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Activation energy from rate constants at two temperatures, or the reverse

This type of question, asked 6 times: 19 Apr 2025 S2, 20 Apr 2025 S1, 25 Apr 2025 S1, 16 Apr 2026 S1, 19 May 2026 S1, 19 May 2026 S2. All 6 questions below, oldest first.

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  1. Question 1

    19 Apr 2025 S2

    88
    Half-life of a first order reaction is 900 minute at 400 K , find its half-life at 300 K ?
    (Ea2.303R=1.3056×103)\left( \frac{E_{a}}{2.303R}= 1.3056 \times10^{3} \right)

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  2. Question 2

    20 Apr 2025 S1

    54
    The rate constant for a first order reaction is 0.58 s−10.58{\text{ }s}^{- 1} at 300 K and 0.026 s−10.026{\text{ }s}^{- 1} at 290 K . What is the energy of activation? ( R=8.314 J K−1 mol−1R = 8.314\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} )

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  3. Question 3

    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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  4. Question 4

    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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  5. Question 5

    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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  6. Question 6

    19 May 2026 S2

    29
    The rate constants for the first order reaction are x s−1x\ \mathrm{s^{-1}} at 300300 K and y s−1y\ \mathrm{s^{-1}} at 320320 K. Calculate the activation energy of the reaction. [R=8.314 J K−1 mol−1R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} and y/x=4y/x = 4]

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