At a given Temperature, the elementary reaction A<--->B in the forward direction is first order in A with a rate constant of 3.90*10^-2s^-1. The reverse reaction is first order in B and the rate constant is 8.20*10^-2s^-1. What is the value at equillibrium for the reaction at this temperature? What is the value of the equillibrium constant for the reaction B<--->A at this temperature?

f = forward

r = reverse
Isn't ratef = kf(A) and
rater = kr(B), then
(rater/ratef) = kr(B)/Kf(A) so
kr/kf = Keq.
For the reverse, K' = 1/Keq

To find the value of equilibrium for the reaction A <--> B, we need to use the rate constants for both the forward and reverse reactions. The equilibrium constant (K) can be calculated using the following formula:

K = (rate constant for forward reaction) / (rate constant for reverse reaction)

Given the rate constants:
rate constant for forward reaction (k1) = 3.90 * 10^-2 s^-1
rate constant for reverse reaction (k2) = 8.20 * 10^-2 s^-1

Substituting these values into the formula, we get:

K = k1 / k2
= (3.90 * 10^-2) / (8.20 * 10^-2)
≈ 0.476

So, the value at equilibrium for the reaction A <--> B is approximately 0.476.

To find the equilibrium constant for the reverse reaction B <--> A, we can use the following formula:

K_reverse = 1 / K_forward

Where K_forward is the equilibrium constant for the forward reaction (A <--> B). Substituting the value of K_forward (0.476) into the formula, we get:

K_reverse = 1 / 0.476
≈ 2.101

So, the equilibrium constant for the reverse reaction B <--> A at this temperature is approximately 2.101.