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
- 414 s
- 552 s
- 690 s
- 276 s
Answer
(3) 690 s
Solution
A constant half-life (independent of concentration) ⇒ first order reaction.
[A]0[A] = 1.280.04 = 32 = 25
⇒ Number of half-lives n = 5
∴ t = 5 138 = 690 s
Answer: (3)