Finding mole fraction in gaseous mixture

Question

For the decomposition of gaseous dinitrogen pentoxide, 2N₂O₅(g) → 4NO₂(g) + O₂(g) the rate constant is k = 2.8 × 10⁻³ s⁻¹ at 60°C. The initial concentration of N₂O₅ is 2.95 mol/L in a closed container at 60°C. After 3.75 min of reaction, what will be the mole fraction of NO₂ in the gaseous mixture?

Answer 1

Rate = k[N₂O₅]

Since k has units of s^-1, we can tell this is a first order reaction. We can write the rate equation as

rate = 2.8x10^-3s^-1 [2.95 M]

rate = 8.26x10^-3 Ms^-1

when t = 3.75 minutes, we can find the [N₂O₅] remaining and the [NO₂] formed

3.75 min x 60 sec / min = 225 sec

Integrated rate equation for 1st order reaction :

ln [A] = -kt + ln [A]_o

ln [A] = -(2.8x10^-3s^-1)(225s) + ln [2.95 M)

ln [A] = -0.630 + 1.08 = -0.452

[A] = 0.636 M

Using an ICE table for simplicity...

2N₂O₅ ===> 4NO₂ + O₂

2.95.................0...................Initial

-2x.................+4x.................Change

2.95-2x.........4x....................Equilibrium

2.95-2x = 0.636

x = 1.16

4x = 4.63

So, at equilibrium

[N₂O₅] = 0.636 M

[NO₂] = 4.63 M

mole fraction NO₂ = 4.63 / 4.63 + 0.636 = 0.88