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The equilibrium constant Kc converts the qualitative language of Le Chatelier's principle into a single quantitative number that pins down exactly where a reversible reaction settles at a given temperature. Where the previous lesson asked which way the equilibrium would shift, this lesson asks by how much — and answers in mol dm⁻³ units that drop out of the balanced equation itself. The work here is threefold: constructing Kc expressions from a balanced equation (including the rule that pure solids and pure liquids are omitted because their activities equal one); calculating Kc from equilibrium concentrations using ICE (Initial-Change-Equilibrium) tables; and deducing the units by inspecting the stoichiometric mole change Δn = (c + d) − (a + b). We also examine how Kc responds (or does not respond) to changes in concentration, pressure, temperature, and catalyst, and how the reaction quotient Q lets us predict the direction of shift before equilibrium is reached.
Spec mapping (AQA 7405): This lesson maps to §3.1.6 (chemical equilibria, Le Chatelier's principle, and Kc). The qualitative groundwork was laid in lesson 4 (Le Chatelier's principle); lesson 6 extends Kc to the gas-phase analogue Kp using partial pressures; lesson 7 anchors Required Practical 7, the experimental determination of Kc for the esterification of ethanol with ethanoic acid. The thermodynamic underpinning ΔG° = −RT ln K is covered in §3.1.4 lesson 5 (free energy and entropy). Refer to the official AQA specification document for the exact wording.
Assessment objectives: AO1 requires you to write the Kc expression directly from a balanced equation and to state and assign its units. AO2 dominates the marks: ICE-table set-up, division of moles by container volume to obtain concentrations, substitution into Kc, and the reverse direction (computing an unknown equilibrium concentration from a known Kc) all feature heavily on Paper 2. AO3 demands interpretation — predicting whether products or reactants are favoured from the magnitude of Kc, predicting the direction of shift from the reaction quotient Q, and explaining why Kc changes with temperature (via ΔG° = −RT ln K) but not with concentration, pressure, or catalyst.
For the general reversible reaction
aA + bB ⇌ cC + dD
the equilibrium constant in terms of concentrations is
Kc = [C]^c [D]^d / ([A]^a [B]^b)
where each square-bracketed term is the equilibrium concentration of that species in mol dm⁻³, raised to the power of its stoichiometric coefficient. The convention is products on top, reactants on the bottom — fixed by the way Guldberg and Waage originally framed the law of mass action in the 1860s.
Three rules of construction matter at A-Level:
Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Kc = [NH₃]² / ([N₂] · [H₂]³)
Contact process step: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]² · [O₂])
Hydrogen iodide equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g)
Kc = [HI]² / ([H₂] · [I₂])
Decomposition of dinitrogen tetroxide: N₂O₄(g) ⇌ 2NO₂(g)
Kc = [NO₂]² / [N₂O₄]
Esterification (homogeneous liquid mixture): CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
Kc = ([CH₃COOC₂H₅] · [H₂O]) / ([CH₃COOH] · [C₂H₅OH])
Thermal decomposition of calcium carbonate (heterogeneous, with a pure solid and a pure gas): CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Kc = [CO₂]
Each expression is constructed by reading the stoichiometric coefficients off the balanced equation; no other information is needed.
The units of Kc are not fixed — they depend on the stoichiometry. To derive them, substitute the unit (mol dm⁻³) into each concentration term and simplify.
Define Δn = (c + d) − (a + b), the change in total moles of (counted) species from reactants to products. The unit of Kc is then (mol dm⁻³)^Δn.
| Reaction | Δn | Units of Kc |
|---|---|---|
| N₂ + 3H₂ ⇌ 2NH₃ | 2 − 4 = −2 | (mol dm⁻³)⁻² = mol⁻² dm⁶ |
| 2SO₂ + O₂ ⇌ 2SO₃ | 2 − 3 = −1 | (mol dm⁻³)⁻¹ = mol⁻¹ dm³ |
| H₂ + I₂ ⇌ 2HI | 2 − 2 = 0 | dimensionless (no units) |
| N₂O₄ ⇌ 2NO₂ | 2 − 1 = +1 | mol dm⁻³ |
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | only 1 gaseous term counted | mol dm⁻³ |
Notice that for H₂ + I₂ ⇌ 2HI the powers cancel — two on top, two on the bottom — so Kc is a pure number. For the Haber synthesis four powers sit on the bottom and only two on top, leaving units of mol⁻² dm⁶. Always work the units out explicitly; do not memorise them.
Exam Tip: When asked for "Kc with units", show the substitution: write Kc = (mol dm⁻³)² / [(mol dm⁻³) · (mol dm⁻³)³] = (mol dm⁻³)⁻² = mol⁻² dm⁶. Marks are often allocated specifically for the working, not only the final unit.
An ICE table organises the bookkeeping required to compute Kc from initial amounts: rows for Initial moles (or concentrations), Change (using stoichiometry), and Equilibrium values. The three steps are:
The single most common error in A-Level Kc questions is forgetting to divide by volume. Moles are not concentrations. If the question states "in a 2.00 dm³ container", every equilibrium mole figure must be divided by 2.00 before going into Kc.
A mixture of 1.00 mol of H₂(g) and 1.00 mol of I₂(g) is sealed in a 2.00 dm³ container and heated to 700 K. At equilibrium, the concentration of HI is found to be 0.780 mol dm⁻³. Calculate Kc.
Step 1 — Initial concentrations. [H₂]ᵢ = 1.00/2.00 = 0.500 mol dm⁻³; [I₂]ᵢ = 0.500 mol dm⁻³; [HI]ᵢ = 0.
Step 2 — ICE table in mol dm⁻³. Let x be the decrease in [H₂].
| H₂ | I₂ | HI | |
|---|---|---|---|
| Initial (mol dm⁻³) | 0.500 | 0.500 | 0 |
| Change (mol dm⁻³) | −x | −x | +2x |
| Equilibrium (mol dm⁻³) | 0.500 − x | 0.500 − x | 2x |
Step 3 — Use the given [HI] to find x. 2x = 0.780, so x = 0.390 mol dm⁻³.
Step 4 — Equilibrium concentrations. [H₂] = 0.500 − 0.390 = 0.110 mol dm⁻³. [I₂] = 0.110 mol dm⁻³. [HI] = 0.780 mol dm⁻³.
Step 5 — Substitute into Kc.
Kc = [HI]² / ([H₂] · [I₂]) = (0.780)² / (0.110 × 0.110) = 0.6084 / 0.0121 = 50.3 (no units, since Δn = 0).
The value 50.3 is the standard A-Level figure for this equilibrium at 700 K. Kc > 1 indicates that products are favoured, though not overwhelmingly — the equilibrium mixture still contains substantial unreacted H₂ and I₂.
Ethanoic acid and ethanol react reversibly to form ethyl ethanoate and water:
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
In a sealed flask, 1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed in a total volume of 0.100 dm³. At equilibrium (room temperature, with a trace of acid catalyst), 0.667 mol of ethyl ethanoate is present. Calculate Kc.
Equilibrium moles:
Kc = ([CH₃COOC₂H₅] · [H₂O]) / ([CH₃COOH] · [C₂H₅OH]) = (6.67 × 6.67)/(3.33 × 3.33) = 44.5 / 11.1 = 4.01 (no units, since Δn = 0).
The standard textbook value is approximately 4. Note that because volume cancels (same number of terms top and bottom), the answer would be identical if we worked in moles directly — for this special case only.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
At 700 K and equilibrium, a 1.00 dm³ vessel contains 0.100 mol N₂, 0.300 mol H₂, and 0.400 mol NH₃. Calculate Kc.
Concentrations are numerically equal to moles (V = 1.00 dm³).
Kc = [NH₃]² / ([N₂] · [H₂]³) = (0.400)² / (0.100 × (0.300)³) = 0.160 / (0.100 × 0.0270) = 0.160 / 0.00270 = 59.3 mol⁻² dm⁶.
Units: (mol dm⁻³)² / [(mol dm⁻³) · (mol dm⁻³)³] = (mol dm⁻³)⁻² = mol⁻² dm⁶. The large Kc indicates that, at 700 K with these initial amounts, ammonia is significantly favoured. (Industrial conditions trade off Kc against rate — the Haber process operates at ~700 K because lower temperatures, although giving larger Kc, are kinetically too slow.)
N₂O₄(g) ⇌ 2NO₂(g), Kc = 0.200 mol dm⁻³ at 350 K.
A 2.00 dm³ flask is charged with 0.500 mol of N₂O₄ and allowed to reach equilibrium. Calculate the equilibrium [NO₂].
ICE table in mol dm⁻³, with x = decrease in [N₂O₄]:
| N₂O₄ | NO₂ | |
|---|---|---|
| Initial | 0.250 | 0 |
| Change | −x | +2x |
| Equilibrium | 0.250 − x | 2x |
Kc = (2x)² / (0.250 − x) = 0.200
4x² = 0.200 × (0.250 − x) = 0.0500 − 0.200x
4x² + 0.200x − 0.0500 = 0
Quadratic formula: x = [−0.200 ± √(0.0400 + 0.800)] / 8 = [−0.200 ± √0.840] / 8 = [−0.200 ± 0.9165] / 8.
Take the positive root: x = 0.7165/8 = 0.0896 mol dm⁻³.
Equilibrium [NO₂] = 2x = 0.179 mol dm⁻³. Equilibrium [N₂O₄] = 0.250 − 0.0896 = 0.160 mol dm⁻³.
Check: Kc = (0.179)²/0.160 = 0.0320/0.160 = 0.200 ✓.
The fundamental statement: Kc depends ONLY on temperature. Concentration, pressure, and catalyst changes do not alter Kc.
| Change | Effect on Kc | Effect on equilibrium position | Reasoning |
|---|---|---|---|
| Concentration change | No change | Position shifts (Le Chatelier) | The same Kc must be restored; concentrations of every species readjust so the ratio is unchanged. |
| Total pressure change (gases) | No change | Position shifts toward the side with fewer moles of gas (if Δn ≠ 0) | The ratio in Kc is restored by a shift, not by a change in Kc itself. For liquid- or solution-phase reactions, modest pressure changes have negligible effect. |
| Temperature increase (exothermic forward) | Kc decreases | Position shifts backward, toward reactants | ΔG° = −RT ln K. For exothermic forward, ΔH° < 0; the van't Hoff isobar d(ln K)/dT = ΔH°/RT² gives a negative slope, so ln K (and hence K) falls as T rises. |
| Temperature increase (endothermic forward) | Kc increases | Position shifts forward, toward products | Same van't Hoff relation with ΔH° > 0 gives a positive slope. |
| Catalyst added | No change | None — equilibrium reached faster but same composition | A catalyst lowers Ea for both forward and reverse equally; both rates increase by the same factor, so the equilibrium ratio (and Kc) is unchanged. |
Key Point: When a question asks "how does Kc change?" the answer is always either "it does not change" (for concentration, pressure, or catalyst) or "it changes because temperature changed". Anything else is wrong.
The thermodynamic root of why temperature alone alters Kc is the relation ΔG° = −RT ln K (covered in §3.1.4 lesson 5). Both ΔG° and T appear, so K depends on T — but no other intensive variable enters. Specifically, the van't Hoff equation in differential form,
d(ln K)/dT = ΔH°/(RT²),
shows directly that the sign of ΔH° fixes the direction of change: ΔH° < 0 (exothermic forward) gives a negative slope (K falls as T rises); ΔH° > 0 (endothermic forward) gives a positive slope.
The numerical value of Kc tells you immediately where the equilibrium lies:
The size of Kc does NOT tell you how fast equilibrium is reached. A reaction can have an enormous Kc but a tiny rate constant — the combustion of paper in air at 298 K is thermodynamically essentially complete (Kc >> 1) but kinetically arrested without ignition. Magnitude of Kc and rate are independent properties; one is set by ΔG°, the other by Ea.
The reaction quotient Q has the same algebraic form as Kc but uses the current (not necessarily equilibrium) concentrations:
Q = [C]^c [D]^d / ([A]^a [B]^b) (current concentrations)
Comparing Q with K at the same temperature predicts the direction of shift required to reach equilibrium:
Worked use of Q. For H₂(g) + I₂(g) ⇌ 2HI(g) with Kc = 50.3 at 700 K, suppose [H₂] = 0.200, [I₂] = 0.200, [HI] = 0.500 mol dm⁻³.
Q = (0.500)² / (0.200 × 0.200) = 0.250/0.0400 = 6.25.
Since Q (6.25) < K (50.3), the reaction must proceed forward to generate more HI, raising Q until it equals K. No new information about temperature is needed — Q vs K is a snapshot test of direction.
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