Mol fraction, first order, 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

Since this is a first order reaction we can use the integrated rate law for a first order reaction

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

ln [A] = ?

k = 2.8x10^-3s^-1

t = 3.75 min x 60 s / min = 225 s

ln [A]_o = ln 2.95 = 1.082

ln [N₂O₅] = (-2.8x10^-3s^-1)(225) + 1.082

ln [N₂O₅] = 0.452 M @ 225 s

[N₂O₅] = 1.57 M @ equilibrium

∆[N₂O₅] = 2.95 - 1.57 = 1.38

2N₂O₅ <===> 4NO₂ + O₂

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

-1.38.................+2.76.....+0.69..Change

1.57...................2.76.......0.69....Equilibrium

At equilibrium we should have the following;

[NO₂] = 2.76 M

[O₂] = 0.69M

[N₂O₅] = 1.57

TOTAL MOLES = 2.76 + 0.69 + 1.57 = 5.02

Mol fraction NO₂ = 2.76 / 5.02 = 0.550