Answer to Question #107407 in General Chemistry for Beverlie

Question

3. Suppose the below reaction in the forward direction is first order in A and the rate constant is 1.90×10-2 s–1. A ⇌ B. The reverse reaction is first order in B and the rate constant is 4.30×10-2 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

Kf=1.9*10^-2 s^-1 forward reaction

Kb=4.30×10^-2 s^-1 reverse reaction

A ⇌ B

$Keq=\frac{Kf}{Kb}$

Keq=(1.9*10^-2)\divide 4.30×10^-2=0.44

Keq=0.44