At 25 C the equilibrium partial pressures for the reaction w

At 25 \'C, the equilibrium partial pressures for the reaction were found to be PA 4.85 bar, PB = 5.77 bar, Pc = 470 bar, and PD 5.06 bar. A(g) + 2 B(g) 4 C(g) + D(g) What is the standard change in Gibbs free energy of this reaction at 25 ? kJ mol A:

Solution

consider the given reaction

A + 2 B ---> 4 C + D

the equilibrium constant is given by

K = [pC]^4 [pD] / [pA] [pB]^2

plug in the given values

K = [4.7]^4 [5.06] / [4.85] [5.77]^2

K = 15.291447

now

dGo = -RT lnK

dGo = - (8.314 J/mol K) x (298 K) x ln 15.291447

dGo = -6757 J/mol

dGo = -6.757 kJ/mol

so

the standard change in gibbs free energy is -6.757 kJ/mol