Answer
-29.09 kJ/mol
Work Step by Step
$\Delta _{r}H^{\circ}=\Sigma n_{p}\Delta _{f}H^{\circ}(products)-\Sigma n_{r}\Delta _{f}H^{\circ}(reactants)$
$=[\Delta_{f}H^{\circ}(CH_{3}OH,l)]-[\Delta _{f}H^{\circ}(CO,g)+2\Delta _{f}H^{\circ}(H_{2},g)]$
$=(-238.66\,kJ/mol)-[(-110.525\,kJ/mol)+2(0)]$
$=-128.14\,kJ/mol$
$\Delta _{r}S^{\circ}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[S^{\circ}(CH_{3}OH,l)]-[S^{\circ}(CO,g)+2S^{\circ}(H_{2},g)]$
$=(126.8\,JK^{-1}mol^{-1})-[(197.674\,JK^{-1}mol^{-1})+2(130.684\,JK^{-1}mol^{-1})]$
$=-332.2\,JK^{-1}mol^{-1}$
$\Delta _{r}G^{\circ}=\Delta _{r}H^{\circ}-T\Delta _{r}S^{\circ}$
$=(-128.14\,kJ/mol)-(298.15\,K)(-0.3322\,kJK^{-1}mol^{-1})$
$=-29.09\,kJ/mol$
$\Delta _{r}G^{\circ}$ is negative verifying that the reaction is product-favored.
