This paper covers the three Unit Assessment areas: chemical calculations (the mole), rate of reaction, and reversible reactions and equilibrium, at full Extended stretch. All questions are new — none repeat Unit Exam 1.
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, Cl = 35.5, Ca = 40, 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.
Question Navigation
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 mass of 0.150 mol of calcium carbonate, CaCO₃? (Mr of CaCO₃ = 100)
2 How many molecules are there in 3.2 g of oxygen gas, O₂? (Mr of O₂ = 32)
3 What is the relative formula mass, Mr, of calcium hydroxide, Ca(OH)₂?
4 0.0500 mol of a solute is dissolved in water and the solution is made up to 200 cm³. What is its concentration?
5 What volume does 0.0150 mol of carbon dioxide occupy at room temperature and pressure?
6 0.060 mol of zinc is added to 0.100 mol of hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. Which reagent is limiting, and why?
7 An oxide of sulfur contains 40% sulfur and 60% oxygen by mass. What is its empirical formula? (Ar: S = 32, O = 16)
8 The theoretical mass of a product is 12.0 g. A student obtains 9.6 g. What is the percentage yield?
9 Limestone decomposes on heating: CaCO₃ → CaO + CO₂. What mass of calcium oxide forms from 25 g of calcium carbonate? (Mr: CaCO₃ = 100, CaO = 56)
10 Which sample contains the most molecules?
11 In an experiment, 90 cm³ of gas is collected in the first 60 seconds. What is the average rate over this period?
12 Raising the temperature increases the rate of a reaction mainly because…
13 A powdered solid reacts faster with acid than a single lump of the same mass because…
14 Which statement about a catalyst in a reversible reaction is true?
15 A graph of gas volume against time is steepest at the very start because…
16 Which statement must be true for a system at dynamic equilibrium?
17 For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the forward reaction is exothermic. Lowering the temperature…
18 Extra nitrogen is pumped into the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at constant temperature. The equilibrium…
19 What is the effect of increasing the pressure on N₂(g) + 3H₂(g) ⇌ 2NH₃(g)?
20 Ammonium salts made from Haber-process ammonia supply which NPK element, and why do plants need it?
Section B · Question 1 — The Mole: Limiting Reagent, Yield and Titration
Total: 13 marks
A student makes magnesium chloride by adding 0.48 g of magnesium ribbon to 50.0 cm³ of 0.600 mol/dm³ dilute hydrochloric acid.
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
(Ar: Mg = 24, Cl = 35.5, 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 magnesium added. [1]
(ii) Calculate the number of magnesium atoms this contains. [1]
Model Answer — 1(a)
n(Mg) = mass ÷ Ar = 0.48 ÷ 24 = 0.0200 mol [1]
The Avogadro constant is the number of atoms in one mole, so multiply by the moles: 0.0200 × 6.02 × 10²³ = 1.20 × 10²² atoms [1]
⚠ If you missed marks here: Moles = mass ÷ Ar, never Ar ÷ mass — 24 ÷ 0.48 = 50 mol would be over a kilogram of magnesium. And the Avogadro constant is per mole, so multiply it by 0.0200 mol, not by the 0.48 g mass — using the mass gives 2.9 × 10²³, which is 24 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]
0.0200 mol of Mg needs 2 × 0.0200 = 0.0400 mol of HCl, but only 0.0300 mol is present, so the hydrochloric acid is the limiting reagent and magnesium is in excess [1]
Mg used = 0.0300 ÷ 2 = 0.0150 mol, so Mg left over = 0.0200 − 0.0150 = 0.0050 mol: 0.0050 × 24 = 0.12 g [1]
⚠ If you missed marks here: Comparing 0.0200 with 0.0300 directly suggests the magnesium runs out first — wrong, because each Mg eats two HCl. Adjust by the ratio before comparing. The other classic slip is leaving the 50.0 cm³ unconverted: 0.600 × 50.0 = 30 mol of acid, an absurd amount that would dissolve two kilograms of magnesium.
(c)[2]
Calculate the volume of hydrogen gas, in cm³, produced at room temperature and pressure. [2]
Model Answer — 1(c)
The product is fixed by the limiting acid: n(H₂) = n(HCl) ÷ 2 = 0.0300 ÷ 2 = 0.0150 mol [1]
V = 0.0150 × 24 000 = 360 cm³ [1]
⚠ If you missed marks here: Using the magnesium's 0.0200 mol gives 480 cm³ — but the last 0.0050 mol of magnesium has no acid left to react with. Every calculation after a limiting-reagent part must run through the limiting reagent; that is why part (b) was asked first.
(d)[2]
After evaporating and drying, the student obtains 1.14 g of anhydrous magnesium chloride, MgCl₂. Calculate the percentage yield. [2]
Model Answer — 1(d)
Mr(MgCl₂) = 24 + (2 × 35.5) = 95, so the theoretical mass = 0.0150 × 95 = 1.425 g [1]
percentage yield = 1.14 ÷ 1.425 × 100 = 80.0% [1]
⚠ If you missed marks here: The theoretical mass again comes from the limiting acid (0.0150 mol of MgCl₂), not from the magnesium — using 0.0200 mol gives 1.90 g and a yield of 60%, punishing you twice for the same slip. And percentage yield is actual ÷ theoretical: inverting the fraction gives 125%, and a yield can never exceed 100%.
(e)[4]
A second batch of hydrochloric acid is too concentrated to titrate directly. The student pipettes 25.0 cm³ of the batch into a volumetric flask and makes it up to exactly 250.0 cm³ with distilled water.
In a titration, 25.0 cm³ of this diluted acid is exactly neutralised by 30.0 cm³ of 0.250 mol/dm³ sodium carbonate solution:
Calculate the concentration, in mol/dm³, of the original batch of acid. [4]
Model Answer — 1(e)
n(Na₂CO₃) = 0.250 × 30.0 ÷ 1000 = 0.00750 mol [1]
Each carbonate needs 2 mol of HCl, so n(HCl) = 2 × 0.00750 = 0.0150 mol in the 25.0 cm³ sample [1]
Concentration of the diluted acid = 0.0150 ÷ 0.0250 = 0.600 mol/dm³ [1]
The batch was diluted 25.0 → 250.0 cm³, a factor of 10, so the original concentration = 0.600 × 10 = 6.00 mol/dm³ [1]
⚠ If you missed marks here: The ratio runs the OTHER way from an acid–alkali titration: here the HCl moles are double the carbonate moles — halving instead gives 0.150 mol/dm³ diluted and 1.50 mol/dm³ at the end. And the original batch was ten times stronger than what was titrated, so multiply by 10 at the last step — dividing gives 0.0600 mol/dm³, an acid too weak to have needed diluting at all.
Section B · Question 2 — Which Variable Changed?
Total: 10 marks
A small measured mass of manganese(IV) oxide catalyst is added to 50.0 cm³ of 0.400 mol/dm³ hydrogen peroxide solution, and the oxygen is collected in a gas syringe. This is Run 1.
2H₂O₂(aq) → 2H₂O(l) + O₂(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. (One mole of gas occupies 24 000 cm³ at r.t.p.)
(a)[3]
Use the two tangents drawn at t = 0 to calculate the initial rate of each run, in cm³/s, and state how the two initial rates compare. [3]
Model Answer — 2(a)
Run 1 tangent passes through (0 s, 0 cm³) and (100 s, 200 cm³): initial rate = 200 ÷ 100 = 2.0 cm³/s [1]
Run 2 tangent passes through (0 s, 0 cm³) and (60 s, 240 cm³): initial rate = 240 ÷ 60 = 4.0 cm³/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 155 ÷ 100 = 1.55 cm³/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 concentration of the hydrogen peroxide was doubled, at the same volume. Student Y: the temperature was increased. Student Z: the same mass of catalyst was used, but in larger lumps.
Use both the initial gradients and the final volumes of oxygen to decide which student is right, and explain why each of the other two must be wrong. [4]
Model Answer — 2(b)
Student Y is right: a higher temperature doubles the initial rate, but it cannot create extra hydrogen peroxide, so the final volume stays at 240 cm³ — exactly what the graph shows [1]
Student X is wrong: doubling the concentration at the same volume would double the moles of H₂O₂, so the final volume would double to 480 cm³ — the graph shows it unchanged [1]
Student Z is wrong: larger lumps of catalyst expose less surface, so Run 2 would start slower than Run 1 — the graph shows it faster [1]
Check: 240 cm³ of O₂ = 240 ÷ 24 000 = 0.0100 mol, needing 0.0200 mol of H₂O₂ (2 : 1), and 0.0200 ÷ 0.0500 dm³ = 0.400 mol/dm³ — the concentration is unchanged, ruling X out by calculation [1]
⚠ If you missed marks here: Neither clue alone settles it — the doubled gradient fits X and Y, and only the unchanged plateau eliminates X. An answer using just one feature of the graph cannot score full marks. The final mark needs the mole arithmetic on the 240 cm³ plateau, not just "the amount of gas is the same".
(c)[3]
A third run, Run 3, uses 25.0 cm³ of the same 0.400 mol/dm³ hydrogen peroxide solution, with everything else exactly as in Run 1. Predict, with reasons, how the initial rate and the final volume of oxygen in Run 3 compare with Run 1. [3]
Model Answer — 2(c)
The initial rate is the same as Run 1: the concentration, temperature and catalyst are unchanged, so collisions happen just as frequently [1]
The final volume is half Run 1's: 25.0 cm³ holds 0.0100 mol of H₂O₂, giving 0.00500 mol of O₂ = 0.00500 × 24 000 = 120 cm³ [1]
So the Run 3 curve starts along Run 1's line but levels off sooner, at 120 cm³ [1]
⚠ If you missed marks here: Half the volume does NOT mean half the rate — the rate depends on the concentration (particles per cm³), which is identical, not on how much solution there is. The amount of solution fixes only the total gas: halving 0.0200 mol of peroxide halves the plateau from 240 to 120 cm³.
Section B · Question 3 — Surface Area, Catalysts and Collision Theory
Total: 4 marks
Powdered calcium carbonate reacts far faster with dilute hydrochloric acid than the same mass of marble lumps, even though the temperature, the acid and the total mass of solid are identical.
(a)[3]
Explain, in terms of particles and collisions, why the powder reacts so much faster, and state clearly what property of the collisions does not change. [3]
Model Answer — 3(a)
Grinding exposes far more surface area — in a lump, most carbonate particles are buried inside where the acid cannot reach them [1]
Acid particles can therefore collide with exposed carbonate particles much more frequently — more collisions per second means more successful collisions per second, so a higher 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: "The powder has more energy" is the classic confusion with temperature — surface area changes how often acid particles hit exposed solid, temperature changes how hard they hit. And "more collisions" without per second misses the rate idea entirely.
(b)[1]
In Question 2 the manganese(IV) oxide was filtered off, dried and weighed after the reaction had finished. State what the balance shows, and why. [1]
Model Answer — 3(b)
The mass is unchanged — a catalyst is not used up by the reaction it speeds up (it is chemically unchanged at the end) [1]
⚠ If you missed marks here: "The catalyst is slowly used up" is exactly the misconception this part tests — a catalyst takes part in the reaction pathway but is regenerated, so the same mass of MnO₂ comes back out. That is also why a tiny amount of catalyst can decompose an unlimited amount of peroxide.
Section B · Question 4 — Reversible Reactions and Equilibrium
Total: 13 marks
Reversible reactions reach equilibrium only under the right conditions, and industry must choose conditions that balance yield, rate and cost.
(a)[2]
Explain why a reversible reaction can reach equilibrium only in a closed system. [2]
Model Answer — 4(a)
In a closed system nothing can enter or escape, so the products stay in contact with the reactants and the reverse reaction can occur; the concentrations then become constant [1]
In an open container a gaseous product escapes, so the reverse reaction cannot happen and the forward reaction simply runs to completion — equilibrium is never reached [1]
⚠ If you missed marks here: The reverse reaction needs the products to still be there — a product that escapes can never turn back. "Closed system" is not decoration in the definition of equilibrium; without it the word equilibrium does not apply at all.
(b)[3]
Colourless N₂O₄ and dark brown NO₂ reach equilibrium in a sealed gas syringe:
N₂O₄(g) ⇌ 2NO₂(g)
The plunger is pushed in, increasing the pressure at constant temperature. Explain what happens to the position of the equilibrium and to the colour of the mixture once it settles. [3]
Model Answer — 4(b)
The equilibrium shifts to the left, towards N₂O₄ [1]
Because the left side has only 1 gas molecule against 2 on the right — shifting left reduces the number of molecules and so opposes the pressure increase [1]
Brown NO₂ is converted into colourless N₂O₄, so the mixture becomes paler [1]
⚠ If you missed marks here: "Pressure shifts it right, towards the products" is the reflex this question punishes — the direction depends only on which side has FEWER gas molecules, and here that is the left. Count the molecules on each side every single time; a student who guesses "right" predicts the mixture darkens, the exact opposite of what happens.
(c)[5]
Sulfur trioxide for sulfuric acid is made by the Contact process:
(i) State the catalyst and the temperature used industrially. [1]
(ii) A lower temperature would give a higher equilibrium yield of sulfur trioxide. Explain why, and explain why a low temperature is nevertheless not used. [2]
(iii) A much higher pressure would also raise the yield. Explain why, and explain why the process actually runs at only about 2 atmospheres. [2]
Model Answer — 4(c)
Vanadium(V) oxide, V₂O₅, at about 450 °C [1]
The forward reaction is exothermic, so lowering the temperature shifts the equilibrium towards SO₃ 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 3 gas molecules on the left and only 2 on the right, so a higher pressure would shift the equilibrium towards SO₃ [1]
But the yield is already very high (about 98%) at just above atmospheric pressure, so the expense and hazard of high-pressure equipment cannot be justified — about 2 atm is enough [1]
⚠ If you missed marks here: Each compromise needs BOTH halves — the benefit and the reason it is given up. Note the pressure argument is the reverse of Haber's: there the high pressure is paid for because the yield is poor without it; here the equilibrium already sits ~98% to the right at 2 atm, so extra pressure would buy almost nothing. Quoting "200 atm" here is a straight Haber/Contact mix-up.
(d)[3]
(i) Ammonia is made industrially from nitrogen and hydrogen. Write the balanced symbol equation, with the equilibrium arrow, and name the catalyst used. [2]
(ii) Explain why most of the ammonia produced is used to make NPK fertilisers. [1]
Model Answer — 4(d)
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) [1]
Catalyst: iron (used at about 450 °C and 200 atm) [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. The catalyst is iron, not vanadium(V) oxide (that is the Contact process — the same swap trap in the other direction). For (ii), "fertilisers help plants grow" says nothing: the mark is for naming ammonia as the source of the N in NPK.
Self-Assessment
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A : 36-41
B : 30-35
C : 24-29
D : 18-23
E : 12-17
U : <12
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