2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    For the reaction of\[{{H}_{2}}\]with\[{{I}_{2}},\]the rate constant is \[2.5\times {{10}^{-4}}d{{m}^{3}}mo{{l}^{-1}}{{s}^{-1}}\]at \[327{}^\circ C\] and \[1.0d{{m}^{3}}mo{{l}^{-1}}{{s}^{-1}}\]at \[527{}^\circ C.\]The activation energy for the reaction, in \[kJ\,mo{{l}^{-1}}\]is: \[\left( R=8.314J\text{ }{{K}^{1}}mo{{l}^{1}} \right)\] [JEE Main 10-4-2019 Afternoon]

    A) 72                               

    B) 166

    C) 150                 

    D) 59

    Correct Answer: B

    Solution :

    \[{{H}_{2}}(g)+{{I}_{2}}(g)\to 2HI(g)\] Apply Arrhenius equation \[\log \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left( \frac{1}{600}-\frac{1}{800} \right)\] \[\log \frac{1}{2.5\times {{10}^{-4}}}=\frac{{{E}_{a}}}{2.303\times 8.31}\left( \frac{200}{600\times 800} \right)\] \[\therefore \]\[{{\text{E}}_{\text{a}}}\approx 166\text{kJ/mol}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner