Dinitrogen tetraoxide partially decomposes according to the following equilibrium:
n2o4(g) = 2no2(g)
a 1.00 l flask is charged with 0.0400 mol of n2o4. at equilibrium at 373k, 0.0055 mol of n2o4 remains. keq for this reaction is
a. 0.87
b. 13
c. 2.2

Keq for this reaction is 0.87 (option a.)

Explanation:

The equilibrium is this:

N₂O₄  ⇄ 2N₂O₂

1 mol of N₂O₄ decomposes to 2 moles N₂O₂

                    N₂O₄  ⇄ 2N₂O₂

Initially       0.04m           -

React               x             2x

Initially I have 0.04 moles of N₂O₄. Some amount (x) reacts to decompose and make 2 moles of N₂O₂; according to reaction, ratio is 1:2.

So if x amount of compound, reacted. I will have the double, as product decomposed.

                    N₂O₄  ⇄ 2N₂O₂

Initially       0.04m           -

React               x             2x

Eq.              0.055m       2x

We have moles of N₂O₄ in equilibrium, so we have to know how many amount has reacted.

(0.04 - 0.0055) = 0.0345

The double of this, is what we have in equilibrium of N₂O₂.

0.0345 .2 = 0.069

Let's make Kc:

[N₂O₂]² / [N₂O₄]

0.069² / 0.0055 = 0.87