# Chemistry

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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?

• Chemistry -

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

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