Answer
(a) -32.74 kJ/mol. The reaction is product-favored.
(b) -21.33 kJ/mol. The reaction is spontaneous.
Work Step by Step
(a) $\Delta _{r}G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(NH_{3},g)]-[\Delta G_{f}^{\circ}(N_{2},g)+3\Delta G_{f}^{\circ}(H_{2},g)]$
$=[2(-16.37\,kJ/mol)]-[(0)+3(0)]$
$=-32.74\,kJ/mol$
As $\Delta _{r}G^{\circ}$ is negative, the reaction is product-favored at equilibrium.
(b) $Q=\frac{P_{NH_{3}}^{2}}{P_{N_{2}}P_{H_{2}}^{3}}=\frac{(0.10)^{2}}{(0.10)(0.10)^{3}}=100$
$\Delta _{r}G=\Delta _{r}G^{\circ}+RT\ln Q$
$=-32.74\,kJ/mol+(8.314\times10^{-3}\,kJmol^{-1}K^{-1})(25+273)K(\ln 100)$
$=-21.33\,kJ/mol$
As $\Delta _{r}G$ is negative, the reaction is spontaneous under these conditions.
