The half life of a substance in a certain enzyme catalyzed reaction is 138s. The time required for the concentration of the substance to fall from `1.28 mg L^(-1) to 0.04 mg L^(-1)` :

Options

A. 690 s
B. 276 s
C. 414 s
D. 552 s

Answer

A

Explanation

Fall of concentration from 1.28 mg `"L"^(-1)` to the requires 5 half-lives.
`therefore` Time required `=5 xx t_(1//2) = 5 xx 138 = 690` s