JEE Main & Advanced Chemistry Chemical Kinetics Question Bank Rate of a reaction

  • question_answer For the reaction \[{{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}\] if \[\frac{\Delta [N{{H}_{3}}]}{\Delta t}=2\times {{10}^{-4}}mol\,{{l}^{-1}}{{s}^{-1}},\] the value of \[\frac{-\Delta [{{H}_{2}}]}{\Delta t}\] would be                                               [MP PMT 2000]

    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 :

               \[{{N}_{2}}+3{{H}_{2}}\]⇌\[2N{{H}_{3}}\]                    \[\frac{-\Delta [{{N}_{2}}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [{{H}_{2}}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N{{H}_{3}}]}{\Delta t}\]                    \[\therefore \frac{\Delta [{{H}_{2}}]}{\Delta t}=\frac{3}{2}\times \frac{\Delta [N{{H}_{3}}]}{\Delta t}=\frac{3}{2}\times 2\times {{10}^{-4}}\]                                 \[=3\times {{10}^{-4}}mol\,\,litr{{e}^{-1}}se{{c}^{-1}}\]

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