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Activation energy when the rate constant doubles for a given rise in temperature

The same question, asked 3 times: 25 Apr 2025 S1*, 16 Apr 2026 S1*, 19 May 2026 S1. Every wording below, oldest first.

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  1. Question 1

    25 Apr 2025 S1

    91
    The rate constant is doubled when temperature increases from 27∘C27^{\circ}C to 37∘C37^{\circ}C. What is activation energy in kJ ?

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  2. Question 2

    16 Apr 2026 S1

    30
    The rate constant of reaction is 1.5×107 s−11.5 \times 10^7\ \mathrm{s^{-1}} at 300 K and 3.0×107 s−13.0 \times 10^7\ \mathrm{s^{-1}} at 330 K. What is the activation energy for the reaction? [ R×2.303=19.15 JK−1mol−1R \times 2.303 = 19.15\ \mathrm{JK^{-1}mol^{-1}}]

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  3. Question 3

    19 May 2026 S1

    32
    Calculate the activation energy when the rate of chemical reaction doubles with an increase of 1010 K in absolute temperature from 298298 K. (R×2.303=19.15 J K−1 mol−1)(R \times 2.303 = 19.15\ \mathrm{J\ K^{-1}\ mol^{-1}}).

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