In a reversible first-order A ⇌ B with forward rate k1 and reverse rate k-1, how is the equilibrium constant defined in terms of rate constants?

In a reversible first-order A ⇌ B with forward rate k1 and reverse rate k-1, how is the equilibrium constant defined in terms of rate constants?

• A. K = k-1 / k1
• B. K = k1 / k-1
• C. [B]/[A]
• D. K = k1 + k-1

Answer: (B)

In a reversible A ⇌ B with forward rate k1 and reverse rate k-1, the rates at equilibrium are equal: k1[A] = k-1[B]. Rearranging gives [B]/[A] = k1/k-1. Since the equilibrium constant defined in terms of rate constants is the ratio of the forward to the reverse rate constants, K = k1/k-1. This matches the concentration ratio Kc = [B]/[A] for this elementary step. The reciprocal k-1/k1 would give the inverse ratio, and summing the rates isn’t related to the equilibrium constant.

In a reversible A ⇌ B with forward rate k1 and reverse rate k-1, the rates at equilibrium are equal: k1[A] = k-1[B]. Rearranging gives [B]/[A] = k1/k-1. Since the equilibrium constant defined in terms of rate constants is the ratio of the forward to the reverse rate constants, K = k1/k-1. This matches the concentration ratio Kc = [B]/[A] for this elementary step. The reciprocal k-1/k1 would give the inverse ratio, and summing the rates isn’t related to the equilibrium constant.