Equilibrium Constant Kc
Learn and apply Equilibrium Constant Kc in the published Chemistry course sequence.
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The core idea
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H1 Equilibrium Constant Kc: Orientation
The equilibrium constant translates a balanced reversible equation into a numerical description of equilibrium composition. H1 uses concentration terms, Kc, and disciplined interpretation.
- Write and interpret Kc expressions using equilibrium concentrations.
- Kp and partial-pressure expressions are outside 8873 Chemical Equilibria.
Definitions (Must Know)
- K_c is the equilibrium constant expressed using equilibrium concentrations for a stated equation at a stated temperature.
- Equilibrium concentration is the constant concentration after dynamic equilibrium has been established.
- A homogeneous equilibrium contains all reacting species in the same phase.
- A heterogeneous equilibrium contains more than one phase; pure solids and pure liquids are omitted from K_c.
Detailed Explanations
A. Expression routine
Balance the equation, identify aqueous or gaseous species, place products over reactants, then apply coefficients as powers.
B. Heterogeneous example
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), both solids are omitted:
K_c = [CO₂]
C. Interpreting magnitude
K_c describes equilibrium composition, not rate. A very large value can belong to a reaction that takes a long time to reach equilibrium.
D. Conditions
Concentration and pressure disturbances change the position until the same K_c ratio is restored.
A catalyst provides a lower-activation-energy pathway for both the forward and the reverse reaction, so it speeds up both reactions by the same factor. Equilibrium is reached sooner, but the equilibrium composition, and therefore K_c, is unchanged.
Temperature changes the value of K_c. Raising the temperature favours the endothermic direction: K_c increases if the forward reaction is endothermic and decreases if it is exothermic.
Worked Examples
Modelled example 1
Writing a Kc expression
Problem
Study the worked solution
Arrange the species
Method
Place the equilibrium concentration of ammonia above those of nitrogen and hydrogen.Reason
The expression follows products over reactants for the balanced equation as written.Working
K_c = [NH₃]^?/[N₂]^?[H₂]^?.Apply coefficient powers
Method
Use each balanced coefficient as its concentration power.Reason
Two ammonia and three hydrogen appear in the equation; nitrogen’s coefficient of 1 need not be written.Working
K_c = [NH₃]²/[N₂][H₂]³.
Guided practice 2
A heterogeneous equilibrium
Problem
Try this before viewing the solution
Hints
Hint 1: audit phases
Hint 2: remove constant activity
View solution step by step
Apply the phase rule
Method
Omit the pure solid NH₄HS.Reason
Its activity is constant while the solid phase is present, so it does not appear as a variable concentration term.Working
Included species: NH₃(g) and H₂S(g) only.Write the expression
Method
Multiply the two product concentration terms.Reason
Each gaseous product has coefficient 1 and there are no variable reactant terms in the denominator.Working
K_c = [NH₃][H₂S].
Common misconception 3
Correct a large-Kc speed claim
Learner claim
Classify the information
View solution step by step
State the valid inference
Method
Interpret 10⁸ as a very large equilibrium constant.Reason
The products-over-reactants concentration ratio is very large at equilibrium.Working
Products predominate once equilibrium is established.Reject the invalid inference
Method
State that no equilibration time follows from K_c.Reason
Rate depends on kinetic factors, including activation energy, which K_c does not specify.Working
The reaction may still take a long time to reach its product-favoured equilibrium.
Challenge 4
Connect temperature, composition and Kc
Problem
Try this before viewing the solution
Hints
Hint 1: place heat
Hint 2: link the new ratio
View solution step by step
Predict the temperature response
Method
Shift equilibrium in the endothermic reverse direction.Reason
The reverse reaction absorbs some of the added heat.Working
The equilibrium product proportion decreases.Connect the new composition to Kc
Method
State that K_c becomes smaller at the higher temperature.Reason
The new equilibrium has a smaller products-to-reactants concentration ratio for the equation as written.Working
T↑ for an exothermic forward reaction ⇒ K_c↓.Contrast concentration
Method
State that changing concentration at fixed temperature does not change K_c.Reason
The system shifts until the original equilibrium ratio is restored.Working
Only the temperature change creates a new K_c value here.
Mind Stretchers
Attempt each unfamiliar application before opening the hint, then compare your reasoning with the solution.
Mind stretcher 1: Equation directionExtension
Question. If K_c = 25 for a forward equation, what is Kc for its reverse?
Show Hint
Reversing an equilibrium inverts the product/reactant ratio.
Show Answer
K_c = 1/25 = 0.040 for the reverse equation.
Mind stretcher 2: Magnitude without kineticsExtension
Question. A reaction has K_c = 10⁸ but takes days to equilibrate. Explain why there is no contradiction.
Show Hint
Kc describes composition; rate depends on activation barriers.
Show Answer
The large Kc means products predominate once equilibrium is reached. It gives no timescale; a high activation barrier can make both directions slow.
Mind stretcher 3: Reading the enthalpy sign from KcExtension
Question. When the temperature of an equilibrium mixture is raised, the value of K_c increases. Deduce whether the forward reaction is exothermic or endothermic.
Show Hint
A larger K_c means a greater proportion of products at the new equilibrium. Which direction does a temperature rise favour?
Show Answer
The forward reaction is endothermic (Δ H positive). The larger K_c shows that the temperature rise has favoured the forward direction, and raising the temperature favours the endothermic direction.