N2O4(g)rightleftharpoons 2NO2(g) is studied in a rigid vessel at constant temperature. When 1 mol of N2O4 is introduced so that the initial pressure…

For initial pressure $P_0$ and degree of dissociation $\alpha$: $p(\mathrm{N_2O_4})=P_0(1-\alpha)$, $p(\mathrm{NO_2})=2P_0\alpha$, total $=P_0(1+\alpha)$. First experiment: $1+\alpha=1.4$, so $\alpha=0.40$ and $K_p=\frac{(2\cdot0.4)^2}{0.6}=1.067\ \mathrm{atm}$. In half the volume at the same $T$ the initial pressure is $P_0'=2.0\ \mathrm{atm}$. Then $K_p=\frac{4P_0'\alpha^2}{1-\alpha}$, i.e. $8\alpha^2+1.067\alpha-1.067=0$, giving $\alpha=\frac{-1.067+\sqrt{1.138+34.13}}{16}=0.3045\approx0.30$. Using $K_p=4\alpha^2P/(1-\alpha^2)$ with the initial pressure in place of the equilibrium total pressure gives $0.34$; assuming $\alpha$ halves gives $0.20$; and $0.40$ ignores the shift altogether. Compression favours fewer moles of gas, so $\alpha$ must decrease, but not to $0.20$.