UNIT 2 CHEMISTRY · PEARSON AAQ · D1.4.4
Calculations Involving Equilibrium Constants
FREE CHEMISTRY REVISION · TOPIC 46
Calculations Involving Equilibrium Constants
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.4.4
What you will learn
Calculate Kc and Kp from equilibrium data, derive units from each expression, and rearrange to find a missing concentration or partial pressure. Work from the balanced equation before touching the calculator. All numerical values below are supplied illustrative teaching data at a fixed stated temperature, not a table of measured constants for these reactions.
1. A reliable calculation method
- Balance the equation, including state symbols.
- Write the expression and omit pure solid/liquid phases.
- Identify equilibrium values; use stoichiometry if only initial amounts and a change are given.
- Convert amount to concentration using [X] = n(X)/V in dm³, or gas amount to partial pressure using p(X) = n(X)Ptotal/ntotal.
- Substitute with brackets, evaluate powers, then divide.
- Derive units separately and round the final answer.
- Check non-negative amounts, conserved atoms, and consistency with the given K.
Keep extra digits during working. Report a suitable number of significant figures, usually matching the least precise supplied measured quantity. Exact stoichiometric coefficients do not limit significant figures. Never round an intermediate concentration so early that it changes the final result.
2. Direct homogeneous Kc calculations
Example 1 — a ratio with cancelling units
At a fixed temperature, H₂(g) + I₂(g) ⇌ 2HI(g) has equilibrium concentrations [H₂] = 0.200, [I₂] = 0.100 and [HI] = 0.800 mol dm⁻³.
Expression: Kc = [HI]²/([H₂][I₂]).
Substitution: Kc = 0.800²/(0.200 × 0.100) = 0.640/0.0200 = 32.0.
Units: (mol dm⁻³)²/(mol dm⁻³)² cancels. The answer has no units; 32.0 has three significant figures.
Example 2 — coefficients and inverse-square units
N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At equilibrium: [N₂] = 0.200, [H₂] = 0.300 and [NH₃] = 0.400 mol dm⁻³.
Kc = [NH₃]²/([N₂][H₂]³) = 0.400²/(0.200 × 0.300³) = 0.160/0.00540 = 29.6296…
Kc = 29.6 dm⁶ mol⁻² (3 s.f.). Units: (mol dm⁻³)2−4. Cubing 0.300 is essential; multiplying it by 3 gives the wrong denominator.
Example 3 — equilibrium moles are not concentrations
N₂O₄(g) ⇌ 2NO₂(g). A 2.00 dm³ vessel contains 0.400 mol N₂O₄ and 0.600 mol NO₂ at equilibrium.
| Species | Equilibrium amount / mol | Equilibrium concentration / mol dm⁻³ |
|---|---|---|
| N₂O₄ | 0.400 | 0.400/2.00 = 0.200 |
| NO₂ | 0.600 | 0.600/2.00 = 0.300 |
Kc = [NO₂]²/[N₂O₄] = 0.300²/0.200 = 0.450 mol dm⁻³. Using amounts directly gives 0.900 mol, which is not Kc. Because the total powers differ, the volume does not cancel.
3. Use a stoichiometric equilibrium table
An initial–change–equilibrium table separates the quantities present before and after the reaction readjusts. The changes must have the balanced mole ratio. Do not put an initial amount directly into K unless it is also the equilibrium amount.
Example 4 — deduce equilibrium composition first
H₂(g) + I₂(g) ⇌ 2HI(g) in a 2.00 dm³ vessel. Initially there are 1.00 mol H₂, 1.00 mol I₂ and no HI. At equilibrium, 0.400 mol I₂ remains.
I₂ consumed = 1.00 − 0.400 = 0.600 mol. The 1:1:2 ratio means 0.600 mol H₂ is consumed and 1.20 mol HI is formed.
| Amount / mol | H₂ | I₂ | HI |
|---|---|---|---|
| Initial | 1.00 | 1.00 | 0 |
| Change | −0.600 | −0.600 | +1.20 |
| Equilibrium | 0.400 | 0.400 | 1.20 |
Divide by 2.00 dm³: [H₂] = [I₂] = 0.200 and [HI] = 0.600 mol dm⁻³. Kc = 0.600²/(0.200 × 0.200) = 9.00, no units.
Check hydrogen atoms in moles: initially 2.00, finally 2(0.400) + 1.20 = 2.00. Iodine gives the same result. Every final amount is non-negative.
4. Heterogeneous Kc and Kp
Example 5 — a heterogeneous concentration constant
CaCO₃(s) ⇌ CaO(s) + CO₂(g), with both pure solids present. At a fixed temperature the equilibrium CO₂ concentration is 0.0120 mol dm⁻³.
Omit both solid factors: Kc = [CO₂] = 0.0120 mol dm⁻³. Do not divide by the mass of CaCO₃. The expression relies on coexistence of both solid phases.
Example 6 — heterogeneous gas pressures with a squared term
C(s) + CO₂(g) ⇌ 2CO(g), with solid carbon present. Equilibrium pressures are p(CO) = 0.600 atm and p(CO₂) = 0.300 atm.
Kp = p(CO)²/p(CO₂) = 0.600²/0.300 = 1.20 atm. Units: atm²/atm. The pure solid is omitted; both gases remain.
5. Calculate Kp from an equilibrium gas mixture
For an ideal gas mixture, the mole fraction of X is n(X)/ntotal and p(X) = x(X)Ptotal. Sum all gas amounts, including inert gases. Solids are not part of the gas total. Use each species’ own partial pressure, never the total pressure in every position.
Example 7 — total pressure, mole fractions and an inert gas
N₂O₄(g) ⇌ 2NO₂(g). At equilibrium there are 0.200 mol N₂O₄, 0.600 mol NO₂ and 0.200 mol inert Ar. Total pressure is 5.00 atm.
| Gas | Amount / mol | Mole fraction | Partial pressure / atm |
|---|---|---|---|
| N₂O₄ | 0.200 | 0.200/1.000 = 0.200 | 1.00 |
| NO₂ | 0.600 | 0.600/1.000 = 0.600 | 3.00 |
| Ar | 0.200 | 0.200/1.000 = 0.200 | 1.00 |
Pressure check: 1.00 + 3.00 + 1.00 = 5.00 atm. Kp = p(NO₂)²/p(N₂O₄) = 3.00²/1.00 = 9.00 atm. Ar enters the mole-fraction denominator but is absent from the reaction expression.
The supplied equilibrium composition need not have been produced from pure N₂O₄; without initial data, do not invent an initial mixture.
6. Find a missing equilibrium quantity
Example 8 — solve for an equilibrium concentration
H₂(g) + I₂(g) ⇌ 2HI(g). At the stated temperature Kc = 16.0, [H₂] = 0.100 and [I₂] = 0.400 mol dm⁻³. Find [HI].
Kc = [HI]²/([H₂][I₂]) ⇒ [HI]² = Kc[H₂][I₂] = 16.0 × 0.100 × 0.400 = 0.640 (mol dm⁻³)².
[HI] = √0.640 = 0.800 mol dm⁻³ (3 s.f.). Use the positive root: a negative concentration is physically impossible.
Back-substitution: 0.800²/(0.100 × 0.400) = 16.0, as required.
Example 9 — solve for a gas partial pressure
N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Kp = 0.250 atm⁻²; equilibrium p(N₂) = 2.00 atm and p(H₂) = 4.00 atm. Find p(NH₃).
p(NH₃)² = Kp × p(N₂) × p(H₂)³ = 0.250 × 2.00 × 4.00³ = 32.0 atm².
p(NH₃) = √32.0 = 5.66 atm (3 s.f.). Units under the root: atm⁻² × atm × atm³ = atm². Back-substitution using the unrounded root gives 0.250 atm⁻².
If these are the only gases, the implied total pressure is 2.00 + 4.00 + 5.65685… = 11.7 atm (3 s.f.). A partial pressure may exceed another gas’s partial pressure, but not the total.
Example 10 — find the equilibrium change from K
H₂(g) + I₂(g) ⇌ 2HI(g). Initially [H₂] = [I₂] = 0.500 mol dm⁻³ and [HI] = 0. At this temperature Kc = 4.00. The volume is constant. Let x mol dm⁻³ of each reactant be consumed.
| Concentration / mol dm⁻³ | H₂ | I₂ | HI |
|---|---|---|---|
| Initial | 0.500 | 0.500 | 0 |
| Change | −x | −x | +2x |
| Equilibrium | 0.500 − x | 0.500 − x | 2x |
4.00 = (2x)²/(0.500 − x)². Physical range: 0 ≤ x < 0.500. Since both numerator and denominator are positive, take the positive square root: 2.00 = 2x/(0.500 − x).
2.00(0.500 − x) = 2x ⇒ 1.00 = 4x ⇒ x = 0.250 mol dm⁻³. Thus [H₂] = [I₂] = 0.250 and [HI] = 0.500 mol dm⁻³.
Check: 0.500²/(0.250²) = 4.00. All concentrations are non-negative and x is within the permitted range. Do not accept an algebraic root that consumes more reactant than was supplied.
7. Check whether the information is sufficient
One equilibrium expression is one relationship. If two independent unknown concentrations remain and no stoichiometric balance, total pressure, total concentration or other relationship is given, a unique answer usually cannot be calculated. State what additional information is needed instead of guessing.
| Problem signal | Check or correction |
|---|---|
| Given moles and volume | Convert each relevant amount to mol dm⁻³. |
| Given equilibrium gas amounts and total pressure | Calculate mole fractions including all gases. |
| Given initial quantities | Find the equilibrium change first. |
| Negative final amount | Reject the root or revisit the stoichiometry. |
| A pressure larger than the stated total | Check mole fractions and arithmetic. |
| Unit power does not match the expression | Recount included product and reactant powers. |
Common misconceptions
- Use 3[H₂] instead of [H₂]³: the coefficient is a power.
- Use initial concentrations in K: determine equilibrium values.
- Cancel volume automatically: it cancels only when total concentration powers match.
- Include the mass of a solid: pure solid activity is omitted.
- Ignore an inert gas when finding mole fractions: it contributes to total gas amount.
- Use a negative square root: concentrations and pressures cannot be negative.
- Use rounded intermediate values: retain working digits until the final line.
Examination tips
- Label every row of an equilibrium table and show signed changes.
- Write a separate unit line.
- Show the rearranged expression before substitution.
- Check the final answer by putting it back into the original expression.
- Quote the supplied temperature, or state that all data refer to the same fixed temperature.
Check your understanding
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