Question

The reaction N_2O_4(g) ⇌ 2NO_2(g) has K_c = 0.140 at 25.0°C. Exactly 0.0245 mol N_2O_4 and 0.0116 mol N_2O are placed into a 2.0 L sealed flask. What is the density of the gas (in g/L) once equilibrium is reached? Do not enter units with your answer.

Answer 1

Good Method

Here, we notice that the mass of the flask must remain conserved before and after the reaction. Volume of flask must also remain constant.

Molar Mass of NO₂ = 46.0055 g/mol

Mass of NO₂ = Moles * Molar Mass = (0.0116 mol) * (46.0055 g/mol)= 0.533 g

Molar Mass of N₂O₄ = 92.011 g/mol

Mass of N₂O₄ = Moles * Molar Mass = (0.0245 mol) * (92.011 g/mol)= 2.254 g

Total Mass = 0.533 g + 2.254 g = 2.787 g

Density = Mass / Volume = 2.787 g / 2.0 L = 1.39 g/L

Up to 2 sig fig, the answer will be 1.4 without units.

Bad Method

We will use ICE Table. Let change in N₂O₄ be -x which means +2x moles of NO₂ will be produced

Reaction	N2O4		2NO2
Initial moles	0.0245		0.0116
Change in moles	-x		+2x
Equilibrium moles	0.0245-x		0.0116+2x

Volume is V = 2.0 L

[N₂O₄] = (0.0245-x)/2 M

[NO₂] = (0.0116+2x)/2 M

Thus, we will use expression for K_c

Now, if x= -0.0986446, then [NO₂] = 0.0116 + 2*(-0.0986446)= -0.1856892 which is not possible.

Thus, x = 0.0170446

Moles of NO₂ = 0.0116 + 2* 0.0170446 = 0.0456892 mol

Mass of NO₂ = Moles * Molar Mass = (0.0456892 mol ) * (46.0055 g/mol)= 2.102  g

Moles of N₂O₄ =0.0245 - (0.0170446) = 0.0074554 mol

Mass of N₂O₄ = Moles * Molar Mass = (0.0074554 mol) * (92.011 g/mol)= 0.686 g

Total Mass = 2.102  g + 0.686 g = 2.788 g

Density = Mass /Volume = 2.788 g /2.0 L = 1.39 g/L