Answer to Question #155151 in General Chemistry for Alex

Question

Calculate the Gibbs energy for the following reaction:

2NH3 (g) + 2.5O2 (g) = 2NO (g) + 3H2O (g)

Answer 1

2NH3 (g) + 2.5O2 (g) = 2NO (g) + 3H2O (g)

Multiply both sides by 2,

4NH3 (g) + 5O2 (g) = 4NO (g) + 6H2O (g)

at 298K,

$\Delta G^\circ = \sum {G^\circ _f {\rm{products}}} - \sum {G^\circ _f {\rm{reactants}}}$

ENTHALPY OF REACTION

${[}4ΔHf(NO (g)) + 6ΔHf(H2O (g)){]} - {[}4ΔHf(NH3 (g)) + 5ΔHf(O2 (g)){]} \\{[}4(90.25) + 6(-241.82){]} - {[}4(-46.11) + 5(0){]}\\ = -905.48 kJ$

ENTROPY CHANGE 

${[}4ΔSf(NO (g)) + 6ΔSf(H2O (g)){]} - {[}4ΔSf(NH3 (g)) + 5ΔSf(O2 (g)){]}\\{[}4(210.65) + 6(188.72){]} - {[}4(192.34) + 5(205.03){]} \\= 180.41 J/K$

FREE ENERGY OF REACTION (AT 298.15 K)

From ΔGf° values:

${[}4ΔGf(NO (g)) + 6ΔGf(H2O (g)){]} - {[}4ΔGf(NH3 (g)) + 5ΔGf(O2 (g)){]}\\{[}4(86.57) + 6(-228.59){]} - {[}4(-16.48) + 5(0){]}\\ = -959.34 kJ (spontaneous)$

FROM $\Delta G^\circ = \sum {G^\circ _f {\rm{products}}} - \sum {G^\circ _f {\rm{reactants}}}$ ,

=-959.27 kJ (spontaneous)