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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 24 January · Morning Shift · Chemistry · Chemical Kinetics

At 27^ C in presence of a catalyst, activation energy of a reaction...

At 27∘Cinpresenceofacatalyst,activationenergyofareactionisloweredby27^{\circ} \mathrm{C} in presence of a catalyst, activation energy of a reaction is lowered by10 kJ\mathrm{ kJ}mol−1\mathrm{ mol}^{-1}. The logarithm of ratio of \frac{k (catalysed) \mathrm{k} \text{ (catalysed) }}{k\mathrm{k} (uncatalysed) \text{ (uncatalysed) }} is.... (Consider that the frequency factor for both the reactions is same)

Options

  1. A1.741
  2. B17.41
  3. C3.482
  4. D0.1741

Correct answer

A. 1.741

Step-by-step solution

Using the Arrhenius equation: k=Ae−Ea/RTk = A e^{-E_a/RT}
The ratio of rate constants is: kcatalysedkuncatalysed=Ae−Eac/RTAe−Eau/RT=e(Eau−Eac)/RT=eΔEa/RT\frac{k_{catalysed}}{k_{uncatalysed}} = \frac{A e^{-E_a^c/RT}}{A e^{-E_a^u/RT}} = e^{(E_a^u - E_a^c)/RT} = e^{\Delta E_a/RT}
Since the catalyst lowers activation energy by 10 kJ/mol: ΔEa=10\Delta E_a = 10 kJ/mol = 10000 J/mol
At T = 27°C = 300 K, R = 8.314 J/(mol·K):
kcku=e10000/(8.314×300)=e10000/2494.2=e4.009\frac{k_c}{k_u} = e^{10000/(8.314 \times 300)} = e^{10000/2494.2} = e^{4.009}
Taking logarithm base 10: log⁡10kcku=4.009×log⁡10(e)=4.009×0.4343=1.741\log_{10}\frac{k_c}{k_u} = 4.009 \times \log_{10}(e) = 4.009 \times 0.4343 = 1.741
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