Answer
55.2 kJ/mol
Work Step by Step
$Q=\frac{P_{COCl_{2}}^{2}}{P_{CO_{2}}P_{CCl_{4}}}=\frac{(0.744)^{2}}{(0.112)(0.174)}=28.4$
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(COCl_{2},g)]-[\Delta G_{f}^{\circ}(CO_{2},g)+\Delta G_{f}^{\circ}(CCl_{4},g)]$
$=[2(-204.9\,kJ/mol)]-[(-394.4\,kJ/mol)+(-62.3\,kJ/mol)]$
$=46.9\,kJ/mol$
$\Delta G=\Delta G^{\circ}+RT\ln Q$
$=46.9\,kJ/mol+(8.314\,Jmol^{-1}K^{-1})(298.15\,K)(\ln 28.4)$
$=46.9\,kJ/mol+(8.3\times10^{3}\,J/mol)$
$=46.9\,kJ/mol+8.3\,kJ/mol$
$=55.2\,kJ/mol$
