Question

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 $L^{-1}$ is :

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

Answer 1

The correct option is C 690 s

Fall of concentration from 1.28 mg L-1 to 0.04 mg L-1 requires 5 half-lives.

Time required is = $5\times t_{\frac{1}{2}}=5\times 138s=690sec$