UNIT 2 CHEMISTRY · PEARSON AAQ · D1.4.3
Equilibrium Constant Expressions
FREE CHEMISTRY REVISION · TOPIC 45
Equilibrium Constant Expressions
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.4.3
What you will learn
Turn a balanced reversible equation into an equilibrium-constant expression, justify every exponent and omitted phase, and derive the units. K describes equilibrium composition at a stated temperature; it does not tell you how quickly equilibrium is reached.
1. What Kc and Kp mean
For a specified reaction at a fixed temperature, an equilibrium constant describes a particular ratio of equilibrium amounts. In this course, Kc uses equilibrium concentrations, usually in mol dm⁻³. Kp uses equilibrium gas partial pressures, in atm for Pearson calculations. The subscript is part of the name, not a power.
| Feature | Kc | Kp |
|---|---|---|
| Quantities entered | Equilibrium concentration [X] | Equilibrium partial pressure p(X) |
| Common units for inputs | mol dm⁻³ | atm |
| Included phases | Gases and dissolved solutes | Gaseous species |
| Pure separate solid/liquid phases | Omitted | Omitted |
| Temperature | State it; changing it can change K | State it; changing it can change K |
These simple expressions use the ideal-gas / sufficiently dilute-solution approximation. Pearson expects units from the concentration or pressure expression where they do not cancel. The more advanced thermodynamic equilibrium constant uses dimensionless activities; do not mix that convention with the units requested here.
At unchanged temperature, adding reactant, compressing a gas or using a catalyst does not change K under this treatment. The mixture can readjust to satisfy the same K. Temperature changes can change K. Kc and Kp need not have the same numerical value.
2. Build the expression from the equation
For aA + bB ⇌ cC + dD, when all species belong in the concentration expression: Kc = ([C]c[D]d)/([A]a[B]b). Multiply product terms in the numerator and reactant terms in the denominator. A coefficient becomes a power on the entire concentration or partial pressure. A missing written coefficient means 1.
- Balance atoms and charge, then retain state symbols.
- Identify which phases are included.
- Put included products above included reactants.
- Use balanced coefficients as exponents.
- Substitute units into exactly the same expression.
Equilibrium exponents follow the balanced equation. This is different from empirical rate-equation orders, which must be found experimentally. Do not infer a multistep reaction’s rate equation from its overall balanced equation.
3. Worked homogeneous derivations
Example 1 — equal total powers
H₂(g) + I₂(g) ⇌ 2HI(g). All gases are included. Product power is 2; each reactant power is 1.
Kc = [HI]² / ([H₂][I₂]). Units: (mol dm⁻³)²/(mol dm⁻³)² = 1, so no units.
Kp = p(HI)² / (p(H₂)p(I₂)). atm²/atm² cancels. K = 1 would mean the product of the denominator terms equals the numerator, not that all three concentrations are equal.
Example 2 — a negative unit power
N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Kc = [NH₃]²/([N₂][H₂]³). The total included product power is 2 and reactant power is 4.
Units = (mol dm⁻³)2−4 = dm⁶ mol⁻². Kp = p(NH₃)²/(p(N₂)p(H₂)³), with units atm⁻².
Example 3 — dissociation with a positive power
N₂O₄(g) ⇌ 2NO₂(g). Kc = [NO₂]²/[N₂O₄]; units (mol dm⁻³)2−1 = mol dm⁻³. Kp = p(NO₂)²/p(N₂O₄), with units atm. Both powers apply to the measured equilibrium values, not to initial values.
Example 4 — homogeneous aqueous equilibrium
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). All three dissolved species enter Kc = [FeSCN²⁺]/([Fe³⁺][SCN⁻]). Units = (mol dm⁻³)1−2 = dm³ mol⁻¹. Charge balances: +3 − 1 = +2. Kp is not an expression for these aqueous ion concentrations.
4. Heterogeneous equilibria and pure phases
Homogeneous means all species are in one phase; heterogeneous means more than one phase is present. The activity of a pure solid or a pure liquid in its standard state is taken as 1. It therefore introduces no variable factor in the equilibrium expression. At this level, this is often explained by its effectively constant concentration/density. Omit the term; do not substitute zero.
This rule applies to a pure phase, not to every substance with an (l) label in a liquid mixture. Dissolved substances must not be omitted just because they are in a liquid solution. Water acting as the bulk solvent in a dilute aqueous equilibrium is usually incorporated into the constant, provided its activity is effectively constant.
Example 5 — two solids and a gas
CaCO₃(s) ⇌ CaO(s) + CO₂(g). Omit both pure solids. Kc = [CO₂], units mol dm⁻³; Kp = p(CO₂), units atm. This does not mean the solids are unimportant: the stated equilibrium requires both solid phases to be present.
Example 6 — pure liquid and vapour
H₂O(l) ⇌ H₂O(g), with pure liquid water present. The liquid activity is 1. Kp = p(H₂O), units atm; Kc = [H₂O(g)], units mol dm⁻³ in the concentration convention. This pressure is the equilibrium vapour pressure at that temperature. Do not omit the gaseous water term.
Example 7 — solid carbon and two gases
C(s) + CO₂(g) ⇌ 2CO(g). Omit C(s) only. Kp = p(CO)²/p(CO₂), units atm. Kc = [CO]²/[CO₂], units mol dm⁻³. The carbon coefficient does not contribute to the unit exponent because its term is absent.
5. Partial pressure and unit shortcuts
A gas’s partial pressure is the pressure it would exert alone in the same volume at the same temperature. For an ideal mixture, p(X) = x(X)Ptotal, where x(X) = n(X)/ntotal. Mole fractions add to 1; partial pressures add to total pressure. An inert gas contributes to total moles and total pressure but has no reaction term in Kp.
Example 8 — choose the correct pressure
A mixture contains 0.200 mol NO₂, 0.300 mol N₂O₄ and 0.500 mol inert gas at total pressure 4.00 atm. Total amount = 1.000 mol. p(NO₂) = 0.200 × 4.00 = 0.800 atm; p(N₂O₄) = 0.300 × 4.00 = 1.20 atm. Use these individual pressures in Kp, not 4.00 atm for each gas.
If all concentration inputs use mol dm⁻³, the units of Kc are (mol dm⁻³)Δν, where Δν = total included product powers − total included reactant powers. For Kp using atm, units are atmΔn(gas). Count only included species.
| Exponent difference | Kc units | Kp units |
|---|---|---|
| 0 | No units | No units |
| +1 | mol dm⁻³ | atm |
| −1 | dm³ mol⁻¹ | atm⁻¹ |
| −2 | dm⁶ mol⁻² | atm⁻² |
6. Interpret the magnitude carefully
For a fixed equation and stated unit convention, a large K means the product-side activity/concentration/pressure terms are favoured relative to the reactant-side terms; a small K favours reactant-side terms. It does not measure rate. A large K can accompany a very slow reaction. K is not a percentage yield, and its value alone does not determine every species concentration.
For the simple ideal A ⇌ B equilibrium, Kc = [B]/[A]: Kc = 100 means [B] = 100[A], whereas Kc = 0.0100 means [B] = 0.0100[A]. For more complex expressions, do not translate K directly into a product:reactant particle ratio. Different powers, total amounts and pressure/volume matter. A dimensional K also depends numerically on the units used, so compare values only using the same equation and convention.
Example 9 — rewrite the reaction, rewrite K
If H₂ + I₂ ⇌ 2HI has K = 25.0, reversing the equation gives K = 1/25.0 = 0.0400. Doubling every coefficient gives K = (25.0)² = 625. The chemistry at that temperature has not changed; the definition of the expression has.
Common misconceptions
- Use initial concentrations: K requires equilibrium values.
- Multiply by the coefficient: 2HI gives [HI]², not 2[HI].
- Include solids as zero: omit their pure-phase factors, which are 1.
- Omit every liquid: only pure liquid phases, or effectively constant solvent activity, qualify.
- K = 1 means equal concentrations: it means numerator and denominator terms are equal.
- Large K means fast: equilibrium composition and reaction rate are separate.
- Every K is unitless in this course: derive the requested Kc/Kp units from the expression.
Examination tips
- State the balanced reaction and temperature.
- Write the symbolic expression before putting in numbers.
- Bracket the full denominator and all powers.
- Show unit cancellation on a separate line.
- Use atm for Pearson partial-pressure calculations.
- Identify the pure phases and say why they are omitted.
Check your understanding
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