For the reaction 2SO2 + O2 ⇌ 2SO3, when the pressure is increased, the forward rate increases and the yield of SO3 increases because the forward reaction reduces the total number of gas molecules. Which statement best explains this observation?

• A. The system shifts toward the side with more moles of gas.
• B. The system shifts toward the side with fewer moles of gas.
• C. There is no effect on equilibrium.
• D. The equilibrium constant increases with pressure.

Answer: (C)

When pressure is increased in a gaseous equilibrium, the system shifts to the side that has fewer gas molecules to reduce the overall pressure. In this reaction, the left side has 3 moles of gas (2 SO2 and O2) while the right side has 2 moles of gas (2 SO3). So the equilibrium moves toward the production of SO3, increasing its yield. The forward rate also speeds up because higher pressure means more frequent collisions, and because it leads to fewer gas molecules on the product side, the overall shift favors the forward direction. Importantly, the equilibrium constant stays the same at a given temperature; increasing pressure changes the position of equilibrium, not the value of K. Therefore, the statement that best explains the observation is that the system shifts toward the side with fewer moles of gas.