At 27 in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ mol-1. The logarithm of ratio of k (catalysed)k (uncatalysed) is.... (Consider that the frequency factor for both the reactions is same)
- 1.741
- 0.1741
- 17.41
- 3.482
Answer
(1) 1.741
Solution
By the Arrhenius equation, k = A e-Ea/RT, and A is the same for both reactions, so
kcatkuncat = e(Ea - Ea')/RT
kcatkuncat = Ea - Ea'2.303RT = 10 1032.303 8.314 300 = 100005744.1 = 1.741
Answer: (1)