Answer
$\Delta S_{univ}=110.5\,JK^{-1}mol^{-1}\gt0$, so the process is product-favored.
Work Step by Step
The chemical equation for the synthesis of ammonia is
$N_{2}(g)+3H_{2}(g)\rightarrow2NH_{3}(g)$
$\Delta H^{\circ}_{rxn}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
$=[2\Delta H_{f}^{\circ}(NH_{3},g)]-[\Delta H_{f}^{\circ}(N_{2},g)+3\Delta H_{f}^{\circ}(H_{2},g)]$
$=[2(-46.11\,kJ/mol)]-[(0)+3(0)]$
$=-92.22\,kJ/mol$
$\Delta S_{surr}=\frac{-\Delta H_{sys}}{T}=\frac{-(-92.22\times10^{3}\,J/mol)}{298.15\,K}=309.3\,JK^{-1}mol^{-1}$
$\Delta S^{\circ}_{rxn}=\Delta S_{sys}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[2S^{\circ}(NH_{3},g)]-[S^{\circ}(N_{2},g)+3S^{\circ}(H_{2},g)]$
$=[2(192.45\,JK^{-1}mol^{-1})]-[(191.61\,JK^{-1}mol^{-1})+3(130.684\,JK^{-1}mol^{-1})]$
$=-198.762\,JK^{-1}mol^{-1}$
$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}$
$=-198.762\,JK^{-1}mol^{-1}+309.3\,JK^{-1}mol^{-1}$
$=110.5\,JK^{-1}mol^{-1}$
As $\Delta S_{univ}\gt0$, the process is product-favored.
