JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • question_answer
    For a given reaction A \[\to \] Product, rate is \[1\times {{10}^{-4}}M{{s}^{-1}}\]when \[[A]=0.01\] M and rate is \[1.41\times {{10}^{-4}}M{{s}^{-1}}\]when \[[A]=0.02\] M. Hence, rate law is:

    A)  \[-\frac{d[A]}{dt}=\frac{k}{4}[A]\]        

    B)  \[-\frac{d[A]}{dt}=k{{[A]}^{1/2}}\]

    C)  \[-\frac{d[A]}{dt}=\frac{k}{4}[A]\]        

    D)  \[-\frac{d[A]}{dt}=k{{[A]}^{1/2}}\]

    Correct Answer: D

    Solution :

    \[A\xrightarrow{\,}\] Product  We know.       Rate \[=K{{[conc.]}^{n}}\] \[1\times {{10}^{-4}}\,=K{{[.01]}^{n}}\]                    ?(1) \[1.41\times {{10}^{-4}}\,=K{{[.02]}^{n}}\]                ?(2) (i)(ii) \[\frac{1}{1.41}={{\left( \frac{1}{2} \right)}^{n}}\] \[n=\frac{1}{2}\] \[\frac{-d(A)}{dt}=K{{[A]}^{1/2}}\]


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