Question

The half-life period 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 mol^–1 to 0.04 mg L^–1 is

Options

• 4145

• 5528

• 6905

• 2765

Answer 1

6905

Explanation:

Amount left after n half-life period ${\displaystyle {\left[A\right]}_0=\frac{{\left[A\right]}_0}{2^n}}$

= ${\displaystyle \frac{{\left[A\right]}_0}{A}=1.28 \text{mg}\frac{1.28 {\text{mg L}}^{-1}}{0.04 {\text{mg L}}^{-1}}}$ 

[2]ⁿ = 82 = [2]⁵

No. of tY2 = peroids = 5

Time required = 138 x 5 = 6905.