The rate constant for the first-order decomposition of N2O5 (g) to NO2 (g) and O2 (g) is 7.48 * 10-3 s-1 at a given temperature.

a. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.100 atm to rise to 0.145 atm.

b. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.100 atm to rise to 0.200 atm.

c. Find the total pressure after 100 s of reaction.

In this problem, we have a first-order decomposition reaction with a given rate constant. The rate law for a first-order reaction like this is

r

a

t

e

=

k

[

A

]

, where k is the rate constant and [A] is the concentration of the reactant (renamed as A, for brevity). To find the dynamics of the reaction with time, we can integrate the rate law to get an expression for [A] (t):

rate = - d[A]dt = k[A]

[A]f = [A]i e-kt

We want the total pressure of the reaction chamber to be 0.145 atm, with a starting reactant pressure of 0.110 atm. To solve for the time this reaction takes, we need the reaction equation:

2N2O5 (g) → 4NO2 (g) + O2 (g)

Using the stoichiometry of the reaction equation, we can determine the final pressure of the reactant. This requires us to rewrite the total pressure equation in terms of the change in pressure of the reactant.

Pf=0.145atm

Pi=0.110atm = pN2O5

iPf = pN2O5

f + pNO2 + pO2pNO2 = 4pO2

This comes from the stoichiometry.

pNO2 = 2 (pN2O5i - pN2O5f)

This comes from the stoichiometry.

pNO2 = - 2ΔpN2O5Pf = (pN2O5i + ΔpN2O5) - 2ΔpN2O 5 - 12ΔpN2O5

0.145atm = (0.110atm + ΔpN2O5) - 2.5ΔpN2O5 = 0.110atm - 1.5ΔpN2O5

ΔpN2O5 = - 0.0233atm

pN2O5f = 0.110atm - 0.0233atm =

0.0867atm

This is our final pressure! Now we can use the integrated rate law.