CET Karnataka Medical CET - Karnataka Medical Solved Paper-2003

  • question_answer
    For a reaction \[A+B\xrightarrow{{}}C+D,\], if concentration of A is doubled without altering that of B, rate doubles. If concentration of B is increased nine times without altering that of A, rate triples. Order of the reaction is:

    A)  \[2\]                

    B)  \[1\]

    C)  \[1\frac{1}{2}\]             

    D)  \[1\frac{1}{3}\]

    Correct Answer: C

    Solution :

    Suppose, the rate of reaction depends upon x power of concentration of A and y power of concentration of B. Hence, \[{{r}_{1}}=k.{{[A]}^{x}}\] \[{{r}_{2}}=k{{[2A]}^{x}}\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{[A]}^{x}}}{{{[2A]}^{x}}}=\frac{1}{2}\] \[\therefore \] \[x=1\] Similarly \[{{r}_{1}}=k{{[B]}^{y}}\] \[{{r}_{2}}=k{{[9B]}^{y}}\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{[B]}^{y}}}{{{[9B]}^{y}}}=\frac{1}{3}\] \[\therefore \] \[{{\left( \frac{1}{3} \right)}^{2y}}=\frac{1}{3}\] or \[y=\frac{1}{2}\] Hence, correct rate-law for the reaction \[=k[A]{{[B]}^{1/2}}\] Hence, order of reaction \[=1+\frac{1}{2}=1\frac{1}{2}\]


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