A bacterial growth follows first order kinetics and it is observed that increase in bacteria is by 10% in 10 min . How long will it take to increase bacteria concentration to double of original.

Question

TestHub · Accepted Answer

Concepts: First-order kinetics, rate constant, integrated rate law, doubling time.

For first-order kinetics, $k = \frac{1}{t} ln \frac{[ A ] _{0}}{[ A ] _{t}}$ .

Given 10% increase in 10 min, so $[A]_{t} = 1.1 [A]_{0}$ .

$k = \frac{1}{10} ln \frac{[ A ] _{0}}{1.1 [ A ] _{0}} = \frac{1}{10} ln (1.1) \approx 0.00953 mi n^{- 1}$ .

For doubling, $[A]_{t} = 2 [A]_{0}$ .

$t = \frac{1}{k} ln \frac{[ A ] _{0}}{2 [ A ] _{0}} = \frac{1}{0.00953} ln (2) \approx 72.9 min$ .

Chemistry - Chemical Kinetics Question with Solution | TestHub

Options:

Doubts & Discussion