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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    Consider the following two reactions, \[A\xrightarrow[{}]{{}}\]Product; \[-\frac{d[A]}{dt}={{k}_{1}}{{[A]}^{0}}\] \[B\xrightarrow[{}]{{}}\]Product; \[-\frac{d[B]}{dt}={{k}_{2}}[B]\] \[{{k}_{1}}\]and\[{{k}_{2}}\]are expressed in terms of molarity (mol\[{{L}^{-1}}\]) and time\[({{s}^{-1}})\]as

    A)  \[{{s}^{-1}},M\,{{s}^{-1}}{{L}^{-1}}\]

    B)  \[M\,{{s}^{-1}}M\,{{s}^{-1}}\]

    C)  \[{{s}^{-1}}{{M}^{-1}}{{s}^{-1}}\]

    D)  \[M\,{{s}^{-1}}{{s}^{-1}}\]

    Correct Answer: D

    Solution :

     Unit of rate constant\[k=\frac{1}{Time}.\frac{1}{{{(conc.)}^{n-1}}}\] For \[{{k}_{1}},n=0,\] Hence, \[k=\frac{1}{t}.\frac{1}{{{(conc.)}^{0-1}}}\] \[=conc.{{t}^{-1}}\equiv M.{{s}^{-1}}\] For \[{{k}_{2}},n=1\] Hence,    \[k=\frac{1}{t}.\frac{1}{{{(conc)}^{-1-1}}}=\frac{1}{t}\equiv {{s}^{-1}}\]


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