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March 28, 2017

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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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