Atkins’ Physical ChemistryEighth Edition

Chapter 7 – Lecture 2Chemical Equilibrium

Copyright © 2006 by Peter Atkins and Julio de Paula

Peter Atkins • Julio de Paula

How equilibria respond to pressure

K ln RTΔGor

• Defined at a single standard pressure, 1 bar

• Therefore K is independent of pressure:

• However, equilibrium composition is

NOT necessarily pressure-independent!

e.g., 3 H2 (g) + N2 (g) 2 NH⇌ 3 (g)

• Recall two ways to pressurize a gas:

0PK

T

Fig 4.11

Two methods of

applying pressure

to a condensed phase

Fig 7.6 Compression of a reaction at equilibrium

Consider: A ⇌ 2B

AA

BB

A

2B

PP

K

Le Chatelier’s principle:

If an external stress is applied to a system at equilibrium, the system will adjust itself in order to minimize the stress

A ← 2B and K remains constant

∴ as system is compressed,

Fig 7.7 Pressure dependence of degree of dissociation

Consider: A ⇌ 2B

Pure A

Pure B

values of K

Example:

3 N2 (g) + H2 (g) ⇌ 2 HN3 (g)

Predict the effect on K of a ten-fold increase in pressure

so to preserve the initial value of K,

LeChaletier’s principle says that K must increase by 100-fold

01.0atm10atm10

atm10

PP

PK

3

2

H3

N

2HN

22

3

How equilibria respond to temperature

• System will shift in endothermic direction as T

• System will shift in exothermic direction as T

Summary:

• Exothermic reactions: Increased T favors reactants

• Endothermic reactions: Increased T favors products

From Le Chatelier’s principle for a system at equilibrium:

How equilibria respond to temperature

The van’t Hoff equation:

2

or

RT

HdT

K ln d

2

or

P T

HdTG/T)( d

The Gibbs-Helmholtz equation (3.53):

K ln RT ΔGor

Since Eqn 7.17 is:

K ln R T

ΔGor

RH

d(1/T)K ln d o

r

2T1

dT(1/T) d

Fig 7.8 Effect of temperature on a chemical equilibrium A ⇌ B

in terms of the Boltzmann distribution

Endothermic Exothermic

B increases at expense of A

A increases at expense of B

Fig 7.9 Plot of ‒ ln K versus 1/T

Ag2CO3 (s) Ag⇌ 2O (s) + CO2 (g)

RH

d(1/T)K ln d o

r

Variation of K with temperature for:

How equilibria respond to temperature

The value of K at different temperatures

To find K2 at T2, given K1 at T1, integrate:

RH

d(1/T)K ln d o

r )T/1(d

RH

K ln dor

2

1

or

2

1

)T/1(dRH

K ln d

Assuming ΔHro is T-independent

12

or

12 T1

T1

RΔH

K lnK ln