Question

For a reversible reaction how rate constant change with temp

Answer 1

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.

Answer 2

Answer:

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.

Explanation: