Answer:
[NO₂] = 0.434 M
[N₂O₄] = 0.0971 M
Step-by-step explanation:
The equilibrum is: N₂O₄(g) ⇆ 2NO₂ (g)
1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.
Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:
2NO₂ (g) ⇆ N₂O₄(g)
Initially 2.20 mol -
React x x/2
X amount has reacted, and the half has been formed, according to stoichiometry.
Eq (2.20-x) / 3.50L (x/2)/ 3.50L
We divide by the volume because we need molar concentrations. Let's make the Kc's expression:
Kc = [N₂O₄] / [NO₂]²
0.513 = ((x/2)/ 3.50L) / [(2.20-x) / 3.50L]
0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]
0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]
0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)
0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50
0.203 - 0.184x + 0.0419x² = x/2 / 3.50
3.50(0.203 - 0.184x + 0.0419x²) = x/2
7 (0.203 - 0.184x + 0.0419x²) - x = 0
1.421 - 2.288x + 0.2933x² = 0 → Quadratic formula
a = 0.2933 ; b = -2.288 ; c = 1.421
(-b +- √(b²-4ac)) / (2a)
x₁ = 7.12
x₂ = 0.68 → We consider this value, so we can have a (+) concentration.
Concentrations in the equilibrium are:
[NO₂] = (2.20-0.68) / 3.50 = 0.434 M
[N₂O₄] = (0.68/2) / 3.50 = 0.0971 M