What is the time required for 99% completion of a first order reaction if rate constant is 23.03 min − 1 ?

Question

TestHub · Accepted Answer

For a first order reaction, the time required is given by the integrated rate law:

$t = \frac{2.303}{k} lo g_{10} (\frac{[ A ] _{0}}{[ A ] _{t}})$

For $99%$ completion, let $[A]_{0} = 100$ . Then $[A]_{t} = 100 - 99 = 1$ .

Given $k = 23.03 min^{- 1}$ .

Substituting the values:

$t = \frac{2.303}{23.03} lo g_{10} (\frac{100}{1})$

$t = 0.1 \times lo g_{10} (100)$

$t = 0.1 \times 2 = 0.2 min$

Answer: 0.2 minute

Chemistry - Chemical Kinetics Question with Solution | TestHub

Options:

Doubts & Discussion