Answer to Question #106853 in Physical Chemistry for mya 18

Question

At a given temperature, the elementary reaction A−⇀B, in the forward direction, is first order in A with a rate constant of 0.0410s−1. The reverse reaction is first order in B and the rate constant is
0.0880s−1.

What is the value of the equilibrium constant for the reaction A−⇀B at this temperature?

What is the value of the equilibrium constant for the reaction B−⇀A at this temperature?

Answer 1

For the reaction A⇀B

the equilibrium constant

K = k_f / k_b , forward (f) and backward (b) processes

K = 0.0410 / 0.0880 = 0.4659

For the reaction B⇀A

K^' = 0.0880 / 0.0410 = 2.1463