Example:
If the initial concentrations in the Haber process were [N2(g)] = 0.90 M, [3H2(g)] = 3.00 M, and [2NH3(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) + 3H2(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: |
| N2(g) | 3H2(g) | 2NH3(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. |
| [N2] = 0.90 – x = 0.90 – 0.011 = 0.89 |
| [H2] = 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
- 1
-
2
Ask a Question
Have questions about this article or topic? AskRelated Questions
See More QuestionsToday on Education.com
WORKBOOKS
May Workbooks are Here!
WE'VE GOT A GREAT ROUND-UP OF ACTIVITIES PERFECT FOR LONG WEEKENDS, STAYCATIONS, VACATIONS ... OR JUST SOME GOOD OLD-FASHIONED FUN!
Get Outside! 10 Playful Activities
Local SAT & ACT Classes
Popular Articles
- Kindergarten Sight Words List
- The Five Warning Signs of Asperger's Syndrome
- What Makes a School Effective?
- Child Development Theories
- Why is Play Important? Social and Emotional Development, Physical Development, Creative Development
- 10 Fun Activities for Children with Autism
- Test Problems: Seven Reasons Why Standardized Tests Are Not Working
- Bullying in Schools
- A Teacher's Guide to Differentiating Instruction
- First Grade Sight Words List
2NH3(g) (Haber process)
cC + dD
2NH3(g)




Celebrate Memorial Day! Worksheets and Activities About American History
7 Parenting Tips to Take the Pressure Off 
Add your own comment