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 2.50×10⁻² mol of N2O5(g) in a volume of 1.8 L .

Part A: How many moles of N2O5 will remain after 4.0 min ?

Part B: How many minutes will it take for the quantity of N2O5 to drop to 1.9×10⁻² mol ?

Part C: What is the half-life of N2O5 at 70∘C?

Answer 1

From first order reaction,

ln{[N2O5]t / [N2O5]o} = - K * t --------(1)

[N2O5]o = Initial concConcentra of N2O5

[N2O5]t = Concentration after time t

K = rate constant

Part A :-

t = 4.0 min = 4.0 * 60 s = 240 s ( 1 min = 60 s)

ln{[N2O5]t / [N2O5]o} = - 6.82 * 10^-3 s^-1 * 240 s

ln{[N2O5]t / [N2O5]o} = - 1.6368

[N2O5]t / [N2O5]o = e^-1.6368 = 0.1946

(N2O5)t = 0.1946 * (N2O5)o = 0.1946 * 2.50 * 10^-2 mol = 0.4865 * 10^-2 mol = 4.865 * 10^-3 mol

Number of moles of N2O5 will remain = 4.865 * 10^-3 mol

Part B

From (1)

ln ( 1.9 * 10^-2 mol / 2.50 * 10^-2 mol) = - K * t

ln 0.76 = - K * t

-0.2744 = - K * t

t = 0.2744 / K = 0.2744 / 6.82 * 10^-3 s^-1 = 40.23 s

= 40.23 min / 60 = 0.67 min

Part C

Half life, t1/2 = 0.693 / K = 0.693 / 6.82 * 10^-3 s^-1 = 101.6 s

Half life = 101.6 s