AP Chemistrymediummcq1 pt

For a simple one-step reversible reaction, how do the forward and reverse rate constants relate to the equilibrium constant at equilibrium?

A.K equals kforward multiplied by kreverse
B.K equals kreverse divided by kforward
C.K equals kforward divided by kreverse
D.K equals kforward minus kreverse

Explanation

Core Concept

At equilibrium the forward and reverse rates match: kforward times the reactant term equals kreverse times the product term. Rearranging gathers the concentrations into the equilibrium expression on one side and kforward over kreverse on the other. The ratio of rate constants therefore equals K. A large K just means the forward rate constant dominates its partner.

Correct Answer

CK equals kforward divided by kreverse

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