Question

2. Calculate the equilibrium constant for the reaction


N¬2O4(g) 2NO2(g)



given that fGo (N2O4(g)) = 97.89 kJ mol-1 and fGo (NO2(g)) = 51.31 kJ mol-1



Calculate the equilibrium constant at 100oC given that fHo (N2O4(g)) = 9.16 kJ mol-1 and fHo (NO2(g)) = 33.18 kJ mol-1.

Answer 1

N2O4(g) = 2 NO2(g)

Equilibrium constant K is:

K = exp(-Δ_rG⁰/RT)

where Δ_rG⁰ = 2*Δ_fG⁰(NO2) - Δ_fG⁰(N2O4) = 2*51.31 - 97.89 = 4.73 kJ,

T = 100 ^oC + 273 = 373 K

K = exp(-4730/(8.31*373)) = 0.217

Given the enthalpy values, the change of entropy can be calculated:

Δ_rG⁰ = Δ_rH⁰ - T*Δ_rS⁰ = 2*Δ_fH⁰(NO2) - Δ_fH⁰(N2O4) - T*Δ_rS⁰

4730 = 2*33180 - 9160 - 373*Δ_rS⁰

Δ_rS⁰ = 140.67 J/K

Answer: equilibrium constant K = 0.217