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NEET Sample Paper NEET Sample Test Paper-37

  • question_answer
    For the reaction, \[{{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}},\]if \[\frac{d\left[ N{{H}_{3}} \right]}{dt}\] \[=2\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}},\]the value of \[\frac{-d[{{H}_{2}}]}{dt}\] would be:

    A) \[1\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]

    B) \[3\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]

    C) \[4\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]

    D) \[6\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]    

    Correct Answer: B

    Solution :

     \[\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}=\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}\] \[\therefore \] \[-\frac{d[{{H}_{2}}]}{dt}=\frac{3}{2}\times \frac{d[N{{H}_{3}}]}{dt}\] \[=\frac{3}{2}\times 2\times {{10}^{-4}}=3\times {{10}^{-4}}\]


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