Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
Learn
In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
R2.3.1 Dynamic equilibrium
Dynamic equilibrium
The state reached by a reversible reaction or process in a closed system when the rates of the forward and backward reactions are equal and not zero. Both reactions continue, so particles are constantly converted in both directions, but the concentrations of all species and the macroscopic properties (colour, pressure, density) remain constant.
Closed system
A system that no matter can enter or leave, although energy can be exchanged with the surroundings. Equilibrium can be established only in a closed system: if a product escapes, as CO₂ does from an open vessel, the backward reaction cannot keep pace and the reaction does not reach equilibrium.
Characteristics of a system at equilibrium
For both physical and chemical systems at equilibrium at constant temperature: the forward and backward processes occur at equal rates; the concentrations of reactants and products are constant but not in general equal; the macroscopic properties are constant; the same equilibrium can be approached from either direction; and equilibrium is reached only in a closed system.
Physical equilibrium
A dynamic equilibrium involving a change of state or a dissolving process, with no chemical reaction. In a stoppered flask containing liquid bromine, Br₂(l) ⇌ Br₂(g): evaporation and condensation occur at equal rates, so the amount of vapour, and its colour and pressure, stay constant.
Students often think At equilibrium the reactions have stopped; nothing is happening because nothing appears to change. In fact Yes. Equilibrium is dynamic: both reactions continue at equal rates, so the concentrations stay constant.
Students often think At equilibrium the concentrations of the reactants and products (or of all the species) are equal. In fact Not in general. At equilibrium the concentrations are constant, but their values depend on K and on the starting amounts; they are equal only by coincidence.
R2.3.2 Homogeneous and heterogeneous equilibria
Homogeneous and heterogeneous equilibria
In a homogeneous equilibrium all the species are in the same phase, for example all gases, as in N₂(g) + 3H₂(g) ⇌ 2NH₃(g). In a heterogeneous equilibrium the species are in different phases, for example CO₂(g) ⇌ CO₂(aq).
Equilibrium law and the equilibrium constant, K
For a homogeneous reaction aA + bB ⇌ cC + dD at a given temperature, the equilibrium law states that K = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), where the square brackets are equilibrium concentrations in mol dm⁻³. The products of the reaction as written go on top and the reactants on the bottom; each concentration is raised to the power of its coefficient in the equation, and the terms are multiplied.
Students often think Each concentration is multiplied by its coefficient, so 2NH₃ in the equation becomes 2[NH₃] in the expression. In fact Each coefficient becomes the power to which the concentration of that species is raised.
Students often think The reactants go on the top of the expression because they come first in the equation. In fact The products of the reaction as written (the right-hand side of the equation) go on top; the reactants go on the bottom.
R2.3.3 K for reactions that are the reverse of each other
K for reactions that are the reverse of each other
At the same temperature, the equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction: K(reverse) = 1/K(forward). If K = 4.0 × 10² for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at a given temperature, then K = 1/(4.0 × 10²) = 2.5 × 10⁻³ for 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) at that temperature.
Magnitude of K and extent of reaction
K ≫ 1: the equilibrium lies far to the right and the reaction goes almost to completion. K > 1: products predominate over reactants at equilibrium. K = 1: neither side is favoured; appreciable amounts of both reactants and products are present. K < 1: reactants predominate. K ≪ 1: the equilibrium lies far to the left and very little reaction occurs. K describes the extent of reaction, not its rate.
Temperature dependence of K
K for a given reaction is constant at a fixed temperature and changes only when the temperature changes. For a reaction whose forward reaction is exothermic, K decreases as the temperature increases; for an endothermic forward reaction, K increases as the temperature increases. Changes in concentration or pressure and the addition of a catalyst do not change K.
Students often think A reaction and its reverse share one equilibrium mixture, so they have the same value of K. In fact No. The equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction: K(reverse) = 1/K(forward).
Students often think Reversing an equation changes the sign of K, just as it changes the sign of ΔH. In fact No. K cannot be negative; reversing the equation gives the reciprocal, 1/K.
R2.3.4 Position of equilibrium
Position of equilibrium
The relative amounts of reactants and products in an equilibrium mixture. A shift of the position of equilibrium to the right increases the proportion of products; a shift to the left increases the proportion of reactants.
Le Châtelier’s principle
If a system at equilibrium is subjected to a change in concentration, pressure or temperature, the position of equilibrium shifts so as to partly oppose the change. The principle predicts the direction of the shift qualitatively; it does not restore the original conditions.
Effect of a change in concentration
Adding a reactant or removing a product at constant temperature causes net forward reaction until equilibrium is re-established; removing a reactant or adding a product causes net backward reaction. The equilibrium composition changes, but K does not.
Effect of a change in pressure
Increasing the pressure of a gaseous equilibrium by reducing the volume shifts the position of equilibrium towards the side with fewer moles of gas; decreasing the pressure shifts it towards the side with more moles of gas. If both sides have the same number of moles of gas, the position is unaffected. K is unchanged.
Effect of a change in temperature
Raising the temperature shifts the position of equilibrium in the endothermic direction; lowering it shifts the position in the exothermic direction. Temperature is the only one of the three changes that alters the value of K.
Catalysts and equilibrium
A catalyst increases the rates of the forward and backward reactions by the same factor, so equilibrium is reached sooner. It changes neither the position of equilibrium nor the value of K.
Gas–solution equilibrium, X(g) ⇌ X(aq)
A heterogeneous equilibrium between a gas and the same substance dissolved in a liquid in a closed container. Increasing the pressure of X(g) above the liquid shifts the equilibrium to the right, so more X dissolves; lowering the pressure, for example by opening a bottle of carbonated water, shifts it to the left and X comes out of solution.
Students often think A catalyst shifts the equilibrium towards the products, increasing the equilibrium yield. In fact No. A catalyst increases the rates of the forward and backward reactions by the same factor, so equilibrium is reached sooner but its composition and K are unchanged.
Students often think Changing the concentration of a reactant or product changes the value of K, so that K follows the new composition of the mixture. In fact No. At constant temperature K is unchanged; the mixture responds by net forward reaction until the concentrations again fit the same value of K.
R2.3.5 Reaction quotient, Q HL
Reaction quotient, Q
The value obtained by substituting the concentrations of reactants and products at a particular time, not necessarily at equilibrium, into the equilibrium constant expression. At equilibrium Q = K.
Direction of reaction from Q and K
If Q < K, the net reaction proceeds forward (to the right) until Q = K. If Q > K, the net reaction proceeds backward (to the left) until Q = K. If Q = K, the system is at equilibrium and there is no net change.
Students often think If Q is less than K the net reaction goes backward, and if Q is greater than K it goes forward. In fact Forward (to the right). Q < K means the ratio of product to reactant terms is too small, so net forward reaction increases the product concentrations until Q = K.
Students often think The amounts in moles can be substituted directly into the expression for Q or K without dividing by the volume. In fact Only if the volume terms cancel. In general each amount must first be divided by the volume to give a concentration in mol dm⁻³.
A calculation in which, for each species, the initial concentration, the change (in the ratio of the coefficients in the equation) and the equilibrium concentration are set out and combined with the equilibrium law, either to find K from equilibrium data or to find an equilibrium concentration from K.
Approximation for a very small K
When K is very small, only a negligible fraction of the reactants is converted, so [reactant]initial ≈ [reactant]eqm can be used. This avoids solving a quadratic or higher-order equation. The approximation is valid only when the change is small compared with the initial concentration.
Students often think The equilibrium concentration of a reactant can always be taken as its initial concentration, whatever the size of K or of the change. In fact No. The approximation [reactant]initial ≈ [reactant]eqm is valid only when K is very small, so that the amount of reactant used up is negligible. Otherwise the change must be subtracted.
Students often think In the change row of an equilibrium calculation every species changes by the same amount, whatever the coefficients. In fact No. The changes are in the ratio of the coefficients: for N₂O₄(g) ⇌ 2NO₂(g), if 0.080 mol dm⁻³ of NO₂ forms, only 0.040 mol dm⁻³ of N₂O₄ reacts.
R2.3.7 Relationship between ΔG⦵ and K HL
Relationship between ΔG⦵ and K
ΔG⦵ = −RT ln K, where ΔG⦵ is the standard Gibbs energy change in J mol⁻¹, R = 8.31 J K⁻¹ mol⁻¹, T is the absolute temperature in K, and ln is the natural logarithm. Rearranged, K = e^(−ΔG⦵/RT).
Sign of ΔG⦵ and position of equilibrium
ΔG⦵ < 0 corresponds to K > 1 (products favoured at equilibrium); ΔG⦵ = 0 corresponds to K = 1; ΔG⦵ > 0 corresponds to K < 1 (reactants favoured). The more negative ΔG⦵, the larger K.
Students often think ΔG⦵ in kJ mol⁻¹ can be used with R = 8.31 J K⁻¹ mol⁻¹ without converting the units. In fact No. ΔG⦵ must first be converted to J mol⁻¹ (multiplied by 1000) so that its units match those of R.
Students often think ln and log₁₀ are interchangeable, so the 'log' key (and 10ˣ to reverse it) can be used in ΔG⦵ = −RT ln K. In fact No. ln is the natural logarithm (base e), so K = e^(−ΔG⦵/RT). Using log₁₀, or 10ˣ to reverse it, gives a wrong value.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Hydrogen and iodine are heated in a sealed flask at constant temperature until the amount of hydrogen iodide no longer changes:
H₂(g) + I₂(g) ⇌ 2HI(g)
Which statement must be true of the mixture at this point?
Answer and reasoning
No HI forms or decomposes now, as both reactions have stopped. — A student who thinks equilibrium is a state of rest picks this. The concentrations are constant because the forward and backward reactions occur at equal rates, not because they have stopped.
HI forms and decomposes at equal rates, so its amount stays the same. — The system is in dynamic equilibrium: in the closed flask the forward and backward reactions continue at equal rates, so the concentrations of H₂, I₂ and HI are constant although the reactions have not stopped.
The concentrations of H₂, I₂ and HI have become equal to each other. — A student who reads 'equilibrium' as 'equal amounts' picks this. It is the rates of the two reactions that are equal; the concentrations are constant but their values depend on K and the starting amounts.
Half of the hydrogen and iodine has been converted into hydrogen iodide. — A student who pictures equilibrium as the halfway point of a reaction picks this. How far the reaction goes is set by K at that temperature, not by a halfway rule; nothing in the stem says half has reacted.
2 Which is the equilibrium constant expression for the following reaction?
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Answer and reasoning
K = 2[NH₃]/([N₂] × 3[H₂]) — A student who multiplies each concentration by its coefficient picks this. The coefficients become powers, not multipliers: 2NH₃ gives [NH₃]² and 3H₂ gives [H₂]³.
K = [N₂][H₂]³/[NH₃]² — A student who puts the reactants on top picks this. This is the expression for the reverse reaction, 2NH₃(g) ⇌ N₂(g) + 3H₂(g), whose value is 1/K.
K = [NH₃]²/([N₂][H₂]³) — Products over reactants, with each concentration raised to the power of its coefficient in the equation: [NH₃]² on top, [N₂] and [H₂]³ multiplied together on the bottom.
K = [NH₃]²/([N₂]+[H₂]³) — A student who carries the '+' sign of the equation into the expression picks this. The reactant terms are multiplied, not added: the denominator is [N₂][H₂]³.
3 For the reaction A(g) ⇌ B(g), K = 1.0 × 10⁻⁸ at 298 K. Which statement about this reaction at 298 K is correct?
Answer and reasoning
At equilibrium, the mixture consists almost entirely of A. — K ≪ 1, so [B]/[A] = 1.0 × 10⁻⁸ at equilibrium: the position of equilibrium lies far to the left and very little A is converted into B.
At equilibrium, A and B are present in similar amounts. — A student who expects every equilibrium to lie roughly halfway picks this. Similar amounts would correspond to K close to 1; here [B]/[A] = 1.0 × 10⁻⁸.
The small value of K shows that equilibrium is reached slowly. — A student who links the size of K with rate picks this. K describes how far the reaction goes, not how fast; a reaction with a small K may reach equilibrium quickly or slowly.
Adding a catalyst would increase the equilibrium yield of B. — A student who thinks a catalyst shifts the equilibrium picks this. A catalyst speeds up the forward and backward reactions by the same factor, so equilibrium is reached sooner, but K and the equilibrium composition are unchanged.
4 The forward reaction in the following equilibrium is exothermic:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
The temperature of the equilibrium mixture is increased at constant pressure. How do the amount of SO₃ and the value of K at the new equilibrium compare with those at the original equilibrium?
Answer and reasoning
A larger amount of SO₃ and a larger value of K. — A student who thinks heating makes every reaction go further picks this. Heating favours the endothermic (backward) direction, so less SO₃ forms and K decreases.
A smaller amount of SO₃ and a smaller value of K. — Raising the temperature favours the endothermic direction, here the backward reaction, so less SO₃ is present at the new equilibrium; K depends on temperature and, for an exothermic forward reaction, decreases.
An unchanged amount of SO₃ and an unchanged K. — A student who treats temperature like a catalyst, speeding up both directions equally, picks this. The endothermic (backward) rate increases more, so the equilibrium shifts to the left and K decreases.
A smaller amount of SO₃ and the same value of K. — A student who applies Le Châtelier’s principle correctly but thinks K is constant at every temperature picks this. K is constant only at a fixed temperature; the shift to the left caused by heating is accompanied by a smaller K.
5 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
At one instant, a sealed 2.00 dm³ vessel contains 0.100 mol N₂, 4.00 mol H₂ and 0.200 mol NH₃. What is the value of the reaction quotient, Q, at this instant? HL
Answer and reasoning
0.667 — A student who multiplies concentrations by their coefficients gets 2(0.100)/(0.0500 × 3(2.00)) = 0.200/0.300 = 0.667. The coefficients are powers: Q = 0.100²/(0.0500 × 2.00³) = 0.0250.
40.0 — A student who puts reactants over products gets (0.0500 × 2.00³)/0.100² = 0.400/0.0100 = 40.0. That is Q for the reverse reaction; Q for the reaction as written is 0.0250.
0.00625 — A student who substitutes the amounts in moles gets 0.200²/(0.100 × 4.00³) = 0.0400/6.40 = 0.00625. Each amount must be divided by 2.00 dm³ first; because the powers on top and bottom differ, the volume does not cancel.
6 N₂O₄(g) ⇌ 2NO₂(g)
N₂O₄ is placed in a sealed vessel at an initial concentration of 0.200 mol dm⁻³ and the mixture is allowed to reach equilibrium at constant temperature. At equilibrium, [NO₂] = 0.080 mol dm⁻³. What is the value of K at this temperature? HL
Answer and reasoning
0.053 — A student who subtracts 0.080 from the N₂O₄ concentration, ignoring the 1 : 2 ratio, gets 0.080²/0.120 = 0.053. Only 0.040 mol dm⁻³ of N₂O₄ reacts to form 0.080 mol dm⁻³ of NO₂.
0.040 — Forming 0.080 mol dm⁻³ of NO₂ uses 0.040 mol dm⁻³ of N₂O₄ (1 : 2 ratio), so [N₂O₄]eqm = 0.200 − 0.040 = 0.160 mol dm⁻³. K = [NO₂]²/[N₂O₄] = 0.080²/0.160 = 0.040.
0.032 — A student who takes [N₂O₄]eqm as the initial 0.200 mol dm⁻³ gets 0.080²/0.200 = 0.032. That approximation is valid only when K is very small; here 20% of the N₂O₄ reacts, so 0.160 mol dm⁻³ must be used.
0.500 — A student who leaves out the power gets 0.080/0.160 = 0.500. The coefficient 2 of NO₂ makes its concentration squared: K = 0.080²/0.160 = 0.040.
7 For a reaction at 227 °C, ΔG⦵ = −8.00 kJ mol⁻¹.
Use ΔG⦵ = −RT ln K, with R = 8.31 J K⁻¹ mol⁻¹.
What is the value of K at this temperature? HL
Answer and reasoning
84.2 — A student who uses 10ˣ instead of eˣ gets 10^1.925 = 84.2. The equation uses the natural logarithm, so K = e^1.925 = 6.86.
1.00 — A student who substitutes ΔG⦵ in kJ mol⁻¹ with R in J K⁻¹ mol⁻¹ gets ln K = 8.00/(8.31 × 500) = 0.00193, so K = 1.00. ΔG⦵ must be converted to −8000 J mol⁻¹.
69.5 — A student who uses 227 instead of 500 K gets ln K = 8000/(8.31 × 227) = 4.241, so K = 69.5. The temperature must be absolute: T = 500 K.
6.86 — T = 227 + 273 = 500 K. ln K = −ΔG⦵/RT = 8000/(8.31 × 500) = 1.925, so K = e^1.925 = 6.86. K > 1, consistent with a negative ΔG⦵.
Working T = 227 + 273 = 500 K; ΔG⦵ = −8000 J mol⁻¹. ln K = −ΔG⦵/(RT) = 8000/(8.31 × 500) = 1.925. K = e^1.925 = 6.86.
8 Some liquid bromine is placed in a flask, which is then stoppered and kept at constant temperature. The brown colour of the vapour above the liquid deepens at first and then stays constant. Which statement about the flask once the colour is constant is correct?
Answer and reasoning
Bromine still evaporates and condenses, and the two processes occur at equal rates. — This is a dynamic physical equilibrium in a closed system: molecules leave the liquid and return to it at equal rates, so the amount of vapour, and hence the colour, stays constant.
Bromine has stopped evaporating, because the vapour above the liquid is saturated. — A student who thinks nothing happens at equilibrium picks this. The vapour is saturated in the sense that its pressure is constant, but evaporation continues; it is balanced by condensation at the same rate.
The system cannot be at equilibrium, because no chemical reaction takes place. — A student who thinks equilibrium applies only to chemical reactions picks this. A change of state in a closed container reaches a dynamic equilibrium with the same characteristics as a chemical one.
The concentration of Br₂ is now the same in the liquid and in the vapour. — A student who believes that concentrations become equal at equilibrium picks this. The liquid is far more concentrated than the vapour; at equilibrium the rates of evaporation and condensation are equal, not the concentrations.
9 Methanol can be made from carbon dioxide and hydrogen in a homogeneous gas-phase equilibrium:
CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g)
Which is the equilibrium constant expression for this reaction?
Answer and reasoning
K = [CH₃OH]/([CO₂][H₂]³) — Water cannot simply be left out. It is omitted only when its concentration is effectively constant: the solvent in dilute aqueous solution, or a pure liquid in a heterogeneous mixture. Here H₂O(g) is a gaseous product whose concentration rises as the reaction proceeds, so [H₂O] belongs in the numerator.
K = [CH₃OH][H₂O]/([CO₂][H₂]) — The coefficient 3 on H₂ is not ignored: each concentration is raised to the power of its stoichiometric coefficient, so the denominator is [CO₂][H₂]³, not [CO₂][H₂].
K = [CH₃OH][H₂O]/([CO₂]·3[H₂]) — A coefficient becomes an exponent, not a multiplier. The equilibrium law raises each concentration to the power of its coefficient, so 3H₂ gives [H₂]³, not 3[H₂].
K = [CH₃OH][H₂O]/([CO₂][H₂]³) — Correct. All four species are gases in a single phase, so every concentration appears: products over reactants, each raised to the power of its coefficient. 3H₂ gives [H₂]³, and H₂O(g) is a gaseous product whose concentration changes, so [H₂O] stays in the numerator.
10 At a certain high temperature, K = 3.16 × 10⁻¹⁰ for the reaction
2H₂O(g) ⇌ 2H₂(g) + O₂(g)
What is the value of K for the following reaction at the same temperature?
2H₂(g) + O₂(g) ⇌ 2H₂O(g)
Answer and reasoning
3.16 × 10⁻¹⁰ — A student who thinks a reaction and its reverse have the same K picks this. They share one equilibrium mixture, but the expression for the reverse is upside down, so its value is 1/K.
−3.16 × 10⁻¹⁰ — A student who reverses the sign, as for ΔH, picks this. K is a ratio of positive concentrations and cannot be negative; reversing the equation gives 1/K.
3.16 × 10¹⁰ — A student who changes the sign of the power of ten without inverting the coefficient picks this. 1/(3.16 × 10⁻¹⁰) = 0.316 × 10¹⁰, which is 3.16 × 10⁹ in standard form.
3.16 × 10⁹ — The second reaction is the reverse of the first, so K = 1/(3.16 × 10⁻¹⁰) = (1/3.16) × 10¹⁰ = 0.316 × 10¹⁰ = 3.16 × 10⁹. The very large value shows that, at this temperature, the formation of water lies far to the right.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
11 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 The following system is at equilibrium in a sealed container:
H₂(g) + I₂(g) ⇌ 2HI(g)
More hydrogen is injected, and the temperature and volume are kept constant. Which statement about the new equilibrium mixture, compared with the original one, is correct?
Answer and reasoning
K has a larger value than in the original mixture. — A student who thinks K follows the composition of the mixture picks this. More HI does form, but at constant temperature K keeps its value.
[H₂] has returned to exactly its original value. — A student who reads Le Châtelier’s principle as 'the change is undone' picks this. The shift only partly opposes the change: the new equilibrium [H₂] is lower than just after the injection but still higher than originally.
[I₂] is lower than in the original mixture. — The added H₂ causes net forward reaction, which uses up I₂ as well as H₂ and forms HI; the volume is constant, so [I₂] falls. K is unchanged because the temperature is unchanged.
[HI] is the same as in the original mixture. — A student who thinks the equilibrium cannot shift while K is unchanged picks this. The added H₂ causes net forward reaction, so [HI] rises while K keeps its value.
2 A gas syringe contains an equilibrium mixture of brown NO₂ and colourless N₂O₄:
2NO₂(g) ⇌ N₂O₄(g)
The intensity of the brown colour depends on [NO₂]. The plunger is pushed in to halve the volume, and the temperature is kept constant. How does the colour of the new equilibrium mixture compare with that of the original mixture?
Answer and reasoning
Paler, since the shift takes [NO₂] below its first value. — A student who applies 'the equilibrium shifts towards N₂O₄' to the original mixture, overlooking that the compression itself first raises every concentration, picks this. The shift removes only part of the extra NO₂ concentration; [NO₂] stays above its original value, so the colour is darker.
The same, since the shift restores [NO₂] to its first value. — A student who thinks Le Châtelier’s principle cancels the change completely picks this. The shift partly opposes the increase in [NO₂]; it does not restore the original concentration.
Darker, since no shift can occur while K is unchanged. — A student who thinks a constant K means the position of equilibrium cannot shift picks this. The mixture is darker, but not because nothing happens: halving the volume doubles every concentration, and the equilibrium then shifts towards N₂O₄, removing part of the extra NO₂ while K keeps its value.
Darker, since the shift removes only part of the extra NO₂. — Halving the volume doubles every concentration at once, so [NO₂] immediately doubles and the mixture darkens. The position of equilibrium then shifts towards N₂O₄, the side with fewer moles of gas, which lowers [NO₂] from its doubled value. The shift only partly opposes the change, so [NO₂] does not fall back to its original value: the new equilibrium mixture is darker than the original.
3 In a sealed bottle of carbonated water at constant temperature, carbon dioxide gas above the liquid is in equilibrium with dissolved carbon dioxide:
CO₂(g) ⇌ CO₂(aq)
When the cap is removed, gas escapes from the space above the liquid and bubbles of CO₂ form in the liquid. Which statement explains the formation of the bubbles?
Answer and reasoning
The pressure of CO₂(g) falls, so the equilibrium shifts to the left. — Opening the bottle lowers the pressure of CO₂(g) above the liquid. The position of the heterogeneous equilibrium shifts to the left to partly oppose this, so dissolved CO₂ comes out of solution as bubbles.
Opening the bottle lowers K, so less CO₂ can stay dissolved. — A student who thinks a pressure change alters K picks this. The temperature is constant, so K is unchanged; it is the lower pressure of CO₂(g) that shifts the position of equilibrium to the left.
The equilibrium was static while sealed, and restarts on opening. — A student who thinks nothing happens at equilibrium picks this. In the sealed bottle CO₂ was dissolving and escaping at equal rates; opening the bottle makes escape faster than dissolving.
CO₂ leaves the liquid until its former gas pressure is restored. — A student who reads Le Châtelier’s principle as 'the change is undone' picks this. A shift only partly opposes a change, and with the bottle open the escaping gas cannot rebuild the former pressure.
4 H₂(g) + I₂(g) ⇌ 2HI(g) K = 50 at temperature T
A sealed vessel at temperature T contains H₂, I₂ and HI, each at a concentration of 0.20 mol dm⁻³. Which statement describes what happens as the mixture reaches equilibrium at temperature T? HL
Answer and reasoning
There is net backward reaction, and [HI] decreases. — A student who has the Q–K rule the wrong way round picks this. Q = 1.0 is less than K = 50, so more product is needed and the net reaction goes forward.
No net reaction occurs, as the concentrations are equal. — A student who thinks equilibrium means equal concentrations picks this. Equal concentrations give Q = 1.0, not K = 50, so the mixture is not at equilibrium.
There is net forward reaction, and [HI] increases. — Q = [HI]²/([H₂][I₂]) = 0.20²/(0.20 × 0.20) = 1.0. Q < K, so the ratio of product to reactant terms is too small: net forward reaction increases [HI] and decreases [H₂] and [I₂] until Q = 50.
K becomes 1.0, and the concentrations stay the same. — A student who thinks K adjusts to the concentrations picks this. At temperature T, K is fixed at 50; it is the concentrations, and hence Q, that change until Q = K.
Working Q = [HI]²/([H₂][I₂]) = (0.20)²/(0.20 × 0.20) = 1.0. Q (1.0) < K (50), so net forward reaction occurs.
5 N₂(g) + O₂(g) ⇌ 2NO(g) K = 1.0 × 10⁻⁵ at a certain high temperature
A sealed vessel at this temperature initially contains N₂ at 0.80 mol dm⁻³ and O₂ at 0.20 mol dm⁻³, and no NO. What is the concentration of NO at equilibrium? HL
Answer and reasoning
1.6 × 10⁻⁶ mol dm⁻³ — A student who leaves out the power on [NO] gets [NO] = K[N₂][O₂] = 1.6 × 10⁻⁶ mol dm⁻³. The coefficient 2 makes the term [NO]², so a square root is needed: [NO] = 1.3 × 10⁻³ mol dm⁻³.
8.0 × 10⁻⁷ mol dm⁻³ — A student who writes 2[NO] instead of [NO]² gets 2[NO] = 1.6 × 10⁻⁶, so [NO] = 8.0 × 10⁻⁷ mol dm⁻³. The coefficient is a power: [NO]² = 1.6 × 10⁻⁶, [NO] = 1.3 × 10⁻³ mol dm⁻³.
1.3 × 10⁻³ mol dm⁻³ — Because K is very small, very little N₂ and O₂ react, so [N₂]eqm ≈ 0.80 and [O₂]eqm ≈ 0.20 mol dm⁻³. Then [NO]² = K[N₂][O₂] = 1.0 × 10⁻⁵ × 0.80 × 0.20 = 1.6 × 10⁻⁶, so [NO] = 1.3 × 10⁻³ mol dm⁻³. Only 6.3 × 10⁻⁴ mol dm⁻³ of each reactant is used, which confirms the approximation.
3.2 × 10⁻³ mol dm⁻³ — A student who adds the reactant concentrations gets [NO]² = 1.0 × 10⁻⁵ × (0.80 + 0.20), so [NO] = 3.2 × 10⁻³ mol dm⁻³. The terms are multiplied: [NO]² = 1.0 × 10⁻⁵ × 0.80 × 0.20.
Working K = [NO]²/([N₂][O₂]). K is very small, so the amounts of N₂ and O₂ that react are negligible: [N₂]eqm ≈ 0.80 mol dm⁻³, [O₂]eqm ≈ 0.20 mol dm⁻³. [NO]² = 1.0 × 10⁻⁵ × 0.80 × 0.20 = 1.6 × 10⁻⁶; [NO] = √(1.6 × 10⁻⁶) = 1.26 × 10⁻³ ≈ 1.3 × 10⁻³ mol dm⁻³. Check: N₂ used = [NO]/2 = 6.3 × 10⁻⁴ mol dm⁻³, less than 0.1% of 0.80 and about 0.3% of 0.20 mol dm⁻³, so the approximation is valid.
6 For a reaction at 298 K, ΔG⦵ = +5.70 kJ mol⁻¹.
Which statement about the position of equilibrium for this reaction at 298 K is correct? HL
Answer and reasoning
K is less than 1, so reactants predominate at equilibrium. — ΔG⦵ = −RT ln K, so a positive ΔG⦵ means ln K is negative and K < 1: K = e^(−5700/(8.31 × 298)) = 0.100. The position of equilibrium lies to the left, but some products are present.
K is greater than 1, so products predominate at equilibrium. — A student who drops the minus sign from ΔG⦵ = −RT ln K picks this. With the minus sign, a positive ΔG⦵ gives a negative ln K, so K = 0.100, which is less than 1.
No products are present, as a positive ΔG⦵ prevents reaction. — A student who reads a positive ΔG⦵ as 'no reaction at all' picks this. K = e^(−ΔG⦵/RT) = 0.100 is small but not zero, so the equilibrium mixture contains some products.
Equilibrium is reached slowly, because ΔG⦵ is positive. — A student who links the position of equilibrium with rate picks this. ΔG⦵ and K describe how far the reaction goes at equilibrium, not how fast equilibrium is reached.
Working K = e^(−ΔG⦵/RT) = e^(−5700/(8.31 × 298)) = e^(−2.30) = 0.100, which is less than 1.
7 For the reaction X(g) ⇌ Y(g), K = 1.0 × 10¹⁵ at 298 K. Which statement about this reaction at 298 K is correct?
Answer and reasoning
At equilibrium, the mixture is made up almost entirely of X. — A student who puts the reactants on top of the expression reads K as [X]/[Y] and so thinks a large K means mostly X. K = [Y]/[X], so a large K means mostly Y.
At equilibrium, almost all of X has been converted into Y. — K ≫ 1, so [Y]/[X] = 1.0 × 10¹⁵ at equilibrium: the position of equilibrium lies far to the right and the reaction goes almost to completion.
At equilibrium, X and Y are present in similar amounts. — A student who expects every equilibrium to lie roughly halfway picks this. Similar amounts would correspond to K close to 1; here [Y]/[X] = 1.0 × 10¹⁵.
The large value of K shows that equilibrium is reached quickly. — A student who links the size of K with rate picks this. K describes how far the reaction goes, not how fast; a reaction with a very large K may still be extremely slow.
8 Pure X is placed in a sealed vessel at constant temperature and the reaction X(g) ⇌ Y(g) takes place. The graph shows how the concentrations of X and Y change with time; three times, t₁, t₂ and t₃, are marked. Which statement about the system is correct?
Answer and reasoning
Equilibrium is first reached at t₁, when the concentrations of X and Y become equal. — At t₁ the curves cross, so [X] = [Y] at that instant, but both concentrations are still changing: [X] goes on falling and [Y] goes on rising after t₁. Equal concentrations are not a condition for equilibrium; constant concentrations are.
Equilibrium is first reached at t₂, when both concentrations have ceased to change. — Correct. A system is at equilibrium once its macroscopic properties no longer change: from t₂ both curves are horizontal, so the forward and backward rates have become equal. Between t₁ and t₂ [X] is still falling and [Y] still rising, so there is still net forward reaction.
The forward and backward reactions both stop at t₂, which is why the curves then flatten. — The concentrations are constant after t₂ because the forward and backward reactions continue at equal rates, not because they have stopped. Equilibrium is dynamic: X still forms Y and Y still forms X at t₃.
The backward reaction only starts at t₂, once the forward reaction has finished. — The backward reaction does not wait for the forward reaction to finish. It begins as soon as any Y is present, and its rate rises as [Y] rises while the forward rate falls; t₂ is simply where the two rates become equal.
9 The graph shows how the concentrations of H₂, I₂ and HI in a reaction vessel change when one change is made to the equilibrium mixture at time t₁:
H₂(g) + I₂(g) ⇌ 2HI(g)
Which change was made at t₁?
Answer and reasoning
Hydrogen was added to the vessel. — Correct. At t₁ only [H₂] changes instantly, jumping to a higher value: that is the signature of adding hydrogen at constant volume. The system then shifts to the right, so [H₂] and [I₂] fall and [HI] rises until a new equilibrium is reached, with [H₂] still above its original value.
The temperature of the vessel was raised. — A change in temperature alters no concentration instantaneously; all three curves would move gradually from t₁ as the position shifted. The graph shows a sudden jump in [H₂] alone, which only the addition of hydrogen produces. The later rise in [HI] is the response to that addition, not evidence of heating.
A catalyst was added to the mixture. — A catalyst added to a mixture already at equilibrium produces no change at all: it speeds up the forward and backward reactions equally, so no curve would move. The graph shows a sudden jump in [H₂] followed by a shift, which a catalyst cannot cause.
The volume of the vessel was decreased. — Decreasing the volume would raise all three concentrations at once by the same factor, and with two gas molecules on each side there would then be no shift. The graph shows a jump in [H₂] only, followed by a shift that lowers [I₂] and raises [HI]: hydrogen was added.
10 The graph shows the percentage of Z in the equilibrium mixture for the gas-phase reaction
X(g) + Y(g) ⇌ Z(g)
plotted against pressure at 400 K and at 600 K. Which statement about the equilibrium constant, K, for this reaction is correct?
Answer and reasoning
K is the same at 400 K and at 600 K, because K is a constant for a given reaction at any pressure. — K is constant only at a fixed temperature. The graph shows that at any pressure the equilibrium mixture contains less Z at 600 K than at 400 K, so the extent of reaction, and K, differ at the two temperatures.
K increases as the pressure is raised, because the percentage of Z rises with pressure at any temperature. — Along each curve the temperature is fixed, so K is fixed too. Raising the pressure shifts the position of equilibrium towards Z (fewer gas molecules), which changes the composition but not the value of K.
K is smaller at 600 K than at 400 K, since whatever the pressure the hotter equilibrium mixture contains less Z. — Correct. At every pressure the 600 K curve lies below the 400 K curve, so the extent of reaction, and hence K, is smaller at the higher temperature. K depends on temperature only, and here it decreases as the temperature rises.
K is larger at 600 K than at 400 K, because equilibrium is reached faster at the higher temperature. — How quickly equilibrium is reached is a matter of rate and says nothing about K. K measures how far the reaction goes, and the graph shows it goes less far at 600 K: at any pressure less Z is present, so K is smaller, not larger.
Working No arithmetic. Read the graph: at every pressure the 600 K curve lies below the 400 K curve, so the equilibrium mixture contains a smaller percentage of Z at 600 K. A smaller extent of reaction at the same pressure means a smaller K, so K decreases as the temperature rises from 400 K to 600 K. Along either curve the temperature is fixed, so K is unchanged by the pressure.
11 X(g) ⇌ 2Y(g)
The graph shows how the concentrations of X and Y change with time after X is placed in a sealed vessel at constant temperature. What is the value of K for this reaction at this temperature? HL
Answer and reasoning
4.0 — This is [Y]/[X] = 0.80/0.20. The coefficient 2 on Y means [Y] is squared in the expression: K = [Y]²/[X] = 0.64/0.20 = 3.2.
8.0 — This is 2[Y]/[X] = 1.60/0.20, treating the coefficient as a multiplier. A coefficient becomes a power: K = [Y]²/[X] = 0.64/0.20 = 3.2.
1.1 — This uses the initial concentration of X, 0.60 mol dm⁻³, read from the start of the X curve. K is defined by equilibrium concentrations only: [X] has fallen to 0.20 mol dm⁻³ by the time the curves are horizontal, so K = 0.64/0.20 = 3.2.
3.2 — Correct. From the horizontal parts of the curves, [X] = 0.20 mol dm⁻³ and [Y] = 0.80 mol dm⁻³ at equilibrium. K = [Y]²/[X] = 0.64/0.20 = 3.2.
Working Equilibrium concentrations are read from the horizontal parts of the curves (from 50 s onwards): [X] = 0.20 mol dm⁻³ and [Y] = 0.80 mol dm⁻³. The initial [X] = 0.60 mol dm⁻³ is not used.
K = [Y]²/[X] = (0.80)²/0.20 = 0.64/0.20 = 3.2 (K has no units).
That was your twenty minutes. Real practice on R2.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
DP Science Cafe