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Manipal Engineering Manipal Engineering Solved Paper-2009

  • 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 :

    \[A\xrightarrow{{}}B\] Rate       \[=k{{[A]}^{n}}\] \[{{(Rate)}_{1}}=k{{(0.05)}^{n}}=2\times {{10}^{-3}}\]                    ... (i) \[{{(Rate)}_{2}}=k{{(0.1)}^{n}}=1.6\times {{10}^{-2}}\]                   ... (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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