Answer:
Kc → 41.9
Explanation:
This is the equilibrium:
2NO₂(g) ⇌ N₂O₄(g)
So the expression for Kc will be:
Kc = [N₂O₄] / [NO₂]²
We prospose the situations:
Initially we have 1.2 moles of NO₂ and N₂O₄
X amount has reacted. As stoichiometry is 2:1, we have produced x/2 of the product during the reaction
Finally In equilibrium we have, 0.38 NO₂
2NO₂(g) ⇌ N₂O₄(g)
Initially 1.2 1.2
React x x/2
Eq (1.2 - x) = 0.38 1.2 + x/2
As we have [NO₂] in the equilibrium, we can determine x (the amount that has reacted) to solve and determine, the [N₂O₄] in the equilibrium
1.2-0.38 = x → 0.82
1.2 + 0.82/2 = 2.02 → [N₂O₄]
For Kc, we need Molar concentration, so we have to divide [N₂O₄] and [NO₂] by the volume
[N₂O₄] → 2.02 mol/3L = 0.673 M
[NO₂] → 0.38 mol/3L = 0.127 M
Now we can replace the Kc expression:
Kc → [N₂O₄] / [NO₂]² → 0.673 / 0.127² = 41.9
Remember that Kc has no UNITS
