Answer
$\Delta S^{\circ}_{rxn}=-429.3\,JK^{-1}mol^{-1}$
$\Delta G^{\circ}_{rxn}=-505\,kJ/mol$
The reaction is spontaneous from left to right.
Work Step by Step
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[S^{\circ}(C_{6}H_{6},l)]-[3S^{\circ}(C_{2}H_{2},g)]$
$=173.4\,JK^{-1}mol^{-1}-[3(200.9\,Jmol^{-1}K^{-1})]$
$=-429.3\,Jmol^{-1}K^{-1}=-0.4293\,kJmol^{-1}K^{-1}$
$\Delta G^{\circ}_{rxn}=\Delta H^{\circ}_{rxn}-T\Delta S^{\circ}_{rxn}$
$=-633.1\,kJ/mol-(298\,K\times-0.4293\,kJ/mol\cdot K)$
$=-505\,kJ/mol$
As $\Delta G^{\circ}$ is negative, the reaction is spontaneous from left to right.
