Answer
$\Delta_{r}G^{\circ}=-100.97\,kJ/mol$
As $\Delta _{r}G^{\circ}$ is negative, the reaction is product-favored under standard conditions.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(C_{2}H_{6},g)]-[\Delta_{f}G^{\circ}(C_{2}H_{4},g)+\Delta_{f}G^{\circ}(H_{2},g)]$
$=(-32.82\,kJ/mol)-[(68.15\,kJ/mol)+(0)]$
$=-100.97\,kJ/mol$
As $\Delta _{r}G^{\circ}$ is negative, the reaction is product-favored under standard conditions.
