Please Answer! Help! Will Give Brainliest. Find Keq for a 3.0 L container with 1.2 moles of both NO2 and N2O4 initially and 0.38 M NO2 at equilibrium. 2NO2(g) ⇌ N2O4(g)

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

the answer is 303.15 or 303

i think   the answer is c hope i'm right