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The half life of a substance in a certain enzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg $L^{-1}$ to 0.04 mg $L^{-1}$ is
A
690 s
B
276 s
C
414 s
D
552 s
Detailed Solution
The reaction is treated as first order, for which the half-life is constant.
$\frac{\text{final concentration}}{\text{initial concentration}} = \frac{0.04}{1.28} = \frac{1}{32} = \left(\frac{1}{2}\right)^5$
Step by step: 1.28 → 0.64 → 0.32 → 0.16 → 0.08 → 0.04, which is five half-lives.
Time required = $5\times t_{1/2} = 5\times138$
= 690 s
$\frac{\text{final concentration}}{\text{initial concentration}} = \frac{0.04}{1.28} = \frac{1}{32} = \left(\frac{1}{2}\right)^5$
Step by step: 1.28 → 0.64 → 0.32 → 0.16 → 0.08 → 0.04, which is five half-lives.
Time required = $5\times t_{1/2} = 5\times138$
= 690 s