Using $\Delta_{\mathrm{f}} H_{298}^{\circ}$ And $S_{298}^{\circ}$ Data Given Below Calculate the Standard Gibbs Energy Change

Using $\Delta_{\mathrm{f}} H_{298}^{\circ}$ and $S_{298}^{\circ}$ data given below:
$\begin{array}{|l|l|l|l|}\hline & \mathrm{NH}_{3}(\mathrm{~g}) & \mathrm{N}_{2}(\mathrm{~g}) & \mathrm{H}_{2}(\mathrm{~g}) \\\hline \Delta_{\mathrm{f}} H_{298}^{\circ} / \mathrm{kJ} \mathrm{mol}^{-1} & -46.1 & 0 & 0 \\\hline S_{298}^{\circ} / \mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1} & 192.5 & 191.6 & 130.7 \\\hline\end{array}$
Calculate the standard Gibbs energy change, $\Delta_{r} G_{298}^{\circ}$ (kJ mol^-1) , for the following reaction:
2 NH₃ (g) → N₂ (g) + 3 H₂ (g)

A) + 151.
B) - 151.
C) + 33.
D) - 59.