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JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    Consider the following two reactions, \[A\xrightarrow{{}}\] Product \[-\frac{d[A]}{dt}={{k}_{1}}{{[A]}^{o}}\] \[B\xrightarrow{{}}\] 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   AIEEE  Solved  Paper-2002

    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 \[k=\frac{1}{\text{Time}}.\frac{1}{{{(\operatorname{co}nc)}^{n-1}}}\] For \[{{k}_{1}}n=0k=\frac{1}{t}.\frac{1}{{{(\operatorname{co}nc)}^{0-1}}}\] hence cone. \[{{t}^{-1}}\equiv M.\,{{\sec }^{-1}}\]    \[{{k}_{2}}\,\,n=1\,\,k=\frac{1}{t}.\frac{1}{{{(\operatorname{c}onc.)}^{1-1}}}=\frac{1}{t}={{s}^{-1}}\]


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