Question

At a particular temperature 8.0 mol NO2 gas is placed in a 1.0-L container. Over time the NO2 decomposes to NO and O2: 2NO2(g) 2NO(g) O2(g) At equilibrium the concentration of NO(g) was found to be 2.9 mol/L. Calculate the value of K for this reaction (using units of mol/L for the concentrations).

Answer 1

Answer:

Kc = 0.7731

Explanation:

Let's write the given reaction:

2NO2 <-------> NO + O2

Now, we know that initially we have 8 mol of NO2, at 1 L (Concentration is 8 M then). At the end of the equilibrium we have letf 2.9 M of NO2 and we need to calculate the value of K, which is the equilibrium constant.

Now, the expression to calculate equilibrium constant is:

Kc = [Product]/[Reactants]

Aplying this to the given reaction we have:

Kc = [NO] [O2] / [NO2]²

But these concentrations must be in equilibrium. To know this, we use an ICE chart in this reaction, and see what we have:

          2NO2 <-------> NO + O2

I:             8                    0        0

C:           -2x               +x        +x

E:           8 - 2x              x        x

According to this, we know that the concentration in equilibrium of NO2 is 8 - 2x; however we know that the total of this is 2.9, so we can solve for x from here:

2.9 = 8 - 2x

2x = 8 - 2.9

2x = 5.1

x = 5.1/2

x = 2.55 mol/L

This means that the concentrations in equilibrium of NO and O2 are 2.55, so the equilibrium constant Kc would be:

Kc = 2.55 * 2.55 / (2.9)²

Kc = 0.7731

And this is the value of Kc.