At a given temperature, the elementary reaction A <--->B in the forward directio

Question

At a given temperature, the elementary reaction A <--->B in the forward direction is first order in A with a rate constant of 2.50*10^-2 s^-1. The reverse reaction is first order in B and the rate constant is 7.30*10^-2 s^-1
What is the value of the equilibrium constant for the reaction A< --->B at this temperature?
What is the value of equilibrium constant for the reaction B<-->A at this temperature?

At a given temperature, the elementary reaction A B in the forward direction is first order in A with a rate constant of 2.50*10^-2 s^-1. The reverse reaction is first order in B and the rate constant is 7.30*10^-2 s^-1 What is the value of the equilibrium constant for the reaction AB at this temperature? What is the value of equilibrium constant for the reaction BA at this temperature?

Explanation / Answer

A < --- > B.

Rate of forward reaction is first order with respect to A. Rate constant for forward reaction would be = 2.50 E-2 s-1.

The reverse reaction rate with respect to B is also first order.

Rate constant for reversible reaction is 7.30 E-2 s-1 .

Solution:

We have to find equilibrium constant For A to B ( forward reaction )

We know equilibrium constant for forward reaction is the ratio ration of rate constant or forward reaction to rate constant of backward reaction.

So equilibrium constant for forward reaction = (kf/kb) = 2.50 E -2 / 7.30 E-2 = 0.3425

Lets calculate equilibrium constant for backward reaction.

Keq (backward reaction) = 1 / K forward reaction = 1 / 0.3425 = 2.91

So the equilibrium constant for backward direction is 2.91

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