Answer to Question #107405 in General Chemistry for Beverlie

Question

1. Suppose the below reaction in the forward direction is first order in A and the rate constant is 1.30×10-2 s–1. A ⇌ B. The reverse reaction is first order in B and the rate constant is 4.10×10-2 s–1 at the same temperature. What is the value of the equilibrium constant for the below reaction at this temperature?
A ⇌ B

Answer 1

The equilibrium constant, K = (rate constant for the forward reaction)/( rate constant for the reverse reaction)

Therefore here k= (1.30×10-2 s–1)/(4.10×10-2 s–1)

=0.317