Question

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 __________.

Answer 1

6.94x10-3

Step-by-step explanation:

Answer 2

Keq for this reaction is 6.94x10⁻³

Step-by-step 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