At a given Temperature, the elementary reaction AB 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 BA at this temperature?

Question

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?

DrBob222 · Answer

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