# An amount of NO2 was initially put into a 5.0L flask. When equilibrium was attained according to the equation: 2NO (g) + O2 (g)--> 2NO2 (g), the concentration of NO was 0.800M. If Keq for this system is 24.0, what was the initial concentration of the NO2?

##### 2 Answers

The answer is

The best way to approach this problem is by using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart). We know that, initially, a concentration of

We must use the reverse reaction, since no amounts of

SInce the equilibrium constant for the reverse reaction is *smaller than 1*, the final mixture will contain *mostly reactants*, which means that the concentration of

.....

**I**.........A.................0...................0

**C**......-2x...............+2x................+x

**E**.......(A-2x)...........2x.................x

Since we know that **0.800 M**, we get

The expression for the equilibrium constant is

Solving for **3.3**.

Therefore,

Notice that the prediction about

The initial concentration of the NO₂ was 3.279 mol/L.

**Step 1.** Write the balanced chemical equation for the equilibrium.

2NO + O₂ ⇌ 2NO₂

**Step 2.** Set up an ICE table:

Let

#2"NO" + "O"_2 ⇌ 2"NO"_2"#

I/mol·L⁻¹: 0; 0;

C/ mol·L⁻¹: +2

E/ mol·L⁻¹: 2

We know that

So

Then

**Step 3.** Write the

**Step 4.** Insert these values into the

**Step 5.** Solve for

**The initial concentration of the NO₂ was 3.279 mol/L.**

**Check**: