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NEET AIPMT SOLVED PAPER MAINS 2011

  • question_answer
    The half-life of a substance in a certain enzyme-catalyses reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg \[{{\text{L}}^{\text{-1}}}\] to 0.04 mg \[{{\text{L}}^{\text{-1}}}\] is                    

    A)  414 s                    

    B)  552 s    

    C)  690 s                    

    D)  276 s

    Correct Answer: C

    Solution :

    Enzyme-catalysed reactions follow first order kinetics. \[\because \]Fall of concentration from 1.28 mg \[{{\text{L}}^{-1}}\]to \[\text{0}\text{.04}\,\text{mg}\,{{\text{L}}^{-1}}\] involves five half-lives. \[1.28\,\xrightarrow[{}]{{{T}_{1/2}}}0.64\xrightarrow[{}]{{{t}_{1/2}}}0.32\xrightarrow[{}]{{{t}_{1/2}}}0.16\] \[\xrightarrow[{}]{{{t}_{1/2}}}0.08\xrightarrow[{}]{{{t}_{1/2}}}0.04\] \[\therefore \]Time required \[=5\times {{t}_{1/2}}\]                                 \[=5\times 138\,s\] = 690 s.


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