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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    For chemical reaction \[{{N}_{2}}+3{{H}_{2}}\xrightarrow[{}]{{}}2N{{H}_{3}}\] \[\frac{d[N{{H}_{3}}]}{dt}=2\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\] The value of\[-\frac{d[{{H}_{2}}]}{dt}\]will

    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 :

     \[{{N}_{2}}+3{{H}_{2}}\xrightarrow[{}]{{}}2N{{H}_{3}}\] \[\frac{d[N{{H}_{3}}]}{dt}=2\times {{10}^{-4}}mol/L-s\] \[-\frac{d[{{N}_{2}}]}{dt}=-\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}=\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\] \[-\frac{d[{{H}_{2}}]}{dt}=\frac{3}{2}\frac{d[N{{H}_{3}}]}{dt}=\frac{3}{2}\times 2\times {{10}^{-4}}\] \[=3\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]


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