Example:
If the initial concentrations in the Haber process were [N_{2}(g)] = 0.90 M, [3H_{2}(g)] = 3.00 M, and [2NH_{3}(g)] = 0 M, what is the equilibrium concentrations? The equilibrium constant is 6.0 * 10^{–5}.
Solution:
1.  Write the balanced equation and equilibrium expression: 
N2(g) + 3H_{2}(g) 2NH3(g) 
2.  Identify the initial concentrations of each substance and determine the shift of the equation. Identify the "change" for each substance using a variable (usually x). Add the initial and change to generate the equilibrium amount. These three steps can be easily handled as follows: 
N_{2}(g)  3H_{2}(g)  2NH_{3}(g)  
Initial  0.90  3.00  0 
Change (per mole)  –x  –x  2x 
Equilibrium  0.90 – x  3.00 – 3x  2x 
The shift is clearly to the right because no products exist initially. The sign of the change reflects whether we are adding or subtracting to a substance. In this example, because the equilibrium is shifting right, the reactants are disappearing (negative x) and the product is appearing (positive x). The coefficients in the balanced equation corresponding to the number of moles are used as coefficients with the variable x. The initial and change are added to yield the equilibrium concentration.
3.  Use the equilibrium expression to solve for x: 
If the K is very small, then the reaction is not expected to proceed far to the right and the "– x" and "– 3x" can be canceled because x will be a small number. This approximation is typical, unless you want to solve the long equation:
4.  Calculate the equilibrium concentration using the solved x value. 
[N_{2}] = 0.90 – x = 0.90 – 0.011 = 0.89 
[H_{2}] = 3.00 – 3x = 3.00 – 3(0.011) = 2.97 
[NH3] = 2x = 2(0.011) = 0.022 
For equations that do not have small equilibrium constants, the quadratic formula must be used to solve for x: 
Le Châtelier's Principle
Le Châtelier's principle states that if an equilibrium system is stressed, the equilibrium will shift in the direction that relieves the stress. For the Haber process at equilibrium, if more nitrogen is added, then the process will shift to the right. If nitrogen is removed, then the equilibrium will shift to the left.
Practice problems for these concepts can be found at  Chemical Equilibrium Practice Questions
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