Secondary 6 Chemistry Practice Paper: Chemical Equilibrium And Kc, Acids And Buffers, Electrochemistry And Electrolysis, Organic Chemistry And Polymers, Industrial And Green Chemistry, Energetics And Rates Of Reaction (100 Marks)

Secondary 6 Chemistry Practice Paper: Chemical Equilibrium and Kc, Acids and Buffers, Electrochemistry and Electrolysis, Organic Chemistry and Polymers, Industrial and Green Chemistry, Energetics and Rates of Reaction (100 marks)

2026-11-14

Chemistry – Secondary 6 – Practice Paper

S6 chemistry practice paper: 24 original parts on equilibrium, buffer pH, electrolysis masses, organic formulae, polymers, atom economy and rates, with full answers.

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Paper details　S6 chemistry mock paper with answers: 24 parts covering equilibrium constants, buffers, electrolysis, polymers and energetics. Free printable DSE S6 practice.

Part 1 – Chemical Equilibrium and the Equilibrium Constant (17 marks)

Answer all parts. Show your working and state the units.

#	Question	Answer and marking notes
1	In a 1.0 dm3 vessel, nitrogen reacts with hydrogen: N2(g) + 3H2(g) gives 2NH3(g). At equilibrium there are 0.50 mol of N2, 0.50 mol of H2 and 0.50 mol of NH3. Calculate Kc at this temperature. (4 marks)	Answer: Concentrations are 0.50, 0.50 and 0.50 mol dm-3. Kc = 0.50 squared divided by (0.50 times 0.50 cubed) = 4.00.
2	In a second experiment at the same temperature, the equilibrium concentrations are 0.50 mol dm-3 of H2 and 1.00 mol dm-3 of NH3. Find the equilibrium concentration of N2. (5 marks)	Answer: From Kc = 4.00, [N2] = 1.00 squared divided by (4.00 times 0.50 cubed) = 2.00 mol dm-3.
3	The temperature is unchanged but the vessel is compressed from 1.0 dm3 to 0.50 dm3. State and explain the effect on the equilibrium yield of ammonia. (4 marks)	Answer: The yield increases. Raising the pressure shifts the position of equilibrium towards the side with fewer moles of gas, and the forward reaction changes 4 mol of gas into 2 mol, so more ammonia forms. Kc itself is unchanged because temperature is fixed.
4	The forward reaction is exothermic. Explain the effect of raising the temperature on Kc and on the equilibrium yield of ammonia. (4 marks)	Answer: Kc decreases and the equilibrium shifts in the endothermic reverse direction, so the equilibrium yield of ammonia falls.

Part 2 – Acids, Bases and Buffer Solutions (17 marks)

At room temperature Kw = 1.0 x 10 to the power minus 14.

#	Question	Answer and marking notes
5	Calculate the pH of 0.020 mol dm-3 hydrochloric acid. (4 marks)	Answer: [H+] = 0.020 mol dm-3, so pH = 1.70.
6	Calculate the pH of 0.050 mol dm-3 sodium hydroxide solution. (4 marks)	Answer: pOH = 1.30, so pH = 14.00 minus pOH = 12.70.
7	A buffer contains 0.10 mol dm-3 ethanoic acid and 0.20 mol dm-3 sodium ethanoate, Ka = 1.8 x 10 to the power minus 5. Calculate its pH. (4 marks)	Answer: [H+] = Ka x (acid divided by salt) = 9.0 x 10 to the power minus 6 mol dm-3, pH = 5.05.
8	What mass of solid sodium hydroxide is needed to neutralise completely 25.0 cm3 of 0.10 mol dm-3 ethanoic acid? (Na 23.0, O 16.0, H 1.0) (5 marks)	Answer: moles of acid = 0.10 x 0.0250 = 0.0025 mol; moles of NaOH are the same, molar mass 40.0 g mol-1, so mass = 0.10 g.

Part 3 – Electrochemistry and Electrolysis (16 marks)

Faraday constant = 96500 C mol-1. Standard electrode potentials: Zn2+/Zn = minus 0.76 V, Cu2+/Cu = plus 0.34 V.

#	Question	Answer and marking notes
9	A cell is made from a zinc electrode and a copper electrode. Calculate its standard cell potential. (4 marks)	Answer: E cell = 0.34 minus (minus 0.76) = 1.10 V.
10	Identify the anode in the cell and state the direction of electron flow. (3 marks)	Answer: Zinc is the anode where oxidation occurs, and electrons flow from zinc through the external circuit to copper.
11	A current of 2.0 A passes through copper sulphate solution for 30.0 minutes. Calculate the mass of copper deposited at the cathode. (Cu 63.5) (5 marks)	Answer: charge = 2.0 x 1800 = 3600 C; moles of electrons = charge divided by 96500 = 0.03731 mol; moles of copper are half of that, so mass = 1.18 g.
12	If 0.50 A is passed through silver nitrate solution for 1.00 hour, calculate the mass of silver deposited. (Ag 107.9) (4 marks)	Answer: charge = 1800 C; moles of electrons = 0.01865 mol; silver and electrons are in 1 to 1 ratio, so mass = 2.01 g.

Part 4 – Organic Chemistry and Polymers (17 marks)

Relative atomic masses: C 12.0, H 1.0, O 16.0. Molar volume of gas at room conditions = 24.0 dm3 mol-1.

#	Question	Answer and marking notes
13	A compound contains 40.0 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen by mass, and has relative molecular mass 180.0. Find its empirical and molecular formulae. (5 marks)	Answer: mole ratio C 3.33 : H 6.7 : O 3.33 gives the empirical formula CH2O with mass 30.0; 180.0 divided by 30.0 = 6, so the molecular formula is C6H12O6.
14	Using that compound, 0.10 mol is burnt completely. Calculate the volume of carbon dioxide produced at room conditions. (4 marks)	Answer: each molecule has 6 carbon atoms, so 0.10 mol gives 0.6 mol of carbon dioxide; volume = 14.4 dm3.
15	Explain the difference between addition polymerisation and condensation polymerisation, giving one example of the monomer and polymer for each. (4 marks)	Answer: in addition polymerisation the monomer contains a carbon to carbon double bond and units join without losing any small molecule, for example ethene to poly(ethene); in condensation polymerisation two monomers with functional groups join and lose a small molecule such as water, for example nylon from a diamine and a diacid.
16	Write the structural formulae of all the structural isomers of C4H10 and name them. (4 marks)	Answer: there are 2 isomers: the straight chain butane and the branched 2-methylpropane.

Part 5 – Industrial and Green Chemistry (17 marks)

Relative atomic masses: S 32.1, H 1.0, O 16.0, Ca 40.1, C 12.0.

#	Question	Answer and marking notes
17	In the Contact process, sulphur is burnt to make sulphuric acid. Assuming 100 percent conversion, calculate the mass of sulphuric acid (in tonnes) obtained from 1.00 tonne of sulphur. (5 marks)	Answer: moles of S = 1000 divided by 32.1 = 31.153 kmol; the same number of moles of acid weighs 31.153 times 98.1 = 3056.1 kg, that is 3.06 tonnes.
18	The actual output is only 2.80 tonnes. Calculate the percentage yield. (4 marks)	Answer: yield = 2.80 divided by 3.06 times 100 = 91.62 percent.
19	Limestone decomposes on heating: CaCO3 gives CaO and CO2. Calculate the atom economy with CaO as the desired product. (4 marks)	Answer: atom economy = 56.1 divided by 100.1 times 100 = 56.04 percent.
20	The Contact process uses a vanadium(V) oxide catalyst at about 450 degrees. Explain why the temperature is not raised further and state the role of the catalyst. (4 marks)	Answer: the forward reaction is exothermic, so a much higher temperature shifts the equilibrium backwards and lowers the yield while wasting energy; the catalyst provides a pathway of lower activation energy so equilibrium is reached faster, without changing its position or the final yield.

Part 6 – Energetics and Rates of Reaction (16 marks)

Specific heat capacity of water = 4.18 J g-1 K-1.

#	Question	Answer and marking notes
21	Burning 0.0100 mol of a fuel raises the temperature of 50.0 g of water by 6.5 K. Calculate the molar enthalpy change of combustion in kJ mol-1. (4 marks)	Answer: heat = 50.0 times 4.18 times 6.5 = 1358.5 J; per mole = 1358.5 divided by 0.0100 = 135.8 kJ, and since heat is released the value is minus 135.8 kJ mol-1.
22	Bond energies in kJ mol-1: hydrogen to hydrogen 436, chlorine to chlorine 243, hydrogen to chlorine 431. Calculate the enthalpy change of H2 + Cl2 giving 2HCl. (4 marks)	Answer: breaking bonds absorbs 436 plus 243 = 679 kJ; forming bonds releases 2 times 431 = 862 kJ; enthalpy change = 679 minus 862 = -183 kJ mol-1.
23	The concentration of a reactant falls from 0.80 mol dm-3 to 0.20 mol dm-3 in 40 seconds. Calculate the average rate over this period. (4 marks)	Answer: the change is 0.60 mol dm-3, so rate = 0.60 divided by 40 = 0.015 mol dm-3 s-1.
24	Explain how a catalyst speeds up a reaction and why it does not change the enthalpy change of the reaction. (4 marks)	Answer: a catalyst provides an alternative pathway with lower activation energy, so more molecules have enough energy and collisions are more successful; the enthalpy change depends only on the energies of reactants and products, which the catalyst does not alter, so the enthalpy change stays the same.

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