Equilibria

Physical Chemistry Weightage: AS · Papers 1, 2 and 3 (assumed in Papers 4 and 5) Topic 7

“Welcome! Equilibria is two topics in one: how far a reversible reaction goes, and how acids and bases behave. Both reward the same habit: say what the system does and why. A prediction with no explanation earns half the marks, and a calculation with no units loses the last one, so we will practise both.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in equilibria

Definitions: dynamic equilibrium needs both rates equal and concentrations constant. Le Chatelier’s principle needs “moves to minimise the change”.
Predict and explain: one mark for the direction of shift, one for the reason (fewer gas molecules, the endothermic direction, removes the added substance).
Calculations: the expression, the concentrations or partial pressures, the value with units. Expression marks are lost for square brackets missing, or for including solids and liquids.
Industrial processes: give the conditions and the compromise: rate against yield, and cost.
Acids and bases (Papers 2 and 3): proton donor and acceptor, the dissociation equation with ⇌, and what to observe to tell strong from weak. In Paper 3, choose an indicator by its range against the vertical section of the curve.

1. Reversible Reactions and Dynamic Equilibrium

A reversible reaction is one in which the products can react to re-form the reactants. It is written with the sign ⇌.

Dynamic equilibrium: the state of a reversible reaction in a closed system in which the rate of the forward reaction equals the rate of the backward reaction, so the concentrations of reactants and products stay constant.

Dynamic: both reactions carry on; molecules are still being converted in both directions.
Equal rates: forward rate = backward rate.
Constant concentrations: they are constant, not necessarily equal.
Closed system: nothing enters or leaves. In an open system a gaseous product escapes, and the reaction can go to completion (CaCO₃(s) → CaO(s) + CO₂(g) in an open tube).
The same equilibrium is reached from either direction.

Approaching dynamic equilibrium in a closed system. The forward rate falls and the backward rate rises until they are equal; after that the concentrations stay constant, but both reactions carry on. The concentrations of reactant and product need not be equal.

Position of equilibrium: the relative amounts of reactants and products at equilibrium. If there is more product, the position is to the right; if more reactant, to the left. It always refers to the equation as written.

⚠️ Examiner Trap: Equal or constant? At equilibrium the rates are equal and the concentrations are constant. Writing “the concentrations are equal” is a common wrong answer. Also say dynamic: the reactions have not stopped.

2. Le Chatelier's Principle

Le Chatelier’s principle: if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise (oppose) that change.

Change	Position moves…	Reason (write this)
Increase a reactant concentration	to the right (products)	removes some of the extra reactant
Increase a product concentration	to the left	removes some of the extra product
Increase the pressure (gases)	towards the side with fewer gas molecules	reduces the pressure
Decrease the pressure	towards the side with more gas molecules	increases the pressure
Increase the temperature	in the endothermic direction	absorbs the added energy
Decrease the temperature	in the exothermic direction	releases energy
Add a catalyst	no change in position	speeds up forward and backward reactions equally

Effect of adding more reactant. The reactant concentration jumps up, then falls as the system removes some of the extra; the product concentration rises. The position of equilibrium moves to the right, and K_c is unchanged because the temperature has not changed.

If the number of gas molecules is the same on both sides, pressure has no effect on the position. Only gases count: solids and liquids are ignored.
A catalyst makes equilibrium arrive sooner but does not change the yield at equilibrium.
Of all the changes, only a change in temperature alters the value of K_c or K_p.

Worked example 1. For 2NO(g) + O₂(g) ⇌ 2NO₂(g), ΔH = −114 kJ mol⁻¹, predict the effect of (i) raising the pressure and (ii) raising the temperature.

(i) There are 3 mol of gas on the left and 2 on the right. Higher pressure moves the position to the right, to reduce the number of gas molecules and so oppose the increase in pressure.
(ii) The forward reaction is exothermic, so the backward reaction is endothermic. Higher temperature moves the position to the left, to absorb the extra energy. K_c decreases.

⚠️ Examiner Trap: Position, rate and K. Pressure and concentration change the position but not K. Temperature changes the position and K. A catalyst changes only the rate. Keep these three separate; questions like “does K_c change?” are asked every year.

3. The Equilibrium Constant, K_c

For aA + bB ⇌ cC + dD, at constant temperature:

K_c = [C]^c[D]^d ÷ [A]^a[B]^b

Square brackets mean the equilibrium concentration in mol dm⁻³. Products go on top, reactants below, each raised to the power of its coefficient.
Solids and pure liquids are left out. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K_c = [CO₂].
A large K_c means the position is well to the right; a small K_c means well to the left.
Units: substitute the units into the expression and cancel. For N₂ + 3H₂ ⇌ 2NH₃: (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = mol⁻² dm⁶.

Worked example 2. 0.80 mol of NO and 0.50 mol of Cl₂ are put in a 2.0 dm³ vessel: 2NO(g) + Cl₂(g) ⇌ 2NOCl(g). At equilibrium 0.40 mol of NOCl is present. Calculate K_c.

NOCl formed = 0.40 mol, so NO used = 0.40 mol and Cl₂ used = 0.20 mol.
At equilibrium: NO = 0.40 mol, Cl₂ = 0.30 mol, NOCl = 0.40 mol.
Concentrations (÷ 2.0): [NO] = 0.20, [Cl₂] = 0.15, [NOCl] = 0.20 mol dm⁻³.
K_c = [NOCl]² ÷ ([NO]²[Cl₂]) = 0.20² ÷ (0.20² × 0.15) = 6.67 dm³ mol⁻¹.

⚠️ Examiner Trap: Moles, not concentrations. Convert the moles at equilibrium to concentrations using the total volume before you substitute (unless the number of moles on both sides is equal, when the volume cancels). Also use the equilibrium amounts: reactant used = initial − equilibrium.

4. Partial Pressures and K_p

For equilibria of gases we can use pressures instead of concentrations.

Mole fraction: x = moles of the gas ÷ total moles of gas. The mole fractions add up to 1.
Partial pressure: the pressure the gas would exert if it alone occupied the container. p = x × total pressure. The partial pressures add up to the total pressure.

For aA(g) + bB(g) ⇌ cC(g): K_p = p(C)^c ÷ ( p(A)^a p(B)^b ). Only gases appear. The units come from the expression: for 2SO₂ + O₂ ⇌ 2SO₃ with pressures in kPa, the units are kPa² ÷ kPa³ = kPa⁻¹.

Route to K_p. Convert the equilibrium amounts to mole fractions, multiply by the total pressure to get the partial pressures, then substitute into the expression. The sum of the mole fractions is 1, and the sum of the partial pressures is the total pressure.

Worked example 3. At equilibrium, N₂O₄(g) ⇌ 2NO₂(g) contains 0.60 mol N₂O₄ and 0.80 mol NO₂ at a total pressure of 200 kPa.

Total moles = 1.40. Mole fractions: N₂O₄ 0.60 ÷ 1.40 = 0.429; NO₂ 0.80 ÷ 1.40 = 0.571.
Partial pressures: p(N₂O₄) = 0.429 × 200 = 85.7 kPa; p(NO₂) = 0.571 × 200 = 114 kPa.
K_p = p(NO₂)² ÷ p(N₂O₄) = 114² ÷ 85.7 = 152 kPa.

⚠️ Examiner Trap: The units of K_p. Use the pressure unit given in the question (kPa, Pa or atm) and raise it to the power from the expression. A K_p of 0.40 with no unit for 2SO₂ + O₂ ⇌ 2SO₃ loses the unit mark. Also note that K_p and K_c for the same reaction are different numbers.

5. The Haber and Contact Processes

	Haber process	Contact process
Equilibrium	N₂(g) + 3H₂(g) ⇌ 2NH₃(g)	2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
ΔH	−92 kJ mol⁻¹ (exothermic)	−197 kJ mol⁻¹ (exothermic)
Gas molecules, left : right	4 : 2	3 : 2
Catalyst	iron	vanadium(V) oxide, V₂O₅
Temperature	about 450 °C (compromise)	about 450 °C (compromise)
Pressure	high (about 200 atm)	about 1–2 atm: K_p is already large, so high pressure is not worth the cost
Removing product	NH₃ liquefied and removed; N₂ and H₂ recycled	SO₃ absorbed in concentrated sulfuric acid

Why a compromise temperature is used for an exothermic equilibrium such as the Haber process. Raising the temperature increases the rate but moves the position to the left, so the yield falls. The plant uses about 450 °C with a catalyst, and a high pressure to help the yield.

Why not a lower temperature? The yield would be higher (the reaction is exothermic), but the rate would be too slow. Why not a higher one? Faster, but the position moves to the left and the yield falls.
Why a high pressure in the Haber process? It moves the position to the right (fewer gas molecules) and increases the rate; the cost is the strong, expensive plant and the energy for compression.
Catalyst: speeds up both reactions equally, allowing a lower temperature; it does not change the yield at equilibrium or K_p.

⚠️ Examiner Trap: Both rate and yield. Every industrial-conditions answer needs a rate statement and a yield statement for each variable, and then the word compromise. Saying “a high temperature gives a higher yield” for an exothermic reaction is wrong.

6. Brønsted–Lowry Acids and Bases

A Brønsted–Lowry acid is a proton (H⁺) donor.
A Brønsted–Lowry base is a proton acceptor.
An acid and its base after loss of H⁺ are a conjugate pair. Every acid–base reaction has two pairs.
Amphoteric species can act as an acid or a base; water is one (H₃O⁺ when it accepts H⁺ and OH⁻ when it donates).

Brønsted–Lowry acids and bases in HCOOH + H₂O ⇌ HCOO⁻ + H₃O⁺. HCOOH donates a proton (acid 1) and becomes its conjugate base HCOO⁻. Water accepts the proton (base 2) and becomes its conjugate acid H₃O⁺. Water is amphoteric: it can act as an acid or as a base.

Common acids: HCl, H₂SO₄, HNO₃ (strong); CH₃COOH (weak). Common alkalis: NaOH, KOH (strong); NH₃ (weak).
Neutralisation: H⁺(aq) + OH⁻(aq) → H₂O(l).
The definition is not limited to water: in HCl + HI ⇌ H₂Cl⁺ + I⁻, HI donates the proton (acid) and HCl accepts it (base).

⚠️ Examiner Trap: Which is which? To decide which reactant is the acid, follow the proton: the species that loses H⁺ is the acid. Do not label by the charge or by the name of the compound. In a conjugate-pair question, name the pair with the same species on both sides of the arrow (for example HCOOH and HCOO⁻).

7. Strong and Weak Acids, pH

pH: acids have pH below 7, alkalis above 7 and neutral solutions exactly 7. A lower pH means a higher H⁺ concentration.
Strong acid: fully dissociated in water (HCl → H⁺ + Cl⁻). Weak acid: only partially dissociated; the equilibrium lies to the left (CH₃COOH ⇌ H⁺ + CH₃COO⁻).
Strong base: fully dissociated (NaOH, KOH). Weak base: partially, e.g. NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

Strong and weak acids of the same concentration. A strong acid is fully dissociated in water, so it has a high concentration of H⁺ ions, a low pH, a high conductivity and reacts quickly with magnesium. A weak acid is only partially dissociated, so its equilibrium lies well to the left.

Telling them apart (same concentration, e.g. 0.1 mol dm⁻³):

Test	Strong acid (HCl)	Weak acid (CH₃COOH)
pH (meter or universal indicator)	about 1	about 3
Electrical conductivity	higher	lower
Rate of reaction with magnesium	fast effervescence	slow effervescence

⚠️ Examiner Trap: Strength is not concentration. “Strong” means fully dissociated; “concentrated” means many moles per dm³. To compare, use the same concentration, and say that the strong acid has a higher H⁺ concentration. Note that the total amount of acid, and so the volume of alkali needed to neutralise it, is the same for equal moles of strong and weak acid.

8. Titration Curves and Indicators

A pH titration curve shows the pH against the volume of alkali added. At the end-point (equivalence) there is a vertical section where the pH changes by several units for a tiny volume.

Sketches of pH against volume of alkali added to 25 cm³ of acid (0.1 mol dm⁻³ each). An indicator is suitable only if its colour-change range lies inside the vertical section of the curve. With a weak acid and a weak base there is no vertical section, so no indicator gives a sharp end-point.

Combination	Start pH	Vertical section (about)	Suitable indicator
Strong acid + strong alkali	about 1	3 to 11	methyl orange, bromothymol blue or phenolphthalein
Strong acid + weak alkali	about 1	3 to 7	methyl orange (or methyl red)
Weak acid + strong alkali	about 3	7 to 11	phenolphthalein
Weak acid + weak alkali	about 3	none	none: no sharp end-point

An indicator changes colour over a pH range of about two units (an indicator is itself a weak acid, HIn, whose two forms are different colours).
Choose an indicator whose colour-change range lies inside the vertical section; a colour change elsewhere is gradual, and the end-point cannot be read to 0.05 cm³.
Equal concentrations and volumes of monoprotic strong or weak acid and alkali give the same volume at equivalence; only the pH at equivalence and the shape differ.

⚠️ Examiner Trap: Curve and indicator. Phenolphthalein is wrong for a strong acid with a weak base, because its range (8.2–10.0) is above the vertical section (about 3 to 7). Methyl orange is wrong for a weak acid with a strong alkali. When you are given a table of indicators, compare each range with the vertical section, and write that comparison.

9. Quick Sheet and Checklist

Idea	Say or write
Dynamic equilibrium	forward rate = backward rate; concentrations constant; closed system
Le Chatelier	the position moves to minimise the change
Pressure	towards fewer gas molecules (if higher pressure)
Temperature	endothermic direction (if higher); K changes
Catalyst	rate only; position and K unchanged
K_c	[products]^powers ÷ [reactants]^powers; no solids or liquids
K_p	partial pressures; p = mole fraction × total pressure
Acid / base	proton donor / proton acceptor
Strong / weak	fully / partially dissociated
Indicator choice	range inside the vertical section

Before you leave the question, check:

Direction of shift and the reason are both written.
K expressions have no solids or liquids, correct powers, and units worked out.
Moles have been converted to concentrations, or mole fractions to partial pressures, before substituting.
Industrial answers mention rate, yield and compromise.
Titration answers name the indicator and compare its range with the vertical section.