

3

f

r

k

2 4 2kCO(g) + 3H (g) CH (g) + H O(g) Which of the following statements is

correct?

1. At equilibrium the reaction stops.

2. At equilibrium the rate constants for the forward and reverse reactions are equal.

3. At equilibrium the rates of the forward and reverse reactions are equal.

4. At equilibrium the rates of the forward and reverse reactions are zero.

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H2

CO

CH4 = H2O

time

mol

es o

f su

bsta

nce

0

1

2

3

f

r

k

2 4 2kCO(g) + 3H (g) CH (g) + H O(g) Which of the following statements is

correct?

1. At equilibrium the reaction stops.

2. At equilibrium the rate constants for the forward and reverse reactions are equal.

3. At equilibrium the rates of the forward and reverse reactions are equal.

4. At equilibrium the rates of the forward and reverse reactions are zero.