Worked example 1
Write Kc for 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g).
Try it first: Place the products on top and turn every coefficient into an exponent.
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Chemical Equilibrium & SolubilityEquilibriumContent level: Core 20 min
Writing Equilibrium Constant Expressions: Kc, Kp and Heterogeneous Systems
What you'll be able to do: Write a correct equilibrium expression for any balanced equation, including heterogeneous systems, and convert between Kc and Kp.
Introduction
Every equilibrium calculation starts with the expression. Get the coefficients, the phases, or the units wrong here and everything downstream is wrong too.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Write Kc and Kp expressions from a balanced equation
- Exclude pure solids and liquids from equilibrium expressions
- Convert between Kc and Kp using Dn
- Explain why pure phases are omitted
Lesson
Products over reactants
For aA + bB ⇌ cC + dD the products go in the numerator and the reactants in the denominator, each raised to the power of its coefficient. The coefficients become exponents, never multipliers.
Kc = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
Kp for gases
When every species is a gas you can use partial pressures instead of concentrations. The structure is identical, only the quantity substituted changes.
Kp = (PCᶜ PDᵈ) / (PAᵃ PBᵇ)
Leaving out pure phases
A pure solid or pure liquid has a fixed density, so its effective concentration does not change as the reaction runs. Anything that is constant is folded into K itself and disappears from the expression. Dilute aqueous solutions treat the water solvent the same way.
Omit (s) and (l). Include (g) and (aq).
Heterogeneous equilibria
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), both solids drop out and the expression is simply Kc = [CO₂]. This is why the pressure of CO₂ above heated limestone in a sealed vessel depends only on the temperature, not on how much limestone is present.
Kc = [CO₂]
Relating Kc and Kp
Substituting P = (n/V)RT into the expression gives a factor of RT for every net mole of gas produced. Dn counts moles of gaseous product minus moles of gaseous reactant. Use R = 0.08206 L atm/mol K with pressures in atm and T in kelvin.
Kp = Kc (RT)Δn
If Dn = 0 then (RT)⁰ = 1 and Kp = Kc.
Pure solid activity is fixed
Pure solids and pure liquids have activity 1 in an equilibrium expression, so their amounts do not appear in K as long as the phase remains present. Adding more of a pure solid does not directly shift equilibrium or change K at fixed temperature. Gases and dissolved species still enter with stoichiometric exponents.
Omit pure phases; do not omit gases just because a solid is present.
Key ideas
Rule
Coefficients become exponents
Each concentration or pressure is raised to the power of its stoichiometric coefficient.
Rule
Omit pure solids and liquids
Their effective concentrations are constant and are absorbed into K.
Definition
Heterogeneous equilibrium
An equilibrium involving more than one phase, so some species are excluded from the expression.
Definition
Dn
Moles of gaseous product minus moles of gaseous reactant in the balanced equation.
Key concept
K is unitless
Equilibrium constants are reported without units because each term is really a ratio to a standard state.
Equation
Concentration equilibrium constant
Kc = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
- [X] = equilibrium concentration in mol/L
Equation
Pressure equilibrium constant
Kp = (PCᶜ PDᵈ) / (PAᵃ PBᵇ)
- PX = equilibrium partial pressure in atm
Equation
Converting between them
Kp = Kc (RT)Δn
- Dn = change in moles of gas
- R = 0.08206 L atm/mol K
- T = temperature in K
Worked examples
Worked example 2
Write Kc for Fe₃O₄(s) + 4 H₂(g) ⇌ 3 Fe(s) + 4 H₂O(g).
Try it first: Cross out every species that is a pure solid or liquid.
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Worked example 3
For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), Kc = 0.50 at 400 C (673 K). Find Kp.
Try it first: Work out Dn from the balanced equation before touching the calculator.
0 of 3 steps revealed.
Worked example 4
For NH₄HS(s) ⇌ NH₃(g) + H₂S(g), write Kp and predict the effect of adding more NH₄HS(s) at fixed T while solid remains.
Try it first: Only gaseous products appear in Kp.
0 of 3 steps revealed.
Common mistakes
Including a solid or pure liquid in the expression.
Why it's wrong: Its effective concentration is constant and already inside K.
Check instead: Scan the phase labels and delete every (s) and (l).
Multiplying by the coefficient instead of raising to it.
Why it's wrong: 2[SO₃] is not [SO₃]².
Check instead: Write the exponent directly from the balanced equation.
Putting reactants on top.
Why it's wrong: It inverts K, giving the reciprocal of the right answer.
Check instead: Products over reactants, every time.
Using Dn as total moles instead of moles of gas.
Why it's wrong: Only gases contribute the RT factor.
Check instead: Count only species labelled (g).
After the lesson · Question bank
Independent practice
Work through new problems on this topic on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
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What you should now know
The equilibrium expression puts products over reactants, each raised to its coefficient. Concentrations in mol/L give Kc; partial pressures give Kp. Pure solids and pure liquids, including the solvent water in dilute solution, are left out because their concentrations do not change as the reaction proceeds. Equilibria containing more than one phase are called heterogeneous, and it is these that most often lose marks for including a solid. Kc and Kp are related by Kp = Kc(RT)Δn, where Dn is the change in the number of moles of gas; when Dn = 0 the two constants are numerically equal.
- Products over reactants, coefficients as exponents
- Kc uses concentrations, Kp uses partial pressures
- Pure solids and liquids are omitted
- Kp = Kc(RT)Δn with Δn counting only gases
- Dn = 0 makes Kp equal to Kc
Sources and further reading
This lesson is original Chem Help content. No external sources were adapted.