Quantitative Aspects of Le Châtelier's Principle
Consider the effect of changing the concentration of a participant in a system that is already at equilibrium.
 
SO2(g) + NO2(g) <=====>  SO3(g) + NO(g)

 
Suppose we add SO2 to the system. This produces a non-equilibrium that is a stress. It must be relieved by a reaction that relieves the stress. The increase in the denominator (of the Keq equation for this reaction) means that the experimental reaction quotient, Q , is less that Keq. In order to re-establish equilibrium the numerator must increase. ie. The reaction must shift to the right and increase to yield of products.
 
At 200oC analysis of the equilibrium mixture shows:

[SO2] = 4.0 M,   [NO2] = 0.50 M

[SO3] = 3.0 M,   [NO] = 2.0 M
 

What is the new equilibrium concentration of NO when 1.5 moles of NO2 is added to the equilibrium mixture above?
 
Solution
This is a new equation. Find the value of Keq first. Since its a constant at this temperature then it will be valid for the equilibrium mixture and the stressed mixture.
 

 
So the value of Keq for this reaction, at this temperature is 3.0
 
Fill in the start and finish values including an 'x' term.
 
              SO2(g)    +      NO2(g) <------>  SO3(g)     +     NO(g)
start       4.0 M              0.5 M                  3.0 M               2.0 M
finish      (4.0-x)         (0.5+1.5-x)            (3.0+x)             (2.0+x)
 
Let's take a second to explain what is happening. When we add NO2 we stress the reaction so that more product is formed. This means we will lose 'x' amount of SO2 more. The NO2 term is 0.5 +1.5-x because we started with 0.5 M, added 1.5 M more to stress it and this amount will then react 'x' amount of itself to produce product. We are lucky here because everything works on a one-to-one basis. This is not always the case. SO3 and NO will each increase by 'x' as more product forms.
 
Now we will fill in these factors in the Keq equation and solve for 'x'.
 

 
3.0 = (3.0 + x)(2.0 + x)
          (4.0 - x)(2.0 - x)
 
After a few multiplications and additions we get:
 
2.0x2 -23x +18 = 0
 
After using the quadratic equation on this formula we get a value for x = 0.75 mol/L
 
Therefore the [NO] = 2.0 + x = 2.75 mol/L
 
Follow-Up Problem
How many moles of NO2 would have to be added to the original equilibrium mixture to increase the equilibrium concentration of SO3 from 3.0 to 4.0 moles at the same temperature