CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    For the decomposition of a compound AB at 600 K, the following data were obtained
    [AB] mol \[d{{m}^{-3}}\] Rate of decomposition of AB in mol \[d{{m}^{-3}}{{s}^{-1}}\]
    0.20 \[2.75\times {{10}^{-8}}\]
    0.40 \[11.0\times {{10}^{-8}}\]
    0.60 \[24.75\times {{10}^{-8}}\]
                    The order for the decomposition of AB is

    A)  1.5                                        

    B)  0

    C)  1                                            

    D)  2

    Correct Answer: D

    Solution :

    \[AB\xrightarrow{{}}\] Product                                 Rate \[=k{{[AB]}^{n}}\]                 \[{{(Rate)}_{1}}=k{{[0.20]}^{n}}=2.75\times {{10}^{-8}}\]              ... (i)                 \[{{(Rate)}_{2}}=k{{[0.40]}^{n}}=11.0\times {{10}^{-8}}\]              ... (ii) Dividing Eq. (ii) byEq. (i),                 \[\frac{{{(Rate)}_{2}}}{{{(Rate)}_{1}}}=\frac{k{{[0.40]}^{n}}}{k{{[0.20]}^{n}}}=\frac{11.0\times {{10}^{-8}}}{2.75\times {{10}^{-8}}}\]                 \[{{2}^{n}}=4\] or            \[{{2}^{n}}={{2}^{2}}\] Hence,  \[n=2\]


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