Question

He rate constant for the forward reaction, k1, is 297 l·mol–1·min–1 and the rate constant for the reverse reaction, k–1, is 393 l·mol–1·min–1 at a given temperature. the activation energy for the forward reaction is 42.1 kj·mol–1, while the activation energy for the reverse reaction is 22.1 kj·mol–1. determine the equilibrium constant, k, of this reaction.

Answer 1

Equilibrium constant of reaction,
 = $\frac{rate.constant.forward.reaction}{rate.constant.of.backward.reaction}$
= $\frac{297}{393}$
=0.7557

Answer 2

The equilibrium constant is a number that shows the extent to which reactants are converted into products. The equilibrium constant of the reaction is 0.76.

What is equilibrium constant?

The equilibrium constant is a number that shows the extent to which reactants are converted into products.

We know that to obtain the equilibrium constant; K = k1/k-1 where;

k-1 = rate constant of reverse reaction

k1 = rate constant of forward reaction.

Hence;

K = 297 l·mol–1·min–1/393 l·mol–1·min–1

K = 0.76

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