Answer
$-29.10\,kJ/mol$
The result obtained here and in the previous part are the same.
Work Step by Step
$\Delta _{r}G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[\Delta G_{f}^{\circ}(CH_{3}OH,l)]-[\Delta G_{f}^{\circ}(CO,g)+2\Delta G_{f}^{\circ}(H_{2},g)]$
$=(-166.27\,kJ/mol)-[(-137.168\,kJ/mol)+2(0)]$
$=-29.10\,kJ/mol$
The result obtained here and in the previous part are the same.
