Question

The first-order rate constant for the decomposition of N2O5,

2N2O5(g)→4NO2(g)+O2(g)

at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.30×10⁻² mol of N2O5(g) in a volume of 1.8 L .

a) How many moles of N2O5 will remain after 6.0 min ?

b) How many minutes will it take for the quantity of N2O5 to drop to 1.6×10⁻² mol ?

c) What is the half-life of N2O5 at 70∘C?

Answer 1

a)

we have:

[N2O5]o = 0.023 mol

t = 6.0 min

k = 6.82*10^-3 min-1

use integrated rate law for 1st order reaction

ln[N2O5] = ln[N2O5]o - k*t

ln[N2O5] = ln(2.3*10^-2) - 6.82*10^-3*6

ln[N2O5] = -3.772 - 6.82*10^-3*6

ln[N2O5] = -3.813

[N2O5] = e^(-3.813)

[N2O5] = 2.21*10^-2 mol

Answer: 2.21*10^-2 mol

b)

we have:

[N2O5]o = 2.3*10^-2 mol

[N2O5] = 1.6*10^-2 mol

k = 6.82*10^-3 min-1

use integrated rate law for 1st order reaction

ln[N2O5] = ln[N2O5]o - k*t

ln(1.6*10^-2) = ln(2.3*10^-2) - 6.82*10^-3*t

-4.135 = -3.772 - 6.82*10^-3*t

6.82*10^-3*t = 0.3629

t = 53.2 min

Answer: 53.2 min

c)

Given:

k = 6.82*10^-3 min-1

use relation between rate constant and half life of 1st order reaction

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(6.82*10^-3)

= 1.016*10^2 min

Answer: 102 min