Answer to Question #128911 in General Chemistry for A

Question

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?

Thanks

Answer 1

A catalyst accelerates both the forward and reverse reactions with no noted change in the equilibrium.

There are changes in both $kf$ and $kr$ with a constant in $kc$ .

Thus $kc=$ $kf\over kr$

$\therefore kr=$ $kf\over kc$

$kr_{cat}=$ $kf_{cat}\over kc$ $=$ $2.93\times10^{-7}s^{-1}\over 7.45\times10^6$

$=3.93s^{-1}$

The units for the forward reaction constant are similar to those of the reverse reaction constant.

The equilibrium reaction has no units.