Answer
Requires work to be done.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(C_{2}H_{4},g)+\Delta_{f}G^{\circ}(H_{2},g)]-[\Delta_{f}G^{\circ}(C_{2}H_{6},g)]$
$=[(68.15\,kJ/mol)+(0\,kJ/mol)]-(-32.82\,kJ/mol)]$
$=100.97\,kJ/mol$
As $\Delta _{r}G^{\circ}$ is positive, work needs to be done for this reaction to occur.
At least 100.97 kJ/mol work is required.
