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Activation energy from the slope of a ln k against 1/T graph

This type of question, asked 2 times: 14 Jun 2022, 28 Jul 2022. All 2 questions below, oldest first.

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  1. Question 1

    14 Jun 2022

    55
    The rate constants for decomposition of acetaldehyde have been measured over the temperature range 00−1000 K00 - 1000\text{ }K. The data has been analysed by plotting ln⁡k\ln k vs 103 T\frac{10^{3}}{\text{ }T} graph. The value of activation energy for the reaction is ____\_\_\_\_ kJ mol−1kJ\ mol^{- 1} (Nearest integer) (Given : R=8.31 J K−1 mol−1R = 8.31\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} )

  2. Question 2

    28 Jul 2022

    148
    For a reaction, given below is the graph of ln⁡k\ln k vs 1T\frac{1}{T}. The activation energy for the reaction is equal to _____ cal mol−1mol^{- 1}. (Nearest integer). (Given: R=2calK−1 mol−1R = 2calK^{- 1}{\text{ }mol}^{- 1} )