For the gas-phase equilibrium A(g)rightleftharpoons 2B(g) at a fixed temperature, Kp = 0.50 atm. Starting with pure A, the degree of dissociation at…

Let A dissociate to a degree $\alpha$. The mole fraction of B is $2\alpha/(1+\alpha)$, so $K_p=4\alpha^2P/(1-\alpha^2)$ and $\alpha=\sqrt{K_p/(K_p+4P)}$. At $P=2$ atm, $\alpha=\sqrt{0.5/8.5}=0.243$, so A is correct. At $P=8$ atm, $\alpha=\sqrt{0.5/32.5}=0.124$, a ratio of 0.511 and not exactly 0.5 (the exact halving holds only for $\alpha\ll1$), so B is wrong. An inert gas added at constant total pressure lowers the partial pressures of A and B, which favours the reaction producing more moles, so $\alpha$ rises and C is correct. At constant volume the partial pressures, and hence $Q_p$, do not change, so the equilibrium does not shift and D is correct.