Physical Chemistry
Lecture 28
Solving Equilibrium Problems
Equilibrium in chemical reactions involving
gases, pure liquids and pure solids
For a chemical reaction occurring at equilibrium
∆Gθ (T )
ln K a (T ) = −
RT
Must find ΔGθ(T) to evaluate Ka(T), and vice versa
K a (T ) = ∏ aνk k
k
Using Ka(T) requires an understanding of ideality and
activity (or fugacity)

Gases:
ai = γi(Pi/Pθ) = fi/Pθ
 γi determined from Newton graphs (use total pressure)

Solids and liquids:
 ai ≅ 1 (unless overpressure is far from Pθ)
 ai = exp(Vmθ(P - Pθ)/RT), assuming incompressibility
Activity of Condensed Phases
Calculated using assumption of
incompressibility
Generally a reasonable assumption
P
Vm
dP ≅
ln a = ∫
RT
Pθ
(
Vm
P − Pθ
RT
)
Example: Zn at 298.15K
Example: Benzene at 298.15K
Vm = 7.1 cm3
Vm = 89 cm3
P (bar)
a
P (bar)
a
1
1.000
1
1.000
10
1.003
10
1.033
50
1.014
50
1.192
100
1.029
100
1.427
Gas equilibrium assuming
ideality
Example: Boudouard reaction
C ( s, gr ) + CO2 ( g ) → 2 CO ( g )
Set up KP
KP
=
2
PCO
PCO2
=
2
X CO
Ptotal
(1 − X CO )
Solve for XCO and use Dalton’s law to find the partial
pressures


KP(1123K) = 14.11 atm
Ptotal = 10 atm; solution gives XCO = 0.676
Gas equilibrium with corrections
for nonideality
Example: Boudouard reaction
KP
=
2
PCO
PCO2
=
2
X CO
Ptotal
(1 − X CO )
=
K a Pθ
Kγ
Calculation of Kγ at 200 atm and 1123 K



γ(CO) = 1.03
γ(CO2) = 1.07
Kγ = 0.99 (Almost ideal at this condition)
Solve for KP under these conditions
KP
Pθ
=
2
PCO
PCO2 Pθ
=
2
X CO
Ptotal
(1 − X CO ) Pθ
=
Ka
Kγ
=
14.11
= 14.25
0.99
Then solve for XCO as previously

XCO = 0.679, not a large change from the ideal situation
Many reactions have substantial changes of K P due to nonideality
Variations on a theme: different
ways to express concentrations
Dissociation reaction: degree of
dissociation, α
Example: Dissociation of Br2
Br2 ( g ) → 2 Br ( g )
KP
=
( PBr / Pθ ) 2
( PBr2 / Pθ )
=
PBr2
( PBr2 )( Pθ )
Solve for concentrations as fraction, α
Ka
Kγ
=
2α
Pt ) 2
1+ α
1−α
Pt ) Pθ
(
1+ α
(
4α 2  Pt 
=


1 − α 2  Pθ 
Summary
To solve for concentrations, obtain KP in gas reactions
For ideal systems (usually at low pressures and temperatures well above
the critical temperature), can be equated to Ka (with proper accounting of
pressures)
For nonideal systems, calculate Kγ from fugacity coefficients of the
components using the total pressure


Solve for KP
Solve expression for the concentrations
Different kinds of problems


Total pressure is known: calculate through mole fractions
Total pressure not known: solve as partial pressures (which are related)
 Ideal case is straightforward
 In nonideal case, must do an iterative solution
Dissociation problems: use dissociation fraction as the measure of
concentration
At sufficiently high pressures, must calculate the activity of even the
condensed phases, which is then used in the equilibrium expression