Answer
$\Delta S^{\circ}_{rxn}=-128.6\,Jmol^{-1}K^{-1}$
$\Delta G^{\circ}_{rxn}=-5.88\,kJ/mol$
The reaction is spontaneous from left to right.
Work Step by Step
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[S^{\circ}(C_{2}H_{5}OH,l)]-[S^{\circ}(C_{2}H_{4},g)+S^{\circ}(H_{2}O,l)]$
$=160.7\,JK^{-1}mol^{-1}-(219.3\,Jmol^{-1}K^{-1}+70.0\,Jmol^{-1}K^{-1})$
$=-128.6\,Jmol^{-1}K^{-1}=-0.1286\,kJmol^{-1}K^{-1}$
$\Delta G^{\circ}_{rxn}=\Delta H^{\circ}_{rxn}-T\Delta S^{\circ}_{rxn}$
$=-44.2\,kJ/mol-(298\,K\times-0.1286\,kJ/mol\cdot K)$
$=-5.88\,kJ/mol$
As $\Delta G^{\circ}$ is negative, the reaction is spontaneous from left to right.
