At a particular temperature, the first-order, gas-phase reaction 2N2O5 →2N2O4+ O2 has a half-life for the disappearance of N2O5 of 3,240 s. If 1.00 atmof N2O5 is introduced into an evacuated 5.00 L flask, what will be the total pressure of the gases in the flask after 1.50 hours?

It is perfectly valid to substitute pressure for concentration in this equation since pressure is directly proportional to concentration in moles /L through the ideal gas law:

PV = nRT⇒ p =(n/V) RT

So we have

Pt / P₀ = e ^ -k t

We dont know the rate constant k but we can compute it from the half life as:

t₁/₂ = 3240 s x 1 hr/3600 s = 0.90 hr

k = 0.693 / t₁/₂ = 0.693 /0.90 hr = 0.77 hr⁻¹

Pt/ P₀ = e ^ - (0.77 hr⁻¹ x 1.5 hr) = 0.32

⇒ Pt = 0.32 x 1 atm (P₀ = 1 atm)

this the parcial pressure of N2O5, therefore pressure reacted = 1-0.32 = 0.68 atm

pressure produced N2O4 is .68 atm (1:1 relation)

pressure produced O2 is 0.34 (2:1)

Ptotal = (0.32 + 0.68 + 0.34) atm = 1.34 atm

aaregbe aaregbe · Answer 2 · 2019-12-12T16:16:24+01:00

Answer:

1.343 atm

Explanation:

For the given compound, its half life is 3240 s.

At time t = 1.50 hours = 1.50*3600 s = 5400 s

Therefore, the number of half lives = 5400/3240 = 1.667

The fraction of compound left after 1.667 half lives = 1/(2^1.667) = 1/3.176 = 0.315

In the given problem, it was stated that the reaction commenced at 1 atm. Thus, there is 0.315 atm of the reactant remaining.

The pressure of the reactant consumed 1.00 atm - 0.315 atm = 0.685 atm. By using the coefficients of the reaction, then 0.685 atm of [tex]N_{2}O_{4}[/tex] was produced and 0.685/2 = 0.3425 atm of [tex]O_{2}[/tex] was also produced.

Therefore, the total pressure is the sum of all the partial pressures:

[tex]P_{total} = P_{N_{2}O_{5} } +P_{N_{2}O_{4} } +P_{O_{2} } = 0.315 + 0.685 + 0.343 = 1.343[/tex] atm