Dinitrogen tetroxide partially decomposes according to the following equilibrium: n2o4 (g) → 2 no2 (g)
a 1.000-l flask is charged with 3.00 × 10-2 mol of n2o4. at equilibrium, 2.36 × 10-2 mol of n2o4 remains.

keq for this reaction is

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

the answer is d. 74.02