Answer to Question #107406 in General Chemistry for Beverlie

Question

1. 2. Suppose the below reaction in the forward direction is first order in A and the rate constant is 1.10×10-2 s–1. A ⇌ B. The reverse reaction is first order in B and the rate constant is 4.70×10-2 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

k₁ is rate constant of forward direction, k₁ = 1.10*10^-2 s^-1

k_-1 is rate constant of backward direction, k_-1 = 4.70*10^-2 s^-1

equilibrium constant K = ${\frac {k_1} {k_{-1}} }$ = ${\frac {1.10*10^{-2}} {4.70*10^{-2}}}$ =0.234 = 2.34*10^-1