Edexcel International A Level Chemistry: Chemical equilibria (YCH11)

This guide teaches outcomes 13.1 to 13.9, which make up Topic 13: Chemical Equilibria in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. Every outcome here falls within the International A2 (IA2) part of the course, in Unit 4: Rates, Equilibria and Further Organic Chemistry. The specification says it assumes and extends the equilibrium work from Unit 2, Topic 9B, and the entropy outcomes of Topic 12 feed straight into 13.8.

Useful pages: the Edexcel chemistry course hub, the printable chemistry checklist, the Topic 13 revision notes, the matching practice questions and the free diagnostics.

Outcomes in this topic

Spec ref	Skill you need
13.1	Deduce a Kc expression for homogeneous and heterogeneous systems
13.2	Deduce a Kp expression, using partial pressures in atm
13.3	Calculate Kc and Kp, with units where appropriate, from experimental data
13.4	Explain how temperature, pressure or a catalyst affects the equilibrium composition
13.5	Know that K is unchanged by concentration, pressure or a catalyst
13.6	Know how temperature changes K for exothermic and endothermic reactions
13.7	Explain temperature shifts through the change in K
13.8	Link temperature, ΔStotal and the size of K through ΔStotal = R lnK
13.9	Use the size of K to predict how far a reaction goes

Writing Kc (13.1)

For the general reaction aA + bB ⇌ cC + dD, the expression is:

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Each bracketed term is a concentration at equilibrium, measured in mol dm⁻³. The right-hand side of the equation forms the numerator, the left-hand side forms the denominator, and every balancing number becomes an index.

In a homogeneous system everything shares one phase: a mixture of gases, say, or a set of species dissolved in the same solution. A heterogeneous system spans two or more phases. In heterogeneous expressions you omit solids. The concentration of a pure solid is set by its density, so it stays constant and is folded into the value of K.

Worked example 1. Iron(II) ions react with silver ions:

Fe²⁺(aq) + Ag⁺(aq) ⇌ Fe³⁺(aq) + Ag(s)

Silver is a solid, so it is omitted:

Kc = [Fe³⁺] / ([Fe²⁺][Ag⁺])

Units: (mol dm⁻³) / (mol dm⁻³ × mol dm⁻³) = (mol dm⁻³)⁻¹ = dm³ mol⁻¹.

To get units, substitute mol dm⁻³ for every term and cancel. When the indices on top and bottom total the same, Kc is a pure number.

Writing Kp (13.2)

With gases you can work in partial pressures. The partial pressure of a gas is the pressure it would exert if it were alone in the vessel. Two relationships do all the work:

• mole fraction, x(A) = amount of A ÷ total amount of gas
• p(A) = x(A) × P(total)

Adding up every partial pressure gives back P(total). The specification asks for partial pressures in atm.

Kp has the same shape as Kc, with p(A) replacing [A]. Only gases are written into Kp. In a heterogeneous system you omit solids, and dissolved species also drop out because they have no partial pressure.

Worked example 2. Iron(II) oxide is reduced by carbon monoxide in a closed furnace:

FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g)

Kp = p(CO₂) / p(CO). The units are atm / atm, so Kp has no units.

Solid ammonium chloride heated in a sealed tube gives NH₄Cl(s) ⇌ NH₃(g) + HCl(g). The solid is omitted: Kp = p(NH₃) × p(HCl), with units atm².

Calculating Kc and Kp from data (13.3)

Lay out an initial / change / equilibrium table. Work out the change line from the stoichiometry of the balanced equation.

Worked example 3 (Kc, ester equilibrium). A sealed flask holds 0.250 mol ethanoic acid and 0.350 mol ethanol with a little acid catalyst. After several days, titration shows that 0.0580 mol ethanoic acid remains. Calculate Kc.

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

	CH₃COOH	C₂H₅OH	CH₃COOC₂H₅	H₂O
Initial / mol	0.250	0.350	0	0
Change / mol	−0.192	−0.192	+0.192	+0.192
Equilibrium / mol	0.0580	0.158	0.192	0.192

All four species share one liquid phase, so the system is homogeneous and water belongs in the expression. Each concentration would be n ÷ V, but with two terms above and two below, every V divides out:

Kc = (0.192 × 0.192) / (0.0580 × 0.158) = 4.02 (no units)

Worked example 4 (Kp from moles). 0.800 mol NO₂ is sealed in a vessel. At equilibrium, 2NO₂(g) ⇌ N₂O₄(g), 0.240 mol NO₂ remains and the total pressure is 1.30 atm. Calculate Kp.

1. NO₂ used = 0.800 − 0.240 = 0.560 mol, so N₂O₄ formed = 0.560 ÷ 2 = 0.280 mol.
2. Total moles of gas = 0.240 + 0.280 = 0.520 mol.
3. p(NO₂) = (0.240 ÷ 0.520) × 1.30 = 0.600 atm; p(N₂O₄) = (0.280 ÷ 0.520) × 1.30 = 0.700 atm.
4. Kp = p(N₂O₄) / p(NO₂)² = 0.700 / 0.600² = 1.94 atm⁻¹.

Worked example 5 (heterogeneous Kc). Equal volumes of solutions are mixed so that, just after mixing, [Fe²⁺] = [Ag⁺] = 0.100 mol dm⁻³. At equilibrium a titration of the remaining silver ions gives [Ag⁺] = 0.0820 mol dm⁻³.

The ratio is 1 : 1 : 1, so [Fe²⁺] = 0.0820 mol dm⁻³ and [Fe³⁺] = 0.100 − 0.0820 = 0.0180 mol dm⁻³.

Kc = 0.0180 / (0.0820 × 0.0820) = 2.68 dm³ mol⁻¹

What changes the composition, and what changes K (13.4, 13.5)

The central idea is that temperature is the only thing that alters K. A change of concentration or pressure, or adding a catalyst, can alter the make-up of the mixture or the time it takes to settle, but the mixture always returns to the same value of K at that temperature.

Concentration. Adding a reactant makes the fraction in the K expression smaller than K. The system responds by converting reactant to product until the fraction equals K again. K itself does not move.

Pressure. Take the mixture in worked example 4 and halve its volume at constant temperature. Every partial pressure doubles at that instant:

p(N₂O₄) / p(NO₂)² becomes 1.40 / 1.20² = 0.972 atm⁻¹

That is half of Kp, because the denominator carries the higher index. To get back to 1.94 atm⁻¹, more NO₂ combines to give N₂O₄, so the mixture moves towards the side with fewer gas moles. When both sides carry the same number of gas moles, as in the FeO and CO system, every term scales equally and the composition stays put.

Catalyst. A catalyst raises the forward and backward rates by the same factor. Equilibrium arrives sooner, but neither the composition nor K changes.

Heterogeneous systems. Reason only with the species that appear in the expression. Adding more FeO or more iron to the furnace changes nothing, because neither solid is in Kp. Removing CO₂ does move the position, because p(CO₂) is in Kp.

Temperature and K (13.6, 13.7)

Forward reaction	Temperature rises	Temperature falls
Exothermic (ΔH negative)	K decreases	K increases
Endothermic (ΔH positive)	K increases	K decreases

The specification wants the position change explained through K. When the temperature rises for an exothermic reaction, K becomes smaller. The existing mixture now has a fraction that is too big for the new K, so products react back to form reactants until the fraction matches the new, smaller K. The equilibrium has moved to the left because K changed, not the other way round.

For 2NO₂ ⇌ N₂O₄ the forward change is exothermic. Warming a sealed syringe of the mixture lowers Kp, so more brown NO₂ forms and the colour deepens. Cooling raises Kp and the colour fades.

Entropy and the equilibrium constant (13.8)

From Topic 12, ΔStotal = ΔSsystem + ΔSsurroundings, and ΔSsurroundings = −ΔH / T. The specification links ΔStotal to K by:

ΔStotal = R lnK

where R = 8.31 J K⁻¹ mol⁻¹. Treat K here as a pure number. A positive ΔStotal gives K greater than 1; a negative ΔStotal gives K less than 1.

Effect of temperature on ΔStotal. As T rises, ΔH / T gets smaller, so ΔSsurroundings shrinks in size. ΔSsystem changes very little.

• Exothermic reaction: ΔSsurroundings is positive and gets smaller, so ΔStotal falls and K falls.
• Endothermic reaction: ΔSsurroundings is negative and gets less negative, so ΔStotal rises and K rises.

Worked example 6. Use these values for 2NO(g) + O₂(g) ⇌ 2NO₂(g): ΔH = −114.2 kJ mol⁻¹ and ΔSsystem = −146.6 J K⁻¹ mol⁻¹. Calculate ΔStotal and K at 298 K and at 800 K.

At 298 K:

• ΔSsurroundings = −(−114 200) / 298 = +383.2 J K⁻¹ mol⁻¹ (ΔH converted to J first)
• ΔStotal = −146.6 + 383.2 = +236.6 J K⁻¹ mol⁻¹
• lnK = 236.6 / 8.31 = 28.5, so K = e^28.5 ≈ 2.3 × 10¹²

At 800 K:

• ΔSsurroundings = 114 200 / 800 = +142.75 J K⁻¹ mol⁻¹
• ΔStotal = −146.6 + 142.75 = −3.85 J K⁻¹ mol⁻¹
• lnK = −3.85 / 8.31 = −0.463, so K = e^−0.463 ≈ 0.63

By 800 K, ΔStotal is slightly negative and K is below 1, as predicted.

Predicting the extent of reaction (13.9)

The size of K tells you how far a reaction goes before equilibrium is reached:

Size of K	Approximate ΔStotal / J K⁻¹ mol⁻¹	What you see
Greater than about 10¹⁰	More positive than about +200	Reaction goes essentially to completion
Between about 10⁻¹⁰ and 10¹⁰	Between about −200 and +200	A true equilibrium mixture with noticeable amounts of both sides
Less than about 10⁻¹⁰	More negative than about −200	Reaction hardly happens at all

These boundaries are a rule of thumb, not sharp cut-offs. Check one: K = 10¹⁰ gives R lnK = 8.31 × 23.0 ≈ +191 J K⁻¹ mol⁻¹. Worked example 6 sits in the top row at 298 K and the middle row at 800 K.

K says nothing about rate. A reaction with a huge K can still be too slow to observe if its activation energy is high. That is kinetic stability, which you met in outcome 12.11.

Further suggested practicals

Know what each of these suggested practicals measures.

• Ethanol and ethanoic acid. Mixtures are left to reach equilibrium, then titrated with sodium hydroxide. Allow for the acid catalyst before finding moles of ethanoic acid. The specification notes this one can be used to show ICT for presenting and analysing data.
• Fe²⁺(aq) + Ag⁺(aq) ⇌ Fe³⁺(aq) + Ag(s). The silver ions left are found by titration with thiocyanate; the other concentrations follow from the 1 : 1 ratio.
• Distribution between two immiscible solvents. Ammonia or iodine is shaken with two solvents; titrating each layer gives the ratio of concentrations, an equilibrium constant with no units.
• Thermal decomposition of ammonium chloride. A heterogeneous gas equilibrium with solid present.
• 2NO₂ ⇌ N₂O₄. Colour changes in a sealed syringe show the effects of temperature and pressure.

Common errors

• Writing a solid into Kc or Kp. Heterogeneous expressions omit solids.
• Plugging amounts in moles straight into Kc when the indices above and below differ. V only divides out when the totals match.
• Forgetting to square a partial pressure when the coefficient is 2.
• Saying that raising the pressure “increases Kp”. Pressure changes the composition, never K.
• Quoting Le Chatelier’s principle on its own for a temperature change when the question wants the reasoning through K.
• Leaving ΔH in kJ mol⁻¹ while ΔSsystem is in J K⁻¹ mol⁻¹.
• Claiming a large K means a fast reaction.

Next steps

Condense this topic with the chemical equilibria revision notes, then test yourself on the chemical equilibria practice questions. Record progress against each outcome on the checklist and find weak spots with the diagnostics.

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.