Question

Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitrogen pentoxide and has a half-life of 2.81 h at 25 ∘C.

If a 1.5-L reaction vessel initially contains 760 torr of N2O5 at 25 ∘C, what partial pressure of O2 is present in the vessel after 225 minutes?

Answer 1

229 torr

Explanation

Initial moles of N2O5 is obtained by ideal gas equation

PV = nRT

number of moles, n = PV/RT

n = 1atm × 1.5L/0.082057(L atm/mol K) ×298.15K

n = 0.06131mol

2N2O5(g) <-------> 4NO2(g) + O2(g)

For first order reaction

rate constant , k = 0.693/t_1/2

k = 0.693/2.81hr = 0.2466hr^-1

Integrated form of first order equation is

2.303log([A]₀/[A]_t) = kt

[A]₀ = initial amount , 0.06131mol

[A]_t = amount at time t , ?

2.303log(0.06131mol/[A]_t) = 0.2466hr^-1 × 3.75 hr

2.303log(0.06131mol/[A]_t) = 0.92475

log(0.06131mol/[A]_t) = 0.40154

0.06131mol/[A]_t = 2.5208

[A]_t = 0.02432mol

moles of N2O5 decomposed = 0.06131mol - 0.02432mol = 0.03699 mol

stoichiometrically, 1moles of O2 formed from 2moles of N2O5

moles of O2 formed = 0.03699mol/2 = 0.018495mol

Applying ideal gas equation

P = nRT/V

Partial pressure of O2 = (0.018495mol× 0.082057(L atm/mol K) × 298.15K / 1.5L

= 0.3017atm

= 229 torr