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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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.
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.
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.
For N₂ + 3 H₂ ⇌ 2 NH₃, Kc = [NH₃]²/([N₂][H₂]³). The exponent three belongs to the whole concentration term [H₂], not to the subscript inside H₂. Use coefficients from the balanced equation and include dissolved ions as well as gases. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), the solid is omitted but both ions remain.
The thermodynamic K is dimensionless and uses activities: ideal gas activity is approximately P/P°, and dilute-solute activity approximately c/c°. Introductory Kc and Kp calculations often suppress those reference factors and use numerical molarities or pressures in specified units. Keep one convention throughout rather than mixing atm with bar or interpreting a dimensionless thermodynamic constant as carrying arbitrary units.
Kc = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
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)
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.
Water is omitted when it is a pure liquid or an effectively constant solvent in the dilute-solution approximation. Water vapor is a gas and must be included. A dissolved substance is not omitted just because it is a liquid in its pure state: ethanol concentration can matter in a solution equilibrium. The pure-phase rule refers to its actual role and phase in the written reaction.
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 is fixed by temperature while the required pure solid phases coexist; complete consumption of a phase changes the constraints.
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), equilibrium fixes the CO₂ pressure at a given T only while the required solid phases coexist. If all CaCO₃ is consumed before that pressure is attained, the assumed heterogeneous equilibrium cannot be maintained. Adding more of an already present pure solid changes neither its activity nor K, but removing an entire phase can invalidate that equilibrium condition. More exposed surface can change the speed of equilibration without changing the equilibrium relation.
Kc = [CO₂]
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.
The classroom ideal-gas conversion Kp = Kc(RT)Δn uses Δn = Σνgas,products − Σνgas,reactants. Count gases only: solids, liquids and aqueous species do not enter Δn. For 2 NO₂(g) ⇌ N₂O₄(g), Δn = 1 − 2 = −1, so Kp = Kc/(RT). For H₂ + I₂ ⇌ 2 HI, Δn = 0 and the corresponding numerical constants agree under matching conventions. Use kelvin, and choose R for the pressure and concentration units supplied.
Kp = Kc (RT)Δn
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.
Each concentration or pressure is raised to the power of its stoichiometric coefficient.
Their effective concentrations are constant and are absorbed into K.
An equilibrium involving more than one phase, so some species are excluded from the expression.
Moles of gaseous product minus moles of gaseous reactant in the balanced equation.
Equilibrium constants are reported without units because each term is really a ratio to a standard state.
Kc = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
Kp = (PCᶜ PDᵈ) / (PAᵃ PBᵇ)
Kp = Kc (RT)Δn
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.
0 of 3 steps revealed.
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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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.
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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.
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| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Including a solid or pure liquid in the expression. | Its effective concentration is constant and already inside K. | Scan the phase labels and delete every (s) and (l). |
| Multiplying by the coefficient instead of raising to it. | 2[SO₃] is not [SO₃]². | Write the exponent directly from the balanced equation. |
| Putting reactants on top. | It inverts K, giving the reciprocal of the right answer. | Products over reactants, every time. |
| Using Dn as total moles instead of moles of gas. | Only gases contribute the RT factor. | Count only species labelled (g). |
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After the lesson · Question bank
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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.
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