For the reaction, $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$, if dinitrogen…
For the reaction, $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$, if dinitrogen tetroxide is $50 \%$ dissociated at $60^{\circ} \mathrm{C}$, the standard free energy change at this temperature and $1 \mathrm{~atm}$ pressure is
$-367.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-763.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-867 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-249 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$
\begin{array}{ccc}
(b) & \mathrm{N}_2 \mathrm{O}_4(g) \longrightarrow & \mathrm{NO}_2(g) \\
t=0 & 1 \mathrm{~mol} & 0 \\
t=t_{\mathrm{cq}} & (1-0.5) & 2(0.5)
\end{array}
$
Total number of moles $=0.5+1=1.5 \mathrm{~mol}$ $p_{\mathrm{N}_2 \mathrm{O}_4}=\frac{0.5}{1.5} \times 1 \mathrm{~atm}=\frac{1}{3} \mathrm{~atm}$ and $p_{\mathrm{NO}_2}=\frac{1}{1.5} \times 1 \mathrm{~atm}=\frac{1}{1.5} \mathrm{~atm}$
According to law of chemical equilibrium,
$
K_p=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{\left(p_{\mathrm{N}_2 \mathrm{O}_4}\right)}=\frac{\left(\frac{1}{1.5}\right)^2}{\left(\frac{1}{3}\right)}=1.33 \mathrm{~atm}
$
Apply, $\Delta G^{\circ}=-2.303 R T \log K_p$
$
\begin{aligned}
\Delta G^{\circ} & =-2.303 \times(8.314) \times(333) \log (1.33) \\
& =-763.8 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$
Hence, the correct option is (2)