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CET Karnataka Engineering CET - Karnataka 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 \[0.1\text{ }mol\text{ }d{{m}^{-3}}\]. 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)                 \[{{(Rate)}_{2}}=k{{(0.01)}^{n}}=1.6\times {{10}^{-2}}\] ... (ii) Dividing the Eq. (ii)byEq. (i).                 \[=\frac{{{(Rate)}_{2}}}{{{(Rate)}_{1}}}\frac{k{{(0.01)}^{n}}}{k{{(0.1)}^{n}}}\frac{1.6\times {{10}^{-2}}}{2\times {{10}^{-3}}}\]                 \[{{(2)}^{n}}=8\] or  \[{{(2)}^{n}}={{2}^{3}}\] \[\therefore \]                  \[n=3\]


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