Q45 / 120 · NEET 2011 (Mains) · Chemistry

The half-life of a substance in a certain enzyme-catalysed reaction is $138 \mathrm{~s}$. The time required for the concentration of the substance to fall from $1.28 \mathrm{mg} \mathrm{L}^{-1}$ to $0.04 \mathrm{mg} \mathrm{L}^{-1}$ is

  • A
    $414 \mathrm{~s}$
  • B
    $552 \mathrm{~s}$
  • Correct answer
    C
    $690 \mathrm{~s}$
  • D
    $276 \mathrm{~s}$

Solution

Enzyme-catalysed reactions follow first order kinetics. $$ \begin{aligned} & 1.28 \stackrel{t_{1 / 2}}{\longrightarrow} 0.64 \stackrel{t_{1 / 2}}{\longrightarrow} 0.32 \stackrel{t_{1 / 2}}{\longrightarrow} 0.16 \\ & \stackrel{t_{1 / 2}}{\longrightarrow} 0.08 \stackrel{t_{1 / 2}}{\longrightarrow} 0.04 \end{aligned} $$ $$ \begin{aligned} \therefore \text { Time required } & =5 \times t_{1 / 2} \\ & =5 \times 138 \mathrm{~s} \\ & =690 \mathrm{~s} . \end{aligned} $$

Hence, the correct answer is option: 'C'.

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