Atkins’ Physical Chemistry Eighth Edition Chapter 7 – Lecture 1 Chemical Equilibrium Copyright © 2006 by Peter Atkins and Julio de Paula Peter Atkins Julio.

Atkins’ Physical ChemistryEighth Edition

Chapter 7 – Lecture 1Chemical Equilibrium

Copyright © 2006 by Peter Atkins and Julio de Paula

Peter Atkins • Julio de Paula

Sections 7.1 - 7.4 only

Homework Set #7Homework Set #7

Atkins & de Paula, 8eAtkins & de Paula, 8e

Chap 7 Chap 7

ExercisesExercises: all part (b) unless noted:: all part (b) unless noted:

2, 3, 4, 5, 7, 9, 10, 122, 3, 4, 5, 7, 9, 10, 12

Objectives:

• Further develop the concept of chemical potential, μ

• Apply μ to account for equilibrium composition of a chemical reaction

• Establish relationship between Gibbs energy and the equilibrium constant, K

• Establish the quantitative effects of pressure and temperature on K

Equilibrium - state in which there are no observable changes with time

Chemical equilibrium - achieved when:

• rates of the forward and reverse reactions are equal and

• concentrations of the reactants and products remain constant

• dynamic equilibriumdynamic equilibrium Physical equilibrium

H2O (l)

Chemical equilibrium

N2O4 (g)

H2O (g)

2NO2 (g)

colorless red-brown

dG = VdP - SdT

dTT

GdP

P

GdG

PT

VP

G

T

ST

G

P

“spontaneous” ⇒ G → min

Fig 3.18 Gibbs energy tends to minimum at ΔT = 0, ΔP =0

Fig 7.1 Plot of Gibbs energy vs. extent of reaction, ξ

ξ pronounced ziReaction Gibbs energy:

P,T

r

GdG

AB

P,T

G

For equilibrium A B⇌

Therefore

ΔGr = μB - μA

Exergonic and endergonic reactions

At constant temperature and pressure:

• ΔGrxn < 0 forward rxn spontaneous (Exergonic)

• ΔGrxn > 0 reverse rxn spontaneous (Endergonic)

• ΔGrxn = 0 rxn at equilibrium (Neither)

Fig 7.2 A strongly exergonic process can drive

a weaker endergonic process

Nonspontaneous,

but!!.....

Q ln RTG

P

Pln RTG

)P ln RT(μ)P ln RT(μ

μ-μΔG

orxn

A

Borxn

AoAB

oB

ABrxn

Description of Equilibrium

For perfect gases: A ⇌ B

At equilibrium: ΔGrxn = 0

K ln RTG0 orxn

K ln RTGΔ orxn

N2O4 (g) 2NO2 (g)

= 4.63 x 10-3 at 25 °CK = [NO2]2

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]bLaw of Mass Action

K > 1

K < 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium will:

Must be caps

Equilibrium constant

Fig 7.3 Plot of Gibbs energy vs. extent of reaction

Hypothetical rxn A(g) → B(g)

Molecular interpretation

of tendency for ΔGr = 0

(i) slope = ΔGr at all times

(ii) from Eqn. 5.18:

)x ln xx ln nRT(xΔG BBAAmix

(iii) curve has minimum

corresponding to

equilibrium composition

Ways of Expressing Equilibrium Constants

Concentration of products and reactants may be expressed in different units, so:

• Heterogeneous equilibria

• Homogeneous equilibria

K = [C]c[D]d

[A]a[B]b Law of Mass Action

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases

CaCO3 (s) ⇌ CaO (s) + CO2 (g)

Kc =[CaO][CO2]

[CaCO3][CaCO3] = constant[CaO] = constant

Kc = [CO2] or Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant

PCO 2= Kp

PCO 2 does not depend on the amount of CaCO3 or CaO

CaCO3 (s) ⇌ CaO (s) + CO2 (g)

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase

N2O4 (g) ⇌ 2NO2 (g)

Kc = [NO2]2

[N2O4]Kp =

NO2P2

N2O4P

In most cases

Kc Kp

aA (g) + bB (g) ⇌ cC (g) + dD (g)

Kp = Kc(RT)n

n = moles of gaseous products – moles of gaseous reactants

= (c + d) – (a + b)

Fig 7.4 Boltzmann distribution of populations of A and B

For endothermic reaction A → B

Bulk of population is species A

Therefore:

A is dominant species at equilibrium

Assumption:

Similar densities of E-levels

Fig 7.5 Plot of energy levels vs. population

For endothermic reaction A → B

Assumption:

Density of E-levels of B >>

density of E-levels of A

Bulk of population is species B

Therefore:

B is dominant species at equilibrium