Introduction
Stoichiometric calculations assume that reactions progress to completion consuming the limiting reactant. However, because many reaction products can also react to again produce the original reactants or because a reaction does not proceed to completion, chemical equilibrium is established. Chemical equilibrium is reached when the amount of products and reactants remain constant.
Equilibrium
Equilibrium is when two opposing reactions occur at the same rate. In chemical equilibrium, the concentrations of the reactants and products remain constant, and no change is observed in the system. Any chemical process will achieve equilibrium over time.
N_{2}(g) + 3H_{2}(g) 2NH_{3}(g) (Haber process)
At equilibrium, there may be greater amounts of products (reaction lies far to the right) or greater amounts of reactants (reaction lies far to the left).
The Equilibrium Constant
For the following general reaction,
aA + bB cC + dD
where A, B, C, and D are chemical species, and a, b, c, and d are their corresponding coefficients, the following equilibrium expression is obtained:
where K is the equilibrium constant.
Example 1:
Write the equilibrium expression for the Haber process.
1. 
Write the balanced equation: 
N_{2}(g) + 3H_{2}(g) 2NH_{3}(g) 
2. 
Write the equilibrium expression: 
What about solids and liquids? Pure liquid and solid concentrations do not vary significantly in chemical processes, and their concentrations are always equal to their standard concentrations (usually one). So pure liquids and solids are omitted from equilibrium expressions. Of course, aqueous species are always included in the equilibrium expression.
If a chemical process has a species that is gaseous and one that is aqueous, solid, or liquid, then the gaseous species is usually listed as a partial pressure.
Example 2:
Write the equilibrium expression for the reaction when gaseous carbon dioxide dissolves in pure water to form carbonic acid.
1. 
Write the balanced equation: 

CO_{2}(g) + H_{2}O(l) H_{2}CO_{3}(aq) 
2. 
Write the equilibrium expression: 



CO_{2}(g): a gas, listed as a partial pressure 

H_{2}O(l): a pure liquid, omitted 

H_{2}CO_{3}(aq): aqueous, listed as its concentration 
Note
H_{2}CO_{3} is a dissolved acid; hence, it is aqueous.
Solving Equilibrium Problems
The equilibrium constant and equation developed the means to calculate equilibrium. Typically, when the amounts of the reactant or products are given, the equilibrium concentrations can be calculated as long as the equilibrium constant is known at the reaction temperature and the direction of the equations shift (left, toward reactants, or right, toward products) can be predicted.
Equilibrium problems can be solved with the six following steps:
1. 
Write the balanced equation and equilibrium expression. 
2. 
Identify the initial concentrations of each substance and determine the shift of the equation. 
3. 
Identify the "change" for each substance using a variable (usually x). 
4. 
Add the initial, and change to generate the equilibrium amount. 
5. 
Use the equilibrium expression to solve for x. 
6. 
Calculate the equilibrium concentration using the solved x value. 
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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