Question #128980
Good day,
For a different reaction, Kc= 7.45×106, kf=6.05×104s−1 and kr=8.12×10−3 s−1
. Adding a catalyst increases the forward rate constant to 2.93×107 s−1
. What is the new value of the reverse reaction constant, after adding catalyst?
I think the answer is 1.28x10^13 but unsure if the units,please help.
Thanks
For a different reaction, Kc= 7.45×106, kf=6.05×104s−1 and kr=8.12×10−3 s−1
. Adding a catalyst increases the forward rate constant to 2.93×107 s−1
. What is the new value of the reverse reaction constant, after adding catalyst?
I think the answer is 1.28x10^13 but unsure if the units,please help.
Thanks
Expert's answer
A catalyst accelerates both forward and reverse reactions, but doesn't affect the equilibrium. So kf and kr change while Kc remains constant.
From a relation between constants Kc=kf/kr:
Kr=kf/Kc,
Krcat=kfcat/Kc=2.93×10^7s-1/7.45×10^6=3.93 s-1.
The units of the constant of reverse reaction is the same as of the constant of the forward reaction, while equilibrium constant is unitless.
