Calculate the standard free energy of reaction at 27°C for the combustion of methane: (g) + 2(g) - (g) + 2(l). delta_f H° (kJ/mol): = -74.5, = 0, = -393.5, = -286.0. S° (J/K-mol): = 186, = 205, = 212

Question

Calculate the standard free energy of reaction at 27°C for the combustion of methane: $CH_4$(g) + 2$O_2$(g) -> $CO_2$(g) + 2$H_2O$(l). delta_f H° (kJ/mol): $CH_4$ = -74.5, $O_2$ = 0, $CO_2$ = -393.5, $H_2O$ = -286.0. S° (J/K-mol): $CH_4$ = 186, $O_2$ = 205, $CO_2$ = 212, $H_2O$ = 70

Prashant Kotian · Accepted Answer

Correct answer: C. δ H° = [-393.5 + 2*(-286)] - (-74.5) = -965.5 + 74.5 = -891.0 kJ/mol. δ S° = [212 + 2*70] - [186 + 2*205] = 352 - 596 = -240 J/(K mol). δ G° = -891.0 - 300*(-0.240) = -891 + 72 = -819 kJ/mol.

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