Chemical Kinetics Practice Test

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For a simple reversible reaction A → B, how is the equilibrium constant related to the forward and reverse rate constants for A → B?

K = k_forward * k_reverse; at equilibrium [B]/[A] = 1/K K = k_forward / k_reverse; at equilibrium [B]/[A] = K K = k_reverse / k_forward; at equilibrium [B]/[A] = 1/K K = k_forward - k_reverse; at equilibrium [B]/[A] = K^2

Correct Answer:

K = k_forward / k_reverse; at equilibrium [B]/[A] = K

Explanation:

In a reversible reaction, equilibrium occurs when the forward and reverse rates balance each other. The forward rate is k_forward times the concentration of A, and the reverse rate is k_reverse times the concentration of B. At equilibrium, k_forward[A] = k_reverse[B]. Rearranging gives [B]/[A] = k_forward/k_reverse. The equilibrium constant for this 1:1 system is defined as K = [B]/[A], so K equals the ratio of the forward to reverse rate constants, and at equilibrium [B]/[A] = K.

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