Law of Mass Action (Equilibrium Constant)
The law of mass action describes chemical equilibrium: the stoichiometrically weighted ratio of the equilibrium concentrations of products to reactants is constant.
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Formula
K_c = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b}Variables & units – Law of Mass Action (Equilibrium Constant)
| Symbol | Meaning | Unit |
|---|---|---|
| Kc | Equilibrium constant (concentration-based) | dimensionless |
| [C],[D] | Equilibrium concentrations of the products | mol/L |
| [A],[B] | Equilibrium concentrations of the reactants | mol/L |
| a,b,c,d | Stoichiometric coefficients | dimensionless |
Derivation & background – Law of Mass Action (Equilibrium Constant)
Guldberg and Waage (1864) formulated the law of mass action. Kc >> 1: equilibrium on the product side. Kc << 1: equilibrium on the reactant side. Le Chatelier's principle: a change in concentration, pressure or temperature shifts the equilibrium so as to compensate.
Exam blueprint
Validity range
Applies to chemical equilibria at fixed temperature; rigorously with activities, in school often with concentrations.
Derivation steps
At equilibrium, forward and reverse reaction rates balance.
- 1Reaction rates depend on concentrations of reactants and products.
- 2The equilibrium ratio becomes the constant K_c.
Rearrangements
Product concentration in a simple equation
Stoichiometric coefficients become exponents.
Task variant
What does K_c >> 1 mean?
The equilibrium lies strongly on the product side.
Common mistakes
Not using coefficients as exponents.
In the law of mass action, stoichiometric factors are exponents.
Exam context
- Often with Le Chatelier, ammonia synthesis and acid-base equilibria.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Chemical equilibrium
Foundation for buffers, Gibbs energy and electrochemistry.
Worked example
N₂ + 3H₂ ⇌ 2NH₃. With [N₂] = 0.5, [H₂] = 0.5, [NH₃] = 0.1 mol/L: Kc = (0.1)² / (0.5·(0.5)³) = 0.01/0.0625 = 0.16.
Applications
The Haber-Bosch process (NH₃ synthesis), acid-base equilibria, buffer calculation, the solubility product
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Which formula describes Law of Mass Action (Equilibrium Constant)?
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How do you rearrange Kc = [C]ᶜ·[D]ᵈ / ([A]ᵃ·[B]ᵇ) for Product concentration in a simple equation?
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Which common mistake happens with Law of Mass Action (Equilibrium Constant)?
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Scientific sources
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Frequently asked questions about Law of Mass Action (Equilibrium Constant)
How do you set up the expression for the equilibrium constant?+
The equilibrium constant K_c is the quotient of the product concentrations and the reactant concentrations, each raised to the power of their stoichiometric coefficients: K_c = [C]^c·[D]^d/([A]^a·[B]^b). The coefficients from the reaction equation thus become exponents. Insert the equilibrium concentrations in mol/L. Example: for N₂ + 3H₂ ⇌ 2NH₃ with [N₂] = 0.5, [H₂] = 0.5 and [NH₃] = 0.1 mol/L you get K_c = (0.1)²/(0.5·(0.5)³) = 0.01/0.0625 = 0.16. Make sure to insert the coefficients correctly as exponents and not as factors; that is the most common mistake.
What does a large or small value of the equilibrium constant mean?+
The equilibrium constant K_c indicates on which side the equilibrium lies. If K_c is much greater than one, the products dominate, so the equilibrium lies far to the right and the reaction proceeds almost completely. If K_c is much smaller than one, the reactants dominate and only little product forms. If K_c is of the order of one, reactants and products are present in comparable amounts. It is important that K_c only describes the position of the equilibrium, not how fast it is reached; that is a question of kinetics. K_c also depends on the temperature.
How does Le Chatelier principle affect the equilibrium?+
Le Chatelier principle states that an equilibrium responds to an external disturbance so as to counteract it. If you increase the concentration of a reactant, the equilibrium shifts to the product side to reduce the excess. If you increase the pressure in gas reactions, it shifts to the side with fewer gas particles. If you change the temperature, for exothermic reactions heating shifts the equilibrium to the reactant side. Importantly: concentration and pressure changes only shift the position but leave K_c unchanged. Only temperature changes the value of K_c itself. Industry uses this principle, for example in ammonia synthesis, to deliberately increase the yield.
What is the difference between the reaction quotient Q and the constant K?+
The reaction quotient Q has the same form as the equilibrium constant K but is calculated with the current, not necessarily equilibrium concentrations. By comparing Q with K you recognize in which direction a reaction will still proceed. If Q is smaller than K, there are still too many reactants and the reaction continues to the right towards the products. If Q is larger than K, there are too many products and the reaction runs backwards. If Q equals K, there is equilibrium and no net change occurs any more. Q is thus a snapshot, K the target value. This comparison is a powerful tool to predict the direction of a reaction.
Why are pure solids and solvents omitted from the law of mass action?+
Pure solids and pure liquids have a constant activity independent of the amount, which by convention is set equal to one. Therefore they do not appear in the expression for K_c, because their concentration does not change in a relevant way during the reaction. The solvent too, for example water in dilute aqueous solutions, is present in such large excess that its concentration stays practically constant and is likewise omitted. Only dissolved substances and gases whose concentration or partial pressure changes noticeably enter the equilibrium constant. A common mistake is to wrongly include a solid or water with a concentration in K_c.
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How do you calculate with Law of Mass Action (Equilibrium Constant)?
Here is how to work through a typical Law of Mass Action (Equilibrium Constant) (Kc = [C]ᶜ·[D]ᵈ / ([A]ᵃ·[B]ᵇ)) task step by step:
- 1
Task
What does K_c >> 1 mean?
Solution path
The equilibrium lies strongly on the product side.