What this chapter covers7.1, 7.2
Most of the reactions met so far have been written with a single arrow, as though they run until something runs out. Very few of them actually do. Left in a closed container, a reaction mixture usually settles into a state where reactants and products sit side by side in fixed proportions and stay there — not because nothing is happening, but because two opposite reactions have arrived at the same speed.
Topic 7 has two halves, and they are closer relatives than they look. The first builds the idea of dynamic equilibrium, gives you a way to predict how a mixture answers a disturbance, and then makes the whole thing quantitative with an equilibrium constant. The second applies exactly that machinery to one particular equilibrium — a proton moving between an acid and a base — and shows what follows when it lies far to the right (a strong acid) or barely moves at all (a weak one).
What topic 7 asks you to do
7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
Understand reversible reactions, dynamic equilibrium and the need for a closed system; define and
use Le Chatelier's principle; deduce K꜀ and Kp expressions; use mole fraction and
partial pressure; carry out calculations with both constants; calculate the quantities present at
equilibrium; say which changes alter the value of an equilibrium constant; and explain the
conditions used in the Haber and Contact processes.
7.2 Brønsted–Lowry theory of acids and bases
Name the common acids and alkalis; describe the Brønsted–Lowry theory; distinguish strong from
weak by the extent of dissociation; relate that to pH, to conductivity and to the rate of reaction
with a metal; understand neutralisation as H⁺(aq) + OH⁻(aq) → H₂O(l), and the formation of salts;
sketch pH curves for the four acid–alkali combinations; and select a suitable indicator from
given data.
Two things are worth saying at the start, because they prevent most of the confusion later.
Position and rate are different questions
"How far does this reaction go?" and "how fast does it get there?" have separate answers, and nothing forces them to agree. An equilibrium constant of 10¹² describes a reaction that would run essentially to completion — and says nothing at all about whether that takes a microsecond or a century. Half the marks lost in this topic come from answering one question with the other: a catalyst, which changes only the rate, is the classic case.
The constant really is constant
Change the concentrations, change the pressure, add a catalyst — the value of K is untouched. The mixture simply rearranges itself until the same quotient is restored, and that rearrangement is what "the equilibrium shifts" means. Only temperature moves the number itself. Keep that sentence somewhere prominent; outcome 7.1.9 is written entirely around it.
What is deliberately not here
Ka, Kb, pKa, Kw, Ksp, buffer calculations and the quantitative treatment of indicators belong to further aspects of equilibria at A Level, and are not in topic 7. This chapter stays inside the AS material: the acid–base half is qualitative, pH values are read from data or a meter rather than calculated, and the syllabus states explicitly that pKa values will not be used when choosing an indicator. Where a model on this page prints a pH, it is doing arithmetic you are not required to reproduce — the number is there so you can see the shape of the curve come out right.
Reversible reactions and dynamic equilibrium7.1.1
A reversible reaction is one whose products can react together to give back the reactants. The double arrow ⇌ says so, and it says nothing else: it is not a statement that the mixture is half and half, or that the reaction is slow, or that it is somehow incomplete.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Mix nitrogen and hydrogen in a sealed vessel and the forward reaction starts fast, because the reactants are at their most concentrated and no ammonia exists yet. As ammonia builds up, two things happen at once. The forward reaction slows, because nitrogen and hydrogen are being used up. The reverse reaction speeds up, because there is more ammonia to decompose. Sooner or later the two rates meet.
Dynamic equilibrium
A system is at dynamic equilibrium when the forward and reverse reactions are proceeding at the same rate, so the concentrations of reactants and products remain constant.
Both halves of that sentence matter. "Constant concentrations" is what you observe; "equal rates" is the reason for it. An examiner who asks you to explain dynamic equilibrium is asking for the rates.
Dynamic is not decoration. Nothing has stopped. Molecules are still crossing in both directions, at a rate that may be enormous; it is only the net change that is zero. Label a few nitrogen atoms with the heavy isotope ¹⁵N and put them in as ammonia, and within minutes the label is spread through the nitrogen gas as well — proof that the reaction has been running the whole time in a mixture that looked entirely static.
The closed system
Equilibrium can only establish itself in a closed system — one that lets energy in and out but keeps matter in. The reason is mechanical rather than philosophical. If a product escapes, it is no longer available to react backwards, so the reverse reaction can never catch up with the forward one, and the reaction runs to completion instead.
The same reaction, open and closed
Heat calcium carbonate in a sealed tube and it reaches an equilibrium:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Carbon dioxide builds up until its pressure stops rising, and most of the carbonate is still there. Heat the same carbonate in an open crucible and the carbon dioxide drifts away; nothing can react back, and given time every last gram decomposes. Identical chemistry, opposite outcome, and the only difference is whether the gas could leave.
Common trap
"At equilibrium the concentrations of reactants and products are equal." They are almost never equal. What is constant is each concentration, not the relationship between them — and how unequal they are is exactly what the equilibrium constant measures.
The position of equilibrium7.1.1
The position of equilibrium is the relative proportions of reactants and products in the equilibrium mixture. If most of the material has ended up as products, the position is said to lie to the right; if hardly any reaction has occurred, it lies to the left.
The model below is not a drawing of that idea — it computes it. It steps two rate equations forward in time, one for the forward reaction and one for the reverse, and lets the concentrations go wherever the arithmetic takes them. The flat portion at the end is a result, not a shape that was drawn in advance.
Two things are worth doing with it. Set the mixture to start from the products instead of the reactants: the curves approach from the other side and finish in exactly the same place. The equilibrium position has no memory of which direction you came from — a fact that is used constantly in calculations, where a problem may start with pure product and be solved with the same expression. Then switch to a single starting mixture and watch the two rate curves: they meet, and from that moment neither concentration changes although both reactions continue.
Where the constant comes from
If the forward rate is kf[A][B] and the reverse rate is kr[C][D], then at equilibrium those two are equal, so
kf[A][B] = kr[C][D] → [C][D] ÷ [A][B] = kf ÷ kr
The ratio of two rate constants is itself a constant at a fixed temperature — and it is the equilibrium constant. That is why K depends on temperature and on nothing else: rate constants do. (The argument above is a simplification, since rate equations are not read off an equation in general, but it gets the right answer here and shows where the expression comes from rather than asking you to accept it.)
Le Chatelier's principle7.1.2
Disturb a system at equilibrium and it responds. The rule that says how is short enough to learn word for word, and the syllabus expects exactly its own wording.
Le Chatelier's principle
If a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.
Not "to oppose", not "to cancel", not "to restore the original conditions" — minimise. The system never fully undoes what you did; it partly offsets it. Add a reactant and some of it is used up, but not all of it: the new equilibrium contains more of that reactant than the old one did.
Used properly, the principle is a fast way of predicting a direction. Used carelessly it becomes a formula recited at a situation it does not fit. Three habits keep it honest:
Exam alert — how to write a Le Chatelier answer
1. Name the change. "The pressure is increased."
2. Name the property of the equation that responds to it. "There are four moles of gas
on the left and two on the right."
3. State the direction and why it minimises the change. "So the equilibrium moves to the
right, which reduces the number of gas molecules and therefore the pressure."
An answer missing the middle step is the one that loses the mark, because that step is the only
part that mentions this particular reaction.
The model below applies the principle to a dozen equilibria. Nothing in it is stored: each prediction is worked out from the number of gas molecules on either side of the equation and the sign of ΔH, which is all the information the principle actually uses.
Changing a concentration7.1.3
Add more of a reactant and the equilibrium moves right; remove a product and it also moves right. Both follow from minimising the change, and both can be seen more precisely through the constant.
Worked example — the iron(III) thiocyanate equilibrium
Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq)
The complex on the right is deep red; the ions on the left are pale. Add iron(III) nitrate and the solution darkens, because the extra Fe³⁺ is partly used up in making more complex. Add sodium hydroxide instead and the colour fades: hydroxide precipitates iron(III) hydroxide, removing Fe³⁺ from the left-hand side, so the equilibrium moves left to replace it.
In both cases K꜀ is the same before and after. What changed was the concentrations, and they changed in exactly the combination that restores the original quotient.
What "shifting" really means
Write the quotient of the current concentrations in the same form as K — products over reactants, each to the power of its coefficient. At equilibrium it equals K. Add a reactant and the bottom of that fraction grows, so the quotient drops below K. The mixture is no longer at equilibrium, and it reacts forwards — using up reactants, making products — until the quotient climbs back to K. "Moving to the right" is the description; restoring the quotient is the mechanism.
Changing the pressure7.1.3
Pressure only matters when gases are involved, and only when the two sides of the equation contain different numbers of gas molecules. Squeezing a mixture raises the pressure; the change is minimised by moving to whichever side has fewer gas molecules, since fewer molecules in the same volume means a lower pressure.
| equilibrium | gas molecules L → R | effect of raising the pressure |
|---|---|---|
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | 4 → 2 | moves right, more ammonia |
| 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) | 3 → 2 | moves right, more SO₃ |
| H₂(g) + I₂(g) ⇌ 2HI(g) | 2 → 2 | no shift at all |
| N₂O₄(g) ⇌ 2NO₂(g) | 1 → 2 | moves left, mixture pales |
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | 0 → 1 | moves left; only the gas is counted |
Common trap
Counting all the species instead of only the gases. In the calcium carbonate equilibrium there is one particle on each side of the equation, which tempts an answer of "no change" — but solids are effectively incompressible and contribute nothing to the pressure. One gas molecule appears on the right and none on the left, so raising the pressure pushes the equilibrium left.
Adding an inert gas
Pump argon into a sealed vessel at constant volume and the total pressure rises, but the partial pressure of every reacting gas is unchanged — each still has the same number of moles in the same volume. The quotient is therefore untouched and the position does not move. Only a change that alters the partial pressures of the reacting species can shift a gas equilibrium.
Changing the temperature7.1.3, 7.1.9
Temperature is the one change that does something different in kind. Raising it is minimised by absorbing heat, so the equilibrium moves in whichever direction is endothermic. Lowering it is minimised by releasing heat, so the equilibrium moves in the exothermic direction.
Worked example — the brown gas equilibrium
N₂O₄(g) ⇌ 2NO₂(g) ΔH = +57 kJ mol⁻¹
Colourless dinitrogen tetroxide on the left, brown nitrogen dioxide on the right. Put a sealed tube of the mixture into hot water and it darkens: heating drives the equilibrium in the endothermic direction, which is forwards. Put it into ice and it pales again. The colour is a direct readout of the position, which is why this is the demonstration everybody uses.
Note what has happened to the constant. More product at the higher temperature means a larger value of K꜀ — the number itself has moved, not just the mixture.
Exam alert
A temperature question needs the sign of ΔH, and ΔH as printed always refers to the forward reaction. If the forward reaction is exothermic, the reverse one is endothermic by the same amount, and heating favours the reverse. Saying "heating always increases the yield" is wrong for every exothermic reaction — which includes both industrial processes in this topic.
Adding a catalyst7.1.3
A catalyst provides an alternative route with a lower activation energy. That route is available to the reverse reaction just as much as to the forward one — it is the same hill, climbed from the other side — so both rate constants are multiplied by the same factor.
What a catalyst does to an equilibrium
It increases the rates of the forward and reverse reactions equally, so equilibrium is reached sooner. The position of equilibrium and the value of the equilibrium constant are unchanged, and so is the yield.
This is the cleanest illustration of the difference between the two questions at the start of the chapter. A catalyst is enormously valuable industrially — without iron, the Haber process would need temperatures at which the yield collapses — and it makes no difference whatsoever to how much ammonia a given mixture can contain at equilibrium.
What changes K, and what only moves the position7.1.9
This outcome is a single short list, and it is tested often because it separates people who have understood the topic from people who have learned the phrase "the equilibrium shifts".
| change | position of equilibrium | value of K꜀ or Kp | rate |
|---|---|---|---|
| increase the concentration of a reactant | moves right | unchanged | faster |
| remove a product as it forms | moves right | unchanged | slower, as the mixture thins |
| increase the total pressure | moves to the side with fewer gas molecules | unchanged | faster |
| add a catalyst | no movement | unchanged | faster, both ways equally |
| increase the surface area of a solid catalyst | no movement | unchanged | faster |
| add an inert gas at constant volume | no movement | unchanged | unchanged |
| raise the temperature | moves in the endothermic direction | changes | faster |
Why only temperature can move the number
K is the ratio of the forward and reverse rate constants, and a rate constant depends on temperature — through the fraction of collisions with enough energy — and on nothing else in this syllabus. Change a concentration or a pressure and you change the quotient; the mixture then reacts until the quotient equals the same old K. Change the temperature and you change K itself, so the mixture must move to a genuinely different composition.
A catalyst is the interesting case: it changes both rate constants, but by the same factor, so their ratio — and therefore K — survives untouched.
Reading ΔH out of a table of constants
A question gives K꜀ for a reaction at two temperatures and asks whether the forward reaction is exothermic or endothermic. There is no need to know any chemistry at all:
| T / K | K꜀ |
|---|---|
| 500 | 0.25 |
| 700 | 4.0 |
K rises as the temperature rises, so heating has pushed the equilibrium towards the products. Heating always pushes an equilibrium in the endothermic direction. Therefore the forward reaction is endothermic, ΔH positive.
Common trap
"Increasing the pressure increases Kp, because the partial pressures go up." Every partial pressure does rise — and they rise in exactly the combination that leaves the quotient at its old value once the mixture has shifted. In the ammonia equilibrium, squaring a larger p(NH₃) on top is offset precisely by the larger p(N₂)p(H₂)³ underneath. That cancellation is not a coincidence; it is what makes Kp useful.
The Haber and Contact processes7.1.10
These two processes are on the syllabus because they are where the whole topic has to be used at once. Nobody chooses industrial conditions from equilibrium alone; the choice is a negotiation between the yield the equilibrium allows, the rate at which that yield arrives, and what the plant costs to build and run.
The Haber process
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
| condition | what equilibrium wants | what rate and cost want | what is used |
|---|---|---|---|
| pressure | as high as possible — 4 gas molecules become 2 | high pressure needs thick vessels and powerful compressors, and is dangerous | about 200 atm |
| temperature | as low as possible — the forward reaction is exothermic | a low temperature makes the reaction hopelessly slow | about 450 °C, a compromise |
| catalyst | no effect on yield | essential: without it, 450 °C is far too cool to be quick | iron |
| the ammonia itself | removing it pulls the equilibrium right | it liquefies well above the temperature at which N₂ and H₂ do | condensed out, the gases recycled |
The Contact process
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJ mol⁻¹
The same reasoning, with one condition coming out differently. Three gas molecules become two, so pressure again favours the product — but the conversion at ordinary pressure is already about 96%, and no plant pays for high-pressure equipment to chase the last few per cent. So the Contact process runs at roughly atmospheric pressure, around 450 °C, over a vanadium(V) oxide catalyst.
Why the two processes choose different pressures
Both have Δn negative, so both are helped by pressure. What differs is where they start. The ammonia equilibrium at 450 °C and one atmosphere gives a yield of a fraction of a per cent — useless — so there is an enormous amount to gain by squeezing, and 200 atm is worth its cost. The sulfur trioxide equilibrium is nearly complete already, and the same investment would buy about three extra per cent. Identical chemistry, opposite economics.
The model below solves each equilibrium properly at whatever temperature and pressure you set: it finds Kp at that temperature and then solves for the composition that satisfies it. The trade-off is not asserted, it falls out.
Exam alert
A question asking why 450 °C is used wants both halves of the compromise, and it wants the direction of each: a lower temperature would give a higher equilibrium yield but an unacceptably slow rate. An answer that gives only the yield argument concludes that the plant should be run cold, which is exactly the error the question is testing for.
Why a real plant beats the equilibrium figure
A single pass through the Haber converter reaches about 36% ammonia at 400 °C and 200 atm, and less at the higher temperature actually used. That is not the plant's overall yield. The ammonia is condensed out and the unreacted nitrogen and hydrogen are fed back in, so the same molecules get repeated attempts; overall conversion approaches 98%. Continuous removal of a product is the one Le Chatelier move that keeps paying, because the mixture never gets to settle.
The equilibrium constant K꜀7.1.4
Le Chatelier gives directions. The equilibrium constant gives numbers, and the same rule builds it for every reaction. For a general equilibrium
aA + bB ⇌ cC + dD
The equilibrium constant in terms of concentration
K꜀ = [C]c[D]d ÷ [A]a[B]b
Products on top, reactants underneath, each concentration raised to the power of its coefficient in the balanced equation, with every concentration measured at equilibrium. Square brackets mean "concentration of, in mol dm⁻³".
Three details decide most of the marks. The expression is written from the equation as it is given — reverse the equation and you invert K, double it and you square K, so an answer must match the equation in the question. Only equilibrium concentrations go in: starting concentrations belong in the first line of a calculation, never in the expression. And the powers come from the coefficients, which is why a 3 in front of H₂ becomes a cube and not a factor of three.
Worked example — writing three expressions
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) K꜀ = [NH₃]² ÷ [N₂][H₂]³
2NH₃(g) ⇌ N₂(g) + 3H₂(g) K꜀′ = [N₂][H₂]³ ÷ [NH₃]² = 1 ÷ K꜀
½N₂(g) + 1½H₂(g) ⇌ NH₃(g) K꜀″ = [NH₃] ÷ [N₂]½[H₂]1½ = √K꜀
All three describe the same chemistry at the same temperature, and all three have different numerical values — including different units. The constant belongs to an equation, not to a reaction in the abstract.
What the size of K tells you
K is a ratio of products to reactants, so a large value means the top dominates and the position lies far to the right; a small value means the opposite. As rough landmarks: K greater than about 10¹⁰ describes a reaction that goes essentially to completion, K below about 10⁻¹⁰ one that barely happens, and anything in between gives a genuine mixture. What K never tells you is how long any of this takes.
What gets left out of the expression7.1.4
Two kinds of species are omitted, and for the same underlying reason: their concentration cannot meaningfully change, so it is absorbed into the constant.
Omit a solid, and omit a liquid that is acting as the solvent
A solid has a fixed number of moles in every cubic decimetre of itself. Doubling the amount of calcium carbonate in the flask doubles the volume of carbonate too, so its "concentration" is the same number as before, whatever you do.
A solvent — water in a dilute aqueous equilibrium — is present in such vast excess that the reaction changes its concentration by a negligible fraction.
| equilibrium | K꜀ expression |
|---|---|
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | K꜀ = [CO₂] |
| Co(H₂O)₆²⁺(aq) + 4Cl⁻(aq) ⇌ CoCl₄²⁻(aq) + 6H₂O(l) | K꜀ = [CoCl₄²⁻] ÷ [Co(H₂O)₆²⁺][Cl⁻]⁴ |
| CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) | K꜀ = [ester][H₂O] ÷ [acid][alcohol] |
Common trap
"Liquids are always left out." The third row above disproves it. In the esterification there is no solvent — the four species are the liquid, all present in comparable amounts, and all four concentrations vary as the reaction proceeds. Water is omitted when it is the solvent, not because it is written (l). The test is whether the concentration can change appreciably, not what state symbol follows it.
The units of K꜀7.1.4
Units are not decoration here; they are marked, and they are easy marks because they follow mechanically from the expression. Put mol dm⁻³ in place of every concentration, raise each to its power, and cancel.
Worked example — A + 2B ⇌ C + D
K꜀ = [C][D] ÷ [A][B]²
a. Insert the units:
(mol dm⁻³)(mol dm⁻³) ÷ (mol dm⁻³)(mol dm⁻³)²
b. Two terms on top, three underneath. Cancel two from each:
1 ÷ (mol dm⁻³)
c. Bring it onto the top by changing the sign of each power:
K꜀ = mol⁻¹ dm³
The shortcut, and why it works
Let Δn be the number of species in the expression on the right minus the number on the left, counting coefficients. Then the units of K꜀ are (mol dm⁻³)Δn, which is written molΔn dm−3Δn.
For N₂ + 3H₂ ⇌ 2NH₃, Δn = 2 − 4 = −2, so the units are mol⁻² dm⁶. For N₂O₄ ⇌ 2NO₂, Δn = +1, so mol dm⁻³. And whenever Δn = 0 — H₂ + I₂ ⇌ 2HI, or the esterification — everything cancels and K꜀ has no units at all. Writing units after such a value is an error, not untidiness.
Exam alert
Deduce the units from the expression every time, even when you think you remember them. The same reaction written in a different but equally valid way — halved coefficients, or reversed — has different units, so a remembered answer is attached to the wrong equation as often as the right one.
Calculating K꜀ from experimental data7.1.7
Almost every K꜀ question has the same shape. You are told what went in, and one thing about what came out. Everything else follows from the balanced equation, because the amounts that react are locked to one another by the coefficients.
The four steps
1. Amounts at equilibrium, in moles, for every species — from the one that was measured.
2. Concentrations, by dividing each amount by the volume in dm³.
3. The K꜀ expression, written from the equation.
4. Substitute, evaluate, and work out the units.
Worked example — steam reforming
1.5 mol of methane and 3.0 mol of steam are heated in a 100 cm³ vessel. At equilibrium the mixture contains 0.25 mol of carbon dioxide.
CH₄(g) + 2H₂O(g) ⇌ CO₂(g) + 4H₂(g)
Step 1. 0.25 mol of CO₂ has formed, so 0.25 mol of CH₄ has reacted (ratio 1 : 1), 0.50 mol of H₂O has reacted (1 : 2), and 1.0 mol of H₂ has formed (1 : 4).
| CH₄ | H₂O | CO₂ | H₂ | |
|---|---|---|---|---|
| start / mol | 1.50 | 3.00 | 0 | 0 |
| change / mol | −0.25 | −0.50 | +0.25 | +1.00 |
| equilibrium / mol | 1.25 | 2.50 | 0.25 | 1.00 |
| ÷ 0.100 dm³ = conc. / mol dm⁻³ | 12.5 | 25.0 | 2.5 | 10.0 |
Step 3 and 4.
K꜀ = [CO₂][H₂]⁴ ÷ [CH₄][H₂O]² = (2.5 × 10.0⁴) ÷ (12.5 × 25.0²) = 3.2 mol² dm⁻⁶
Δn = 5 − 3 = +2, hence mol² dm⁻⁶. Note the 100 cm³ converted to 0.100 dm³ before anything else happened — the single commonest arithmetic slip in this topic.
The model below runs those four steps on whatever you give it. Change the measured amount or the volume of the vessel and watch which parts of the answer move — the concentrations always, the value of K꜀ always, the units never.
Common trap
Putting moles into the expression instead of concentrations. Sometimes you get away with it: when Δn = 0 the volumes cancel and the answer is right by accident. When Δn is not zero the answer is wrong by a factor of VΔn — a plausible-looking number with no obvious sign of trouble. Divide by the volume every time, and the habit costs nothing on the occasions when it was not needed.
Working through the harder cases7.1.7, 7.1.8
Three variations account for nearly all the awkward questions.
When the measured species is a reactant
0.56 mol of nitrogen dioxide decomposes in a 500 cm³ vessel; at equilibrium 0.48 mol of nitrogen monoxide is present.
2NO₂(g) ⇌ 2NO(g) + O₂(g)
0.48 mol of NO has formed, so 0.48 mol of NO₂ has reacted (2 : 2) leaving 0.08 mol, and 0.24 mol of O₂ has formed (2 : 1). Dividing by 0.500 dm³ gives 0.16, 0.96 and 0.48 mol dm⁻³.
K꜀ = [NO]²[O₂] ÷ [NO₂]² = (0.96² × 0.48) ÷ 0.16² = 17.3 mol dm⁻³
When products are present at the start
Nothing changes except the first row of the table. If a vessel already holds 1 mol of HI along with 1 mol each of H₂ and I₂, the HI row starts at 1 rather than 0, and the equilibrium amount is (1 + 2x) rather than 2x. The mixture may then move either way — whichever direction brings the quotient to K — and a negative x simply means it went backwards.
When you are told a percentage
"At equilibrium, 40% of the phosphorus(V) chloride has dissociated" is a statement about the change row, not the equilibrium row. Starting from 1.00 mol, 0.40 mol has reacted, so 0.60 mol of PCl₅ remains and 0.40 mol each of PCl₃ and Cl₂ have appeared. Reading the percentage as the amount left is a favourite way of losing every mark that follows.
Finding the composition when K꜀ is known7.1.8
The other direction: the constant is given and the equilibrium amounts are wanted. The algebra looks worse than it is, because the syllabus guarantees you will never need to solve a quadratic. What it needs instead is the willingness to call the unknown x and fill in the table with expressions rather than numbers.
Worked example — the ester equilibrium
1.0 mol of ethanoic acid and 1.0 mol of ethanol are left with a trace of acid catalyst until no further change occurs. K꜀ = 4.0 at 298 K.
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Let x be the number of moles of acid that react. At equilibrium: acid (1 − x), alcohol (1 − x), ester x, water x. Dividing each by the volume V:
K꜀ = (x/V)(x/V) ÷ [(1−x)/V][(1−x)/V] = x² ÷ (1−x)²
Every V cancels — Δn = 0 — so the volume is never needed, which is fortunate, because the question does not give it. Both sides are perfect squares, so take the square root:
√4.0 = x ÷ (1 − x) → 2.0 − 2.0x = x → x = ⅔
At equilibrium: 0.67 mol each of ester and water, 0.33 mol each of acid and alcohol.
Exam alert — is the answer sensible?
Taking a square root gives two answers, and one of them is usually nonsense: an amount larger than what you started with, or a negative amount. Check every answer against the starting amounts before writing it down. x = ⅔ passes because ⅔ is less than the 1.0 mol available; x = 2 would not.
Why the square root trick keeps working
It works whenever the expression is a perfect square on both top and bottom — which happens when the two reactants start in equal amounts and the coefficients match, as they do in the ester equilibrium, or when a squared product sits over two equal reactants, as in H₂ + I₂ ⇌ 2HI. Those two shapes cover essentially every question this syllabus sets. If you ever find yourself facing a genuine quadratic, re-read the question: something has been misread.
Mole fraction and partial pressure7.1.5
For an equilibrium between gases it is more natural to work in pressures than in concentrations, because pressure is what a gauge on the vessel actually reads. Two definitions carry the whole idea.
Mole fraction and partial pressure
The mole fraction of a gas is the fraction of all the molecules present that are molecules of that gas:
x(A) = n(A) ÷ n(total)
The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the container — found by sharing out the total pressure in proportion:
p(A) = x(A) × p(total)
Two consequences are worth holding onto, because they are the quickest check on any answer. Mole fractions are fractions of a whole, so they add up to exactly 1. Partial pressures therefore add up to the total pressure. And since a mole fraction has no units, a partial pressure carries whatever unit the total pressure was given in — usually kPa or Pa in this syllabus, sometimes atm.
Worked example
A vessel at 200 kPa contains 0.90 mol N₂, 2.70 mol H₂ and 0.40 mol NH₃.
Total: 0.90 + 2.70 + 0.40 = 4.00 mol.
x(N₂) = 0.90/4.00 = 0.225 x(H₂) = 0.675 x(NH₃) = 0.100
p(N₂) = 45.0 kPa p(H₂) = 135 kPa p(NH₃) = 20.0 kPa
The fractions add to 1.000 and the pressures add to 200 kPa. Both checks take two seconds and catch almost every slip.
Common trap
Reading mole fractions off the balanced equation. The coefficients tell you the ratio in which the gases react, not the amounts present in the flask. A mixture at equilibrium is whatever the reaction left behind, and the only time its composition matches the equation is when a question has arranged it that way. Always total the amounts actually given.
Kp expressions and units7.1.6
Kp is built exactly like K꜀, with a partial pressure wherever a concentration would have gone.
The equilibrium constant in terms of partial pressure
For aA(g) + bB(g) ⇌ cC(g) + dD(g):
Kp = p(C)c p(D)d ÷ p(A)a p(B)b
Gases only. A solid or a liquid has no partial pressure, so it never appears — and an equilibrium with no gases in it has no Kp at all.
| equilibrium | Kp | Δn | units |
|---|---|---|---|
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | p(NH₃)² ÷ p(N₂)p(H₂)³ | −2 | kPa⁻² |
| 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) | p(SO₃)² ÷ p(SO₂)²p(O₂) | −1 | kPa⁻¹ |
| H₂(g) + I₂(g) ⇌ 2HI(g) | p(HI)² ÷ p(H₂)p(I₂) | 0 | no units |
| PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) | p(PCl₃)p(Cl₂) ÷ p(PCl₅) | +1 | kPa |
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | p(CO₂) | +1 | kPa |
Common trap
Attaching a unit to a Kp that has none. For H₂ + I₂ ⇌ 2HI there are two pressure terms on top and two underneath, so every kPa cancels and the answer is a bare number: 49.1, not 49.1 kPa. The habit of always writing a unit after a value is right nearly everywhere else in chemistry, which is precisely why this one slips through.
Calculating Kp7.1.7
One extra step compared with K꜀, and it goes in the middle: amounts → total amount → mole fractions → partial pressures → Kp.
Worked example — methanol synthesis
2.0 mol of hydrogen and 1.0 mol of carbon monoxide are mixed. At equilibrium the mixture contains 0.175 mol of methanol, at a total pressure of 1.75 × 10⁴ kPa.
2H₂(g) + CO(g) ⇌ CH₃OH(g)
| H₂ | CO | CH₃OH | total | |
|---|---|---|---|---|
| start / mol | 2.000 | 1.000 | 0 | 3.000 |
| equilibrium / mol | 1.650 | 0.825 | 0.175 | 2.650 |
| mole fraction | 0.623 | 0.311 | 0.066 | 1.000 |
| partial pressure / kPa | 10 900 | 5440 | 1160 | 17 500 |
Kp = p(CH₃OH) ÷ p(H₂)²p(CO) = 1155 ÷ (10902.5² × 5442.5) = 1.79 × 10⁻⁹ kPa⁻²
Δn = 1 − 3 = −2, hence kPa⁻². Notice that the total number of moles falls as the reaction proceeds — three molecules become one — so the mole fractions must be worked out from the equilibrium total, 2.650, and not from the 3.000 that went in.
Change the measured amount in the model and every row of the table moves with it. Change only the total pressure and something more interesting happens: for this reaction Kp changes, because the mixture as described is no longer at equilibrium — which is the numerical face of the fact that pressure shifts an equilibrium whenever Δn is not zero. Try the same experiment on the hydrogen iodide problem, where Δn = 0, and Kp does not budge.
Exam alert
Keep the two constants apart. K꜀ takes concentrations and needs the volume; Kp takes partial pressures and needs the total pressure and the total number of moles. Mixing a concentration into a Kp expression, or quoting mol dm⁻³ as the unit of a Kp, is treated as a chemistry error rather than a slip of the pen.
Acids and bases as proton transfer7.2.1, 7.2.2, 7.2.3
The second half of topic 7 is one particular equilibrium: a hydrogen ion moving from one species to another. Everything in this part is that reaction, seen from different angles.
The Brønsted–Lowry theory
An acid is a proton donor.
A base is a proton acceptor.
A hydrogen ion is a bare proton — a hydrogen atom that has lost its only electron — so "proton" and "H⁺" are the same thing here.
This definition is broader than the older one about producing H⁺ in water, and the extra breadth earns its keep. When ammonia gas meets hydrogen chloride gas, a white smoke of ammonium chloride forms instantly. No water is involved anywhere, and yet a proton has moved from HCl to NH₃ — an acid–base reaction by the Brønsted–Lowry definition and by no earlier one.
NH₃(g) + HCl(g) → NH₄Cl(s)
Water plays both parts
Dissolve hydrogen chloride in water and water accepts the proton, acting as a base:
HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
Dissolve ammonia in water and water donates one, acting as an acid:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
The hydrated proton H₃O⁺ is often written simply as H⁺(aq), and this syllabus accepts either unless a question specifically asks for one. H⁺(aq) is the shorter and the more common in mark schemes.
| acids | formula | alkalis | formula |
|---|---|---|---|
| hydrochloric acid | HCl | sodium hydroxide | NaOH |
| sulfuric acid | H₂SO₄ | potassium hydroxide | KOH |
| nitric acid | HNO₃ | ammonia | NH₃ |
| ethanoic acid | CH₃COOH |
Three of those acids are strong and one — ethanoic — is weak. Two of the alkalis are strong and ammonia is weak. That pattern is the raw material for everything that follows, including which pH curve you draw and which indicator you choose.
Strong and weak7.2.4, 7.2.6
Strong and weak
A strong acid or base is fully dissociated in aqueous solution. A weak one is only partially dissociated.
HCl(aq) → H⁺(aq) + Cl⁻(aq) one arrow: every molecule
CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq) two arrows: an equilibrium, lying far left
Common trap — strength is not concentration
"Concentrated" counts molecules per cubic decimetre; "strong" asks what fraction of them have given up a proton. The two are independent. Concentrated ethanoic acid is a weak acid; very dilute hydrochloric acid is a strong one. A question that says "equal concentrations" is inviting you to compare strengths; a question that says "equal volumes of the same concentration" usually wants you to notice that the amount of acid is the same even though the [H⁺] is not.
The model below solves the dissociation equilibrium for whichever acid and concentration you pick and reports what fraction of the molecules have actually given up a proton.
Three ways to tell them apart in the laboratory
| observation | what you see | why |
|---|---|---|
| pH meter or universal indicator | strong acid gives a much lower pH — about 0 against about 2.4 at 1.0 mol dm⁻³ | [H⁺] is hundreds of times higher when every molecule has dissociated |
| electrical conductivity | strong acid conducts far better | current is carried by ions, and the weak acid is mostly un-ionised molecules at any instant |
| reaction with magnesium | strong acid fizzes faster | rate depends on [H⁺], which is much larger |
| total hydrogen given off, excess magnesium | the same | it depends on the moles of acid, and those are equal |
| volume of alkali needed to neutralise | the same | same reason — every molecule of the weak acid is eventually neutralised |
Why a weak acid still reacts completely
This is the part that feels wrong at first. If only one ethanoic acid molecule in two hundred has ionised, how can the acid react with all of the magnesium?
Because it is an equilibrium. As H⁺ is consumed, the quotient falls below K, and more molecules dissociate to replace it. The reservoir of un-ionised molecules keeps topping up the supply until it is exhausted. A weak acid is slower, not smaller: strength controls the rate, the amount of acid controls the outcome.
The pH scale7.2.5
pH is a compressed way of reporting hydrogen ion concentration. At AS you are not asked to calculate it, but you are expected to read it correctly.
What the numbers mean
Pure water has pH 7. Acidic solutions have pH below 7. Alkaline solutions have pH above 7. The scale is logarithmic: each whole number down the scale means ten times more hydrogen ions, so pH 2 is a hundred times more acidic than pH 4 — not twice.
| solution | pH | why |
|---|---|---|
| strong acid, e.g. HCl | 0 – 1 | [H⁺] equals the concentration of the acid |
| weak acid, e.g. CH₃COOH | 2 – 3 | only a small fraction has dissociated |
| pure water | 7 | [H⁺] = [OH⁻] |
| weak alkali, e.g. aqueous ammonia | 11 – 12 | partially ionised, so [OH⁻] is well below 1 mol dm⁻³ |
| strong alkali, e.g. NaOH | 14 | fully ionised |
Alkali, base, and which word to use
A base is any proton acceptor. An alkali is a base that dissolves in water. Copper(II) oxide is a base — it neutralises acids perfectly well — but it is not an alkali, because it will not dissolve. Every alkali is a base; most bases are not alkalis.
Neutralisation and salts7.2.7, 7.2.8
Write out the neutralisation of hydrochloric acid by sodium hydroxide in full ionic form and most of it turns out to be spectators:
H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)
Sodium and chloride ions start as separated aqueous ions and finish as separated aqueous ions. Strike them out and what is left is the reaction itself:
Neutralisation
H⁺(aq) + OH⁻(aq) → H₂O(l)
This is the ionic equation for the neutralisation of any strong acid by any soluble strong base. The salt is what is left in solution afterwards: the metal ion from the base and the anion from the acid.
It also explains a result from the energetics topic. The enthalpy of neutralisation of a strong acid by a strong alkali is close to −57 kJ mol⁻¹ whichever pair you choose, because in every case the reaction being measured is the same one.
| acid | anion | salt with NaOH |
|---|---|---|
| hydrochloric, HCl | chloride, Cl⁻ | sodium chloride, NaCl |
| nitric, HNO₃ | nitrate, NO₃⁻ | sodium nitrate, NaNO₃ |
| sulfuric, H₂SO₄ | sulfate, SO₄²⁻ | sodium sulfate, Na₂SO₄ |
| ethanoic, CH₃COOH | ethanoate, CH₃COO⁻ | sodium ethanoate, CH₃COONa |
Exam alert
For a weak acid the ionic equation is not H⁺ + OH⁻ → H₂O, because the acid is not present as ions to begin with. Write the molecule:
CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)
That distinction matters at the end of the next section too: it is the reason a weak acid titration does not finish at pH 7.
pH titration curves7.2.9
Run alkali from a burette into acid in a flask, following the pH with a meter, and plot pH against the volume added. The syllabus asks for the sketch, for all four combinations of strong and weak, and what it is really asking for is three features of each curve.
The three things that define a curve
Where it starts. Low for a strong acid (about pH 1), higher for a weak one (about pH 3),
because the weak acid is barely ionised.
Where it ends. High for a strong alkali (approaching pH 13), lower for a weak one (about
pH 11).
The vertical section — how long it is and where it sits. This is the part that decides
everything practical.
The curves below are not drawn; each point is computed by solving the acid–base equilibrium for the mixture in the flask at that moment. Switch between the four combinations and watch the vertical section shrink.
| titration | starting pH | pH at the end point | vertical section |
|---|---|---|---|
| strong acid + strong alkali | 1.0 | 7.0 | very long, roughly pH 3 to 11, centred on 7 |
| weak acid + strong alkali | 2.9 | 8.7 | shorter, and on the alkaline side |
| strong acid + weak alkali | 1.0 | 5.3 | shorter, and on the acidic side |
| weak acid + weak alkali | 2.9 | 7.0 | essentially none — the pH drifts through |
Why the end point is not always pH 7
At the end point the flask contains the salt, and nothing else but water. For NaCl that is the end of the story: neither ion reacts with water, so the solution is neutral.
Sodium ethanoate is different. The ethanoate ion is the conjugate base of a weak acid, so it takes protons back from water — CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ — leaving the solution alkaline. Ammonium chloride does the mirror image: the ammonium ion releases a proton to water, so the solution is acidic. The end point pH follows from which partner was weak, every time.
Exam alert — sketching one of these
A sketch is marked on shape and on three or four specific points, not on artistry. Show: a sensible starting pH; a gentle rise; a near-vertical section at the correct volume; the end-point pH clearly above, below or at 7 as appropriate; and a levelling off towards the pH of the alkali being added. Mark the end-point volume on the axis. A curve that rises vertically at the wrong volume loses more than one that is drawn untidily.
The half-way point
Half way to the end point of a weak acid titration, exactly half the acid has been converted to its anion, so the two are present in equal amounts. That point is a buffer, and its pH stays remarkably steady — which is why the weak-acid curve has a flat shoulder before the rise where the strong-acid curve has none. Buffers themselves are A Level material; the flat shoulder is simply worth recognising when you see it.
Choosing an indicator7.2.10
An acid–base indicator is itself a weak acid whose two forms have different colours. It changes over a range of roughly two pH units, and which range that is depends on the indicator.
The rule for choosing
The indicator must change colour within the vertical section of the curve.
Inside that section, a single drop of alkali takes the pH right through the indicator's range, so the colour change is sharp and marks the end point to within a drop. Outside it, the pH creeps through the range over several cubic centimetres and there is no end point to see.
The table below is not a list of remembered answers. Each verdict is produced by computing the curve, finding the volume at which the indicator is half way through its colour change, and comparing that with the true end point.
| titration | suitable indicator | why |
|---|---|---|
| strong acid + strong alkali | methyl orange or phenolphthalein | the vertical section spans pH 3 to 11, so both ranges lie inside it |
| weak acid + strong alkali | phenolphthalein | the vertical section is on the alkaline side; methyl orange would change far too early |
| strong acid + weak alkali | methyl orange | the vertical section is on the acidic side; phenolphthalein never gets there at all |
| weak acid + weak alkali | none | there is no vertical section, so no indicator can give a sharp end point |
How much error an unsuitable indicator causes
Titrating 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid with 0.100 mol dm⁻³ sodium hydroxide, the true end point is at 25.0 cm³. Methyl orange changes colour around pH 3.7, which this titration passes through after roughly 2 cm³ — an error of over 90%. Phenolphthalein changes around pH 9.2, which is reached within a small fraction of a drop of the true end point. The choice is not a refinement; it is the difference between a titration and a guess.
Common trap
"Phenolphthalein is for acids, methyl orange is for alkalis" — or any other version that ties an indicator to a substance. The indicator is chosen by the curve, which depends on both partners. Phenolphthalein is right for a weak acid with a strong alkali and wrong for a strong acid with a weak alkali, and the acid is strong in exactly the case where phenolphthalein fails.
On pKa
The syllabus states explicitly that pKa values will not be used when selecting an indicator at this level. You will be given pH ranges, as in the table above, and asked to match one to a curve. The quantitative theory of why an indicator changes over about two pH units around its pKa belongs to the A Level equilibria unit.
Self-test7.1, 7.2
Thirty-two questions across the whole of topic 7, each with the reasoning rather than just the answer. Several are built around the specific errors described earlier — the catalyst that is supposed to improve a yield, the units written after a constant that has none, the mole fractions read off the balanced equation, the indicator chosen by habit.
Definitions to learn7.1, 7.2
These are the wordings that earn the mark. Each is short enough to write out under pressure and specific enough that a paraphrase usually drops something an examiner is looking for.
| term | definition |
|---|---|
| reversible reaction | a reaction in which the products can react together to re-form the reactants |
| dynamic equilibrium | the state in which the forward and reverse reactions proceed at equal rates, so the concentrations of reactants and products remain constant |
| closed system | one in which no substance is added or lost, although energy may be transferred in or out |
| position of equilibrium | the relative proportions of reactants and products in the equilibrium mixture |
| Le Chatelier's principle | if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change |
| K꜀ | the equilibrium constant in terms of concentrations: products over reactants, each raised to the power of its coefficient, at equilibrium |
| mole fraction | the number of moles of a component divided by the total number of moles present |
| partial pressure | the pressure a gas in a mixture would exert if it alone occupied the container; mole fraction × total pressure |
| Kp | the equilibrium constant in terms of partial pressures, for the gases only |
| catalyst | a substance that increases the rate of a reaction by providing an alternative route of lower activation energy, and is unchanged at the end |
| Brønsted–Lowry acid | a proton donor |
| Brønsted–Lowry base | a proton acceptor |
| strong acid or base | one that is fully dissociated in aqueous solution |
| weak acid or base | one that is only partially dissociated in aqueous solution |
| alkali | a base that is soluble in water |
| neutralisation | the reaction H⁺(aq) + OH⁻(aq) → H₂O(l), leaving a salt in solution |
| end point of a titration | the volume at which the acid and alkali are present in exactly the proportions of the equation |
Three phrasings that lose marks
"At equilibrium the reaction has stopped." It has not; both reactions continue at equal rates. The word dynamic is in the definition precisely to exclude this answer.
"Le Chatelier's principle says the equilibrium opposes the change." Cambridge wants minimise. The system offsets part of what you did — it never cancels it.
"A strong acid is one with a high concentration of acid." Strength is the fraction of molecules dissociated. Say fully or partially dissociated, and the mark is safe.
Data used on this page7.1, 7.2
Every model on this page computes its answers from the values below rather than storing conclusions, so it is worth being able to see them.
These are reference values, not the official data booklet
Sources disagree in the last digit or two, and equilibrium constants in particular vary with how a source defines its standard state. In an examination, use the figures printed in the question or in the data booklet supplied with the paper. The values here are internally consistent, which is what matters for the models to behave correctly.
| equilibrium | ΔH |
|---|---|
| N₂ + 3H₂ ⇌ 2NH₃ | −92 |
| 2SO₂ + O₂ ⇌ 2SO₃ | −196 |
| H₂ + I₂ ⇌ 2HI | −10 |
| N₂O₄ ⇌ 2NO₂ | +57 |
| PCl₅ ⇌ PCl₃ + Cl₂ | +88 |
| 2H₂ + CO ⇌ CH₃OH | −90 |
| CH₄ + 2H₂O ⇌ CO₂ + 4H₂ | +165 |
| CaCO₃ ⇌ CaO + CO₂ | +178 |
| CO + H₂O ⇌ CO₂ + H₂ | −41 |
The industrial model needs one more thing: a value of Kp to start from. For each process, Kp is fixed at the temperature the plant actually uses so that the model reproduces the conversion usually quoted for it — 36% ammonia from a 1 : 3 mixture at 400 °C and 200 atm, and 96% conversion of sulfur dioxide at 450 °C and 1 atm. The van 't Hoff equation then carries that value to other temperatures using the ΔH above. Treating ΔH as constant over a 300 K range is an approximation, so read the curves as showing the shape of the trade-off rather than as plant data.
| acid | Ka / mol dm⁻³ | pKa |
|---|---|---|
| hydrochloric, nitric, sulfuric (first proton) | treated as fully dissociated | — |
| methanoic acid, HCOOH | 1.78 × 10⁻⁴ | 3.75 |
| ethanoic acid, CH₃COOH | 1.74 × 10⁻⁵ | 4.76 |
| propanoic acid, CH₃CH₂COOH | 1.35 × 10⁻⁵ | 4.87 |
| carbonic acid, H₂CO₃ (first proton) | 4.3 × 10⁻⁷ | 6.37 |
| ammonia, NH₃ (as Kb) | 1.78 × 10⁻⁵ | pKb 4.75 |
The ionic product of water is taken as 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. Titration curves are computed from the charge balance of the mixture rather than from an approximate formula, which is why they stay correct at the end point and in dilute solution.
| indicator | pH range | colour change, acid → alkali |
|---|---|---|
| methyl orange | 3.1 – 4.4 | red → yellow |
| methyl red | 4.2 – 6.3 | red → yellow |
| bromothymol blue | 6.0 – 7.6 | yellow → blue |
| phenolphthalein | 8.3 – 10.0 | colourless → pink |
| thymolphthalein | 9.3 – 10.5 | colourless → blue |
An indicator is counted as suitable when the middle of its colour change falls within half a per cent of the true end-point volume and the whole change is complete within two per cent of it — about a drop or two on a 25 cm³ titre. Both figures are measured on the computed curve for that particular pair of reagents, which is why the same indicator is accepted for one titration and rejected for another.