7. Equilibria

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7.1 Chemical equilibria and dynamic equilibrium

A reversible reaction can proceed from reactants to products and from products back to reactants under the stated conditions. It is represented by opposing half-arrows, ⇌.

At dynamic equilibrium, the forward and reverse reactions continue but have equal rates. Reactant and product concentrations therefore remain constant with time, even though particles keep reacting in both directions.

A closed system prevents reactants or products from escaping or being added. Without that material boundary, loss of a gaseous product or continuing feed/removal changes the composition and prevents the two rates from establishing a stable equilibrium state.

For N₂O₄(g) ⇌ 2NO₂(g) in a sealed vessel at constant temperature, the colour becomes constant when the two rates are equal. Individual N₂O₄ and NO₂ particles still interconvert.

Equal rates do not mean equal concentrations, and constant concentration does not mean the reaction has stopped. ‘Closed’ refers to no transfer of matter; it does not require the system to be thermally insulated.

If a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.

The position of equilibrium describes the relative equilibrium amounts of reactants and products. A move to the right produces more products and consumes some reactants; a move to the left does the reverse.

The response establishes a new dynamic equilibrium under the changed conditions. It reduces the imposed effect but does not normally restore every concentration, pressure or temperature to its original value.

The principle predicts the direction of composition change, not how fast equilibrium is reached or the numerical final amounts. Those applications and catalyst effects are handled by the next objective.

Change imposed	Direction favoured	Boundary
add a reactant or remove a product	forward, to the right	responds to concentrations of species in the equilibrium
remove a reactant or add a product	reverse, to the left	the change is only partly opposed
increase pressure by decreasing volume	side with fewer moles of gas	no shift if gaseous mole totals are equal; ignore solids and liquids in the count
decrease pressure by increasing volume	side with more moles of gas	applies to gaseous equilibria
increase temperature	endothermic direction	identify the sign of the forward ΔH first
decrease temperature	exothermic direction	temperature changes both position and K
add a catalyst	no shift	both directions speed up, so equilibrium is reached sooner

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH < 0: adding O₂ shifts right; increasing pressure shifts right because three gas moles become two; increasing temperature shifts left because the reverse direction is endothermic; a catalyst does not change the final composition.

State how pressure was changed. Adding an inert gas at constant volume leaves reacting-gas partial pressures unchanged and does not shift the equilibrium, even though the total pressure rises.

$\mathrm{for}\ aA+bB\rightleftharpoons cC+dD,\qquad K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Use equilibrium concentrations in mol dm⁻³. Put product terms above reactant terms and turn every stoichiometric coefficient in the balanced equation into a power.

$\mathrm{N_2+3H_2\rightleftharpoons2NH_3},\qquad K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$

Include gases and dissolved species whose concentrations can change. Omit pure solids and pure liquids because their effective concentrations are constant; their presence in the balanced equation does not create a Kc term.

An equilibrium expression is derived from the equation as written. Reversing the equation reciprocates Kc and multiplying every coefficient by a factor raises Kc to that factor; never invent powers from measured concentrations.

$x_i=\frac{n_i}{n_{total}}\qquad\sum x_i=1$

$p_i=x_iP_{total}\qquad\sum p_i=P_{total}$

The partial pressure of gas i is its contribution to the total pressure: for an ideal mixture it equals the pressure that the same amount of that gas would exert alone at the mixture's temperature and volume.

A mixture contains 2.00 mol N₂ and 1.00 mol O₂ at 300 kPa total pressure. x(N₂) = 2.00/3.00 = 0.667, so p(N₂) = 0.667 × 300 = 200 kPa; p(O₂) = 100 kPa, and the partial pressures sum to 300 kPa.

Mole fraction is dimensionless and is not mass fraction. Use all gaseous components in n(total), keep one pressure unit throughout, and verify both sums after calculating.

$\mathrm{for}\ aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g),\qquad K_p=\frac{p_C^c p_D^d}{p_A^a p_B^b}$

Use each gas's equilibrium partial pressure, not the total pressure. Products go in the numerator, reactants in the denominator and balanced-equation coefficients become powers.

$\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)},\qquad K_p=\frac{p_{NO_2}^2}{p_{N_2O_4}}$

Only gaseous species appear in Kp. Pure solids, pure liquids and aqueous concentrations are omitted from this partial-pressure expression. The unit, where requested, follows from the net pressure power in the particular expression.

Do not convert Kp into Kc: their relationship is outside this syllabus outcome. The task is to deduce Kp directly from the gas equation and the given partial pressures.

Write the expression before substituting, confirm that every value is an equilibrium concentration or partial pressure, apply the coefficient powers and derive the unit from the uncancelled concentration or pressure powers.

For N₂O₄(g) ⇌ 2NO₂(g):

Constant	Equilibrium data	Substitution	Result
Kc	[N₂O₄] = 0.200, [NO₂] = 0.300 mol dm⁻³	(0.300)² / 0.200	0.450 mol dm⁻³
Kp	p(N₂O₄) = 80.0 kPa, p(NO₂) = 40.0 kPa	(40.0)² / 80.0	20.0 kPa

The numerical result describes the equilibrium composition for this equation at the stated temperature. It does not measure reaction rate, and a Kc value cannot be substituted into a Kp expression.

Initial values are not equilibrium values unless the system initially happens to be at equilibrium. Keep brackets and powers intact, and do not report a unit copied from the input without cancelling the expression's powers.

Start from a balanced equation and one reaction extent x. Write initial, change and equilibrium amounts in the stoichiometric ratio, convert amounts to concentrations or partial pressures if required, substitute into K, solve the permitted non-quadratic relationship, then reject any value that gives a negative amount.

For H₂(g) + I₂(g) ⇌ 2HI(g) in 1.00 dm³, initially 1.00 mol H₂, 1.00 mol I₂ and no HI, with Kc = 49.0:

Stage / mol dm⁻³	H₂	I₂	HI
initial	1.00	1.00	0
change	−x	−x	+2x
equilibrium	1.00−x	1.00−x	2x

$49.0=\frac{(2x)^2}{(1.00-x)^2}\Rightarrow 7.00=\frac{2x}{1.00-x}\Rightarrow x=0.7778$

The equilibrium quantities are H₂ = 0.222 mol, I₂ = 0.222 mol and HI = 1.56 mol in the 1.00 dm³ vessel. Substitution returns Kc ≈ (1.56)²/(0.222)² = 49.0 after rounding.

Equal forward and reverse rates do not make equilibrium amounts equal. Apply coefficients to the change row—not independently guessed changes—and use the actual vessel volume whenever it is not 1.00 dm³.

Change	Equilibrium position/composition	Value of Kc or Kp
temperature	changes; higher T favours the endothermic direction	changes
concentration	shifts until the equilibrium ratio is restored	unchanged at constant T
pressure	may shift a gaseous equilibrium	unchanged at constant T
catalyst	reaches the same equilibrium faster	unchanged

For an exothermic forward reaction, increasing temperature shifts left and decreases K; decreasing temperature shifts right and increases K. For an endothermic forward reaction, these K changes reverse.

Concentration or pressure disturbances temporarily make the reaction ratio inconsistent with K, so composition changes until the same constant is restored. A catalyst lowers activation energy for both directions and changes neither the ratio nor K.

Always hold temperature constant when claiming that pressure or concentration leaves K unchanged. A pressure change caused by heating also changes temperature, so separate the variables before drawing the conclusion.

Process and equilibrium	Industrial conditions	Equilibrium and rate explanation
Haber: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹	about 450 °C; about 200 atm; iron catalyst; NH₃ removed and unreacted gases recycled	lower T and higher pressure favour NH₃, but very low T is slow and very high pressure is costly and hazardous; catalyst raises rate without changing yield
Contact: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH ≈ −197 kJ mol⁻¹	about 450 °C; near atmospheric pressure; V₂O₅ catalyst; oxygen available and SO₃ removed for conversion to sulfuric acid	lower T and higher pressure favour SO₃, but 450 °C gives useful rate and the equilibrium yield is already high near 1 atm, so extra pressure is not worth its cost

Both forward reactions are exothermic and reduce gaseous mole count, so Le Chatelier predicts low temperature and high pressure for maximum equilibrium yield. Industry instead optimises production rate, energy and compression cost, safety, catalyst performance, separation and recycling.

Removing product shifts each operating equilibrium toward further product formation. In the Haber process ammonia is cooled and condensed; in the Contact process sulfur trioxide is removed from the gas stream and absorbed during sulfuric-acid manufacture.

A catalyst never increases equilibrium yield. The two processes also do not use the same pressure compromise: Haber gains enough yield and rate to justify high pressure, whereas Contact obtains little extra benefit from compression.

7.2 Brønsted–Lowry theory of acids and bases

Common acid	Formula
hydrochloric acid	HCl
sulfuric acid	H₂SO₄
nitric acid	HNO₃
ethanoic acid	CH₃COOH

Match the whole formula to the name. Ethanoic acid is the only listed organic acid and is commonly written CH₃COOH to keep its acidic carboxyl group visible.

This recall objective is limited to these four acids. Do not omit ethanoic acid, change sulfuric acid to sulfurous acid, or infer acid strength solely from the number of H atoms in a formula.

Common alkali	Formula
sodium hydroxide	NaOH
potassium hydroxide	KOH
ammonia	NH₃

An alkali is a base that dissolves in water. Sodium hydroxide and potassium hydroxide contain OH⁻ directly; ammonia is still an alkali in aqueous solution because it reacts with water to generate OH⁻.

$\mathrm{NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)}$

Do not add OH to ammonia's formula: ammonia is NH₃, not NH₄OH for this syllabus recall. ‘Alkali’ is not a synonym for every base; solubility in water matters.

In the Brønsted–Lowry model, an acid is a proton, H⁺, donor and a base is a proton acceptor. An acid–base reaction is therefore identified by tracking transfer of one H⁺ between species.

$\mathrm{HA+B\rightleftharpoons A^-+HB^+}$

HA donates H⁺ and becomes its conjugate base A⁻; B accepts H⁺ and becomes its conjugate acid HB⁺. Each conjugate pair differs by exactly one H⁺.

In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, NH₃ accepts H⁺ and is the base, while H₂O donates H⁺ and is the acid. The pairs are NH₃/NH₄⁺ and H₂O/OH⁻.

A base need not contain OH⁻: NH₃ qualifies through proton acceptance. Acid/base labels can depend on the reaction partner; identify the transferred proton rather than relying only on a memorised substance label.

Type	Behaviour in aqueous solution	Equation convention
strong acid	fully or essentially fully dissociated	HA(aq) → H⁺(aq) + A⁻(aq)
weak acid	partially dissociated	HA(aq) ⇌ H⁺(aq) + A⁻(aq)
strong base	fully or essentially fully dissociated into ions	MOH(aq) → M⁺(aq) + OH⁻(aq)
weak base	partially reacts with water to form ions	B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

HCl, HNO₃ and H₂SO₄ are strong acids in this context; CH₃COOH is weak. NaOH and KOH are strong bases, while NH₃ is weak.

Strength is not concentration. A strong solution may be dilute and a weak solution may be concentrated: strength describes the fraction dissociated, while concentration describes amount per volume.

Aqueous solution at standard conditions	pH	Relative H⁺ and OH⁻ concentrations
acidic	below 7	[H⁺] > [OH⁻]
pure water / neutral	7	[H⁺] = [OH⁻]
alkaline	above 7	[OH⁻] > [H⁺]

Moving to a lower pH means a higher hydrogen-ion concentration and greater acidity; moving to a higher pH means greater alkalinity. The classification applies to aqueous solutions under the stated standard conditions.

pH 7 is the standard-condition neutral reference used by this syllabus. Do not confuse ‘neutral’ with ‘contains no ions’: water contains equal small concentrations of H⁺ and OH⁻.

Compare strong and weak acids at the same concentration and temperature. The strong acid is much more fully dissociated, so it has a higher concentration of mobile H⁺ and counter-ions than the weak acid.

Observation	Strong acid compared with an equally concentrated weak acid	Explanation
pH meter	lower pH	higher [H⁺]
universal indicator	colour further toward the acidic/red end	responds to lower pH
conductivity	higher	more mobile ions carry charge
reaction with the same reactive metal	faster initial bubbling / hydrogen production	more frequent effective collisions involving H⁺

$\mathrm{Mg(s)+2H^+(aq)\rightarrow Mg^{2+}(aq)+H_2(g)}$

The comparison requires controlled concentration, volume, temperature and metal surface area. Faster initial reaction does not necessarily mean more final H₂: if equal stoichiometric amounts of two monoprotic acids react completely with excess metal, both can eventually supply the same total protons.

$\mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)}$

In acid–alkali neutralisation, H⁺ from the acid and OH⁻ from the alkali combine in a 1:1 ratio to form water. Other unchanged ions are spectators in the net ionic equation.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). Removing spectator Na⁺ and Cl⁻ leaves the net ionic equation above.

Neutralisation describes the reaction, not a guarantee that the final mixture has pH 7. Excess reagent or a weak-acid/weak-base conjugate ion can make the final solution acidic or alkaline.

Neutralisation	Salt formed	Ion sources
HCl + NaOH → NaCl + H₂O	sodium chloride	Na⁺ from alkali; Cl⁻ from acid
HNO₃ + KOH → KNO₃ + H₂O	potassium nitrate	K⁺ from alkali; NO₃⁻ from acid
CH₃COOH + NaOH → CH₃COONa + H₂O	sodium ethanoate	Na⁺ from alkali; CH₃COO⁻ from acid
HCl + NH₃ → NH₄Cl	ammonium chloride	NH₄⁺ formed when NH₃ accepts H⁺; Cl⁻ from acid

Name the salt from the base-derived cation followed by the acid-derived anion: hydrochloric acid gives chlorides, nitric acid nitrates, sulfuric acid sulfates and ethanoic acid ethanoates.

Balance charges when writing a salt formula. Salt formation does not require the final solution to be pH 7, and neutralisation by ammonia forms ammonium ions without producing OH⁻ as a written reactant.

Put pH on the vertical axis and volume of titrant added on the horizontal axis. Mark the initial pH, a smooth approach to the stoichiometric equivalence volume, the steep or shallow inflection region, and the final pH set by excess titrant.

Assuming alkali is added to acid:

Acid + alkali	Initial region	Equivalence region	After excess alkali
strong + strong	very low initial pH; gradual rise	large near-vertical jump centred at pH 7	approaches high pH of strong alkali
weak + strong	higher initial pH; broad gradual/buffer-like rise	vertical jump with equivalence pH above 7	approaches high pH of strong alkali
strong + weak	very low initial pH; gradual rise	smaller vertical jump with equivalence pH below 7	approaches the lower pH of weak alkali
weak + weak	moderately acidic start and gradual rise	no sharp vertical section; equivalence pH depends on relative strengths	approaches the pH of excess weak alkali

If acid is added to alkali, reverse the curve vertically: it starts alkaline and falls as volume is added. The equivalence volume is fixed by stoichiometric moles, not by where pH happens to equal 7.

Do not force every equivalence point to pH 7 or draw a sharp vertical jump for weak acid–weak alkali. This syllabus asks for qualitative sketches; pKa is not needed to place the required curve features.

Given a titration curve and indicator transition ranges, choose an indicator whose complete colour-change range lies within the steep section around the equivalence point. A small added volume then carries the indicator through its range, minimising endpoint error.

Titration	Suitable choice from common indicators	Reason
strong acid + strong alkali	methyl orange (pH 3.1–4.4) or phenolphthalein (pH 8.3–10.0)	the wide vertical section spans both ranges
strong acid + weak alkali	methyl orange	acidic transition range lies within the vertical section
weak acid + strong alkali	phenolphthalein	alkaline transition range lies within the vertical section
weak acid + weak alkali	neither	no sufficiently steep section gives a sharp indicator endpoint

The endpoint is the observed indicator colour change; the equivalence point is the stoichiometric point. A suitable range makes them close, but the two terms are not definitions of the same event.

Select from the supplied transition data rather than memorising an indicator name in isolation. No indicator pKa calculation is required, and a range merely containing the equivalence pH is insufficient if it extends outside the steep section.

7. Equilibria

Exam analysis

Most tested topics

7.1 Chemical equilibria and dynamic equilibrium

Dynamic equilibrium needs reversible reactions in a closed system

Le Chatelier's principle predicts how equilibrium position responds

Classify the disturbance before predicting an equilibrium shift

Build Kc from equilibrium concentrations and equation coefficients

Mole fraction converts a gas amount into its partial pressure

Build Kp using only equilibrium partial pressures of gases

Calculate Kc or Kp only from equilibrium values

Use stoichiometric change and K to recover every equilibrium quantity

Only temperature changes the equilibrium constant for a fixed reaction

Haber and Contact conditions balance equilibrium yield, rate, cost and safety

7.2 Brønsted–Lowry theory of acids and bases

Know the four syllabus acids by name and formula

Know the three syllabus alkalis by name and formula

A Brønsted–Lowry acid donates H⁺ and a base accepts H⁺

Strength is the extent of dissociation in aqueous solution

At standard conditions, pH 7 is neutral

At equal concentration, stronger acids show more ions and faster initial reactions

Neutralisation forms water from aqueous H⁺ and OH⁻

The acid supplies the anion and the base supplies the salt cation

Sketch titration curves from acid/base strength and addition direction

Choose an indicator range wholly inside the steep pH change