# Question 7f455

Feb 21, 2016

$\text{0.81 M}$

#### Explanation:

You're dealing with a first-order reaction that takes the form

$\textcolor{b l u e}{\text{A " -> " products}}$

In such cases, the rate of the reaction depends exclusively and linearly on the concentration of $\text{A}$

color(blue)("rate" = -(d["A"])/dt = k * ["A"])" ", where

$k$ - the rate constant for the reaction at a given temperature

That expression represents the differential rate law, which establishes a relationship between the rate of the reaction and the change in the concentration of the reactant.

In order to determine how the rate of the reaction varies with time, you need to integrate the differential rate law. This will give you the integrated rate law, which takes the form

color(blue)(ln( (["A"])/(["A"_0])) = - k * t)" ", where

$\left[\text{A}\right]$ - the concentration of $\text{A}$ after a period of time $t$
${\left[\text{A}\right]}_{0}$ - the initial concentration of $\text{A}$

Now, you know that your first-order reaction has a rate constant equal to

$k = 1.4 \cdot {10}^{- 4} {\text{s}}^{- 1}$

It's important to notice that the time of reaction is given to you in minutes, which means that you will have to convert it to seconds in order to get the units to match those used by the rate constant.

25 color(red)(cancel(color(black)("min"))) * "60 s"/(1color(red)(cancel(color(black)("min")))) = "1500 s"

Remember that you have

${e}^{\ln \left(x\right)} = x$

which means that you can rewrite the integrated rate law as

${e}^{\ln} \left(\left(\left[{\text{A"])/(["A}}_{0}\right]\right)\right) = {e}^{- k \cdot t}$

This will get you

$\left(\left[{\text{A"])/(["A}}_{0}\right]\right) = {e}^{- k t}$

Rearrange to solve for $\left[\text{A}\right]$

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

Plug in your values to get

["A"] = "1.0 M" * "exp"(-1500color(red)(cancel(color(black)("s"))) * 1.4 * 10^(-4)color(red)(cancel(color(black)("s"^(-1)))))#

$\left[\text{A"] = color(green)("0.81 M}\right)$

The answer is rounded to two sig figs.