If Ke for the reaction 2SO3 (g) ⇌ 2SO2 (g) + O2 (g) is 2.3 x 10^-7, what is the equilibrium constant for the reverse reaction?

• A. 2.3 x 10^-7
• B. 1.0 x 10^7
• C. 1.0 x 10^-7
• D. 4.3 x 10^6

Answer: (D)

The key idea is that the equilibrium constant for a reaction and for its reverse are reciprocals of each other. For the forward reaction, K_forward is defined as the product of the activities of the products raised to their stoichiometric powers divided by the activities of the reactants raised to their powers. Here, K_forward = 2.3 × 10^-7.

If you reverse the reaction, the expression switches to the reactants becoming products and vice versa, which makes K_reverse = 1 / K_forward. So K_reverse = 1 / (2.3 × 10^-7) ≈ (1/2.3) × 10^7 ≈ 4.3 × 10^6.

This large value for the reverse constant means the reverse reaction is strongly favored to form SO3 from SO2 and O2, consistent with the forward reaction not proceeding far to products.