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Chemical Kinetics question

2011 · Q63
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Chemical Kinetics question

2011 · Q63

NEETChemistryChemical KineticsMCQ+4 / −1
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
  1. A
    414 s
  2. B
    552 s
  3. C
    690 s
  4. D
    276 s
View written solutionFree

Correct answer: C

We know, N=N0(12)nN = {N_0}{\left( {{1 \over 2}} \right)^n}N=N0​(21​)n

Give N0(original amount) = 1.28 mg L–1

N (amount of substance left after time T) = 0.04 mg L–1

∴\therefore∴ 0.041.28=(12)n{{0.04} \over {1.28}} = {\left( {{1 \over 2}} \right)^n}1.280.04​=(21​)n

⇒\Rightarrow⇒ 132=(12)n{1 \over {32}} = {\left( {{1 \over 2}} \right)^n}321​=(21​)n

⇒\Rightarrow⇒ n = 5

∴\therefore∴ Time required = 5 × t1/2 = 5 × 138 = 690 s

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