Answer to Question #128601 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

The equilibrium constant of any reversible reaction system is the ratio of forward and reverse rate constants:

K_c = k_f/k_r

Adding a catalyst speeds up the forward and reverse reactions equally, so it has no effect on the equilibrium constant:

K_c = 7.45×10⁶

k_f = 2.93×10⁷ s⁻¹

k_r = k_f/K_c

k_r = 2.93×10⁷ / 7.45×10⁶ = 3.93 s⁻¹

Answer: 3.93 s⁻¹