Answer
Requires work to be done.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(N_{2},g)+3\Delta_{f}G^{\circ}(F_{2},g)]-[2\Delta_{f}G^{\circ}(NF_{3},g)]$
$=[(0\,kJ/mol)+3(0\,kJ/mol)]-[2(-83.2\,kJ/mol)]$
$=166.4\,kJ/mol$
As $\Delta _{r}G^{\circ}$ is positive, work needs to be done for this reaction to occur.
At least 166.4 kJ/mol work is required.
