Use the following equations to find $\Delta G_{\mathrm{f}}^{0}$ for $\mathrm{HBr}(\mathrm{g})$ at ${ }^{25^{\circ} \mathrm{C}} .$
$\begin{array}{ll}\mathrm{Br}_{2}(\ell) \longrightarrow \mathrm{Br}_{2}(\mathrm{~g}) & \Delta G^{0}=3.14 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{HBr}(\mathrm{g}) \longrightarrow \mathrm{H}(\mathrm{g})+\mathrm{Br}(\mathrm{g}) & \Delta G^{0}=339.09 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Br}(\mathrm{g}) & \Delta G^{0}=161.7 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{g}) & \Delta G^{0}=406.494 \mathrm{~kJ} / \mathrm{mol}\end{array}$