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 $1.28 m g L^{- 1}$ to $0.04 m g L^{- 1}$ is:

414 s

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552 s

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690 s

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276 s

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Solution

The correct option is A $690 s$
Enzyme catalysed reactions are initially follow first order kinetics when concentration decreases $1.28 m g L^{- 1}$ to $0.04 m g L^{- 1}$ .

$\frac{1.28 m g L^{- 1}}{0.04 m g L^{- 1}} = 32 = 2^{5}$ .

Then five half life completed.
Number of half lives $= 5$

So, times required $= 5 \times 138 = 690 s$

The time required for the concentration of the substance to fall from $1.28 m g L^{- 1}$ to $0.04 m g L^{- 1}$ , is 690s.

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The half-life of a substance in a certain enzyme-catalysed reaction is 138s. The time required for the concentration of the substance to fall from 1.28 mgL−1 to 0.04 mgL−1 is:

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 $1.28 m g L^{- 1}$ to $0.04 m g L^{- 1}$ is: