Answer:
Δ[tex]G=-37.7kJ/mol[/tex]
Explanation:
Hello,
Gibbs free energy for this reaction is defined as:
Δ[tex]G=[/tex]Δ[tex]G^0+RTln(Q)[/tex]
Now, [tex]Q[/tex] is computed via the law of mass action:
[tex]Q=\frac{P_{NH_3}^2}{P_{N_2}P_{H_2}^3} =\frac{0.78^2}{1.3*1.4^3}=0.17[/tex]
Finally:
Δ[tex]G=-33.3kJ/mol+298K*8.314x10^{-3}\frac{kJ}{mol*K}*ln(0.17)=-37.7kJ/mol[/tex]
Best regards.