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CMC Medical CMC-Medical Ludhiana Solved Paper-2011

  • question_answer
    For a chemical reaction \[A\to B,\]the rate of the reaction is\[2\times {{10}^{-3}}mol\,d{{m}^{-3}}{{s}^{-1}},\]when the initial concentration is \[0.05\,mol\,d{{m}^{-3}}\]. The rate of the same reaction is\[1.6\times {{10}^{-2}}\,mol\,d{{m}^{-3}}\,{{s}^{-1}}\] when the initial concentration is \[0.1\,mol\,d{{m}^{-3}}.\] The order of the reaction is

    A)  2                                            

    B)  0

    C)  3                                            

    D)  1

    Correct Answer: C

    Solution :

                    \[A\xrightarrow{{}}B\] \[\text{Rate}=k\,{{[A]}^{n}}\] \[{{\text{(Rate)}}_{1}}=k{{(0.05)}^{n}}=2\times {{10}^{-3}}\]       ?(i) \[{{\text{(Rate)}}_{2}}=k{{(0.01)}^{n}}=1.6\times {{10}^{-2}}\]   ?(ii) Dividing the Eq. (ii) by Eq. (i). \[\frac{{{(Rate)}_{2}}}{{{(Rate)}_{1}}}=\frac{k{{(0.1)}^{n}}}{k{{(0.05)}^{n}}}=\frac{1.6\times {{10}^{-2}}}{2\times {{10}^{-3}}}\] \[{{(2)}^{n}}=8\]               or            \[{{(2)}^{n}}={{2}^{3}}\] \[\therefore \]  \[n=3\]


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