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

Answer

C

Explanation

Amount left after n half lives `=([A]_(0))/(2^(n))`
`0.04 = (1.28)/(2^(n)) or 2^(n) = 32 = 2^(5) therefore n = 5`
Time taken ` = 5 ` half lives `= 5 xx 138 s = 690 s.`