How Does Adding NO2 Affect N2 Concentration in Equilibrium?

[terryds]

Homework Statement

At certain pressure and temperature in a 2 L closed flask, equilibrium reaction undergoes such as :

N₂ (g) + O₂ (g) ↔ 2NO (g)

Each concentration of the substances in equilibrium is 0.8 M.
If we add 2 mole of NO₂ gas into the flask, then the concentration of N₂ gas in the new equilibrium is ...

A. 1.13 M
B. 1.80 M
C. 2.20 M
D. 2.60 M
E. 2.80 M

Homework Equations

[/B]
Kc = products of right molarity / products of left molarity

The Attempt at a Solution

[/B]
At initial equilibrium
Kc = (0.8)^2 / (0.8)(0.8) = 1

Change in mole = 2 mole => Change in molarity = 2 mole / 2 liter = 2 M

At final equilibrium
Kc = (1.8)^2 / (x^2)
1 = (1.8)^2 / (x^2)
x = 1.8 M

However, the answer key is A. 1.13 M
Please help what I missed

 

[DrClaude]

If we add 2 mole of NO₂ gas into the flask, then the concentration of N₂ gas in the new equilibrium is ...

Do you mean NO?

 

[Borek]

Kc = (1.8)^2 / (x^2)

Nope, concentration of NO at equilibrium is not 1.8.

This is best done using an ICE table.

 

[terryds]

Do you mean NO?

NO₂ is what written in the problem. Maybe, it's a typo from the problem-maker.

Nope, concentration of NO at equilibrium is not 1.8.

This is best done using an ICE table.

Okay, I'll try

Kc = (1.8-2x)^2/(0.8+x)^2
1 = (1.8-2x)^2/(0.8+x)^2

Using calculator, x = 0.3333 M
N2 concentration = 0.8+0.33 = 1.13 M

Thanks for all your help! ICE table is really a reliable method!

 

[Borek]

Remember ICE is just an easy and convenient way of keeping track of the stoichiometry, it doesn't add anything new to the way we solve equilibrium problems.