Question

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

2N2O5(g)?4NO2(g)+O2(g),
at 70?C is 6.82×10?3s?1. Suppose we start with 2.90×10^?2 mol of N2O5(g) in a volume of 1.5 L .

a) How many moles of N2O5 will remain after 5.0 min ?

b) How many minutes will it take for the quantity of N2O5 to drop to 2.0×10^?2 mol?

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

Answer 1

For the given first-order reaction

a) [N2O5]i = 2.90 x 10^-2 mol

[N2O5]t = after 5.0 min

t = 5 min = 30 s

ln[N2O5]t = ln(2.90 x 10^-2) - 6.82 x 10^-3 x 30

[N2O5]t remained after 5 min = 3.75 x 10^-3 mol

b) [N2O5]i = 2.90 x 10^-2 mol

[N2O5]t = 2.0 x 10^-2 mol

So,

t = [ln(2.90 x 10^-2) - ln(2.0 x 10^-2)]/6.82 x 10^-3 = 54.5 s = 0.91 min

time taken = 0.91 min

c) half-life t1/2 = ln2/k

                        = ln(2)/6.82 x 10^-3

                        = 101.63 s