atkins’ physical chemistry eighth edition chapter 7 – lecture 1 chemical equilibrium copyright...
Post on 20-Dec-2015
367 views
TRANSCRIPT
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