This paper covers the three Unit Assessment areas: chemical calculations (the mole), rate of reaction, and reversible reactions and equilibrium, at full Extended stretch.
Section A — 20 multiple-choice questions, 20 marks, about 25 minutes. Choose one option per question; it is marked automatically when you submit.
Section B — 4 structured questions, 40 marks. Answer in the spaces provided and show all working — method marks are available even if the final answer is incorrect.
Take all graph readings from the printed graphs, not by guessing round numbers.
Relative atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Zn = 65. One mole of any gas occupies 24 dm³ (24 000 cm³) at room temperature and pressure. Avogadro constant = 6.02 × 10²³ per mol.
Your answers will be automatically graded when you submit.
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Section A — Multiple Choice
20 questions · 20 marks · about 25 minutes · choose ONE option per question · auto-marked when you submit
1 What is the number of moles in 8.8 g of carbon dioxide, CO₂? (Mr of CO₂ = 44)
2 How many water molecules are there in 0.25 mol of water?
3 What is the relative formula mass, Mr, of ammonium sulfate, (NH₄)₂SO₄?
4 2.0 g of sodium hydroxide, NaOH (Mr = 40), is dissolved in water and made up to 250 cm³ of solution. What is the concentration of the solution?
5 What is the number of moles of hydrogen gas in 600 cm³ measured at room temperature and pressure?
6 0.30 mol of magnesium is added to 0.50 mol of hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Which reagent is limiting, and why?
7 A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. What is its empirical formula?
8 The theoretical mass of a product is 4.0 g. A student obtains 3.0 g. What is the percentage yield?
9 Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. What mass of magnesium oxide forms from 6.0 g of magnesium? (Ar: Mg = 24; Mr of MgO = 40)
10 Which sample contains the same number of molecules as 12 dm³ of carbon dioxide at r.t.p.?
11 An excess of large marble chips reacts with dilute hydrochloric acid. Which change increases the total volume of carbon dioxide given off but leaves the initial rate unchanged?
12 In an experiment, 36 cm³ of gas is collected in the first 30 seconds. What is the average rate over this period?
13 Why does the rate of a reaction between marble and acid fall as the reaction proceeds?
14 A catalyst increases the rate of a reaction because it…
15 The same mass of zinc is used twice with an excess of the same acid: once as powder, once as a single lump. Which quantity is the same in the two experiments?
16 Which statement must be true for a reaction at dynamic equilibrium?
17 For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the forward reaction is exothermic. What happens to the equilibrium when the temperature is raised?
18 Increasing the pressure has no effect on the position of which equilibrium?
19 Why is a temperature of about 450 °C used in the Haber process?
20 Which catalyst is used in the Contact process for making sulfuric acid?
Section B · Question 1 — The Mole: Limiting Reagent, Yield and Titration
Total: 13 marks
A student makes zinc sulfate by adding 3.25 g of zinc powder to 40.0 cm³ of 1.00 mol/dm³ dilute sulfuric acid.
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
(Ar: Zn = 65, S = 32, O = 16, H = 1. One mole of gas occupies 24 000 cm³ at r.t.p. Avogadro constant = 6.02 × 10²³ per mol.)
(a)[2]
(i) Calculate the number of moles of zinc added. [1]
(ii) Calculate the number of zinc atoms this contains. [1]
Model Answer — 1(a)
n(Zn) = mass ÷ Ar = 3.25 ÷ 65 = 0.0500 mol [1]
The Avogadro constant is the number of atoms in one mole, so multiply by the number of moles: 0.0500 × 6.02 × 10²³ = 3.01 × 10²² atoms [1]
⚠ If you missed marks here: Moles = mass ÷ Ar, never Ar ÷ mass — 65 ÷ 3.25 = 20 mol would mean 1.3 kg of zinc. And the Avogadro constant is per mole, so it is multiplied by 0.0500 mol, not by 3.25 g; multiplying by the mass gives an answer 65 times too large.
(b)[3]
Show by calculation which reactant is the limiting reagent, and calculate the mass of the other reactant left over when the reaction stops. [3]
The equation ratio is 1 : 1, and 0.0400 mol < 0.0500 mol, so the sulfuric acid is the limiting reagent and zinc is in excess [1]
Zinc left over = 0.0500 − 0.0400 = 0.0100 mol: 0.0100 × 65 = 0.65 g [1]
⚠ If you missed marks here: The limiting reagent is decided by comparing moles adjusted by the equation ratio, never masses — here 3.25 g of zinc is less mass than the acid solution, yet the zinc is the one in excess. The other classic slip is forgetting to convert 40.0 cm³ to 0.0400 dm³: leaving it in cm³ gives 40 mol of acid, which would dissolve half a kilogram of zinc.
(c)[2]
Calculate the volume of hydrogen gas, in cm³, produced at room temperature and pressure. [2]
Model Answer — 1(c)
The amount of product is fixed by the limiting reagent: n(H₂) = n(H₂SO₄) = 0.0400 mol (1 : 1) [1]
V = 0.0400 × 24 000 = 960 cm³ [1]
⚠ If you missed marks here: Using the zinc's 0.0500 mol gives 1200 cm³ — but the last 0.0100 mol of zinc has no acid left to react with. Every calculation after a limiting-reagent part must run through the limiting reagent; that is the whole reason part (b) was asked first.
(d)[2]
After evaporating and drying, the student obtains 5.15 g of anhydrous zinc sulfate, ZnSO₄. Calculate the percentage yield. [2]
Model Answer — 1(d)
Mr(ZnSO₄) = 65 + 32 + (4 × 16) = 161, so the theoretical mass = 0.0400 × 161 = 6.44 g [1]
percentage yield = 5.15 ÷ 6.44 × 100 = 80.0% [1]
⚠ If you missed marks here: The theoretical mass again comes from the limiting acid (0.0400 mol), not the zinc — using 0.0500 mol gives 8.05 g and a yield of 64%, punishing the student for an error made two parts earlier. And percentage yield is actual ÷ theoretical: inverting the fraction gives 125%, which should ring an alarm bell — a yield can never exceed 100%.
(e)[4]
A second batch of sulfuric acid is too concentrated to titrate directly. The student pipettes 10.0 cm³ of the batch into a volumetric flask and makes it up to exactly 100.0 cm³ with distilled water.
In a titration, 25.0 cm³ of this diluted acid is exactly neutralised by 20.0 cm³ of 0.500 mol/dm³ sodium hydroxide:
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
Calculate the concentration, in mol/dm³, of the original batch of acid. [4]
Model Answer — 1(e)
n(NaOH) = 0.500 × 20.0 ÷ 1000 = 0.0100 mol [1]
The equation needs 2 mol of NaOH per mol of acid, so n(H₂SO₄) = 0.0100 ÷ 2 = 0.00500 mol in the 25.0 cm³ sample [1]
Concentration of the diluted acid = 0.00500 ÷ 0.0250 = 0.200 mol/dm³ [1]
The batch was diluted 10.0 → 100.0 cm³, a factor of 10, so the original concentration = 0.200 × 10 = 2.00 mol/dm³ [1]
⚠ If you missed marks here: Two directions to get wrong. The ratio: each acid takes two NaOH, so the acid moles are half the NaOH moles — multiplying by 2 gives 0.800 mol/dm³ at the end. The dilution: the original was ten times stronger than what was titrated, so multiply by 10 at the last step — dividing gives 0.0200 mol/dm³, an acid so weak it would never have needed diluting.
Section B · Question 2 — Which Variable Changed?
Total: 10 marks
An excess of large marble chips (calcium carbonate) is added to 40.0 cm³ of 1.0 mol/dm³ dilute hydrochloric acid in a flask standing on a balance. A cotton-wool plug lets the carbon dioxide escape, so the loss in mass equals the mass of CO₂ released. This is Run 1.
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
The experiment is repeated as Run 2 with exactly one condition changed. Both runs are plotted below, with a tangent drawn at t = 0 for each. (Mr of CO₂ = 44.)
(a)[3]
Use the two tangents drawn at t = 0 to calculate the initial rate of each run, in g/s, and state how the two initial rates compare. [3]
Model Answer — 2(a)
Run 1 tangent passes through (0 s, 0 g) and (100 s, 1.0 g): initial rate = 1.0 ÷ 100 = 0.010 g/s [1]
Run 2 tangent passes through (0 s, 0 g) and (100 s, 2.0 g): initial rate = 2.0 ÷ 100 = 0.020 g/s [1]
The initial rate of Run 2 is exactly double the initial rate of Run 1 [1]
⚠ If you missed marks here: Reading the curve at 100 s instead of the tangent gives 0.64 ÷ 100 = 0.0064 g/s for Run 1 — too low, because the reaction has already slowed by then. The tangent at t = 0 is steeper than every later part of the curve; that is precisely why it is drawn.
(b)[4]
Three students suggest what was changed in Run 2:
Student X: the temperature was increased. Student Y: the volume of acid was doubled to 80.0 cm³, at the same concentration. Student Z: the concentration of acid was doubled to 2.0 mol/dm³, at the same volume.
Use both the initial gradients and the final losses in mass to decide which student is right, and explain why each of the other two must be wrong. [4]
Model Answer — 2(b)
Student Z is right: doubling the concentration at the same volume doubles the initial rate and doubles the moles of acid, so it explains both the doubled gradient and the doubled final loss [1]
Student X is wrong: a higher temperature would make the initial gradient steeper, but it cannot create extra acid, so the final loss would stay at 0.88 g — the graph shows it doubled to 1.76 g [1]
Student Y is wrong: doubling the volume at the same concentration would double the final loss, but the concentration of acid particles is unchanged, so the initial gradient would be the same as Run 1 — the graph shows it doubled [1]
Check: 1.76 g of CO₂ = 1.76 ÷ 44 = 0.040 mol, needing 0.080 mol of HCl in the same 40.0 cm³, which is 2.0 mol/dm³ — exactly Student Z's claim [1]
⚠ If you missed marks here: Neither clue alone settles it — the doubled gradient fits X and Z, and the doubled plateau fits Y and Z. Only using both together isolates Z, which is why an answer that mentions only the gradient (or only the plateau) cannot score full marks here. The final mark needs the mole arithmetic, not just "more acid means more gas".
(c)[3]
Explain, in terms of particles, why doubling the concentration of the acid doubles the initial rate, and state clearly what property of the collisions does not change. [3]
Model Answer — 2(c)
At double concentration there are twice as many acid particles in the same volume of solution [1]
So acid particles collide with the surface of the marble twice as frequently — twice as many collisions per second means twice as many successful collisions per second, doubling the rate [1]
The energy of the collisions is unchanged: the temperature is the same, so the particles move no faster and the activation energy is unchanged — only the collision frequency rises [1]
⚠ If you missed marks here: "More particles so more collisions" without per second misses the rate idea — over the whole of Run 2 there are more collisions in total anyway, simply because more acid reacts. And crediting concentration with "more energetic collisions" is the standard confusion with temperature: concentration changes how often particles hit, temperature changes how hard they hit.
Section B · Question 3 — Temperature, Catalysts and Collision Theory
Total: 4 marks
When sodium thiosulfate solution reacts with dilute hydrochloric acid, increasing the temperature from 20 °C to 30 °C roughly doubles the rate of reaction, even though the particles move only slightly faster.
(a)[3]
Explain, in terms of collisions, why this small rise in temperature produces such a large increase in rate. Your answer must deal with both the frequency of the collisions and their energy. [3]
Model Answer — 3(a)
The particles gain kinetic energy and move faster, so they collide slightly more frequently (more collisions per second) [1]
Far more importantly, a much greater proportion of the colliding particles now have energy greater than or equal to the activation energy [1]
So the number of successful collisions per second rises much faster than the collision frequency itself — which is why a 10 °C rise can double the rate [1]
⚠ If you missed marks here: "The particles move faster so they collide more often" is only a third of the answer — the slight speed increase alone cannot double the rate. The doubling comes from the energy condition: the fraction of collisions beating the activation energy grows steeply with temperature. An answer that never mentions activation energy cannot score the last two marks.
(b)[1]
A catalyst also increases the rate of a reaction, but by a completely different mechanism. State how a catalyst produces its effect. [1]
Model Answer — 3(b)
It provides an alternative reaction pathway with a lower activation energy, so a greater proportion of the existing collisions are successful [1]
⚠ If you missed marks here: A catalyst gives the particles no energy at all — the temperature, the collision frequency and the collision energies are all unchanged. Temperature raises the particles towards a fixed barrier; a catalyst lowers the barrier towards fixed particles. Muddling those two directions is the most examined catalyst error there is.
Section B · Question 4 — Reversible Reactions and Equilibrium
Total: 13 marks
Many industrial reactions are reversible. In a closed system a reversible reaction reaches equilibrium, and industry must then choose conditions that balance yield, rate and cost.
(a)[2]
Explain what is meant by a dynamic equilibrium. [2]
Model Answer — 4(a)
The forward and reverse reactions are both still occurring, at equal rates [1]
So the concentrations of reactants and products remain constant (in a closed system) — constant, not necessarily equal [1]
⚠ If you missed marks here: "The reaction has stopped" is exactly what dynamic rules out — both reactions run continuously, their effects cancelling. And "the concentrations are equal" is the other classic slip: an equilibrium can sit 99% on one side; what matters is that the concentrations stop changing.
(b)[3]
Hydrogen and iodine reach equilibrium in a sealed container:
A student says: "Increasing the pressure will shift the equilibrium to the right and increase the yield of hydrogen iodide."
Explain why the student is wrong, and state one change that would increase the equilibrium yield of HI. [3]
Model Answer — 4(b)
Changing the pressure has no effect on the position of this equilibrium [1]
Because there are equal numbers of gas molecules on the two sides (1 + 1 = 2 on the left, 2 on the right), so neither direction reduces the pressure [1]
To raise the yield: decrease the temperature (the forward reaction is exothermic, so the equilibrium shifts right) — or add more H₂ or I₂, or remove HI as it forms [1]
⚠ If you missed marks here: "More pressure pushes it right" only works when the right-hand side has fewer gas molecules — the rule is to count them, every time, before answering. Here both sides count 2, so the position cannot move (the rate does increase, but the yield does not). This equal-moles case is the deliberate trap examiners set for students who apply the pressure rule without counting.
(i) State the catalyst and the temperature and pressure used industrially. [1]
(ii) A lower temperature would give a higher equilibrium yield of ammonia. Explain why, and explain why a low temperature is nevertheless not used. [2]
(iii) A much higher pressure would also give a higher yield. Explain why, and explain why about 200 atm is used instead. [2]
Model Answer — 4(c)
Iron catalyst, about 450 °C and about 200 atm [1]
The forward reaction is exothermic, so lowering the temperature shifts the equilibrium towards ammonia and raises the yield [1]
But at a low temperature the rate is far too slow — equilibrium would take too long to reach — so 450 °C is a compromise between yield and rate [1]
There are 4 gas molecules on the left and only 2 on the right, so a higher pressure shifts the equilibrium towards ammonia (and also speeds the reaction up) [1]
But very high pressures need extremely strong, thick-walled vessels and pumping, which are expensive and hazardous — so 200 atm is a compromise between yield and cost/safety [1]
⚠ If you missed marks here: Reciting "450 °C, 200 atm, iron" is one mark; the other four are the compromise argument, and each half needs both sides — the benefit AND the reason it is given up. "High temperature gives a better yield" is flatly wrong for an exothermic forward reaction: the temperature is raised despite the yield, for the sake of the rate. And the pressure limit is economics and safety, not chemistry — the equilibrium itself would love 1000 atm.
(d)[3]
(i) Sulfuric acid is made by the Contact process. Write the balanced symbol equation, with the equilibrium arrow, for its reversible step, and name the catalyst used. [2]
(ii) Most of the ammonia from the Haber process is used to make NPK fertilisers. Explain why ammonia matters for these fertilisers. [1]
Model Answer — 4(d)
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) [1]
Catalyst: vanadium(V) oxide, V₂O₅ (used at about 450 °C) [1]
Ammonia supplies the nitrogen — it is converted into ammonium salts (or urea), and nitrogen is one of the three essential elements (N, P, K) plants need, for making proteins and for growth [1]
⚠ If you missed marks here: The equation must balance and carry the reversible arrow — a one-way arrow loses the mark, because the whole point of the Contact conditions is that this step is an equilibrium. The catalyst is vanadium(V) oxide, not iron (that is Haber's). For (ii), "fertilisers help plants grow" says nothing: the mark is for naming ammonia as the source of the N in NPK.
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A : 36-41
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D : 18-23
E : 12-17
U : <12
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