Consider the gas-phase equilibrium $A(g) + B(g) \rightleftharpoons 2C(g)$ at constant temperature. If the volume of the container is HALVED (i.e. the pressure is doubled), what happens to the equilibrium?

1. Apply Le Chatelier for pressure: the equilibrium shifts toward the side with FEWER moles of gas to relieve the increased pressure. 2. Count moles of gas on each side: Left side: $1\,(A) + 1\,(B) = 2$ mol. Right side: $2\,(C) = 2$ mol. 3. $\Delta n = 2 - 2 = 0$. Both sides have the same total moles of gas. 4. With $\Delta n = 0$, halving the volume increases ALL concentrations by the SAME factor. The reaction quotient $Q = [C]^2/([A][B])$ remains unchanged because both numerator and denominator scale by $2^2 = 4$ → cancel. 5. So $Q$ is still equal to $K$, and there is NO net shift — option C. 6. Options A and B incorrectly apply the "shift to side with fewer moles" rule without first checking $\Delta n$. _Source: NCERT Class 11 Chemistry, Ch 6, §6.8.1 (Effect of pressure, depends on $\Delta n$), p. 19–20._