Edexcel International A Level Chemistry: Chemical equilibria (YCH11) – Revision Notes

These notes are the short version. Every idea here is explained step by step, with longer examples, in the chemical equilibria study guide.

They cover Topic 13: Chemical Equilibria of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 13.1 to 13.9. Everything here is IA2 content from Unit 4: Rates, Equilibria and Further Organic Chemistry. It extends the Unit 2 work in Topic 9B. Pair the notes with the practice questions, then use the chemistry checklist, the Edexcel course hub and our diagnostics to plan what to revise next.

Key definitions

Term	Meaning
Homogeneous equilibrium	Every reactant and product is in one phase
Heterogeneous equilibrium	The reacting species are spread over more than one phase
Kc	K written with equilibrium concentrations, each in mol dm⁻³
Kp	K written with equilibrium partial pressures, each in atm
Mole fraction, x(A)	Moles of gas A ÷ total moles of gas
Partial pressure, p(A)	x(A) × total pressure; the pressure A would exert alone
Position of equilibrium	How much of the mixture is on the reactant side and how much on the product side

Formulae at a glance

Quantity	Expression
Kc for aA + bB ⇌ cC + dD	[C]^c [D]^d / ([A]^a [B]^b)
Kp for the same reaction (all gases)	as Kc, but with p(X) in place of every [X]
Partial pressure	p(A) = x(A) × P(total)
Sum of partial pressures	p(A) + p(B) + … = P(total)
Entropy of surroundings	ΔSsurroundings = −ΔH / T
Total entropy	ΔStotal = ΔSsystem + ΔSsurroundings
Link to K	ΔStotal = R lnK, with R = 8.31 J K⁻¹ mol⁻¹

13.1 and 13.2: building expressions

Method in steps

1. Write the balanced equation with state symbols.
2. Cross out every solid. For Kp, keep only gases.
3. Products on top, reactants underneath, coefficients as powers.
4. Units: replace each term with mol dm⁻³ (Kc) or atm (Kp) and cancel.

Quick reminders

• N₂(g) + 3H₂(g) ⇌ 2NH₃(g): four gas moles go to two, so Kc has units dm⁶ mol⁻² and Kp has units atm⁻².
• Same number of gas moles on each side: no units.
• Cu(s) + 2Ag⁺(aq) ⇌ Cu²⁺(aq) + 2Ag(s): Kc = [Cu²⁺] / [Ag⁺]², units dm³ mol⁻¹.
• PbCO₃(s) ⇌ PbO(s) + CO₂(g): Kp = p(CO₂), units atm.

13.3: calculating K from data

Method in steps

1. Set up an initial / change / equilibrium table in moles.
2. Use the reaction ratio for the change row.
3. For Kc, divide by volume (unless it cancels). For Kp, find total moles, then mole fractions, then partial pressures.
4. Substitute, calculate, give units.

Small worked reminder (Kp). C₂H₄(g) + H₂(g) ⇌ C₂H₆(g). At equilibrium p(C₂H₄) = 0.40 atm, p(H₂) = 0.25 atm and p(C₂H₆) = 1.10 atm.

Kp = 1.10 / (0.40 × 0.25) = 11.0 atm⁻¹

Small worked reminder (partial pressure). A gas with mole fraction 0.35 in a mixture at 4.0 atm has p = 0.35 × 4.0 = 1.4 atm.

Small worked reminder (Kc with a volume). For 2X(g) ⇌ Y(g), a 2.0 dm³ flask holds 0.40 mol X and 0.15 mol Y at equilibrium. [X] = 0.40 / 2.0 = 0.20 mol dm⁻³ and [Y] = 0.15 / 2.0 = 0.075 mol dm⁻³. Kc = 0.075 / 0.20² = 1.9 dm³ mol⁻¹. Using moles directly would give 0.94, which is wrong.

When V drops out of Kc. If the indices above and below add to the same total, amounts in moles can go straight in. The ester equilibrium, with two species each side, is the standard case. Otherwise, convert every amount to mol dm⁻³ first.

13.4 and 13.5: changing conditions at constant temperature

Change	Composition	Value of K
Add a reactant	Shifts towards products	Unchanged
Remove a product	Shifts towards products	Unchanged
Increase total pressure	Shifts to the side with fewer gas moles	Unchanged
Increase pressure, equal gas moles each side	No change	Unchanged
Add a catalyst	No change; equilibrium reached sooner	Unchanged
Add more of a solid	No change	Unchanged

How to explain it with K. After the change, the fraction in the K expression no longer equals K. The mixture reacts until it does. For example, doubling the pressure on 2A(g) ⇌ B(g) halves the fraction p(B) / p(A)², so more A turns into B.

13.6 and 13.7: temperature changes K

Forward reaction	Raise T	Lower T
Exothermic	K smaller; equilibrium moves left	K larger; equilibrium moves right
Endothermic	K larger; equilibrium moves right	K smaller; equilibrium moves left

The order of the explanation matters: temperature changes K first, and the composition then changes so that the fraction matches the new K.

13.8: entropy, temperature and K

• ΔSsurroundings = −ΔH / T, so its size falls as T rises.
• Exothermic: ΔSsurroundings positive, gets smaller with rising T → ΔStotal falls → K falls.
• Endothermic: ΔSsurroundings negative, gets less negative → ΔStotal rises → K rises.
• ΔStotal = R lnK, so K = e^(ΔStotal / R).

Small worked reminder (temperature). For a reaction with ΔH = −40 kJ mol⁻¹, ΔSsurroundings = 40 000 / 400 = +100 J K⁻¹ mol⁻¹ at 400 K but only 40 000 / 500 = +80 J K⁻¹ mol⁻¹ at 500 K. ΔStotal is 20 J K⁻¹ mol⁻¹ lower at the higher temperature, so K is smaller.

Small worked reminder (finding K). ΔStotal = +45 J K⁻¹ mol⁻¹ gives lnK = 45 / 8.31 = 5.42, so K ≈ 2.2 × 10².

13.9: extent of reaction

K	ΔStotal / J K⁻¹ mol⁻¹ (approx.)	Extent
> 10¹⁰	> +200	Goes essentially to completion
10⁻¹⁰ to 10¹⁰	−200 to +200	Equilibrium mixture
< 10⁻¹⁰	< −200	Barely reacts

These are rough guides. A large K does not mean a fast reaction: kinetic stability (outcome 12.11) can stop a thermodynamically favoured change.

Must-know distinctions

• Position vs constant. Concentration and pressure move the position; only temperature changes K.
• Homogeneous vs heterogeneous. One phase or more than one. Solids never appear in K.
• Kc vs Kp. Concentrations in mol dm⁻³ vs partial pressures in atm. Kp uses gases only.
• Moles vs concentrations. Kc needs concentrations unless the volume cancels.
• Extent vs rate. K tells you how far, not how fast.
• ΔSsystem vs ΔStotal. Only ΔStotal is linked to K by R lnK.

Further suggested practicals: one-line summaries

• Ethanol + ethanoic acid: titrate the equilibrium mixture with NaOH; allow for the acid catalyst.
• Fe²⁺ + Ag⁺: titrate the silver ions left with thiocyanate.
• Distribution of ammonia or iodine between immiscible solvents: titrate both layers; the ratio of concentrations is constant.
• Ammonium chloride: solid in equilibrium with two gases.
• 2NO₂ ⇌ N₂O₄: colour in a syringe shows temperature and pressure effects.

Quick self-test

1. Write Kc for 2NH₃(g) ⇌ N₂(g) + 3H₂(g) and give its units.
2. Write Kp for CuSO₄·5H₂O(s) ⇌ CuSO₄·3H₂O(s) + 2H₂O(g) and give its units.
3. A gas has mole fraction 0.20 in a mixture at 2.5 atm. What is its partial pressure?
4. For A(g) ⇌ 2B(g), [A] = 0.50 mol dm⁻³ and [B] = 0.30 mol dm⁻³ at equilibrium. Calculate Kc with units.
5. A catalyst is added to an equilibrium mixture. State the effect on Kc.
6. An equilibrium has an exothermic forward reaction. How does heating it affect Kp?
7. ΔStotal = −12 J K⁻¹ mol⁻¹ at a certain temperature. Calculate K.
8. K = 1 × 10⁻¹⁵ for a reaction. Calculate ΔStotal and state the extent of reaction.
9. For an endothermic reaction, state how raising the temperature affects ΔSsurroundings and ΔStotal.
10. Why does adding more Fe(s) not affect the equilibrium 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)?

Answers

1. Kc = [N₂][H₂]³ / [NH₃]²; units mol² dm⁻⁶.
2. Kp = p(H₂O)²; units atm².
3. 0.20 × 2.5 = 0.50 atm.
4. Kc = 0.30² / 0.50 = 0.18 mol dm⁻³.
5. No effect; equilibrium is reached sooner.
6. Kp decreases.
7. K = e^(−12 / 8.31) = 0.24.
8. ΔStotal = 8.31 × ln(1 × 10⁻¹⁵) = −287 J K⁻¹ mol⁻¹; the reaction barely happens.
9. ΔSsurroundings becomes less negative, so ΔStotal increases (and K increases).
10. Fe is a solid, so it does not appear in Kp; adding more does not change the fraction.

Where marks are usually lost

• Including solids, or including aqueous ions in a Kp expression.
• Putting equilibrium amounts in moles into Kc when the V terms do not divide out.
• Using initial amounts instead of equilibrium amounts.
• Forgetting the 2 : 1 ratio when working out moles of N₂O₄-type products.
• Finding mole fractions but never multiplying by total pressure.
• Giving units when there are none, or leaving off units when there are some.
• Writing “Kp increases with pressure”.
• Explaining a temperature effect without saying that K changes.
• Mixing kJ and J in ΔSsurroundings.
• Treating a large K as proof that a reaction is fast.

Next steps

Work through the practice questions, return to the study guide for anything that is still shaky, and mark progress on the checklist. Revisit your Topic 12 entropy notes if ΔSsystem and ΔSsurroundings feel rusty.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Topic 13: Chemical Equilibria (outcomes 13.1 to 13.9), in Unit 4: Rates, Equilibria and Further Organic Chemistry. Published by Pearson Education Limited.