# Question #5c648

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

As you know, the rate of a **first-order reaction** depends linearly on the **concentration** or **partial pressure** of a single reactant.

In your case, the decomposition of *dinitrogen pentoxide*,

The [rate law](https://socratic.org/chemistry/chemical-kinetics/rate-law

#"rate" = k * ["N"_2"O"_5]#

The *integrated rate law* for a **first-order reaction** looks like this

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

For a **first-order reaction**, the half-life,

#color(blue)(t_"1/2" = ln(2)/k)#

In your case, the half-life of the reaction will be equal to

#t_"1/2" = ln(2) * 1/(3.38 * 10^(-5)"s"^(-1)) = "20,507.3 s"#

Rounded to three sig figs, the number of sig figs you have for the rate constant, the answer will be

#t_"1/2" = color(green)(2.05 * 10^4"s")#

Now, you can relate the concentration of the *reactant* and its *partial pressure* by using the ideal gas law equation and the **assumption** that the volume of the reaction vessel **remains constant** and that the reaction takes place at constant temperature.

As you know, you have

#color(blue)(PV = nRT)#

Rearrange to get

#P = n * overbrace((RT)/V)^(color(purple)("constant")) <=> P prop n#

This means that you can say

#ln ( (["N"_2"O"_5])/(["N"_2"O"_5]_0)) = ln( n/color(red)(cancel(color(black)(V))) * (color(red)(cancel(color(black)(V))))/n_0) = ln(n/n_0) = ln( P/color(red)(cancel(color(black)("constant"))) * color(red)(cancel(color(black)("constant")))/P_0) = ln(P/P_0)#

This is of course equal to

#ln(P/P_0) = - k * t#

To find the pressure of the gas at

#e^ln(P/P_0) = e^(-kt)#

#P/P_0 = e^(-kt) implies color(blue)(P = P_0 * e^(-kt))#

Finally, plug in your values to get

#P_1 = "500.0 torr" * e^(-3.38 * 10^(-5) color(red)(cancel(color(black)("s"^(-1)))) * 10color(red)(cancel(color(black)("s"))))#

#P_1 = color(green)("499.8 torr")#

and

#P_2 = "500.0 torr" * e^(-3.38 * 10^(-5) color(red)(cancel(color(black)("s"^(-1)))) * 600color(red)(cancel(color(black)("s"))))#

#P_2 = color(green)("490.0 torr")#

I'll leave the answers rounded to four sig figs, just for good measure.