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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    For a reaction, consider the plot of In k versus 1/r given in the figure. If the rate constant of this reaction at 400 K is \[{{10}^{-5}}{{s}^{-1}},\]then the rate constant at 500 K [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[{{10}^{-4}}{{s}^{-1}}\]                 

    B) \[4\times {{10}^{-4}}{{s}^{-1}}\]

    C) \[{{10}^{-6}}{{s}^{-1}}\]                 

    D)   \[2\times {{10}^{-4}}{{s}^{-1}}\]

    Correct Answer: A

    Solution :

    From Arrhenius equation, In\[k=\ln \,A-\frac{{{E}_{a}}}{RT}\] Slope \[=\frac{-{{E}_{a}}}{R}=-4606K\] or,\[\ln =\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left( \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right)\] \[\ln =\frac{{{k}_{2}}}{{{10}^{-5}}}=4606\left( \frac{500-400}{500\times 400} \right)\] \[\ln \frac{{{k}_{2}}}{{{10}^{-5}}}=2.303;\frac{{{k}_{2}}}{{{10}^{-5}}}=anti\ln (2.303)\] \[{{k}_{2}}=1\times {{10}^{-4}}{{s}^{-1}}\]


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