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For the reversible first-order reaction A <-> B with forward rate constant kf and backward rate constant kb, the equilibrium concentration of B, [B]e, in terms of the equilibrium concentration of A, [A]e, is given by:

Kc * [A]e⁻¹
(kf/kb) * [A]e⁻¹
(kf/kb) * [A]e
kf * kb * [A]e⁻¹

Correct answer: (kf/kb) * [A]e

Solution

At equilibrium kf[A]e = kb[B]e, so [B]e = (kf/kb)[A]e, which equals Kc[A]e since Kc = kf/kb.

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