IB Chemistry Reactivity 2.3: why 'how far' is a number, not a feeling

The IB Chemistry Reactivity 2.3 sub-topic, formally titled 'How far? The extent of chemical change', is the part of the IB Diploma Programme syllabus where learners move from asking whether a reaction occurs to asking how completely it occurs. It is the first sustained encounter with reversible reactions, the equilibrium constant, and Le Chatelier's principle, and it sits inside the second of the three organising themes of the IB Chemistry guide. For most candidates, the marks lost in this sub-topic are not lost on the ideas themselves; they are lost on the calculation scaffolding, the units, and on knowing which paper rewards which kind of answer.

Reactivity 2.3 appears in both Standard Level and Higher Level, but the HL extension adds quantitative treatment of Kp, the relationship between Kc and Kp, and more demanding application questions. Understanding this split shapes a preparation strategy that targets the right marks on the right paper, and avoids the common trap of rehearsing SL-style descriptive answers when the HL Paper 2 wants an ICE table, a Kp expression, and a partial-pressure calculation. The rest of this article walks through the conceptual core, the calculation scaffolds, the common pitfalls, and the question types that decide a level 5 from a level 7.

The conceptual core of Reactivity 2.3: reversible reactions, dynamic equilibrium, and the meaning of 'how far'

Reactivity 2.3 begins by establishing that many reactions do not go to completion. The syllabus requires students to recognise that a reversible reaction can reach a state in which the forward and reverse rates are equal, and that the concentrations of reactants and products then remain constant. This state is dynamic, meaning that both forward and reverse processes continue at the molecular level; it is not a state in which all reaction has stopped. Examiners reward precise language here, and a candidate who writes 'the reaction has stopped' will be marked down against a candidate who writes 'the forward and reverse reactions proceed at equal rates, so macroscopic concentrations remain constant'.

From there, the sub-topic introduces the equilibrium constant, Kc, as a ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient. The syllabus is explicit that this expression is written in terms of concentrations, not amounts, and the units that result from the expression must be handled. A common HL question asks the candidate to deduce the units of Kc from the balanced equation; for a reaction of the form A + 2B ⇌ C + D, the units of Kc are mol dm⁻³, because the denominator contains (mol dm⁻³)³ and the numerator contains (mol dm⁻³)². A candidate who simply writes 'no units' because the equation looks balanced loses a mark that another candidate collects by writing the algebra through.

The phrase 'how far' in the sub-topic title is the operational definition that examiners use. A large value of Kc means the position of equilibrium lies to the right, the products dominate the mixture, and the reaction 'goes further'. A small value of Kc means the reactants dominate. The numerical value of Kc does not, by itself, say anything about the rate at which equilibrium is reached; the rate question belongs to Reactivity 2.4, not Reactivity 2.3. Candidates frequently conflate the two, and a Paper 2 question that tests the rate-equilibrium distinction can be settled with this single conceptual move.

Kc, Kp, and the calculation scaffolds that examiners actually reward

Reactivity 2.3 expects fluency with three calculation scaffolds: an equilibrium concentration table (the ICE table), a Kc expression evaluation, and, at HL, a Kp expression using partial pressures. Each scaffold is worth practising until the candidate can write it under exam pressure in under three minutes, because on a 90-minute Paper 2 section, equilibrium calculations are bundled with acid-base, electrochemistry and organic questions, and the equilibrium item is typically worth 4 to 6 marks. Running out of time on an ICE table is one of the cleanest ways to drop a boundary.

The ICE table scaffold

The ICE table is shorthand for Initial, Change, Equilibrium. For a generic reaction aA + bB ⇌ cC + dD, the candidate writes the starting concentrations (or moles, if the volume is given later), the change in terms of a single variable x, and the equilibrium row as the algebraic sum. Two practical habits raise the mark reliably: writing the change row using the stoichiometric ratio of the balanced equation, and using the equilibrium row directly in the Kc expression. A student who writes 'A: initial 0.20, change -x, equilibrium 0.20 - x' for a 1:1 stoichiometry is more accurate than a student who invents a new variable for each species. Consistency of variable is what examiners reward because it allows the algebraic step to be checked.

Units of Kc and why they are tested

Units are not decorative. For a general reaction, the units of Kc are derived from the expression by tracking (mol dm⁻³) raised to the power (sum of product coefficients minus sum of reactant coefficients). At SL, the most common forms are: Kc with no units (e.g. H₂ + I₂ ⇌ 2HI), Kc in mol dm⁻³ (e.g. PCl₅ ⇌ PCl₃ + Cl₂), and Kc in dm³ mol⁻¹ (e.g. esterification). At HL, candidates must be able to derive the units for less symmetric cases. A candidate who loses one mark on units in a 6-mark calculation question is also losing a mark per additional question that tests the same idea, so the cost compounds across the paper.

Kp at Higher Level

Kp is the gas-phase analogue of Kc, written using partial pressures in place of concentrations. The partial pressure of a gas is its mole fraction multiplied by the total pressure, and candidates at HL must be able to convert between mole fractions and partial pressures inside a single calculation. A common Paper 2 shape is: 'a mixture of 0.40 mol of N₂O₄ in a 2.0 dm³ flask at 100 kPa is allowed to reach equilibrium; the degree of dissociation is α. Calculate Kp.' The expected scaffold is: write the ICE table in moles, compute the total moles at equilibrium as a function of α, find the mole fractions, multiply each by total pressure to get partial pressures, write the Kp expression, and substitute. Candidates who skip the mole-fraction step and write partial pressures directly from the ICE row usually miss the total-mole correction and lose 2 to 3 marks.

For SL candidates, Kp is not on the syllabus, and Paper 1 will not test it. SL candidates should still recognise that the partial pressure of a gas in a mixture is its mole fraction times the total pressure, because that idea occasionally appears as a context in a stoichiometry question. The risk for SL is the reverse: an HL candidate who has not internalised the Kp scaffold will leave 4 to 6 marks on the table on Paper 2 Section B.

Le Chatelier's principle: how the rubric awards marks beyond the obvious

Le Chatelier's principle is often the first thing a candidate thinks of when they see a Reactivity 2.3 question, and that reflex is both useful and dangerous. The principle is qualitative: when a system at equilibrium is subjected to a change, the position of equilibrium shifts in the direction that opposes the change. The danger is in applying it without tracking what the Kc value does. Kc is a constant at a given temperature; a shift in equilibrium position changes the concentrations, but Kc itself does not change unless the temperature changes. Examiners write questions precisely to catch this confusion, and the wording is usually unambiguous: 'State and explain the effect of increasing the temperature on the value of Kc.' A candidate who answers 'Kc increases because the equilibrium shifts to the right' has answered a different question.

Concentration, pressure, and catalyst changes

Changing the concentration of a reactant or product shifts the equilibrium position but does not change Kc. The candidate is expected to describe the shift in terms of the rate of the forward and reverse reactions in the moment after the change, and to connect that to the new equilibrium position. For pressure changes, only gases with different numbers of moles on each side matter; if the moles of gas are equal on both sides, a pressure change has no effect on the position of equilibrium. Catalysts have no effect on the position of equilibrium; they speed up both the forward and reverse rates equally, so the system reaches the same equilibrium position in a shorter time. This is one of the most common marks lost on Paper 1, where a multiple-choice item might offer 'catalyst shifts equilibrium to the right' as a distractor.

Temperature and the sign of ΔH

Temperature is the only change that alters Kc. The syllabus is explicit that for an exothermic reaction, increasing the temperature decreases Kc, and for an endothermic reaction, increasing the temperature increases Kc. The justification is that the equilibrium constant is related to the Gibbs free energy change, ΔG° = -RT ln K, but at the level of the IB Chemistry exam, the candidate is expected to recall the direction without deriving it. A common HL question links the van 't Hoff equation conceptually: 'explain, with reference to the sign of ΔH, why the value of Kc for the exothermic reaction decreases as temperature increases.' The expected answer combines Le Chatelier's principle (heat is a product for an exothermic reaction, so removing heat by cooling shifts equilibrium to the right and increases Kc) with the energy profile interpretation.

Reacting masses, the mole, and the bridge between Reactivity 2.1 and 2.3

Reactivity 2.3 sits immediately after Reactivity 2.1 (Stoichiometric relationships) and Reactivity 2.2 (Competing reactions and yield), and the calculation questions often require the candidate to combine these earlier ideas with the equilibrium framework. A typical 6-mark HL Paper 2 question might give a starting mass of one reagent, ask the candidate to convert to moles, set up an ICE table, solve for x, and then back-calculate the percentage yield or the mass of product at equilibrium. The marks are distributed across the steps, and losing one early step costs marks on the dependent steps as well.

The advice for preparation is to drill the conversion chain: mass → moles → concentration (using volume), then equilibrium calculation, then back-conversion. Most candidates who lose marks on Reactivity 2.3 do not lose them on the equilibrium algebra; they lose them on the unit conversion at the start. A candidate who writes the equilibrium expression using grams instead of moles, or who omits the volume when calculating concentration, will arrive at a numerically plausible but dimensionally wrong Kc. The IB mark scheme typically awards a single 'answer mark' at the end and several 'process marks' along the way, so the dimensional error does not just lose one mark; it loses the chain.

For SL candidates, the syllabus does not require Kp or advanced gas calculations, but it does require the same conversion discipline at the Kc level. The simplest preparation strategy is to keep a worked folder of equilibrium problems organised by starting information: some give masses and volumes, some give concentrations, some give Kc and ask the candidate to find an unknown concentration. Each is a different first step, and recognising which first step is required is itself a mark.

Question types, paper split, and where the marks actually live

The IB Chemistry exam has three papers. Paper 1 is a multiple-choice paper with 30 marks at SL and 40 marks at HL, drawn from the entire syllabus. Paper 2 is a written paper with structured and extended-response questions totalling 50 marks at SL and 72 marks at HL. Paper 3 is the internal-assessment-equivalent written component, with a Section A on experimental techniques and a Section B on options and an HL-only question on a short topic. Reactivity 2.3 appears on Paper 1 in roughly two to three questions at both levels, and on Paper 2 in a structured question worth 6 to 10 marks on average. At HL, an additional 4 to 6 marks on Paper 2 Section B test the Kp extension.

Paper 1 shapes

On Paper 1, the typical Reactivity 2.3 question is a single multiple-choice item testing a qualitative idea. For example: 'Which change increases the value of Kc for an exothermic reaction: A increasing temperature, B decreasing temperature, C adding a catalyst, D increasing the pressure?' The correct answer is B, and the distractor list is constructed to test each common misconception. Candidates preparing for Paper 1 should build a list of qualitative equilibrium statements and check each against the syllabus, because a paper like this rewards recall under time pressure more than it rewards deep calculation.

Paper 2 shapes

On Paper 2, the typical Reactivity 2.3 question is a structured calculation worth between 4 and 10 marks. The mark scheme typically awards marks for: the equilibrium expression in terms of concentration, the ICE table with the change row written in terms of a single variable, the correct substitution into the Kc expression, the algebraic solution, the units of Kc, and a final answer with the correct number of significant figures. HL versions extend this to Kp, the relationship between Kc and Kp, and an additional conceptual part that tests the rate-equilibrium distinction.

A second common Paper 2 shape is a Le Chatelier question linked to an industrial process, such as the Haber process or the contact process. The candidate is given the equation, an enthalpy change, and asked to discuss the conditions that maximise yield. A complete answer addresses temperature (favourable for yield at low temperature for the exothermic Haber process, but unfavourable for rate, so a compromise is used), pressure (high pressure favours the side with fewer moles of gas, but engineering costs), and the role of a catalyst. The marks here are split between stating the principle, applying it to the specific reaction, and connecting it to the compromise. A candidate who states the principle without connecting to the specific reaction loses marks.

Common pitfalls and how to avoid them in Reactivity 2.3

The marks lost in Reactivity 2.3 cluster around five traps, and an aware candidate can pre-empt all of them. Listing them as a tactical block is the most efficient way to internalise them before the exam.

• Conflating position of equilibrium with the value of Kc. Concentration and pressure changes shift the position; only temperature changes the value of Kc. A common Paper 2 phrase is 'explain the effect on the position of equilibrium and on the value of Kc', and a candidate who answers only the first half leaves a mark on the table.
• Omitting pure solids and pure liquids from the Kc expression. The activity of a pure solid or pure liquid is defined as 1, so they are not included. A common trap is a heterogeneous equilibrium such as CaCO₃(s) ⇌ CaO(s) + CO₂(g), where Kc = [CO₂]. A candidate who writes Kc = [CaO][CO₂]/[CaCO₃] has misread the syllabus.
• Forgetting the stoichiometric coefficients when raising to a power. For 2A ⇌ B + C, the Kc expression is [B][C]/[A]². Candidates who write [B][C]/[A] lose marks on every equilibrium question they meet until the habit is corrected.
• Using moles instead of concentrations in the Kc expression. Kc is defined in terms of concentrations, not moles. If the volume of the container is given, the candidate must divide moles by volume before substituting. A candidate who works in moles throughout the calculation and writes the final Kc value in mol dm⁻³ has by accident a wrong answer.
• Confusing Kc with Kp and the rate constant k. The three constants are unrelated beyond their names. A Paper 1 distractor might ask the candidate to identify 'the equilibrium constant for a reaction at 298 K' from a list of four symbols, and a candidate who picks k is a candidate who has not yet separated the ideas.

Worked example: a typical HL Kc calculation question

To consolidate the scaffolds, consider a typical 6-mark HL Paper 2 item: 'Hydrogen and iodine react according to H₂(g) + I₂(g) ⇌ 2HI(g). 0.50 mol of H₂ and 0.50 mol of I₂ were placed in a 1.0 dm³ vessel and allowed to reach equilibrium at 700 K. At equilibrium, the concentration of HI was 0.78 mol dm⁻³. Calculate the value of Kc at this temperature.'

The expected scaffold is: write the equilibrium expression Kc = [HI]²/([H₂][I₂]); build an ICE table where the change in H₂ is x, the change in I₂ is x, and the change in HI is 2x; use the equilibrium concentration of HI to find 0.50 - 2x = 0.78, which gives 2x = -0.28, so x = -0.14, and the equilibrium concentrations of H₂ and I₂ are 0.64 mol dm⁻³. Substitute into the Kc expression: Kc = (0.78)²/((0.64)(0.64)) ≈ 1.5. The units cancel because the moles of gas are equal on both sides, so Kc is dimensionless. The 6 marks are distributed across: the expression (1 mark), the ICE table (2 marks), the algebra (1 mark), the substitution and calculation (1 mark), and the units statement (1 mark).

A second worked example illustrates Kp at HL. 'A mixture of 0.10 mol of PCl₅ was placed in a 2.0 dm³ vessel at 250 °C and allowed to reach equilibrium according to PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium, the total pressure was 200 kPa. Calculate Kp.' The expected scaffold is: write the ICE table in moles, with initial PCl₅ = 0.10, change -x, and equilibrium moles of PCl₃ = x, Cl₂ = x, PCl₅ = 0.10 - x; the total moles at equilibrium are 0.10 + x; the mole fractions are x/(0.10 + x) for both products and (0.10 - x)/(0.10 + x) for PCl₅; multiply each by the total pressure to get partial pressures; substitute into Kp = (P_{PCl₃} × P_{Cl₂})/P_{PCl₅}. Candidates who skip the total-mole correction usually produce a Kp value that is wrong by a factor of two or more, costing 2 to 3 marks on a 4 to 6 mark item.

Preparation strategy: turning Reactivity 2.3 into a 7-band score

For an IB Diploma candidate aiming at a 7 in Chemistry, Reactivity 2.3 is best approached as a calculation-discipline unit rather than a memorisation unit. The list of facts to recall is short: Kc is constant at constant temperature, the position of equilibrium is a separate concept, catalysts do not change Kc, and only gases and aqueous species appear in the equilibrium expression. The bulk of the marks, however, live in the calculation scaffolds, and the difference between a 6 and a 7 is usually a chain of small marks rather than one big idea.

The most efficient preparation strategy is a six-week cycle. Weeks one and two: build the ICE table scaffold by working six to ten past-paper items, marking them against the mark scheme, and noting which process mark is most often lost. Weeks three and four: extend the scaffold to Kp at HL, and to a heterogeneous equilibrium to internalise the rule about pure solids. Week five: switch to a Paper 1 mode and work through a bank of multiple-choice items, focusing on the qualitative Le Chatelier questions and the Kc-versus-rate distinctions. Week six: complete a full Paper 2 under timed conditions, marking and reviewing. The point of this cycle is that the calculation chains are not built in the final week; they are built in the first two, and the remaining weeks are spent transferring the scaffold to new question shapes.

For SL candidates, the strategy is similar but shorter; the Kp extension is removed, and the ICE table scaffold is built only for Kc questions. The SL Paper 2 typically contains one structured equilibrium item worth 4 to 6 marks, and the preparation goal is to bank all of them. A useful self-test is to take a past Paper 2 item, cover the mark scheme, and write the answer; if the candidate cannot reproduce a complete Kc expression, ICE table, and correct answer with units in under eight minutes, the scaffold is not yet internalised.

How Reactivity 2.3 connects to the rest of the IB Chemistry syllabus

Reactivity 2.3 is the foundation for the rest of the Reactivity theme. The acid-base equilibria of Reactivity 2.5 and 2.6 reuse the Kc idea with water as a special case, the solubility product in Reactivity 2.7 is a Kc expression in disguise, and the electrochemistry of Reactivity 2.9 connects to equilibrium through the Nernst equation and the equilibrium constant. A candidate who has internalised the Reactivity 2.3 scaffolds carries them into every later sub-topic, which is why a single weak unit in 2.3 can have a multiplier effect on the final IB mark.

This is also why an IB Chemistry tutor working at the 6-to-7 boundary will often start a student on Reactivity 2.3 even when the student's stated concern is acid-base or electrochemistry. The equilibrium scaffolding is the prerequisite, and a student who can produce a clean ICE table in under three minutes is a student who can transfer that scaffold to Ka, Kb, Ksp and the Nernst equation with relatively little extra effort. The IB Diploma rewards fluency, and fluency is built in the prerequisite units.

For most candidates reading this, the practical next step is to take one past Paper 2 equilibrium item, work through the ICE table scaffold with the mark scheme open, and then repeat with the mark scheme closed. Doing this twice a week for three weeks turns the scaffold from a recalled procedure into an automatic one, and that is the move that separates a 6 from a 7 on a boundary paper. IB Courses' one-to-one IB Chemistry programme drills each Reactivity 2.3 calculation shape against the official rubric until the student can produce a complete Kc or Kp answer with units and a stated assumption in under eight minutes, which is the time budget that a Paper 2 equilibrium item actually allows.

Comparison: SL Reactivity 2.3 versus HL Reactivity 2.3 at a glance

Conclusion and next steps

Reactivity 2.3 is the equilibrium unit of IB Chemistry, and it is the unit where the IB Diploma's preference for calculation discipline is most clearly on display. A candidate who can write a Kc expression without hesitation, build an ICE table in under three minutes, handle units, and state the temperature dependence of Kc with the correct sign convention is a candidate who has done most of the work for the rest of the Reactivity theme. The remaining work is the HL extension into Kp, and a focused three-week drill of the Kp scaffold against past Paper 2 items closes that gap.

For the IB Chemistry candidate aiming at a 7, the next concrete step is to set up a six-week Reactivity 2.3 study plan: weeks one and two on the Kc ICE table, weeks three and four on Kp and heterogeneous equilibria, week five on Paper 1 qualitative items, week six on a timed Paper 2 mock. Each session should end with a written reflection on which process mark was lost, because the mark scheme is the syllabus, and the rubric is the answer key. IB Courses' one-to-one IB Chemistry HL coaching programme marks every Reactivity 2.3 calculation against the official rubric and turns the scaffold into an automatic habit, which is the move that turns a 6 into a 7 on a boundary paper.

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