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Ratio of rate constants for two activation energies with the same pre-exponential factor

This type of question, asked 3 times: 26 Jun 2022, 21 Jan 2026 S1, 24 Jan 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    26 Jun 2022

    148
    Catalyst AA reduces the activation energy for a reaction by 10 kJ mol−110\text{ }kJ{\text{ }mol}^{- 1} at 300 K300\text{ }K. The ratio of rate constants, kT,CatalysedkT,Uncatalysed\dfrac{k_{T,\text{Catalysed}}}{k_{T,\text{Uncatalysed}}} is exe^{x}. The value of xx is [nearest integer] [Assume that the pre-exponential factor is same in both the cases. Given R=8.31 J K−1 mol−1R = 8.31\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} ]

  2. Question 2

    21 Jan 2026 S1

    46
    Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol−120\text{ }kJ{\text{ }mol}^{- 1}. If k1k_{1} and k2k_{2} are the rate constants of first and second reaction respectively at 300 K , then In k2k1\frac{k_{2}}{k_{1}} will be (nearest integer) [R=8.3 J K−1 mol−1]\left\lbrack R = 8.3\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} \right\rbrack

  3. Question 3

    24 Jan 2026 S1

    41
    At 27∘C27^{\circ}C in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ mol−110\text{ }kJ{\text{ }mol}^{- 1}. The logarithm ratio of k (catalysed) k (uncatalysed) \frac{k\text{~(catalysed)~}}{k\text{~(uncatalysed)~}} is .... (Consider that the frequency factor for both the reactions is same)

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