For PCl5(g)rightleftharpoonsPCl3(g)+Cl2(g), pure PCl5 is found to be 50% dissociated at equilibrium at 500 K when the total pressure is 1.0 atm. Take…

For one mole of $\mathrm{PCl_5}$ with degree of dissociation $\alpha$, the moles are $1-\alpha,\ \alpha,\ \alpha$ (total $1+\alpha$), so $K_p=\dfrac{\alpha^2}{1-\alpha^2}P$. At $P=1\ \mathrm{atm}$ and $\alpha=0.5$, $K_p=\dfrac{0.25}{0.75}=\dfrac13$. Then $\Delta G^\circ=-RT\ln K_p=RT\ln3=8.314\times500\times1.10\approx+4.57\ \mathrm{kJ\,mol^{-1}}$ (positive because $K_p<1$; using the exact $\ln3=1.0986$ gives the same value to two decimals). At $P=2\ \mathrm{atm}$, $\dfrac{\alpha^2}{1-\alpha^2}\times2=\dfrac13$, so $\alpha^2=\dfrac{1/3}{1/3+2}=\dfrac17$ and $\alpha=0.378\approx0.38$. Higher pressure favours the side with fewer moles, so $\alpha$ falls below $0.50$; halving it ($0.25$) wrongly treats $K_p$ as proportional to $1/P$ in $\alpha$ itself. Also, $\Delta G^\circ$ refers to the standard state and does not change with the total pressure used in the experiment.