7.1 Gibbs Energy Changes and the Equilibrium Constant 309
Exercise 7.2
Carry out the steps to obtain Eqs. (7.1-16) and (7.1-17).If it is possible to keep the activity of every substance fixed during a chemical
reaction, the value ofQis fixed and the change in Gibbs energy for 1 mol of reaction
can be written∆G(fixed composition)(
∂G
∂ξ)
T,P∣
∣
∣
∣
∣
f ixed comp∫ 1
0∆G◦+RTln(Q)dξ∆G◦+RTln(Q)∫ 1
0dξ∆G(fixed composition)∆G◦+RTln(Q) (7.1-18)EXAMPLE 7.2
Calculate the value of∆G(fixed composition) at 298.15 K for the reaction of Exercise 7.1 if
the partial pressure of PCl 5 is maintained equal to 0.0100 bar, the partial pressure of PCl 3
is maintained equal to 0.00600 bar, and the partial pressure of Cl 2 is maintained equal to
0.00300 bar. Assume that the gases are ideal.
SolutionQa(PCl 3 )a(Cl 2 )
a(PCl 5 )
(0.00600)(0.00300)
(0.0100)
0. 00180(
∂G
∂ξ)T,P 20 .661 kJ mol−^1 +(8.3145 J K−^1 mol−^1 )(298.15 K) ln(0.00180) 20 .661 kJ mol−^1 +(−15667 J mol−^1 )
4994 J mol−^1 4 .994 kJ mol−^1Exercise 7.3
Repeat the calculation of the previous example if the partial pressure of PCl 5 is maintained equal
to 0.00600 bar, the partial pressure of PCl 3 is maintained equal to 0.0100 bar, and the partial
pressure of Cl 2 is maintained equal to 0.00300 bar.At equilibrium, (∂G/∂ξ)T,P0, and Eq. (7.1-16) becomes0 ∆G◦+RTln(Qeq) (7.1-19)