Five teaching hours at both levels, outcomes 2.3.1 to 2.3.4, and four additional higher level hours, outcomes 2.3.5 to 2.3.7.
Guiding question. How can the extent of a reversible reaction be influenced?
Reactivity 2.3 · How far? The extent of chemical change
1Dynamic equilibrium 2.3.1 SL + HL
The syllabus statement: A state of dynamic equilibrium is reached in a closed system when the rates of forward and backward reactions are equal. The skill: describe the characteristics of a physical and chemical system at equilibrium.
Seal some colourless dinitrogen tetroxide in a glass tube at room temperature and it slowly turns brown as nitrogen dioxide forms. The colour deepens, then stops changing — but not because all of the N2O4 has gone. Start instead with pure brown NO2 and the colour fades, as NO2 molecules pair up, until it reaches exactly the same shade. Both tubes arrive at the same mixture from opposite directions:
The reaction is reversible: it can proceed in both directions under the same conditions, shown by the half-arrows ⇌. Reactivity 2.3 is about the mixture that such a reaction produces — how far it goes, and how that can be changed.
What happens at the particle level
At the start only N2O4 is present, so only the forward reaction can occur, and its rate is at its highest. As N2O4 is used up its concentration falls, and so does the forward rate. Meanwhile NO2 builds up, and the reverse reaction, 2NO2 → N2O4, starts and speeds up. Eventually the two rates become equal. From then on, N2O4 is being formed exactly as fast as it is being used, and the same is true of NO2. Both reactions continue, but there is no further net change in the concentrations. This is dynamic equilibrium.
Dynamic equilibrium: the state of a reversible reaction in a closed system in which the rates of the forward and backward reactions are equal, so that the concentrations of reactants and products remain constant.
- Closed system: no matter enters or leaves (energy may be exchanged).
- Rates of forward and backward reactions are equal; both reactions continue at the particle level.
- Concentrations are constant — but not, in general, equal to each other.
- Macroscopic properties are constant: colour, pressure, density, pH.
- Equilibrium can be reached from either direction.
What evidence shows that the reactions have not stopped? If a little radioactive sodium iodide is added to a saturated solution of ordinary sodium iodide in contact with undissolved solid, the amount of solid and the concentration of the solution stay the same, yet radioactivity soon appears in the solid. Iodide ions are continually leaving the solid and others are joining it at the same rate. The same experiment with isotopically labelled hydrogen in H2 + I2 ⇌ 2HI shows the label spreading to every hydrogen-containing species at equilibrium.
Physical and chemical equilibria
The guide asks for both kinds. In a physical equilibrium a substance changes state or dissolves without chemical change. In a sealed flask partly filled with bromine, liquid bromine evaporates and bromine vapour condenses; when the two rates are equal, the depth of the brown colour above the liquid stops changing: Br2(l) ⇌ Br2(g). A saturated solution in contact with undissolved solute, and a gas in contact with its aqueous solution, X(g) ⇌ X(aq), are also physical equilibria. In a chemical equilibrium, bonds are broken and formed, as in N2O4 ⇌ 2NO2 or N2 + 3H2 ⇌ 2NH3.
A closed system is essential. In an open beaker, bromine vapour escapes, the reverse process cannot keep pace, and all of the liquid eventually evaporates. Carbonated water in an open bottle loses its carbon dioxide for the same reason. A reaction whose product escapes, or one that is thermodynamically so favourable that the reverse is negligible, goes to completion instead; the combustion of methane is written with → and not ⇌.
Equilibrium is not the same as “nothing happening”. A diamond left on a table is not at equilibrium with graphite, although nothing visible happens; it is kinetically stable because the rate of conversion is immeasurably slow, and there is no reverse reaction occurring at an equal rate. A dynamic equilibrium, by contrast, is a balance between two processes that are both happening.
Exam focus · what the published papers show
On a recall question, although a recall question less than 50% got the mark as they did not state that forward and backward reactions went at the same rate. The mark scheme's answer is «in a closed system» the rate of the forward reaction equals the rate of the reverse reaction. “The concentrations are equal” is wrong; “the concentrations are constant” is right but is a second characteristic, not the definition.
On a rate question in which a gas pressure stopped rising, some candidates mistook it for a system at equilibrium when the pressure stops changing (although a straight arrow is shown in the equation). The reaction had simply finished. Constant properties are necessary for equilibrium, not sufficient: the reaction must be reversible and the system closed.
2The equilibrium law and the expression for K 2.3.2 SL + HL
The syllabus statement: The equilibrium law describes how the equilibrium constant, K, can be determined from the stoichiometry of a reaction. The skill: deduce the equilibrium constant expression from an equation for a homogeneous reaction.
Experiments on many equilibrium mixtures of the same reaction, started with different amounts of reactants and products, give different equilibrium concentrations each time. But one combination of those concentrations is always the same at a given temperature. For a general homogeneous reaction
the equilibrium law states that
where the square brackets are equilibrium concentrations in mol dm−3 and K (often written Kc to show that it uses concentrations) is the equilibrium constant. Products go on the top, reactants on the bottom; each concentration is raised to the power of its coefficient in the equation; the terms are multiplied, never added. A homogeneous equilibrium is one in which all species are in the same phase — all gases, or all in the same solution — and only these are assessed for K expressions. The guide does not require units for K, and K values in this chapter are written without them.
| Haber process | N2(g) + 3H2(g) ⇌ 2NH3(g): K = [NH3]2 ÷ ([N2][H2]3) |
| Contact process | 2SO2(g) + O2(g) ⇌ 2SO3(g): K = [SO3]2 ÷ ([SO2]2[O2]) |
| Esterification | CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l), all in one liquid phase: K = [CH3COOC2H5][H2O] ÷ ([CH3COOH][C2H5OH]) |
| Hydrogen iodide | H2(g) + I2(g) ⇌ 2HI(g): K = [HI]2 ÷ ([H2][I2]) — the coefficient 2 becomes a power, not a multiplier: [HI]2, not [2HI] |
Question. A 1.00 dm3 vessel at 400 °C contains an equilibrium mixture of 0.70 mol N2, 2.10 mol H2 and 0.60 mol NH3. Calculate K.
| Concentrations | Volume is 1.00 dm3, so [N2] = 0.70, [H2] = 2.10, [NH3] = 0.60 mol dm−3. In any other volume, divide each amount by the volume first. |
| Substitution | K = 0.602 ÷ (0.70 × 2.103) = 0.36 ÷ (0.70 × 9.261) |
| Answer | K = 0.36 ÷ 6.48 = 0.056 |
| Check | K < 1 at this temperature: the mixture contains more reactants than product, consistent with the numbers. Two significant figures, as in the data. |
Writing K expressions is one of the most reliable marks in the course: 89% of the candidates chose the correct Kc expression on one item, it was the easiest question on the paper with 93% of answers correct on another, and reports list Writing the equilibrium constant expression among the well-prepared areas. The slips are specific: a few had the expression upside down and several separated the molecules into atoms; some included [2NH3] instead of [NH3]2; and some forgot to include the [HI] coefficient as a power to its concentration term. When a structured item asks for the expression, some calculated the value of Kc, rather than giving the expression.
3What the value of K tells you 2.3.3 SL + HL
The syllabus statement: The magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium and is temperature dependent. The skill: determine the relationships between K values for reactions that are the reverse of each other at the same temperature, including the extent of reaction for K ≪ 1, K < 1, K = 1, K > 1 and K ≫ 1.
Because products are on the top of the expression, a large K means that at equilibrium the product concentrations are large compared with the reactant concentrations: the position of equilibrium lies to the right, and the reaction has gone far towards completion. A small K means that the position lies to the left. The value of K says nothing about how fast equilibrium is reached — that is kinetics.
| Value of K | Composition at equilibrium | Description |
|---|---|---|
| K ≫ 1 | Almost entirely products | Reaction goes almost to completion (for example K ≈ 1010 or more) |
| K > 1 | Products predominate | Position of equilibrium lies to the right |
| K = 1 | Appreciable amounts of both; neither favoured | Position in the middle (the concentration term ratio equals one) |
| K < 1 | Reactants predominate | Position of equilibrium lies to the left |
| K ≪ 1 | Almost entirely reactants | Reaction hardly proceeds (for example K ≈ 10−10 or less) |
“Almost to completion” is not the same as “complete”. However large K is, some reactant remains at equilibrium; a report notes that some candidates said that the reaction went to completion, and were not awarded the mark, as they did not acknowledge the presence of equilibrium, and another that a common error being to declare the reaction ‘complete’ rather than ‘almost complete’.
K for related equations
The value of K belongs to an equation as written. For the reverse reaction, the expression is turned upside down, so K(reverse) = 1 ÷ K(forward). If every coefficient is multiplied by n, every power in the expression is multiplied by n, so K is raised to the power n: doubling the coefficients squares K; halving them takes the square root.
Question. For 2SO2(g) + O2(g) ⇌ 2SO3(g), K = 400 at a certain temperature. Calculate K, at the same temperature, for (a) 2SO3(g) ⇌ 2SO2(g) + O2(g) and (b) SO3(g) ⇌ SO2(g) + ½O2(g).
| (a) Reversed | K = 1 ÷ 400 = 2.5 × 10−3 |
| (b) Reversed and halved | K = (1 ÷ 400)½ = √(2.5 × 10−3) = 0.050 |
| Check | Write the expression for (b): [SO2][O2]½ ÷ [SO3]; squaring it gives the expression for (a). Halving an equation does not halve K. |
Temperature is the only condition that changes K
At a fixed temperature, K is constant: changing concentrations or pressure moves the system away from equilibrium temporarily, but it returns to a mixture with the same value of K. Changing the temperature changes K itself. For an exothermic forward reaction, K decreases as temperature increases; for an endothermic forward reaction, K increases as temperature increases. The reason is developed in 2.3.4 and, at higher level, through ΔG⦵ = −RT ln K in 2.3.7.
Relationships between K values are consistently weaker than writing expressions. A report lists Kc values for reactions which are multiples or reverses of one another as difficult; on an item in which an equation was reversed and halved, quarter of the students selected A as the answer, instead of C, demonstrating a poor understanding of the effect on Kc when an equation is reversed and halved; on another, answered correctly by 44 %, almost the same number of candidates stated that doubling the coefficients in a balanced equation means one should double the constant as correctly squared it. On a structured item, approximately 60% of the candidates just repeated the Kc value given when the reverse reaction was asked for.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Ammonia is manufactured by the Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH⦵ = −92.0 kJ mol−1.
(i) Outline what is meant by dynamic equilibrium. [1]
(ii) Deduce the Kc expression for the reaction. [1]
PCl5(g) and Cl2(g) were placed in a sealed flask and allowed to reach equilibrium at 200 °C. The enthalpy change for the decomposition of PCl5(g) is positive.
(i) Deduce the equilibrium constant expression, Kc, for the decomposition of PCl5(g). [1]
(ii) Deduce, giving a reason, the factor responsible for establishing the new equilibrium after 14 minutes. [2]
Solutions and mark-scheme guidance · Set R2H
R2H.1 D
Products over reactants, coefficients as powers: [N2O][H2O]3 ÷ ([NH3]2[O2]2). 89 % correct.
R2H.2 D
[NO2F]2 ÷ ([NO2]2[F2]). Terms are multiplied, not added, and coefficients are powers.
R2H.3 C
K = [B][C]3 ÷ [A]2 = 2 × 23 ÷ 22 = 16 ÷ 4 = 4. Almost 85 % correct.
R2H.4 [2]
(i) (In a closed system) the rate of the forward reaction equals the rate of the reverse reaction ✔. Fewer than half gained this recall mark. (ii) Kc = [NH3]2 ÷ ([N2][H2]3) ✔.
R2H.5 [3]
(i) PCl5 ⇌ PCl3 + Cl2: Kc = [PCl3][Cl2] ÷ [PCl5] ✔. The report notes that many calculated a value instead of giving the expression.
(ii) A decrease in temperature ✔; the forward reaction is endothermic AND the equilibrium shifts to the left (PCl5 increases while PCl3 and Cl2 decrease, gradually, with no sudden jump) ✔. “Temperature change” alone is not accepted.
4Le Châtelier’s principle 2.3.4 SL + HL
The syllabus statement: Le Châtelier’s principle enables the prediction of the qualitative effects of changes in concentration, temperature and pressure to a system at equilibrium. The skill: apply Le Châtelier’s principle to predict and explain responses to changes of systems at equilibrium, including the effects on the value of K and on the equilibrium composition, and heterogeneous equilibria such as X(g) ⇌ X(aq).
When a system at equilibrium is subjected to a change in conditions, the position of equilibrium shifts in the direction that tends to oppose (minimize) the change.
The principle predicts the direction of the shift; it does not say the change is fully reversed. If more of a reactant is added, the equilibrium shifts to use some of it up, but the new equilibrium mixture still contains more of that reactant than before. Each prediction can be explained in two further ways: in terms of rates (what happens to the forward and reverse rates immediately after the change), and in terms of K (the concentration ratio must return to the value of K). The explanation in terms of K is the one that also shows what happens to K itself.
Changing a concentration
Iron(III) ions and thiocyanate ions form a blood-red complex: Fe3+(aq) + SCN−(aq) ⇌ [FeSCN]2+(aq). Adding more Fe3+ (as a few drops of concentrated iron(III) chloride solution) deepens the red colour. Adding more of a reactant increases the rate of the forward reaction; more product forms until the rates are equal again. In terms of K: immediately after the addition, [Fe3+] on the bottom of the expression is larger, so the ratio is smaller than K; the system shifts right, raising the numerator and lowering the denominator, until the ratio equals K again. Removing a product has the same effect. K is unchanged.
Changing the pressure of a gas-phase equilibrium
Increasing the pressure by compressing the mixture increases the concentration of every gas. The system opposes the rise in pressure by shifting towards the side with fewer moles of gas, because fewer gas particles exert a lower pressure. For N2(g) + 3H2(g) ⇌ 2NH3(g), there are 4 mol of gas on the left and 2 on the right, so high pressure shifts the equilibrium to the right and increases the yield of ammonia. If both sides have the same number of moles of gas, as in H2(g) + I2(g) ⇌ 2HI(g), pressure has no effect on the position. Only gases are counted: solids, liquids and dissolved species are ignored.
In terms of K, halving the volume doubles every concentration. In the Haber expression the numerator increases by 22 = 4 but the denominator by 24 = 16, so the ratio becomes smaller than K and the system shifts right to restore K. Adding an unreactive gas such as argon at constant volume increases the total pressure but changes no concentration, so it has no effect on the position of equilibrium.
Changing the temperature
Temperature is different from the other changes because it changes K. If the temperature is raised, the equilibrium shifts in the direction that absorbs heat — the endothermic direction — so as to oppose the rise. For the exothermic Haber reaction (ΔH = −92 kJ mol−1), raising the temperature shifts the equilibrium to the left, lowers the yield of ammonia, and decreases K. For the endothermic dissociation N2O4 ⇌ 2NO2, warming the tube makes the mixture a deeper brown and increases K; cooling it in ice makes it paler.
The cobalt(II) chloride equilibrium shows the same idea in solution: [Co(H2O)6]2+(aq) + 4Cl−(aq) ⇌ [CoCl4]2−(aq) + 6H2O(l) is endothermic in the forward direction, so the pink solution turns blue when heated and pink again when cooled. The size of the effect depends on the size of ΔH: an equilibrium with a small enthalpy change is shifted only a little by temperature.
Adding a catalyst
A catalyst lowers the activation energies of the forward and reverse reactions by the same amount (2.2.5) and increases both rates by the same factor. It therefore has no effect on the position of equilibrium or on K. It allows equilibrium to be reached faster, which in industry means more product per hour from the same plant.
| Change | Position of equilibrium | Value of K | Rate of reaching equilibrium |
|---|---|---|---|
| Add reactant / remove product | Shifts right | Unchanged | — |
| Increase pressure (gases) | Shifts to the side with fewer moles of gas; no shift if equal | Unchanged | Faster (higher concentrations) |
| Increase temperature | Shifts in the endothermic direction | Increases if forward is endothermic; decreases if exothermic | Faster |
| Add a catalyst | No shift | Unchanged | Faster |
| Add an inert gas at constant volume | No shift | Unchanged | — |
How far? Le Châtelier at work — N2O4 ⇄ 2NO2
The classic demonstration, integrated rather than asserted. Squeeze the vessel or heat it and watch the mixture move to a new position — and watch the colour follow it.
Note what the temperature does that nothing else does: it changes K itself. Volume and added substance only move the system along a fixed K, which is why they never change the constant.
Concentrations against time, integrated step by step. After any disturbance Q returns to K — the same K, unless you changed the temperature.
The gas mixture. NO2 is brown and N2O4 is colourless, so the depth of colour is a direct readout of the position of equilibrium.
Equilibria in industry: the Haber process
Ammonia, the starting point for most fertilizers, is made from nitrogen and hydrogen. Le Châtelier's principle predicts the conditions for the highest yield: low temperature (exothermic forward reaction) and high pressure (fewer moles of gas on the right). But at low temperature the rate is too slow, and very high pressure is expensive and hazardous. The conditions actually used are a compromise.
| Condition | Typical choice | Reasoning |
|---|---|---|
| Temperature | About 450 °C | A lower temperature would give a higher equilibrium yield but too slow a rate; 450 °C gives an acceptable yield at an acceptable rate. |
| Pressure | About 200 atm (≈ 20 MPa) | High pressure increases both yield and rate; higher still would cost more in plant strength and energy for compression, and increase the hazard. |
| Catalyst | Iron | Increases the rate so that equilibrium is approached quickly; no effect on yield. |
| Removing product | Ammonia liquefied and removed; unreacted gases recycled | Removing product shifts the equilibrium right; recycling means that little reactant is wasted even though each pass converts only a fraction. |
Heterogeneous equilibria: a gas and its solution
The guide extends Le Châtelier's principle to heterogeneous equilibria such as X(g) ⇌ X(aq). A sealed bottle of carbonated drink contains the equilibrium CO2(g) ⇌ CO2(aq). Opening the bottle lowers the pressure of CO2 above the liquid; the equilibrium shifts to the left, and dissolved carbon dioxide comes out of solution as bubbles. Dissolving a gas is usually exothermic, so gases are less soluble in warm water — a warm drink goes flat faster, and warm rivers hold less dissolved oxygen for fish.
Several equilibria linked together respond in the same way. In hot weather, hens pant and lose more carbon dioxide from their blood; the loss shifts the equilibria CO2(aq) + H2O(l) ⇌ H2CO3(aq) ⇌ H+(aq) + HCO3−(aq) ⇌ 2H+(aq) + CO32−(aq) to the left, less carbonate is available, and the eggshells are thinner. In the blood, carbon monoxide competes with oxygen for haemoglobin; because its binding is so much more favourable, a small concentration of CO shifts the oxygen equilibrium well to the left, which is why carbon monoxide poisoning is treated with a high concentration of oxygen.
Question. For 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −198 kJ mol−1, predict and explain the effect on the position of equilibrium and on K of (a) increasing the pressure, (b) increasing the temperature.
| (a) Position | Shifts to the right: there are 3 mol of gas on the left and 2 on the right, so the shift to fewer gas moles opposes the increase in pressure. |
| (a) K | No change: K depends only on temperature. |
| (b) Position | Shifts to the left: the forward reaction is exothermic, so the endothermic (reverse) direction absorbs heat and opposes the rise in temperature. |
| (b) K | Decreases, because the forward reaction is exothermic. |
Each part has a prediction and a reason, and the question asks separately about position and about K. The reports show that the mark for K is the one most often missed.
Exam focus · what the published papers show
Direction of shift is generally well answered — reports list Applying Le Chatelier’s principle to the position of equilibrium among the strengths and note that 77% of the candidates applied Le Chatelier’s Principle correctly on one item. The effect on K is the weakness, recorded in almost every session: the fact that Kc remains constant at fixed temperatures was less well known; about half of the candidates forgot that pressure/concentration has no effect on the equilibrium constant value; many forgot that Kc only changes with temperature; and marks were missed because candidate answered about shift of equilibrium and not value of equilibrium constant.
With temperature, the direction of the change in K must be stated and linked to ΔH: some correctly identified the forward reaction as exothermic but then forgot to say Kc would decrease, weaker candidates thought that Kc would shift to the left or right confusing it with the equilibrium position, and some stated that as T decreased so did Kc for an exothermic reaction. The two most common mistakes on a later item were discussing the shift in equilibrium position without identifying the effect on the value of Kc and stating that Kc increases without any reasoning.
On one item 33% of the candidates identified the equilibrium that would shift left with an increase in pressure; the most popular distractor ignored the state symbols. On a structured item, some candidates stated there was no shift in the equilibrium as the number of moles is the same on both sides of the equation, not acknowledging that only gaseous substances need to be considered. Count moles of gas, from the state symbols, on each side.
The heterogeneous case is newly explicit in the guide. On an item involving a sparingly soluble salt, a report calls it one of the poorest answered questions and notes that in the new syllabus starting 2025 Le Chatelier’ s Principle will be applied to both homogeneous and heterogeneous equilibria.
Questions set in an unfamiliar context — cleaning products that release chlorine, a peroxyacid sold in solution, lead dissolving in acidic water — are marked on whether the answer uses the equilibrium given. On the chlorine item, most students gained at least one mark for stating that ‘chlorine gas will be produced’ but couldn’t link it to equilibrium ideas; on another, most candidates did not refer to equilibrium (2), as directed by the question, and hence could not gain any marks. A report summarizes the weakness as Applying Le Chatelier’s Principle in unfamiliar situations. Identify which species is added or removed, say which way the named equilibrium shifts, and state the consequence.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
H2(g) + I2(g) ⇌ 2HI(g), ΔH⦵ < 0, was allowed to reach equilibrium at 761 K. Outline the effect, if any, of each change on the position of equilibrium, giving a reason: increasing the volume at constant temperature; increasing the temperature at constant pressure. [2]
2NO2(g) ⇌ N2O4(g). At 100 °C, Kc for this reaction is 0.0665.
(i) Outline what this indicates about the extent of this reaction. [1]
(ii) Calculate the value of Kc at 100 °C for the equilibrium N2O4(g) ⇌ 2NO2(g). [1]
A solution of bleach can be made by reacting chlorine gas with sodium hydroxide solution: Cl2(g) + 2NaOH(aq) ⇌ NaOCl(aq) + NaCl(aq) + H2O(l). Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar (dilute ethanoic acid) and bleach together as cleaners. [3]
Urea can be made by the direct combination of ammonia and carbon dioxide: 2NH3(g) + CO2(g) ⇌ (H2N)2CO(g) + H2O(g), ΔH < 0. Predict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased. [1]
Solutions and mark-scheme guidance · Set R2I
R2I.1 C
Doubling every coefficient squares the expression, so K becomes K2.
R2I.2 B
The new equation is the reverse of the given one, halved: K = 1 ÷ √(7.3 × 1034).
R2I.3 B
Reverse (1/4.0 = 0.25) and halve (√0.25 = 0.50).
R2I.4 A
The forward reaction is endothermic and produces more gas moles: higher temperature and lower pressure both shift right. A catalyst has no effect on the position.
R2I.5 C
A catalyst speeds up both directions equally: no change in position or in K.
R2I.6 A
K increases with temperature, so the forward reaction is endothermic and is favoured at higher temperature. At equilibrium the forward and reverse rates are equal, so C is wrong.
R2I.7 B
Count gas moles only. In B, 1 mol gas → 2 mol gas, so higher pressure shifts left. In D, water is a liquid: 7 mol gas → 2 mol gas, which shifts right. Only 33 % answered correctly; more chose D.
R2I.8 C
Solid NaCl dissolves, raising [Cl−]; the equilibrium shifts right to remove it. Removing solid AgCl or changing pressure has no effect; adding water lowers the ion concentrations and shifts left.
R2I.9 [2]
Volume: no effect AND the same number of gas moles on both sides ✔. Temperature: moves to the left AND the forward reaction is exothermic ✔. [1 max] if both effects are correct without reasons.
R2I.10 [2]
(i) The reaction hardly proceeds / the equilibrium lies to the left / reactants are at a greater concentration than products at equilibrium ✔. (ii) Kc = 1 ÷ 0.0665 = 15.0 ✔. About 60 % simply repeated 0.0665.
R2I.11 [3]
Ethanoic acid (vinegar) reacts with NaOH (or produces H+ ions) ✔; this removes NaOH, so the equilibrium moves to the left (reactant side) ✔; chlorine gas is released, and Cl2 is toxic ✔. An equation for the overall reaction that does not refer to the equilibrium is not accepted for the second mark.
R2I.12 [1]
Kc decreases AND the reaction is exothermic (ΔH negative / the reverse, endothermic reaction is favoured) ✔.
Additional higher level · outcomes 2.3.5 to 2.3.7
5The reaction quotient, Q 2.3.5 HL only
The syllabus statement: The reaction quotient, Q, is calculated using the equilibrium expression with non-equilibrium concentrations of reactants and products. The skill: calculate the reaction quotient Q from the concentrations of reactants and products at a particular time, and determine the direction in which the reaction will proceed to reach equilibrium.
The expression for K can be evaluated with the concentrations present at any moment, not only at equilibrium. The result is the reaction quotient, Q. For aA + bB ⇌ cC + dD,
At equilibrium, Q = K. Away from equilibrium, comparing Q with K shows which way the reaction must go to get there. Because the system always moves towards Q = K:
- Q < K: there is too little product relative to reactant; the reaction proceeds in the forward direction (net), increasing Q until it equals K.
- Q = K: the system is at equilibrium; no net change.
- Q > K: there is too much product; the reaction proceeds in the reverse direction (net), decreasing Q.
Q gives Le Châtelier's principle a quantitative basis. Adding a reactant increases the denominator of Q, so Q falls below K and the reaction proceeds forwards. Changing the temperature is different: Q is unchanged at the instant of the change, but K changes, and the system moves to the new K.
Question. For H2(g) + I2(g) ⇌ 2HI(g), K = 50 at a certain temperature. A mixture at that temperature contains 0.10 mol dm−3 H2, 0.10 mol dm−3 I2 and 1.0 mol dm−3 HI. Deduce the direction in which the reaction proceeds.
| Expression | Q = [HI]2 ÷ ([H2][I2]) |
| Substitution | Q = 1.02 ÷ (0.10 × 0.10) = 100 |
| Comparison | Q = 100 > K = 50 |
| Conclusion | The reaction proceeds in the reverse direction (to the left): HI decomposes until Q falls to 50. |
The comparison is always with K. “Q is greater than 1” tells you nothing.
The reaction quotient is regularly reported as difficult: Q and Kc values comparison in equilibrium problems and Reaction quotient meaning / interpretation appear in lists of weak areas. On one item many omitted the calculation for Q or considered it the Kc value and almost half the candidates who calculated the Q value correctly then reached the incorrect conclusion that the forward reaction is favoured. On another, only about a third of the candidates used the reaction quotient to determine the direction the equilibrium proceeds, incorrect answers often failed to compare Q with Kc but rather they stated Q > 1, and there were many attempts that had incorrect powers in the expression. A third report describes candidates who correctly calculated the value of Qc (= 20.8) and recognized it as less than Kc (= 280) but could not say what followed.
6Equilibrium calculations 2.3.6 HL only
The syllabus statement: The equilibrium law is the basis for quantifying the composition of an equilibrium mixture. The skill: solve problems involving values of K and initial and equilibrium concentrations of the components of an equilibrium mixture. The approximation [reactant]initial ≈ [reactant]eqm when K is very small should be understood; quadratic equations are not expected; only homogeneous equilibria are assessed.
Equilibrium problems are organized with an ICE table: Initial concentrations, the Change in each concentration as the system reaches equilibrium, and the Equilibrium concentrations. The changes are linked by the mole ratio of the equation — this is where Reactivity 2.1 meets Reactivity 2.3 — and the equilibrium concentrations are then substituted into the expression for K.
1. Work in concentrations (mol dm−3): divide amounts by the volume first. 2. The changes are in the ratio of the coefficients: if HI increases by 2x, H2 and I2 each decrease by x. 3. Only equilibrium concentrations go into K — never initial ones. 4. Check the answer: substitute back, and make sure no concentration is negative.
A report explains a lost mark exactly: many lost a mark in the calculation of Kc as they used the initial concentrations of nitrogen and hydrogen. On a multiple-choice item, 62% of the candidates were able to deduce the equilibrium concentration of IBr and calculate the equilibrium constant correctly. The most commonly chosen distractor was B where the stoichiometric ratio was not taken into account.
Question. H2 and I2, each at an initial concentration of 0.500 mol dm−3, are heated in a sealed vessel. At equilibrium, [HI] = 0.780 mol dm−3. Calculate K.
| H2(g) | + I2(g) | ⇌ 2HI(g) | |
|---|---|---|---|
| Initial / mol dm−3 | 0.500 | 0.500 | 0 |
| Change / mol dm−3 | −0.390 | −0.390 | +0.780 |
| Equilibrium / mol dm−3 | 0.110 | 0.110 | 0.780 |
| Ratio | 2 mol HI form from 1 mol H2, so each reactant falls by 0.780 ÷ 2 = 0.390 |
| Substitution | K = 0.7802 ÷ (0.110 × 0.110) = 0.6084 ÷ 0.0121 |
| Answer | K = 50.3 |
Small K: the approximation
When K is very small, very little reactant is converted at equilibrium, so the equilibrium concentration of each reactant is almost the same as its initial concentration. Treating them as equal removes the need to solve a quadratic (or cubic) equation, which is not expected. The approximation is justified when the change x turns out to be a very small fraction of the initial concentration — this should be checked after the calculation.
Question. For N2(g) + O2(g) ⇌ 2NO(g), K = 1.0 × 10−5 at a high temperature. Air-like mixture: initially [N2] = [O2] = 0.50 mol dm−3 and no NO. Calculate [NO] at equilibrium.
| ICE | Change: N2 −x, O2 −x, NO +2x. Equilibrium: [N2] = [O2] = 0.50 − x; [NO] = 2x |
| Approximation | K is very small, so x ≪ 0.50 and 0.50 − x ≈ 0.50 |
| Substitution | 1.0 × 10−5 = (2x)2 ÷ (0.50 × 0.50) = 4x2 ÷ 0.25 |
| Solve | x2 = 6.25 × 10−7; x = 7.9 × 10−4 |
| Answer | [NO] = 2x = 1.6 × 10−3 mol dm−3 |
| Check | x is 0.16 % of 0.50, so the approximation is excellent. Note the square: forgetting that 2x is squared, or taking a square root too early, are the arithmetic errors reported. |
The same approximation is the basis of weak-acid pH calculations in Reactivity 3.1, where the acid dissociation constant is small and the equilibrium concentration of undissociated acid is taken as its initial concentration.
The ICE table, solved exactly
Initial, change, equilibrium. The change is one unknown multiplied by the coefficients, and K then fixes it — here by solving the equation rather than by assuming x is small.
Initial against equilibrium concentration. The height lost by the reactants and the height gained by the products are in the ratio of the coefficients — that is the whole content of the middle row of the table.
Reports record that many candidates had difficulty in calculating the equilibrium concentrations of each component present in the mixture, that the calculation of equilibrium mole concentrations was more testing, particularly that for [O2] (a species with a coefficient different from the others), and, on a recent item, that the most common errors were arithmetical, though some students also failed to square terms in the equilibrium expression. By contrast, substituting given equilibrium concentrations is secure: almost 85% of the candidates correctly calculated the value of the equilibrium constant from the given equilibrium concentrations.
7The equilibrium constant and Gibbs energy 2.3.7 HL only
The syllabus statement: The equilibrium constant and Gibbs energy change, ΔG, can both be used to measure the position of an equilibrium reaction. The skill: calculations using ΔG⦵ = −RT ln K, which is given in the data booklet.
Reactivity 1.4 used the Gibbs energy change to decide whether a reaction is spontaneous. Equilibrium shows what that means in practice. As a reaction mixture changes from pure reactants towards products, its total Gibbs energy falls, reaches a minimum, and would rise again if the reaction went further. The minimum is the equilibrium mixture. At equilibrium, ΔG = 0: there is no further tendency to change in either direction. The standard Gibbs energy change, ΔG⦵, which refers to the complete conversion of reactants in their standard states to products in their standard states, determines where the minimum lies — that is, the value of K:
ΔG⦵ in J mol−1 in this equation (R = 8.31 J K−1 mol−1); T in K; ln is the natural logarithm; K is the equilibrium constant at T.
- ΔG⦵ < 0 ⇔ ln K > 0 ⇔ K > 1: products favoured at equilibrium.
- ΔG⦵ = 0 ⇔ K = 1.
- ΔG⦵ > 0 ⇔ ln K < 0 ⇔ K < 1: reactants favoured.
Because the relationship is logarithmic, modest values of ΔG⦵ correspond to very large or very small values of K. At 298 K, ΔG⦵ = −57 kJ mol−1 corresponds to K ≈ 1010, a reaction that goes almost to completion, and +57 kJ mol−1 to K ≈ 10−10. The more negative ΔG⦵, the larger K. The equation also explains the effect of temperature on K: since ΔG⦵ = ΔH⦵ − TΔS⦵, changing T changes ΔG⦵ and therefore K, in the direction Le Châtelier's principle predicts.
The link between ΔG, Q and K can be written ΔG = ΔG⦵ + RT ln Q. When Q < K, ΔG is negative and the forward reaction is favoured; when Q > K, ΔG is positive and the reverse is favoured; when Q = K, ΔG = 0 and the equation reduces to ΔG⦵ = −RT ln K. This form is included for understanding and is not required for calculation.
Question. (a) For a reaction at 298 K, ΔG⦵ = −10.0 kJ mol−1. Calculate K. (b) For another reaction at 500 K, K = 1.0 × 10−3. Calculate ΔG⦵ in kJ mol−1.
| (a) Units | ΔG⦵ = −10 000 J mol−1; T = 298 K |
| (a) Rearrange | ln K = −ΔG⦵ ÷ RT = 10 000 ÷ (8.31 × 298) = 4.04 |
| (a) Answer | K = e4.04 = 56.7 (K > 1, as expected for negative ΔG⦵) |
| (b) Substitute | ΔG⦵ = −8.31 × 500 × ln(1.0 × 10−3) = −8.31 × 500 × (−6.91) |
| (b) Answer | ΔG⦵ = +2.87 × 104 J mol−1 = +28.7 kJ mol−1 (positive, as expected for K < 1) |
Exam focus · what the published papers show
A report lists them together: using an incorrect value for T (500°C instead of 773K), not carry out the ln of Kc or not dividing by 1000 to convert J (in R value) to kJ, as the answered required. Others add calculation error in converting ln Kc into Kc value and many did not change G from kJ to J in the equation for Kc. Signs are the fourth trap: a negative ΔG⦵ must give K > 1, which is a quick check on any answer.
Understanding what is special about equilibrium is also tested: only 59% of candidates selected the appropriate entropy and free energy values (maximum or minimum) at equilibrium, and on a structured item many candidates used the equation, ∆Gᶱ = -nFEᶱ to try and calculate a value for ∆G, rather than realizing that the reaction is at equilibrium and thus ∆G = 0.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
A mixture of 1.00 mol SO2(g), 2.00 mol O2(g) and 1.00 mol SO3(g) is placed in a 1.00 dm3 container and allowed to reach equilibrium: 2SO2(g) + O2(g) ⇌ 2SO3(g).
(a) Distinguish between the terms reaction quotient, Q, and equilibrium constant, Kc. [1]
(b) Kc is 0.282 at temperature T. Deduce, showing your work, the direction of the initial reaction. [2]
2SO2(g) + O2(g) ⇌ 2SO3(g). SO2, O2 and SO3 are mixed and allowed to reach equilibrium at 600 °C.
| SO2 | O2 | SO3 | |
|---|---|---|---|
| Initial concentration / mol dm−3 | 2.00 | 1.50 | 3.00 |
| Equilibrium concentration / mol dm−3 | 1.50 |
Determine the value of Kc at 600 °C. [2]
2NO2(g) ⇌ N2O4(g). At 100 °C, Kc is 0.21.
(i) At a given time, the concentrations of NO2(g) and N2O4(g) were 0.52 and 0.10 mol dm−3 respectively. Deduce, showing your reasoning, if the forward or the reverse reaction is favoured at this time. [2]
(ii) Comment on the value of ΔG when the reaction quotient equals the equilibrium constant, Q = K. [2]
Carbon disulfide undergoes gas-phase hydrolysis: CS2(g) + 2H2O(g) ⇌ CO2(g) + 2H2S(g). An earlier part of the question gives Kc = 1.45 × 104 at 500 K. The concentrations at equilibrium are CS2: 0.0400 mol dm−3; H2O: 0.100 mol dm−3; CO2: x mol dm−3; H2S: 2x mol dm−3. Calculate the numerical value of x. [2]
Solutions and mark-scheme guidance · Set R2J
R2J.1 A
The smallest Q has the smallest ratio of products to reactants.
R2J.2 B
Q (4.5) < K (6.2), so the net reaction is forward: the forward rate is greater than the reverse rate until equilibrium is reached.
R2J.3 C
I2 fell by 0.10, so IBr rose by 0.20: K = 0.202 ÷ (0.10 × 0.10) = 4. The popular distractor B used x = 0.10, ignoring the 1 : 2 ratio; 62 % correct.
R2J.4 D
0.80 mol IBr formed from 0.40 mol each of I2 and Br2, leaving 0.10 mol each. Volumes cancel (equal powers top and bottom): K = 0.802 ÷ (0.10 × 0.10) = 64.
R2J.5 B
N2 fell by 0.2, so H2 fell by 0.6 (to 0.4) and NH3 rose by 0.4 (to 1.4).
R2J.6 A
Spontaneous (products favoured): ΔG⦵ negative, so ln K > 0 and K > 1.
R2J.7 C
The reverse reaction is favoured when K < 1, which corresponds to a positive ΔG⦵.
R2J.8 [3]
(a) Q uses non-equilibrium concentrations (at any time) AND Kc uses equilibrium concentrations ✔.
(b) Q = 1.002 ÷ (1.002 × 2.00) = 0.500 ✔; Q > Kc (0.500 > 0.282), so the reverse reaction is favoured / the reaction proceeds to the left ✔.
R2J.9 [2]
SO2 fell by 0.50, so O2 fell by 0.25 and SO3 rose by 0.50: [O2] = 1.25 AND [SO3] = 3.50 mol dm−3 ✔. Kc = 3.502 ÷ (1.502 × 1.25) = 4.36 ✔.
R2J.10 [4]
(i) Qc = 0.10 ÷ 0.522 = 0.37 ✔; Q > Kc, so the reaction proceeds to the left / the reverse reaction is favoured ✔ (no second mark without the calculation). Almost half of those who calculated Q correctly concluded that the forward reaction was favoured.
(ii) ΔG = 0 ✔; the reaction is at equilibrium / the forward and reverse rates are equal / macroscopic properties are constant ✔. Many tried to use ΔG⦵ = −nFE⦵ instead.
R2J.11 [2]
Kc = [CO2][H2S]2 ÷ ([CS2][H2O]2) = x(2x)2 ÷ (0.0400 × 0.1002) = 4x3 ÷ 4.00 × 10−4 ✔; x3 = 1.45, x = 1.13 mol dm−3 ✔. (The scheme also accepts error carried forward from the earlier value.) The report notes that some failed to square the H2S term.
Review · Reactivity 2.3
8Misconceptions, the examiner’s view, and the question types
- “At equilibrium the reactions stop.” Both continue at equal rates; only the net change stops.
- “At equilibrium the concentrations are equal.” They are constant, and usually very different.
- “[2NH3]” in a K expression. The coefficient becomes a power: [NH3]2.
- “A large K means the reaction is fast” or “complete”. K measures extent, not rate; even a very large K leaves some reactant.
- “Halving the equation halves K.” It takes the square root; reversing gives 1/K.
- “Changing concentration or pressure changes K.” Only temperature changes K.
- “K shifts to the left.” The position shifts; K increases, decreases or stays the same.
- “Pressure has no effect because the total number of moles is equal.” Count moles of gas only.
- “A catalyst increases the yield.” It speeds up both directions equally: no change in position or K.
- HL “Q > 1 means the forward reaction is favoured.” Compare Q with K: Q > K means the reverse reaction proceeds.
- HL Using initial concentrations in K. Only equilibrium concentrations go into K.
- HL ΔG⦵ in kJ, T in °C. R is in J K−1 mol−1; T must be in kelvin.
Evidence base: the IB Diploma chemistry subject reports quoted in this chapter.
Answered well: writing K expressions (89 % and 93 % on two items; routinely listed as a strength); calculating K from given equilibrium concentrations (almost 85 %); predicting the direction of shift for simple changes of concentration, pressure or temperature (77 % and over 70 % on items); recognizing that a catalyst does not affect the position of equilibrium; interpreting K values at different temperatures to decide whether the forward reaction is exothermic.
Found difficult: (1) defining dynamic equilibrium with equal rates (fewer than half on one item); (2) the effect on K, as distinct from the position — forgetting that only temperature changes K, or saying that K “shifts”; (3) relationships between K values for reversed and scaled equations (44 % on one item; about 60 % simply repeating K on another); (4) counting only gaseous species for pressure effects (33 % on one item); (5) heterogeneous equilibria and unfamiliar contexts; (6) extent of reaction — “complete” rather than “almost complete”; HL (7) using Q: comparing with K rather than with 1, and drawing the right conclusion; (8) stoichiometric ratios in ICE tables and using initial concentrations in K; (9) ΔG⦵ = −RT ln K — T in kelvin, J versus kJ, and converting ln K to K; (10) recognizing that ΔG = 0 at equilibrium.
What successful answers did: stated equal rates and constant concentrations; wrote powers from coefficients; answered the position question and the K question separately, each with a reason; counted moles of gas from the state symbols; named the specific equilibrium in unfamiliar contexts; HL calculated Q, compared it with K, and stated the direction; built ICE tables with stoichiometric changes; and checked the sign of ΔG⦵ against the size of K.
Eight question types cover the sub-topic.
| If the question asks… | …then |
|---|---|
| Describe dynamic equilibrium | Closed system; forward rate = reverse rate; concentrations (macroscopic properties) constant. |
| Write the K expression | Products over reactants; coefficients as powers; multiply terms; homogeneous only. |
| Calculate K from equilibrium concentrations | Convert amounts to concentrations; substitute; check the powers. |
| Interpret the size of K; K for a related equation | ≫ 1 almost complete, ≪ 1 hardly proceeds; reverse = 1/K; × n → Kn. |
| Predict and explain a shift | Direction + reason (fewer gas moles; endothermic direction; uses added species); then K separately. |
| HL Which way will it go? | Calculate Q; compare with K; Q < K forward, Q > K reverse. |
| HL Equilibrium composition | ICE table with stoichiometric changes; approximation for small K; check. |
| HL ΔG⦵ and K | ΔG⦵ = −RT ln K; J and K units; K = e−ΔG⦵/RT; sign check. |
9Quick check
- State two characteristics of a system in dynamic equilibrium.
- Write K for 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g).
- K = 2.0 × 1012 for a reaction. What can be said about the composition at equilibrium?
- K = 64 for A + B ⇌ 2C. What is K for C ⇌ ½A + ½B?
- For CO(g) + 2H2(g) ⇌ CH3OH(g), ΔH < 0, state the effect of increasing the pressure on the yield, and of increasing the temperature on K.
- Why does opening a bottle of fizzy drink produce bubbles?
- HL For the reaction in question 4, a mixture has [A] = [B] = 0.50 and [C] = 2.0 mol dm−3. Which way does it proceed?
- HL What is the sign of ΔG⦵ for a reaction with K = 0.020? What is ΔG at equilibrium?
- Any two: forward and reverse rates equal; concentrations constant; closed system; macroscopic properties constant.
- K = [NO]4[H2O]6 ÷ ([NH3]4[O2]5).
- The position lies far to the right: almost entirely products; the reaction goes almost to completion.
- Reversed: 1/64; halved: square root: K = 1/8 = 0.125.
- Higher pressure increases the yield (3 mol gas → 1 mol gas); higher temperature decreases K (exothermic forward reaction).
- Opening lowers the pressure of CO2; CO2(g) ⇌ CO2(aq) shifts left, so dissolved CO2 comes out of solution.
- Q = 2.02 ÷ (0.50 × 0.50) = 16 < K = 64: forward (to the right).
- K < 1, so ln K < 0 and ΔG⦵ is positive. At equilibrium ΔG = 0.
10Summary and knowledge organiser
Essential knowledge
- Dynamic equilibrium: closed system; forward and reverse rates equal; concentrations and macroscopic properties constant; reached from either direction; physical and chemical examples.
- K = [products]coefficients ÷ [reactants]coefficients for homogeneous equilibria, using equilibrium concentrations.
- K ≫ 1 almost complete; K ≪ 1 hardly proceeds; K(reverse) = 1/K; coefficients × n → Kn.
- Le Châtelier: shift opposes the change. Concentration and pressure change the position, not K; temperature changes both; catalyst changes neither.
- Increase T: K increases for endothermic forward reactions, decreases for exothermic ones.
- Heterogeneous X(g) ⇌ X(aq): lower gas pressure or higher temperature (usually) drives gas out of solution.
- HL Q from non-equilibrium concentrations; Q < K forward, Q > K reverse.
- HL ICE tables with stoichiometric changes; [reactant]initial ≈ [reactant]eqm when K is very small.
- HL ΔG⦵ = −RT ln K; ΔG⦵ < 0 ⇔ K > 1; ΔG = 0 at equilibrium.
Examination checklist
- Definition: “rate of forward reaction = rate of reverse reaction”.
- Powers, not multipliers; products on top.
- Answer position and K as two separate statements, each with a reason.
- Count only gaseous moles for pressure.
- HL Compare Q with K, never with 1; state the direction.
- HL T in K, ΔG⦵ in J in the equation; ex to undo ln.
Knowledge organiser
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Dynamic equilibrium 2.3.1 | Closed system; rate forward = rate reverse; constant concentrations. | Constant ≠ equal; reactions continue; a finished reaction is not an equilibrium. |
| Equilibrium law 2.3.2 | K = [C]c[D]d/[A]a[B]b; homogeneous. | [NH3]2 not [2NH3]; expression, not value, when asked. |
| Magnitude of K 2.3.3 | ≫ 1 right, ≪ 1 left; reverse 1/K; × n → Kn; depends on T only. | “Almost complete”; halving → √K. |
| Le Châtelier 2.3.4 | Opposes change; pressure → fewer gas moles; heat → endothermic direction. | K changes only with T; catalyst: no shift; gases only; heterogeneous X(g) ⇌ X(aq). |
| Q 2.3.5 HL | Same expression, any concentrations; Q < K forward; Q > K reverse. | Compare with K, not 1. |
| Calculations 2.3.6 HL | ICE; changes in coefficient ratio; small-K approximation. | Equilibrium, not initial, concentrations in K; square the terms. |
| ΔG⦵ and K 2.3.7 HL | ΔG⦵ = −RT ln K; ΔG = 0 at equilibrium. | T in K; J not kJ; K = e−ΔG⦵/RT. |