# Question #7f455

##### 1 Answer

#### Explanation:

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

#color(blue)("A " -> " products")#

In such cases, the **rate of the reaction** depends *exclusively* and *linearly* on the concentration of

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

*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

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

#k = 1.4 * 10^(-4)"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(x)) = x#

which means that you can rewrite the integrated rate law as

#e^ln( (["A"])/(["A"_0])) = e^(-k * t)#

This will get you

#(["A"])/(["A"_0]) = e^(-kt)#

Rearrange to solve for

#["A"] = ["A"_0] * e^(-kt)#

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)))))#

#["A"] = color(green)("0.81 M")#

The answer is rounded to two **sig figs**.