Question

If a 1.4-L reaction vessel initially contains 760 torr of N2O5 at 25 ∘C, what partial pressure of O2 is present in the vessel after 220 minutes?

Answer 1

2N2O5(g) àà 4NO2(g) + O2(g)

d[N2O5]/dt = - k [N2O5] is the first order rate reaction, where [N2O5] = concentration of N2O5 in moles/L at time t, and k = first order rate constant.

integrating this you get:

INT(d[N2O5] / [N2O5]) = - INT (k dt), where INT means integral of

ln [N2O5] = - k t + C where C = constant

when t = 0, [N2O5] = [N2O5]₀, where [N2O5]₀ = initial concentration of N2O5

ln [N2O5] = - k t + ln [N2O5]₀

[N2O5] = [N2O5]₀ × e^{-k t}

T1/2 = 2.81 hr. where T1/2 = half-life
So [N2O5] / [N2O5]₀ = 1/2 = e^{-k 2.81}
k = 0.2467 hr^-1

Therefore the rate equation is:

[N2O5] = [N2O5]₀ × e^-0.2467t

Now we can use equation PV=nRT

Initially; N2O5: (760 torr) × (1.4 L) = n × (62.36 L - torr K-¹mole^-1) × (25 +273 K)

By solving this equation we get; n = 0.05725 moles of N2O5 initially

This means, [N2O5]₀ = 0.05725 moles / 1.4 L = 0.04089 moles / L
Therefore [N2O5] = 0.04089 × e^-0.2467t is the complete rate equation

At 220 minutes = 3.666 hr., [N2O5] = 0.04089 × e^{-(0.2467*3.666)} = 0.01654 moles / L

This corresponds to (0.01654 moles / L) × (1.4 L) = 0.02316 moles of N2O5

Since 2N2O5(g) àà4NO2(g) + O2(g), and we started with 0.04089 moles of N2O5,

that means at 220 minutes we have 0.02316 moles of N2O5,

(0.04089 - 0.02316)(4/2) = 0.03546 moles of NO2, and

(0.04089 - 0.02316)(1/2) = 0.00886 moles of O2

Substituting back to determine partial pressure
P=nRT/V = 0.00886 × 62.36 L - torr K-¹mole^-1 × 298 K /1.4 L = 0.302 atm = 117 tor

Partial pressure of O2 is 117 tor