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
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
