S6 Chemistry Practice Paper: Equilibrium, Acids And Buffers, Electrolysis, Rates And The Haber Process

S6 Chemistry Practice Paper: Equilibrium, Acids and Buffers, Electrolysis, Rates and the Haber Process

2026-11-12

S6 Chemistry – Equilibrium, Electrolysis and Rates

S6 Chemistry practice paper with model answers on equilibrium, acids and buffers, electrolysis, rates, catalysis and the Haber process, plus titration calculations.

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Paper details　S6 Chemistry practice paper: equilibrium, acids and buffers, electrolysis, rates, catalysis and the Haber process, with full model answers.

Part 1: Chemical equilibrium (20 marks)

Four questions. Show working, with units and suitable significant figures.

#	Question	Answer and marking notes
1	At 298 K, 0.200 mol of N2O4 is put into a closed 1.0 dm3 vessel. At equilibrium the concentration of N2O4 is 0.140 mol dm-3. The reaction N2O4 to 2NO2 is reversible. Find the equilibrium concentration of NO2 and calculate Kc.	N2O4 used = 0.200 – 0.140 = 0.060 mol dm-3; NO2 = 2 x 0.060 = 0.120 mol dm-3; Kc = (0.120)2 / 0.140 = 0.103 mol dm-3 (3 s.f.).
2	Calculate the degree of dissociation of N2O4 at this temperature, as a percentage.	0.060 / 0.200 = 0.30, that is 30.0 per cent.
3	Given Kp = Kc(RT)^n, where n is the change in moles of gas and R = 0.0821 dm3 atm K-1 mol-1, find Kp at 298 K.	n = 2 – 1 = 1; RT = 0.0821 x 298 = 24.5; Kp = 0.103 x 24.5 = 2.52 atm.
4	If the vessel is compressed to half its volume at constant temperature, how does the position of equilibrium shift? Explain also the effect of raising the temperature on Kc, given that the forward reaction is exothermic.	Compression doubles all gas concentrations, so the equilibrium shifts towards the side with fewer moles of gas, that is towards N2O4, and the mole fraction of NO2 falls. As the forward reaction is exothermic, a higher temperature shifts the equilibrium backwards and Kc decreases.

Part 2: Acids, pH and buffer solutions (18 marks)

Four questions. State any approximation used and show your working.

#	Question	Answer and marking notes
5	At 25 C the Ka of a 0.100 mol dm-3 ethanoic acid solution is 1.8 x 10^-5. Calculate the pH, assuming the change in acid concentration is negligible.	Ka = x2 / 0.100, so x = sqrt(1.8e-5 x 0.100) = 1.34e-03 mol dm-3; pH = -log(1.34e-03) = 2.87.
6	A buffer is made from 0.100 mol ethanoic acid and 0.100 mol sodium ethanoate made up to 1.0 dm3. Find its pH, then the new pH after 0.010 mol of hydrochloric acid is added (volume change negligible).	pKa = -log(1.8e-5) = 4.74; acid and salt are equimolar, so pH = pKa = 4.74. After adding strong acid the acid is 0.110 mol and the salt 0.090 mol, so pH = 4.74 + log(0.090 / 0.110) = 4.66.
7	25.0 cm3 of the ethanoic acid solution is titrated with 0.100 mol dm-3 sodium hydroxide. Find the volume of alkali needed to reach the equivalence point, explain why the pH there is above 7, and calculate that pH (Kb = Kw / Ka, Kw = 1.0 x 10^-14).	Acid and alkali react 1:1, so 25.0 cm3 of NaOH is needed. At the equivalence point only sodium ethanoate remains, at 0.100 x 25.0 / 50.0 = 0.0500 mol dm-3; Kb = 1.0e-14 / 1.8e-5 = 5.56e-10; [OH-] = sqrt(Kb x concentration) = 5.27e-06 mol dm-3; pOH = 5.28 and pH = 8.72. The ethanoate ion hydrolyses to give OH-, so the solution is alkaline.
8	Explain why phenolphthalein is a suitable indicator for this titration while methyl orange is not.	The equivalence pH is about 8.7. Phenolphthalein changes over roughly 8.2 to 10.0, which covers this value, so the end point is close to the equivalence point. Methyl orange changes over about 3.1 to 4.4, far below the equivalence pH, so it changes colour before enough alkali is added and the end point volume is too small.

Part 3: Electrolysis and electrochemical cells (18 marks)

Four questions. Write the half-equations and quote the values used.

#	Question	Answer and marking notes
9	A current of 2.00 A is passed through copper(II) sulphate solution for 30.0 minutes using platinum (inert) electrodes. One faraday = 96500 C mol-1 and the relative atomic mass of copper is 63.5. Find the mass of copper deposited at the cathode.	Charge = 2.00 x 1800 = 3600 C; moles of electrons = 3600 / 96500 = 0.0373 mol; the cathode reaction Cu2+ + 2e- gives Cu, so copper = 0.0187 mol; mass = 0.0187 x 63.5 = 1.18 g.
10	In the same electrolysis, find the volume of oxygen released at the anode at room temperature (molar gas volume = 24.0 dm3 mol-1).	Anode reaction: 2H2O gives O2 + 4H+ + 4e-; moles of oxygen = 0.0373 / 4 = 9.33e-03 mol; volume = 9.33e-03 x 24.0 = 0.224 dm3, about 224 cm3.
11	A Daniell cell is made from a copper electrode (Cu2+/Cu, standard electrode potential +0.34 V) and a zinc electrode (Zn2+/Zn, standard electrode potential -0.76 V). Find the standard cell e.m.f. and state the direction of electron flow in the external circuit.	Standard e.m.f. = 0.34 – (-0.76) = 1.10 V. Zinc has the lower reduction potential and is the anode, where Zn gives Zn2+ + 2e-, so electrons flow from the zinc electrode through the external circuit to the copper electrode (cathode).
12	Explain the function of the porous separator (or salt bridge) in a cell and state the consequence of removing it.	The separator lets ions pass so that each half cell stays electrically neutral, and it keeps copper(II) ions from touching the zinc metal directly. If it is removed the two solutions mix, zinc transfers electrons straight to copper(II) ions, oxidation and reduction occur at the same place, and the cell gives no steady external current.

Part 4: Reaction rate and catalysis (14 marks)

Four questions using initial rate data at one temperature. Justify each order.

#	Question	Answer and marking notes
13	At one temperature the initial rates for A with B are: experiment 1, [A] = 0.10 and [B] = 0.10 mol dm-3, rate = 2.0 x 10^-4 mol dm-3 s-1; experiment 2, [A] = 0.20 and [B] = 0.10, rate = 8.0 x 10^-4; experiment 3, [A] = 0.10 and [B] = 0.20, rate = 4.0 x 10^-4. Find the order with respect to A and to B and write the rate equation.	Compare experiments 1 and 2: [A] doubles while [B] is fixed and the rate rises from 2.0e-4 to 8.0e-4, a factor of 4, so the order in A is two. Compare experiments 1 and 3: [B] doubles while [A] is fixed and the rate rises by a factor of 2, so the order in B is one. Rate = k[A]2[B], overall order three.
14	Use these data to calculate the rate constant k and give its units.	k = 2.0e-4 / (0.10 x 0.10 x 0.10) = 0.20; the units are dm6 mol-2 s-1, obtained by balancing the concentration powers in the rate equation against mol dm-3 s-1.
15	Another reaction is first order with k = 0.0231 s-1. Find its half-life.	For a first order reaction the half-life is ln2 / k = 0.693 / 0.0231 = 30.0 s and is independent of the initial concentration.
16	A catalyst lowers the activation energy of a reaction by 20.0 kJ mol-1. By what factor is the rate increased at 298 K? (R = 8.314 J K-1 mol-1)	From the Arrhenius equation the rate ratio is exp(Ea / RT) = exp(20000 / (8.314 x 298)) = exp(8.07), about 3205, that is roughly 3.2 x 10^3 times faster.

Part 5: Industrial processes and green chemistry (18 marks)

Four questions on the Haber process. Use N = 14.0 and H = 1.0.

#	Question	Answer and marking notes
17	The Haber process combines nitrogen and hydrogen (mole ratio 1:3) at 450 C and 200 atm over an iron catalyst: N2 + 3H2 gives 2NH3, a reversible exothermic reaction. If the conversion per pass is 15 per cent, find the moles of each gas leaving the bed when the feed is 100 mol of N2 and 300 mol of H2.	N2 reacted = 15 mol, H2 reacted = 3 x 15 = 45 mol, NH3 formed = 2 x 15 = 30 mol. Exit gas: N2 = 85 mol, H2 = 255 mol, NH3 = 30 mol, total 370 mol.
18	For the exit gas above, find the mole fraction of ammonia and the mass of ammonia formed.	Mole fraction = 30 / 370 = 0.0811, that is 8.1 per cent. The molar mass of ammonia is 14.0 + 3 x 1.0 = 17.0 g mol-1, so the mass is 30 x 17.0 = 510 g.
19	Calculate the atom economy of this reaction (mass of ammonia formed relative to the total mass of reactants) and explain why the unreacted gases are recycled industrially.	Mass of reactants = 15 x 28.0 + 45 x 2.0 = 510 g, equal to the 510 g of product, so the atom economy is 100 per cent: every atom ends up in the target product. However the conversion per pass is only 15 per cent, so recycling raises the overall use of raw materials and cuts emissions and cost.
20	Explain why industry uses 450 C rather than a lower temperature, and give two advantages of using a high pressure.	The forward reaction is exothermic and the moles of gas fall from 4 to 2, so low temperature and high pressure are favoured thermodynamically. But at too low a temperature the rate is very slow and the iron catalyst is not active enough, so 450 C balances yield against rate. High pressure raises the equilibrium yield and also raises gas concentrations, which speeds up the reaction.

Part 6: Mixed calculations (12 marks)

Four questions. Show full working with units and correct significant figures.

#	Question	Answer and marking notes
21	2.50 g of calcium carbonate (molar mass 100.1 g mol-1) is added to excess 1.00 mol dm-3 hydrochloric acid and reacts completely. Find the volume of carbon dioxide formed at room temperature (molar gas volume = 24.0 dm3 mol-1).	Moles of CaCO3 = 2.50 / 100.1 = 2.50e-02 mol; from CaCO3 + 2HCl giving CaCl2 + CO2 + H2O the moles of CO2 are the same, so the volume is 2.50e-02 x 24.0 = 0.599 dm3, about 599 cm3 (3 s.f.).
22	For the reaction above, find the minimum volume of 1.00 mol dm-3 hydrochloric acid required.	From the stoichiometry the moles of HCl = 2 x 2.50e-02 = 5.00e-02 mol; volume = 5.00e-02 / 1.00 = 0.0500 dm3, about 50 cm3.
23	25.0 cm3 of 0.100 mol dm-3 sodium hydroxide exactly neutralises 20.0 cm3 of a sulphuric acid solution. Find the concentration of the acid.	Moles of NaOH = 0.0250 x 0.100 = 2.50e-03 mol; sulphuric acid is diprotic, so half as many moles are needed, 1.25e-03 mol; concentration = 1.25e-03 / 0.0200 = 0.0625 mol dm-3.
24	To how many significant figures should the answer to question 1 be given? State the units that must appear in the working.	Both 2.50 g and 100.1 g mol-1 have 3 significant figures, so the answer should be given to 3 s.f., that is 0.599 dm3. The working must show the unit mol for the amount of substance and dm3 (or cm3) for the volume.

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