The effect of pressure on equilibrium constant What will happen to the $K_\text{eq}$ in the following reaction if volume is increased?

$$\ce{2NO(g) + O2(g) &lt;=&gt; 2NO2(g) + energy}$$

From le Chatelier's principle I know that it must be constant unless temperature is changed, however it doesn't make sense because if

$$K_\text{eq}= \frac{[\ce{NO2}]^2}{[\ce{O2}][\ce{NO}]^2},$$

then by increasing the volume we know that equilibrium will shift left and favor reactants. Hence concentration of reactants would increase while concentration of products would decrease. If we plug that back into the equilibrium constant equation we see that if numerator decrease and denominator increase that must yield a smaller number than the initial number. Hence the equilibrium is not constant. Why this occurs?
First step: Don't use Kc... use Kp!

You are focusing on concentration, but as we're working with gases it becomes easier to work with partial pressures.

$K_\text{p}= \frac{\ce{p(NO2)}^2}{\ce{p(O2)}\ce{p(NO)}^2}$

Now, if we remember that partial pressure is pressure $\times$ mole fraction ($p_{a}=P \times \chi_{a}$), and substitute these in we get (and cancelling out the pressures):

$K_\text{p}= \frac{\chi_{(NO_{2})}^2}{\chi_{(O_{2})}\chi_{(NO)}^{2}P}$

Now, if we consider increasing the volume. By doing so (and assuming that temperature is kept constant), then we have decreased the pressure (ideal gas law: PV=nRT).

Now we can see that although the number of moles changes (and hence the mole fractions), our equilibrium constant will remain unchanged thanks to the effect of pressure.

A good guide is here.