The forward rate of a reaction from S --> P is 10^-9 per second and the...
The forward rate of a reaction from S --> P is 10^-9 per second and the reverse reaction is 10^-12 per second. Addition of an enzyme enhances this reaction by 100 fold. What will the reverse rate constant be? Show all your work and explanation please.
Homework Answers
Keq=rate constant of the forward rection/rate constant of the backward reaction
Rate of a reaction is directly proportional to the rate contant
Rate for forward reaction
k1
10-9
k1
Rate for reverse reaction
k2
Therefore, 10-12
k2
Therefore, keq (uncatalyzed)
10-9/10-12=1000
Therefore, k1=k2*1000
Now, for catalyzed, k2*100=k1*100/1000
Therefore,k2(catalyzed)=100*10-12=10-10
Keq for catalyzed and uncatalyzed reaction is the same as it is not changed by the use of a catalyst
Thank You!
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