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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2009

  • question_answer
    For a chemical reaction \[A\to B\], the rate of the reaction is\[2\times {{10}^{-3}}\text{ }mol\text{ }d{{m}^{-3}}\text{ }{{s}^{-1}}\], when the initial concentration is \[0.05\text{ }mol\text{ }d{{m}^{-3}}\]. The rate of the same reaction is \[1.6\times {{10}^{-2}}\text{ }mol\text{ }d{{m}^{-3}}\text{ }{{s}^{-1}}\] 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) \[{{(Rate)}_{2}}=k{{(0.1)}^{n}}=1.6\times {{10}^{-2}}\]     ??(ii) Dividing the Eq. (ii) by Eq. (i). \[\frac{{{(Rate)}_{2}}}{{{(Rate)}_{1}}}=\frac{k{{(0.1)}^{n}}}{k{{(0.05)}^{n}}}=\frac{1.6\times {{10}^{-2}}}{2\times {{10}^{-2}}}\] \[{{(2)}^{n}}=8\]   or   \[{{(2)}^{n}}={{2}^{3}}\] \[\therefore \] \[n=3\]


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