Question
At 27^ C 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)
JEE Main
Chemistry
Chemical Kinetics
2026
JEE Main 2026 (24 January Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. 1.741
- B. 17.41
- C. 3.482
- D. 0.1741
Answer
A. 1.741
Step-by-step solution
Using the Arrhenius equation: k = A e^ -E_a/RT The ratio of rate constants is: k_ catalysed k_ uncatalysed = A e^ -E_a^c/RT A e^ -E_a^u/RT = e^ (E_a^u - E_a^c)/RT = e^ E_a/RT Since the catalyst lowers activation energy by 10 kJ/mol: E_a = 10 kJ/mol = 10000 J/mol At T = 27°C = 300 K, R = 8.314 J/(mol·K): k_c k_u = e^ 10000/(8.314 300) = e^ 10000/2494.2 = e^ 4.009 Taking logarithm base 10: _ 10 k_c k_u = 4.009 _ 10 (e) = 4.009 0.4343 = 1.741
Related: Chemistry — Chemical Kinetics · All PYQ Banks