5.111 Lecture Summary #18 Wednesday, October 22, 2014
Reading for Today: Sections 10.1-10.5, 10.9 (Sections 9.1-9.4 in 4th ed.)
Reading for Lecture #19: Sections 10.9-10.13 (Section 9.4-9.5 in 4th ed.)
Topics: Chemical Equilibrium
I. Nature of Chemical Equilibrium
II. Meaning of K
III. External Effects on K
Chemical reactions reach a state of dynamic equilibrium in which the rates of forward and
reverse reactions are equal and there is no net change in composition.
Consider: N2(g) + 3H2(g) 2NH3(g) ∆G° = -32.90 kJ/mol
Free energy, G
Concentration
Pure Pure
reactants products
time
Progress of reaction
When the reaction mixture has not produced enough products to have reached
equilibrium, the spontaneous direction of change is toward more products
(∆Gforward reaction 0).
When excess products are present (ex. pure ammonia), the reverse reaction is spontaneous
(∆Gforward reaction 0).
The reaction free energy (∆G) changes as the proportion of reactants and products
.
∆G = ∆G° + RT ln Q Where
∆G = reaction free energy at any definite, fixed composition of the reaction mixture.
∆G° = is the difference in free energy of the products and reactants in their standard states.
R = universal gas constant, T = Temperature, and Q = reaction quotient
1
, For aA + bB cC + dD
In gaseous phase in solution
(PC/Pref)c (PD/Pref)d ([C]/Cref)c ([D]/Cref)d
∆G = ∆G° + RT ln ∆G = ∆G° + RT ln
(PA/Pref)a (PB/Pref)b ([A]/Cref)a ([B]/Cref)b
Q Q
Pref = 1 bar Cref = 1 M and [C] in M
Q = PCc PDd Q = [C]c [D]d
PAa PBb [A]a [B]b
At equilibrium ∆G = 0 and Q = K (the equilibrium constant),
0 = ∆G° + RT ln K
∆G° = -RT ln K
K = is the equilibrium constant. It has the same form as , but only uses the
amounts of products and reactants at equilibrium.
We can rewrite ∆G = ∆G° + RT ln Q as
∆G = -RT ln K + RT ln Q or
∆G = RT ln (Q/K)
Relationship between K and Q:
If Q < K, ∆G is and the forward reaction will occur
If Q > K, ∆G is and the reverse reaction will occur
Example: N2(g) + 3H2(g) 2NH3(g)
If K=1.9 x 10-4 at 400°C, and PN2 = 5.5 bar PH2=2.2 bar PNH3 = 1.1 bar at 400°C, which
direction will the reaction go?
Q=
2
Reading for Today: Sections 10.1-10.5, 10.9 (Sections 9.1-9.4 in 4th ed.)
Reading for Lecture #19: Sections 10.9-10.13 (Section 9.4-9.5 in 4th ed.)
Topics: Chemical Equilibrium
I. Nature of Chemical Equilibrium
II. Meaning of K
III. External Effects on K
Chemical reactions reach a state of dynamic equilibrium in which the rates of forward and
reverse reactions are equal and there is no net change in composition.
Consider: N2(g) + 3H2(g) 2NH3(g) ∆G° = -32.90 kJ/mol
Free energy, G
Concentration
Pure Pure
reactants products
time
Progress of reaction
When the reaction mixture has not produced enough products to have reached
equilibrium, the spontaneous direction of change is toward more products
(∆Gforward reaction 0).
When excess products are present (ex. pure ammonia), the reverse reaction is spontaneous
(∆Gforward reaction 0).
The reaction free energy (∆G) changes as the proportion of reactants and products
.
∆G = ∆G° + RT ln Q Where
∆G = reaction free energy at any definite, fixed composition of the reaction mixture.
∆G° = is the difference in free energy of the products and reactants in their standard states.
R = universal gas constant, T = Temperature, and Q = reaction quotient
1
, For aA + bB cC + dD
In gaseous phase in solution
(PC/Pref)c (PD/Pref)d ([C]/Cref)c ([D]/Cref)d
∆G = ∆G° + RT ln ∆G = ∆G° + RT ln
(PA/Pref)a (PB/Pref)b ([A]/Cref)a ([B]/Cref)b
Q Q
Pref = 1 bar Cref = 1 M and [C] in M
Q = PCc PDd Q = [C]c [D]d
PAa PBb [A]a [B]b
At equilibrium ∆G = 0 and Q = K (the equilibrium constant),
0 = ∆G° + RT ln K
∆G° = -RT ln K
K = is the equilibrium constant. It has the same form as , but only uses the
amounts of products and reactants at equilibrium.
We can rewrite ∆G = ∆G° + RT ln Q as
∆G = -RT ln K + RT ln Q or
∆G = RT ln (Q/K)
Relationship between K and Q:
If Q < K, ∆G is and the forward reaction will occur
If Q > K, ∆G is and the reverse reaction will occur
Example: N2(g) + 3H2(g) 2NH3(g)
If K=1.9 x 10-4 at 400°C, and PN2 = 5.5 bar PH2=2.2 bar PNH3 = 1.1 bar at 400°C, which
direction will the reaction go?
Q=
2
