Answer:
Keq for this reaction is 6.94x10⁻³
Explanation:
The equilibrium equation is this one:
N₂O₄ (g) ⇄ 2NO₂ (g)
Initially we have 0.03 moles from the dinitrogen tetroxide and nothing from the dioxide.
In the reaction, some amount of compound (x) has reacted.
As ratio is 1:2, we have double x in products.
Finally in equilibrium we have:
N₂O₄ (g) ⇄ 2NO₂ (g)
0.03 - x 2x
And we know [N₂O₄] in equilibrium so:
0.03 - x = 0.0236
x = 0.03 - 0.0236 → 6.4x10⁻³
As this is the amount that has reacted, in equilibrium I have produced:
6.4x10⁻³ .2 = 0.0128 moles of NO₂
This is the expression for K,
[NO₂] ² / [N₂O₄]
0.0128² / 0.0236 = 6.94x10⁻³
