A gas-phase reaction experiences increased pressure. Which side of the equation is favored?

• A. The side with more moles of gas.
• B. The side with fewer moles of gas.
• C. The solid phase is favored.
• D. Temperature is the sole determinant and pressure does not affect the equilibrium.

Answer: (B)

Increasing pressure in a gas-phase equilibrium shifts the balance to reduce the number of gas molecules, so the system favors the side with fewer moles of gas. This is Le Châtelier’s principle in action: by moving toward the smaller gas-number, the overall pressure is lowered.

Think of a reaction where reactants add up to more gas moles than products; raising pressure nudges the equilibrium toward the product side with fewer gas molecules. For example, N2(g) + 3H2(g) ⇌ 2NH3(g) has 4 moles of gas on the left and 2 on the right, so increasing pressure shifts toward NH3.

If the two sides had the same number of gas moles, pressure changes wouldn’t favor one side over the other. The idea that a solid phase is universally favored or that temperature is the sole determinant is inaccurate—pressure specifically affects gas-phase equilibria by reducing gas-mole count.