Question

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?

Answer 1

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 $N_{2}O_{4}$ was produced and 0.685/2 = 0.3425 atm of $O_{2}$ was also produced.

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

$P_{total} = P_{N_{2}O_{5} } +P_{N_{2}O_{4} } +P_{O_{2} } = 0.315 + 0.685 + 0.343 = 1.343$ atm

Answer 2

Answer:

1.34 atm

Explanation:

For a first order reaction:

(At)/(Ao) = e ^ -kt    where  (At)and (A₀)  are final and initial molarities  respectively,  k is the rate constant and t is the time.

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