# Question a42c0

Mar 3, 2016

After $22.45 \text{s}$

#### Explanation:

This is a 1st order reaction of the form:

$A \rightarrow \text{products}$

As the reaction is 1st order the following expression applies:

${\left[A\right]}_{t} = {\left[A\right]}_{0} {e}^{- k t}$

${\left[A\right]}_{0}$ is the initial concentration of $A$.

${\left[A\right]}_{t}$ is the concentration of $A$ after time $t$.

$k$ is the rate constant.

Taking natural logs of both sides gives:

$\ln {\left[A\right]}_{t} = \ln {\left[A\right]}_{0} - k t$

[A]_0=100%

[A]_t=11%#

$\therefore \ln 11 = \ln 100 - k t$

$k t = \ln 100 - \ln 11$

$t = \frac{4.6 - 2.4}{0.098}$

$t = 22.45 \text{s}$