# Question #f0861

Jan 16, 2017

The rate constant $k$ is $3.39 \ast {10}^{-} 4$ ${s}^{-} 1$.

#### Explanation:

We are given that the reaction is a first order reaction, meaning that it can be expressed through the first order reaction integrated rate law:

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

Let us assume a sample size of 1 M. Therefore, ${\left[A\right]}_{0}$ = 1

40%, or .4 of the sample remains after 45 minutes. Therefore,
.4 = $\ln {\left[A\right]}_{t}$

While we are given minutes, I prefer to use seconds for rate law calculations.
$45$ m x $\frac{60 s}{1 m}$ = 2700 seconds = $t$

Now, we can plug in the numbers to get this equation:

$\ln \left[.4\right]$ = -$k \ast 2700$ + $\ln \left[1\right]$

Solving for $k$ in a calculator gives a value of $3.39 \ast {10}^{-} 4$. Since this is a first order reaction, the units are ${s}^{-} 1$.