Paper 2: short and extended responses

How Paper 2 questions are built and marked: setting out calculations, writing equations, explanations that earn every mark, and three fully worked questions with markschemes.

IB Chemistry
Exam skills
SL and HL

16 min readLast reviewed 14 Oct 2026

Paper 2 consists of structured questions, each built around one substance or one context: a fuel, an industrial process, an element and its compounds, a drug molecule. A question opens with a short stem and moves through parts labelled (a), (b), (c) and sub-parts (i), (ii), each with its mark allocation in square brackets. Several shorter questions are followed by one or two longer ones that range widely across the course.

The parts are small. About half are worth 1 mark and most of the rest 2 marks, so a typical part is one fact, one link in a chain of reasoning, or one step of a calculation. The skill being tested is the ability to give exactly what each part asks for, quickly and precisely, and then move on.

How a Paper 2 question is built

The stem names a substance or a situation in one or two lines, often with a structure, a table of data, a graph or a diagram of apparatus. The parts then visit different areas of chemistry that all involve that substance. A question about ethanol might ask for its Lewis formula, why it mixes with water, its combustion equation, the energy released by a given mass, the assumptions behind that calculation, the product of its oxidation and the atom economy of making it from ethene: six subtopics in one question.

New information between parts. Lines such as "The enthalpy of combustion of methanol is …" or "The equilibrium constant at 450 °C is …" appear between parts, unlabelled. They are there because a later part needs them.
Easy parts after each new lead-in. Within a question, parts often restart at simple recall (write, state) after new information arrives, then rise to calculation and explanation. A hard part does not mean the rest of the question is hard.
Follow-on parts. Phrases such as "using your answer to (b)(i)" mean the method is marked even if the earlier answer was wrong. Use your own earlier value, even if you doubt it.
Labelled answer lines. Some parts print labels for a two-part answer ("Oxidized: … Reduced: …", "Similarity: … Difference: …", "Number of signals: … Ratio: …"). Write one answer on each line.
Answers on figures. "Sketch on the axes", "complete the diagram" and "draw an arrow, labelled X" mean the answer goes on the printed figure.

Setting out a calculation

Calculation parts are marked one step at a time. A typical 3-mark calculation gives one mark for the amount in moles, one for applying the mole ratio, and one for the final quantity. The markscheme usually adds "Award [3] for correct final answer", so a correct answer alone earns full marks, but a wrong answer earns only the marks for steps that are visible.

Write what each line calculates. " $n (N a O H)$ =" or "mass of water =" before the number lets the examiner follow the method and award marks for it.
Substitute, then calculate. Write the equation with the numbers in it, then the result. If the arithmetic slips, the substitution can still earn the mark.
Keep unrounded values in the calculator. Round only the answers you write down, and use the stored value in the next step, so rounding errors do not build up.
Give the final answer with a unit and a sensible number of significant figures, usually the same as the least precise data (often three).
Check the size and the sign. An enthalpy change of combustion is negative; a pH is usually between 0 and 14; a percentage yield above 100 % means an error.

(Paper 2 style) (a) Calculate the amount, in mol, of magnesium in $0.486 g$ of magnesium ribbon. [1] (b) The magnesium reacts completely with excess hydrochloric acid: $M g (s) + 2 H C l (a q) \overset{}{\to} M g C l_{2} (a q) + H_{2} (g)$ . Calculate the volume of hydrogen produced, in ${cm}^{3}$ , at STP. [2] A student wrote: (a) $n = 0.486 \times 24.31 = 11.8 mol$ ; (b) $n (H_{2}) = 11.8 mol$ , $V = 11.8 \times 22.7 = 268 {dm}^{3} = 268 000 {cm}^{3}$ . How is this answer marked?

Show how it is marked

(a) scores 0. The mass was multiplied by the molar mass instead of divided: $n = \frac{0.486}{24.31} = 0.0200 mol$ .

(b) scores 2 by ECF. The student used the correct 1 : 1 mole ratio and the molar volume correctly on their own value from (a), and converted to ${cm}^{3}$ . Both marks in (b) are for method, so both are awarded. The correct answer is $0.0200 \times 22.7 = 0.454 {dm}^{3} = 454 {cm}^{3}$ .

The absurd size of 11.8 mol from a strip of ribbon should have been a warning. Checking the size of an answer in (a) would have saved that mark too.

Writing equations

"Write an equation" or "formulate an equation" always means a balanced symbol equation. A word equation scores zero, and so does an unbalanced one, even if every formula is right. Where an equation is worth 2 marks, one mark is usually for the correct species and one for the balancing, or one for the equation and one for its state symbols.

State symbols are ignored unless the question asks for them. When it does, use the state at the conditions given: water is $H_{2} O (l)$ under standard conditions, ionic compounds in solution are $(aq)$ , precipitates are $(s)$ .
Integer coefficients. "Using integer coefficients" or "using only whole numbers" rules out $\frac{1}{2} O_{2}$ . Multiply the whole equation through instead. Equations that define a standard enthalpy change (of formation or combustion) may need fractions, because they refer to one mole of one substance.
Ionic equations leave out the spectator ions: $A g^{+} (a q) + C l^{-} (a q) \overset{}{\to} A g C l (s)$ . Charges must balance as well as atoms.
Half-equations show electrons, and the electrons go on the left for a reduction and on the right for an oxidation. Balance atoms other than O and H first, then O with $H_{2} O$ , then H with $H^{+}$ , then charge with electrons.
Organic equations may use condensed structural formulas ( $C H_{3} C H_{2} O H$ ) so that the product is unambiguous; molecular formulas such as $C_{2} H_{6} O$ do not show which isomer is meant.

Explanations that earn every mark

An explain part is marked as a chain of cause and effect, and each link is usually one mark. The examiner checks the answer against a list of mark points; a point that is implied but not written does not score. Most chemistry explanations have a standard skeleton, and learning the skeletons for the common ones means the right links appear in the right order under exam pressure.

Three chains of three boxes joined by arrows. Boiling point: name the force in each substance, state which is stronger, more energy needed to overcome it. Rate: more kinetic energy, larger proportion with E ≥ Ea, more successful collisions per unit time. — Three common explanations written as chains. Each box is one mark point, and the green box is the conclusion the question asks about. Leaving out a middle link usually costs a mark even when the conclusion is right.

The same habit works for unfamiliar questions. Before writing, decide how many separate points the marks require, then make sure each one appears as its own sentence or clause. A part worth 2 marks that gets one long, accurate sentence containing a single idea scores 1.

Explanation asked for	The points the markscheme looks for
Why first ionization energy increases across a period	greater nuclear charge (more protons); electrons added to the same main energy level, so similar shielding; outer electron held more strongly / smaller radius
Why an ionic solid conducts when molten but not when solid	in the solid the ions are held in fixed positions in the lattice; when molten (or in solution) the ions are free to move and carry charge
Why a catalyst increases the rate	provides an alternative pathway (mechanism); with a lower activation energy; so a larger proportion of collisions have energy ≥ Ea
Why a weak acid has a higher pH than a strong acid of the same concentration	the weak acid only partially dissociates; so the concentration of H⁺ ions is lower
Why the boiling points of the alkanes increase with chain length	more electrons / larger surface area of contact; stronger London forces; more energy needed to separate the molecules
Why a substance with a giant covalent structure has a high melting point	many strong covalent bonds throughout the structure; all must be broken to melt it; this needs a lot of energy

Questions in unfamiliar contexts

Some parts deliberately use a compound or a situation that the course never mentions: a drug molecule, an unusual salt, a new battery. The chemistry needed is always within the course. The approach is to find the familiar feature: a functional group in the structure, a type of bonding, an equilibrium, a redox change. Then apply the ordinary rule to it, and say in the answer which feature you used.

Worked question 1: hydrogen peroxide

A short question of the kind that opens a paper. It moves from structure to stoichiometry, kinetics, intermolecular forces and energetics around one substance. Try each part before opening the markscheme.

Hydrogen peroxide, $H_{2} O_{2}$ , is used as a bleach and a disinfectant. It decomposes slowly at room temperature: $2 H_{2} O_{2} (a q) \overset{}{\to} 2 H_{2} O (l) + O_{2} (g)$

(a) Draw the Lewis formula of hydrogen peroxide. [1]

(b) Deduce the oxidation state of oxygen in hydrogen peroxide. [1]

(c) Calculate the volume of oxygen, in ${cm}^{3}$ at STP, released by the complete decomposition of $25.0 {cm}^{3}$ of $1.50 {mol dm}^{- 3}$ hydrogen peroxide solution. [2]

(d) Outline how adding manganese(IV) oxide increases the rate of decomposition. [1]

(e) Explain why hydrogen peroxide is very soluble in water. [2]

(f) Determine the enthalpy change, in $kJ$ , for $2 H_{2} O_{2} (g) \overset{}{\to} 2 H_{2} O (g) + O_{2} (g)$ , using bond enthalpies from the data booklet. [3]

Show the markscheme

(a)

Lewis formula of hydrogen peroxide. Each oxygen atom has two bonding pairs and two lone pairs, completing its octet.

H–O–O–H with single bonds AND two lone pairs on each O ✓

Accept any combination of dots, crosses and lines.

(b)

−1 ✓

Do not accept "1−" or "−2".

(c)

$n (O_{2}) = \frac{0.0250 \times 1.50}{2} = 0.01875$ «mol» ✓
$V = 0.01875 \times 22.7 = 0.426 {dm}^{3} = 426$ «cm³» ✓

Award [2] for correct final answer. Award [1 max] for 851 cm³ (the 2 : 1 ratio not applied) or for 0.426 cm³ (no unit conversion).

(d)

provides an alternative pathway/mechanism with a lower activation energy ✓

(e)

hydrogen peroxide has O–H bonds/H bonded to O AND lone pairs on O ✓
so it forms hydrogen bonds with water molecules ✓

Do not accept "it is polar" alone for M2, or "like dissolves like".

(f)

bonds broken: $2 (O - O) + 4 (O - H) = 2 (144) + 4 (463) = 2140$ «kJ» ✓
bonds formed: $4 (O - H) + O = O = 4 (463) + 498 = 2350$ «kJ» ✓
$Δ H = 2140 - 2350 = - 210$ «kJ» ✓

Award [3] for correct final answer. Accept the shortcut that the O–H bonds cancel: $2 (144) - 498 = - 210$ . Award [2 max] for $+ 210$ .

Worked question 2: methanol as a fuel

A longer synoptic question. It visits combustion, calorimetry, experimental error, specific energy, incomplete combustion, equilibrium and intermolecular forces, all through one compound. The new information between parts is needed for the parts after it.

Methanol, $C H_{3} O H$ , is a liquid fuel that can be made from carbon monoxide and hydrogen.

(a) Write a balanced equation for the complete combustion of methanol. [1]

A student burns methanol in a spirit burner under a copper can containing $200.0 g$ of water. The mass of the burner decreases by $0.650 g$ and the temperature of the water rises from $20.5^{\circ} C$ to $34.0^{\circ} C$ .

(b) Calculate the enthalpy change of combustion of methanol, in ${kJ mol}^{- 1}$ . [3]

(d) Calculate the specific energy of methanol, in ${kJ g}^{- 1}$ , using the data booklet value. [1]

(e) Write an equation for the incomplete combustion of methanol to form carbon monoxide and water. [1]

(f) State one reason why carbon monoxide is harmful to health. [1]

Methanol is manufactured by the reaction $C O (g) + 2 H_{2} (g) \overset{}{⇌} C H_{3} O H (g)$ , $Δ H = - 91 kJ$ .

(g) Predict, giving a reason, the effect of increasing the temperature on the equilibrium yield of methanol. [2]

(h) Methanol (boiling point $65^{\circ} C$ ) and ethane (boiling point $- 89^{\circ} C$ ) have similar molar masses. Explain the difference in their boiling points. [3]

Show the markscheme

(a)

$2 C H_{3} O H (l) + 3 O_{2} (g) \overset{}{\to} 2 C O_{2} (g) + 4 H_{2} O (l)$ ✓

Accept $C H_{3} O H + 1.5 O_{2} \overset{}{\to} C O_{2} + 2 H_{2} O$ . Ignore state symbols.

(b)

$Q = 200.0 \times 4.18 \times 13.5 = 11 286$ «J» ✓
$n (C H_{3} O H) = \frac{0.650}{32.05} = 0.02028$ «mol» ✓
$Δ H_{c} = - \frac{11.29}{0.02028} = - 556$ «kJ mol⁻¹» ✓

Award [3] for correct final answer. Accept answers from $- 555$ to $- 557$ . Award [2 max] for $+ 556$ . Award [2 max] if the mass of methanol is used in $Q = m c Δ T$ .

(c)

heat lost to the surroundings/air ✓
incomplete combustion ✓
heat absorbed by the copper can ✓
evaporation of methanol from the wick ✓

Any two. [2 max] Do not accept "human error" or "faulty thermometer".

(d)

$\frac{726}{32.05} = 22.7$ «kJ g⁻¹» ✓

(e)

$C H_{3} O H + O_{2} \overset{}{\to} C O + 2 H_{2} O$ ✓

Accept $2 C H_{3} O H + 2 O_{2} \overset{}{\to} 2 C O + 4 H_{2} O$ .

(f)

it binds to haemoglobin, reducing the transport of oxygen in the blood ✓

Accept "toxic/poisonous" only if linked to oxygen transport.

(g)

yield decreases/equilibrium shifts to the left ✓
the forward reaction is exothermic «so the endothermic reverse reaction is favoured» ✓

Do not award M1 for an answer based on the rate of reaction.

(h)

methanol forms hydrogen bonds between its molecules ✓
ethane has only London (dispersion) forces between its molecules ✓
hydrogen bonds are stronger, so more energy is needed to separate methanol molecules ✓

Do not award M3 for breaking covalent bonds.

Worked question 3: an equilibrium mixture (HL)

An HL question on equilibrium calculations, the reaction quotient and the link between the equilibrium constant and Gibbs energy. HL papers mix parts like these into questions that also contain SL parts.

Dinitrogen tetroxide dissociates into nitrogen dioxide: $N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)$ , $Δ H = + 57 kJ$ .

(a) Write the expression for the equilibrium constant, $K$ . [1]

A sealed flask initially contains $0.100 {mol dm}^{- 3}$ of $N_{2} O_{4}$ and no $N O_{2}$ . At equilibrium at 298 K, $[N O_{2}] = 0.0200 {mol dm}^{- 3}$ .

(b) Calculate the value of $K$ at 298 K. [2]

(c) A different mixture at 298 K contains $[N O_{2}] = 0.0300 {mol dm}^{- 3}$ and $[N_{2} O_{4}] = 0.150 {mol dm}^{- 3}$ . Deduce, using the reaction quotient, $Q$ , the direction in which the reaction proceeds. [2]

(d) Calculate the standard Gibbs energy change, $Δ G^{⊖}$ , in ${kJ mol}^{- 1}$ , at 298 K, using section 1 of the data booklet. [2]

(e) State and explain the effect of increasing the temperature on the value of $K$ . [2]

Show the markscheme

(a)

$K = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$ ✓

(b)

$[N_{2} O_{4}]_{eqm} = 0.100 - \frac{0.0200}{2} = 0.0900$ «mol dm⁻³» ✓
$K = \frac{{0.0200}^{2}}{0.0900} = 4.44 \times 10^{- 3}$ ✓

Award [1 max] for $4.00 \times 10^{- 3}$ (the initial $[N_{2} O_{4}]$ used).

(c)

$Q = \frac{{0.0300}^{2}}{0.150} = 6.00 \times 10^{- 3}$ ✓
$Q > K$ , so the reaction proceeds to the left/towards $N_{2} O_{4}$ «until $Q = K$ » ✓

Accept ECF from (b).

(d)

$Δ G^{⊖} = - R T \ln K = - 8.31 \times 298 \times \ln (4.44 \times 10^{- 3})$ ✓
$= + 13 400$ «J mol⁻¹» $= + 13.4$ «kJ mol⁻¹» ✓

Accept ECF from (b). Award [1 max] for $- 13.4$ .

(e)

$K$ increases ✓
the forward reaction is endothermic/ $Δ H$ is positive, so a higher temperature shifts the equilibrium to the right ✓

Common ways marks are lost

What happens	How to avoid it
The answer addresses a different quantity: molecular for empirical formula, lengths for strengths, observation for conclusion	Underline the quantity and the qualifier before writing
A name is given when a formula was asked for, or the reverse	Give exactly what is asked; if both are given and one is wrong, the mark may be lost
Equations are unbalanced or written in words	Always count atoms and charges
The sign of an enthalpy change is missing or wrong	Exothermic means negative; write the sign every time
A 2-mark explanation contains one idea	Count the points against the marks before moving on
A bare final answer is wrong, so no method marks are possible	Show each step, labelled
Vague words: clear, more collisions, enough energy, human error, pollution	Use the precise term and name the substance or process
An oxidation state is written as "2+" instead of "+2"	Sign first for oxidation states, sign after the number for ionic charges

Self-check

To do:
I can follow a Paper 2 question across several topics around one substance and use the new information given between parts.
To do:
I can set out a calculation with labelled steps, unrounded intermediate values and a final answer with a unit.
To do:
I can explain how error carried forward protects later marks.
To do:
I can write balanced symbol, ionic and half-equations, with state symbols when asked.
To do:
I can write explanations as chains of cause and effect, with one point per mark.
To do:
I can use the precise vocabulary that markschemes accept.
To do:
I can apply familiar principles to an unfamiliar compound or context and say which feature I used.
To do:
I can budget time in a long question by marks rather than by parts.

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