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Arrhenius equation at two temperatures: find activation energy, rate constant or temperature

This type of question, asked 5 times: 2021 Paper 21, 27 Jun 2022, 29 Jun 2022, 31 Jan 2023, 2 Apr 2026 S1. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 21

    51
    The first order rate constant for the decomposition of CaCO3CaCO_{3} at 700 K700\text{ }K is 6.36×10−3 s−16.36 \times10^{- 3}{\text{ }s}^{- 1} and activation energy is 209 kJ mol−1209\text{ }kJ{\text{ }mol}^{- 1}. Its rate constant (in s −1\ ^{- 1} ) at 600 K600\text{ }K is x×10−6x \times10^{- 6}. The value of xx is (Nearest integer) [Given R=8.31 J K−1 mol−1;log⁡6.36×10−3=−2.19R = 8.31\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1};\log6.36 \times10^{- 3}= - 2.19, 10−4.79=1.62×10−510^{- 4.79}= 1.62 \times10^{- 5} ]

  2. Question 2

    27 Jun 2022

    148
    It has been found that for a chemical reaction with rise in temperature by 9 K9\ K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K300\ K, the value of activation energy is found to be _____ kJ mol−1kJ\ mol^{-1}. [nearest integer] (Given ln⁡10=2.3\ln 10 = 2.3, R=8.3 J K−1 mol−1R = 8.3\ J\ K^{-1}\ mol^{-1}, log⁡2=0.30\log 2 = 0.30)

  3. Question 3

    29 Jun 2022

    55
    The activation energy of one of the reactions in a biochemical process is 53261 J mol−153261\text{ }J\text{ }mol^{-1}. When the temperature falls from 310 K310\text{ }K to 300 K300\text{ }K, the change in rate constant observed is k300=x×10−3k310k_{300} = x \times 10^{-3}k_{310}. The value of xx is ______ [Given: ln⁡10=2.3\ln10 = 2.3, R=8.3 J K−1 mol−1R = 8.3\text{ }J\text{ }K^{-1}\text{ }mol^{-1}]

  4. Question 4

    31 Jan 2023

    56
    A→BA \rightarrow B The rate constants of the above reaction at 200 K200\text{ }K and 300 K300\text{ }K are 0.03 min⁡−10.03{\text{ }\min}^{- 1} and 0.05 min⁡−10.05{\text{ }\min}^{- 1} respectively. The activation energy for the reaction is ____\_\_\_\_ J mol−1mol^{- 1} (Nearest integer) (Given : ln⁡10=2.3\ln10 = 2.3
    R=8.3 J K−1 mol−1log⁡5=0.70log⁡3=0.48log⁡2=0.30)\begin{matrix} & R = 8.3\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} \\ & \log5 = 0.70 \\ & \log3 = 0.48 \\ & \log2 = 0.30) \end{matrix}

  5. Question 5

    2 Apr 2026 S1

    49
    For reaction A→PA \rightarrow P, rate constant k=1.5×103 s−1k = 1.5 \times 10^{3}{\text{ }s}^{- 1} at 27∘C27^{\circ}C. If activation energy for the above reaction is 60 kJ mol−160\text{ }kJ{\text{ }mol}^{- 1}, then the temperature (in  ∘C\ ^{\circ}C ) at which rate constant, k=4.5×103 s−1k = 4.5 \times 10^{3}{\text{ }s}^{- 1} is ____\_\_\_\_ . (Nearest integer) Given : log⁡2=0.30,log⁡3=0.48,R=8.3 J K−1 mol−1\log2 = 0.30,\log3 = 0.48,R = 8.3\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1}, ln⁡10=2.3\ln10 = 2.3