Question #5c648

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, #"N"_2"O"_5#, will depend exclusively on the concentration of the reactant.

The [rate law](https://socratic.org/chemistry/chemical-kinetics/rate-law#), which allows you to express the rate of the reaction in terms of a *reaction constant*, #k#, and the concentration of the reactant, will look like this

#"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

#["A"]# - the concentration of the reactant after the passing of a time #t#
#["A"_0]# - the initial concentration of the reactant

For a first-order reaction, the half-life, #t_"1/2"#, which expresses the amount of time needed in order to have #["A"] = 1/2 * ["A"_0]#, is equal to

#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 #t_1 = "10 s"# and at #t_2 = "10 min" = "600 s"#, use the exponential form of the rate law

#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.