A reaction is run under conditions in which the equilibrium constant is known to be 2.5. $\times$ 10². At equilibrium, the forward rate is determined to be 5.0 $\times$ 10^-4 mol L^-1 s^-1. What is the rate of the reverse reaction under these conditions?

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The Answer of A reaction is run under conditions in which the equilibrium constant is known to be 2.5. $\times$ 10². At equilibrium, the forward rate is determined to be 5.0 $\times$ 10^-4 mol L^-1 s^-1. What is the rate of the reverse reaction under these conditions?

A Reaction Is Run Under Conditions in Which the Equilibrium ×\times×

A Reaction Is Run Under Conditions in Which the Equilibrium $\times$