For a reversible reaction A ⇌ B with forward rate kf[A] and backward rate kb[B], which statement is true at equilibrium? (Choose the correct relation at equilibrium.)

• A. [B]eq/[A]eq = kb/kf
• B. [B]eq/[A]eq = kf/kb
• C. [A]eq = [B]eq
• D. K_eq = kf + kb

Answer: (A)

When a reversible reaction reaches equilibrium, the forward and backward rates balance each other: kf[A]eq = kb[B]eq. From that equality, you get the ratio of product to reactant concentrations at equilibrium as [B]eq/[A]eq = kf/kb. This ratio defines the equilibrium constant for this simple two-species system and depends only on temperature, not on how the system was prepared. It means if the forward rate constant is larger than the backward one, you’ll have relatively more B than A at equilibrium, and vice versa. The ratio is not the sum of the rate constants, and it’s not generally equal concentrations unless the two rate constants are the same.