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Chemistry · Chemical Kinetics

JEE Main 2026 — 24 January, Morning Shift — Question 59

At 27∘C27^{\circ} \mathrm{C} in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ mol−110 \mathrm{~kJ} \mathrm{~mol}^{-1}. The logarithm ratio of k (catalysed) k (uncatalysed) \frac{\mathrm{k} \text { (catalysed) }}{\mathrm{k} \text { (uncatalysed) }} is .... (Consider that the frequency factor for both the reactions is same)

  1. Option A:

    17.41

  2. Option B:

    1.741

    Correct
  3. Option C:

    3.482

  4. Option D:

    0.1741

Answer: B

Step-by-step solution

Kcatalyst Kuncatalyst =eΔEaRT\frac{\mathrm{K}_{\text {catalyst }}}{\mathrm{K}_{\text {uncatalyst }}}=\mathrm{e}^{\frac{\Delta \mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}} ln⁡Kcatalyst Kuncatalyst =ΔEaRT\ln \frac{\mathrm{K}_{\text {catalyst }}}{\mathrm{K}_{\text {uncatalyst }}}=\frac{\Delta \mathrm{E}_{\mathrm{a}}}{\mathrm{RT}} log⁡Kcatalyst Kuncatalyst =ΔEa2.303RT\log \frac{\mathrm{K}_{\text {catalyst }}}{\mathrm{K}_{\text {uncatalyst }}}=\frac{\Delta \mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} =10×10002.303×8.314×300=\frac{10 \times 1000}{2.303 \times 8.314 \times 300} log⁡Kcatalyst Kuncatalyst =1.741\log \frac{\mathrm{K}_{\text {catalyst }}}{\mathrm{K}_{\text {uncatalyst }}}=1.741

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Arrhenius Equation

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