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Activation energy from a given Arrhenius-form equation for k

This type of question, asked 5 times: 2021 Paper 23, 27 Jun 2022, 29 Jun 2022, 22 Jan 2026 S2, 6 Apr 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 23

    57
    For the reaction A→BA \rightarrow B, the rate constant k(k\left( \right. in s   −1)\left. \ \ ^{- 1} \right) is given by
    log⁡10k=20.35−(2.47×103)T\log_{10}k = 20.35 -\frac{\left( 2.47 \times10^{3} \right)}{T}
    The energy of activation in kJmol−1kJmol^{- 1} is (Nearest integer) [Given: R=8.314 J K−1 mol−1R = 8.314\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} ]

  2. Question 2

    27 Jun 2022

    56
    The rate constant for a first order reaction is given by the following equation:
    ln⁡k=33.24−2.0×104 K T\ln k = 33.24 -\frac{2.0 \times10^{4}\text{ }K}{\text{ }T}
    The Activation energy for the reaction is ____\_\_\_\_ kJ mol−1kJ\ mol^{- 1}. (In Nearest integer) (Given: R=8.3 J K−1 mol−1R = 8.3\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} )

  3. Question 3

    29 Jun 2022

    145
    The equation
    k=(6.5×1012 s−1)e−26000K/Tk =\left( 6.5 \times10^{12}{\text{ }s}^{- 1} \right)e^{- 26000K/T}
    is followed for the decomposition of compound A. The activation energy for the reaction is ____\_\_\_\_ kJ mol−1kJ\ mol^{- 1}. [nearest integer] (Given: R=8.314 J K−1 mol−1R = 8.314\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} )

  4. Question 4

    22 Jan 2026 S2

    50
    Consider A→k1  BA\overset{k_{1}\ }{\rightarrow}\text{ }B and C→k2 DC\overset{k_{2}\ }{\rightarrow}D are two reactions. If the rate constant (k1)\left( k_{1} \right) of the A→BA \rightarrow B reaction can be expressed by the following equation log⁡10k=14.34−1.5×104 T/K\log_{10}k = 14.34 - \frac{1.5 \times 10^{4}}{\text{ }T/K} and activation energy of C→DC \rightarrow D reaction (Ea2)\left( {Ea}_{2} \right) is 15\frac{1}{5} th of the A→BA \rightarrow B reaction (Ea1)\left( {Ea}_{1} \right), then the value of (Ea2)\left( {Ea}_{2} \right) is ____\_\_\_\_ kJmol−1kJ{mol}^{- 1}. (Nearest Integer)

  5. Question 5

    6 Apr 2026 S2

    50
    Decomposition of a hydrocarbon follows the equation k=(5.5×1011 s−1)e−28000 K Tk = \left( 5.5 \times 10^{11}{\text{ }s}^{- 1} \right)e^{\frac{- 28000\text{ }K}{\text{ }T}}. The activation energy of reaction is ____\_\_\_\_ kJmol−1kJ{mol}^{- 1}. (Nearest Integer) Given : R=8.3JK−1 mol−1R = 8.3{JK}^{- 1}{\text{ }mol}^{- 1}