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)

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