Question

The first-order rate constant for the decomposition of N2O5, given below, at 70°C is 6.82 10-3 s-1. Suppose we start with 0.0550 mol of N2O5(g) in a volume of 3.5 L. 2 N2O5(g) → 4 NO2(g) + O2(g) (a) How many moles of N2O5 will remain after 3.0 min? mol (b) How many minutes will it take for the quantity of N2O5 to drop to 0.005 mol? min (c) What is the half-life of N2O5 at 70°C? min

Answer 1

Integrated first rate law for this first-order reaction is
ln[N₂O₅] = -k∙t + ln[N₂O₅]₀

Initial concentration is
[N₂O₅]₀ = 0.0550mol / 3.5L = 0.0157 M

Solve for integrated rate law for [A]:
[N₂O₅] = e^(-k∙t + ln[N₂O₅]₀) = e^(ln[N₂O₅]₀)∙e^(-k∙t) = [N₂O₅]₀∙e^(-k∙t)
=>
[N₂O₅] = 0.0157M∙e^( - 6.82×10⁻³s⁻¹ ∙ 3∙60s)
= 0.0157M∙e^( - 1.2276)
= 0.0046
=>
n(N₂O₅) = 0.0088M ∙ 3.5L = 0.0161mol


b)
Solve this integrated rate law t and you get the time elapsed until concentration has dropped to a specified level:
t = (ln[N₂O₅]₀ - ln[N₂O₅])/k = ln( [N₂O₅]₀/[N₂O₅] )/k

[N₂O₅] = 0.005mol/3.5L = 0.001428M
=>
t = ln( 0.0157M/0.001428M) / 6.82×10⁻²s⁻¹ = 351s


c)
For a first order reaction:
t½ = ln(2)/k
=>
t½ = ln(2)/6.82×10⁻³s⁻¹ = 101.6s