The reaction $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ is at equilibrium in a cylinder at constant temperature. The volume is suddenly halved. Immediately afterwards, the reaction quotient $Q_{c}$ compared with $K_{c}$ is

Answer: B

Halving the volume doubles every concentration.

Q_{c} = (2 [NO X_{2}])^{2} / (2 [N X_{2} O X_{4}]) = 2 K_{c}

. Since

Q_{c} > K_{c}

the reaction goes backward, toward the side with fewer moles of gas. Forgetting that the denominator also doubles gives the incorrect factor 4.