Answer
$11.9\,kJ/mol$
Work Step by Step
$Q=\frac{P_{CO}P_{H_{2}}^{2}}{P_{CH_{3}OH}}=\frac{(0.125)(0.183)^{2}}{0.855}=0.004896$
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$\Delta G^{\circ}=[\Delta G_{f}^{\circ}(CO,g)+2\Delta G_{f}^{\circ}(H_{2},g)]-[\Delta G_{f}^{\circ}(CH_{3}OH,g)]$
$=[(-137.2\,kJ/mol)+2(0)]-(-162.3\,kJ/mol)$
$=25.1\,kJ/mol$
$\Delta G=\Delta G^{\circ}+RT\ln Q$
$=25.1\,kJ/mol+(8.314\,Jmol^{-1}K^{-1})(298.15\,K)(\ln 0.004896)$
$=25.1\,kJ/mol+(-13.2\times10^{3}\,J/mol)$
$=25.1\,kJ/mol-13.2\,kJ/mol$
$=11.9\,kJ/mol$
