# Homework Help: Le Chatelier and Equilibrium

1. Apr 11, 2016

### terryds

1. The problem statement, all variables and given/known data
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

2. Relevant equations

Kc = products of right molarity / products of left molarity

3. The attempt at a solution

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

2. Apr 11, 2016

### Staff: Mentor

Do you mean NO?

3. Apr 11, 2016

### Staff: Mentor

Nope, concentration of NO at equilibrium is not 1.8.

This is best done using an ICE table.

4. Apr 11, 2016

### terryds

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

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!

5. Apr 11, 2016

### Staff: Mentor

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.