Answer
$−59,260\text{ kJ/kmol}$
Work Step by Step
From Tables A-18 and A-26, at 1 atm pressure,
$$
\begin{aligned}
\bar{g} *(600 \mathrm{~K}, 1 \mathrm{~atm}) & =\bar{g}_f^{\circ}+\Delta\left[\bar{h}(T)-T s^{\circ}(T)\right] \\
& =0+(17,563-600 \times 212.066)-(8669-298 \times 191.502) \\
& =-61,278 \mathrm{~kJ} / \mathrm{kmol}
\end{aligned}
$$ The partial pressure of $\mathrm{N}_2$ is
$$
P_{\mathrm{CO}}=y_{\mathrm{N} 2} P=(0.30)(5 \mathrm{~atm})=1.5 \mathrm{~atm}
$$ The Gibbs function of $\mathrm{N}_2$ at $600 \mathrm{~K}$ and $1.5 \mathrm{~atm}$ is
$$
\begin{aligned}
\bar{g}(600 \mathrm{~K}, 1.5 \mathrm{~atm}) & =\bar{g} *(600 \mathrm{~K}, 1 \mathrm{~atm})+R_u T \ln P_{\mathrm{CO}} \\
& =-61,278 \mathrm{~kJ} / \mathrm{kmol}+(8.314 \mathrm{~kJ} / \mathrm{kmol})(600 \\ \mathrm{~K}) \ln (1.5 \mathrm{~atm}) \\
& =-\mathbf{5 9}, \mathbf{2 6 0}\ \mathbf{k J} / \mathbf{k m o l}
\end{aligned}
$$
