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

Answer

D

Explanation

`1.28 to 0.64 to 0.32 to 0.16 to 0.08 to 0.04`
No. of half lifes (n) = 5
`5=("Total time")/(138)`
time required = `5xx138`
=690s