Question

± Free Energy and Chemical Equilibrium

The equilibrium constant of a system, K, can be related to the standard free energy change, ΔG, using the following equation:

ΔG∘=−RTlnK

where T is standard temperature in kelvins and R is equal to 8.314 J/(K⋅mol).

Under conditions other than standard state, the following equation applies:

ΔG=ΔG∘+RTlnQ

In this equation, Q is the reaction quotient and is defined the same manner as K except that the concentrations or pressures used are not necessarily the equilibrium values.

Part A)

At 25 ∘C the reaction from Part A has a composition as shown in the table below.

Substance	Pressure (atm)
C2H2(g)	5.35
H2(g)	4.85
C2H6(g)	3.25×10−2

What is the free energy change, ΔG, in kilojoules for the reaction under these conditions?

Express your answer numerically in kilojoules.

Answer 1

C2H2 + 2H2 -----------------> C2H6

4.75      3.45                          4.25 x 10^-2

Qp = P_C2H6 / P_C2H2 P_H2²

      = 3.25 x 10^-2 / (5.35) (4.85)^2

       = 2.58 x 10^-4

R = 8.314 x 10^-3 kJ/ K mol

T = 273 + 25 = 298 K

G   = Go + RT lnQp

G   = Go + 2.303 RT log Qp

        = -241.9 + 2.303 x 8.314 x 10^-3 x 298 x log (2.58 x 10^-4)

=-241.9 -20.47 kJ

        = -262.37 kJ

G = -262.4 kJ