The following picture represents the equilibrium state for the reaction A2 + B2 ⇌ 2AB (A = white; B = gray). What is the relationship between the rate constant for the forward reaction, kf, and the rate constant for the reverse reaction kr at equilibrium?

• A. kf/kr = Kc
• B. kf > kr
• C. kf = kr
• D. kf < kr

Answer: (A)

Verified step by step guidance

1. Understand that at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This is a fundamental concept in chemical equilibrium.
2. The equilibrium constant, Kc, is defined as the ratio of the concentration of products to the concentration of reactants, each raised to the power of their stoichiometric coefficients.
3. For the given reaction A2 + B2 ⇌ 2AB, the equilibrium constant expression is: $\frac{c^2}{c}$, where [AB] is the concentration of AB, and [A2] and [B2] are the concentrations of A2 and B2, respectively.
4. At equilibrium, the rate of the forward reaction (kf[A2][B2]) equals the rate of the reverse reaction (kr[AB]^2). Therefore, kf/kr = [AB]^2 / ([A2][B2]).
5. This relationship shows that kf/kr = Kc, which is the equilibrium constant for the reaction. This is the correct relationship between the rate constants at equilibrium.