Question

for a reversible reaction the rate constant for forward and backward reaction are 2.38 * 10 ^ -4 and 8.15 * 10 ^-5 respectively. The equilibrium constant for the reaction is

Answer 1

Kc=Kf/Kb
So,Kc=2.38×10^-4/8.15×10^-5
= 0.29×10
= 2.9
~3

Answer 2

We know very well that forward reaction is usually more thermodynamically favorable compared with the reverse reaction and thus the relationship k1>=k2 maintains between the rate constants.

Given that:

forward reaction Kf = 2.38 X 10^-4

and Kb =8.15 X 10^-5;  

Kc= Kf/Kb.

Thus Kc = 2.38 X 10^-4/8.15 X 10^-5 = 2.9~=3.