Question

For a different reaction, Kc = 1.39, kf=12.6s−1 , and kr= 9.08 s−1 . Adding a catalyst increases the forward rate constant to 4.28×103 s−1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

Answer 1

Answer:

        $\large\boxed{\large\boxed{K_r=3.08\times 10^3s^{-1}}}$

Explanation:

The equilibrium constant is equal to the quotient of the forward rate to the reverse rate.

          $K_c=\dfrac{K_f}{K_r}$

Adding a catalyst increases both the forward rate and the reverse rate in the same proportion without net effect on the equilibrium position.

Then, the equilibrium constant before and after adding the catalyst is the same.

Thus, knowing the forward rate constant and the equilibrium constant you can determine the reverse rate constant:

               $K_r=\dfrac{K_f}{K_c}\\ \\ \\ K_r=\dfrac{4.28\times 10^{3}s^{-1}}{1.39}\\ \\ \\ K_r=3.08\times 10^3s^{-1}$