In a reversible reaction A B at a particular temperature, the rate constant for the reverse reaction is twice the rate constant for the forward reaction. At equilibrium, how are the concentrations of A and B related?

Question

In a reversible reaction A  B at a particular temperature, the rate constant for the reverse reaction is twice the rate constant for the forward reaction. At equilibrium, how are the concentrations of A and B related?

Quizplus · Accepted Answer

At equilibrium, the ratio of concentrations of products to reactants is given by the equilibrium constant (K).
K = [B] / [A]
Since the rate constant for the reverse reaction is twice that of the forward reaction, we can write the equilibrium constant as:
K = k(reverse) / k(forward) = 2
Therefore, we have:
[B] / [A] = 2
or
[A] = [B] / 2
This relationship is represented by choice C.

In a Reversible Reaction a B at a Particular ×\times×

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