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NEET Sample Paper NEET Sample Test Paper-14

  • question_answer
    The first order reaction\[2{{N}_{2}}O(g)\to 2{{N}_{2}}(g)+{{O}_{2}}(g)\]has a rate constant of \[1.3\times {{10}^{-11}}{{s}^{-1}}\] at \[207{}^\circ C\] and \[4.5\times {{10}^{-10}}{{s}^{-1}}\]at \[350{}^\circ C\]. What is the activation energy for this reaction?

    A)  15 kJ                          

    B)  30 kJ

    C)  68 kJ                          

    D)  124 kJ

    Correct Answer: D

    Solution :

    \[Lo{{g}_{10}}\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\] \[R=8.314\,J{{K}^{-1}}\text{mo}{{\text{l}}^{-1}}.\] \[{{T}_{1}}=(273+270)K=543\,K.\] \[{{T}_{2}}=(273+350)K=623\,K.\] \[{{k}_{2}}=4.5\times {{10}^{-10}}{{s}^{-1}},{{k}_{1}}=1.3\times {{10}^{-11}}\,{{s}^{-1}}.\] \[{{\log }_{10}}\frac{4.5\times {{10}^{-10}}}{1.3\times {{10}^{-11}}}=\frac{{{E}_{a}}}{2.303\times 8.314}\left( \frac{1}{543}-\frac{1}{623} \right)\] \[1.534=\frac{{{E}_{a}}}{19.1471}\left( \frac{80}{543\times 623} \right)\] \[{{E}_{a}}=\frac{1.534\times 19.1471}{2.364\times {{10}^{-4}}}\] \[=124.2\,kJ\] Hence, the correct option is [d].


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