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:

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 C $690 s$
Fall of concentration from $1.28 {mg L}^{- 1}$ to $0.04 {mg L}^{- 1}$ requires $5$ half-lives.
$∴$
Time required $= 5 \times t_{\frac{1}{2}} = 5 \times 138 = 690 s$

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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:

The correct option is C 690 sFall of concentration from 1.28 mg L−1 to 0.04 mg L−1 requires 5 half-lives. ∴ Time required =5×t12=5×138=690 s

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:

The correct option is C $690 s$
Fall of concentration from $1.28 {mg L}^{- 1}$ to $0.04 {mg L}^{- 1}$ requires $5$ half-lives.
$∴$
Time required $= 5 \times t_{\frac{1}{2}} = 5 \times 138 = 690 s$