S6 Chemistry Practice Paper: Equilibrium, Acids And Electrochemistry (100 Marks)

S6 Chemistry Practice Paper: Equilibrium, Acids and Electrochemistry (100 marks)

2026-11-12

Chemistry – Equilibrium, Acids and Electrochemistry

S6 Chemistry paper: 6 questions, 26 parts, 100 marks, 60 minutes. Equilibrium, acids, cells, rates, enthalpy and volumetric analysis, with full answers.

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Question 1 – Chemical Equilibrium (18 marks)

H2(g) + I2(g) = 2HI(g) proceeds at 700 K in a closed vessel of volume 1.0 dm3. Initially 1.00 mol of H2 and 1.00 mol of I2 are placed in the vessel. At equilibrium the vessel contains 1.20 mol of HI.

#	Question	Answer and marking notes
1	(a) Find the equilibrium concentration of H2 and the percentage conversion of H2.	Answer: Let x mol of H2 react. Since 1.20 mol of HI is formed, x = 1.20 / 2 = 0.60 mol. At equilibrium H2 remains 1.00 – 0.60 = 0.40 mol, so its concentration is 0.40 mol dm-3. The conversion is 0.60 / 1.00 x 100 = 60 per cent. Marking: x value 2, concentration 2, conversion 2.
2	(b) Calculate the equilibrium constant Kc at 700 K and state its unit.	Answer: At equilibrium [H2] = [I2] = 0.40 mol dm-3 and [HI] = 1.20 mol dm-3. Kc = [HI]2 / ([H2][I2]) = 1.44 / 0.16 = 9.0. Since the number of gas moles is unchanged, Kc has no unit. Marking: expression 2, value 1, unit 1.
3	(c) If the vessel is compressed to 0.50 dm3, what happens to the equilibrium yield of HI? Explain.	Answer: The equilibrium yield of HI stays unchanged. The total number of gas moles is the same on both sides, so a change in pressure does not shift the position of equilibrium; the amounts stay constant and only the concentrations double. Marking: conclusion 2, equal moles 2.
4	(d) A second experiment shows that Kc falls to 4.0 at 800 K. Decide whether the forward reaction is exothermic or endothermic and justify your answer.	Answer: The forward reaction is exothermic. Raising the temperature lowers Kc from 9.0 to 4.0, so the equilibrium shifts backwards, meaning the reverse direction is favoured at higher temperature. By Le Chatelier principle heating favours the endothermic direction, so the reverse reaction is endothermic and the forward one is exothermic. Marking: decision 2, Kc evidence 1, principle 2.

Question 2 – Weak Acids and Acid-Base Titration (17 marks)

An ethanoic acid solution of concentration 0.10 mol dm-3 has Ka = 1.8 x 10-5 mol dm-3 at 298 K. A 25.0 cm3 portion of this solution is titrated with 0.10 mol dm-3 sodium hydroxide solution.

#	Question	Answer and marking notes
5	(a) Calculate the hydrogen ion concentration and the pH of the ethanoic acid solution.	Answer: [H+] = square root of (Ka x c) = square root of (1.8 x 10-5 x 0.10) = 1.34 x 10-3 mol dm-3. pH = -log(1.34 x 10-3) = 2.87. Marking: square root approximation 2, [H+] 2, pH 2.
6	(b) Find the volume of sodium hydroxide solution needed to reach the equivalence point.	Answer: Ethanoic acid and sodium hydroxide react in a 1 to 1 ratio, so 0.10 x 25.0 = 0.10 x V and V = 25.0 cm3. Marking: equal moles 2, volume 2.
7	(c) Find the pH of the mixture after 12.5 cm3 of sodium hydroxide solution has been added.	Answer: Half of the equivalence amount of alkali has been added, so equal concentrations of ethanoic acid and ethanoate ion are present. Then pH = pKa = -log(1.8 x 10-5) = 4.74. Marking: half equivalence identified 2, pH equals pKa 1, value 1.
8	(d) Explain why the pH at the equivalence point of this titration is above 7.	Answer: At the equivalence point the solute is sodium ethanoate. The ethanoate ion hydrolyses, CH3COO- + H2O = CH3COOH + OH-, producing hydroxide ions, so the solution is alkaline and the pH is above 7. Marking: sodium ethanoate 2, hydrolysis giving OH- 2.

Question 3 – Electrochemical Cells (17 marks)

A cell uses zinc and copper electrodes. The zinc electrode dips into 1.0 mol dm-3 zinc nitrate solution and the copper electrode into 1.0 mol dm-3 copper(II) nitrate solution; the two solutions are joined by a salt bridge at 298 K. Standard electrode potentials: Zn2+/Zn = -0.76 V and Cu2+/Cu = +0.34 V.

#	Question	Answer and marking notes
9	(a) Write the half equations at both electrodes and state which electrode is the anode.	Answer: Anode (oxidation): Zn(s) -> Zn2+(aq) + 2e-. Cathode (reduction): Cu2+(aq) + 2e- -> Cu(s). The zinc electrode is the anode. Marking: oxidation half equation 2, reduction half equation 1, anode named 1.
10	(b) Calculate the standard electromotive force of the cell.	Answer: E cell = E cathode – E anode = 0.34 – (-0.76) = 1.10 V. Marking: formula 1, substitution 1, answer 1.
11	(c) Describe the direction of electron flow in the external circuit and the movement of ions in the solutions when the cell discharges.	Answer: Electrons flow from the zinc electrode through the external circuit to the copper electrode. In solution cations such as Zn2+ and Cu2+ move towards the copper cathode while anions such as NO3- move through the salt bridge towards the zinc anode, keeping the solutions electrically neutral. Marking: electron direction 2, cations 1, anions 1.
12	(d) If 0.020 mol of electrons passes through the external circuit, find the change in mass of the copper electrode.	Answer: From Cu2+ + 2e- -> Cu, 2 mol of electrons deposit 1 mol of copper, so moles of copper = 0.020 / 2 = 0.010 mol and the mass increases by 0.010 x 63.5 = 0.635 g. Marking: stoichiometry 1, moles 1, mass 1.
13	(e) Explain the function of the salt bridge. What happens if it is removed?	Answer: The electrolyte in the salt bridge allows ions to migrate and neutralises the charge building up in each half cell, keeping both solutions neutral so the cell can keep discharging. Without it the charge cannot be balanced, the internal circuit is broken and no current flows. Marking: ion migration 1, neutrality 1, consequence 1.

Question 4 – Rates of Reaction (16 marks)

Hydrogen peroxide decomposes in the presence of a catalyst: 2H2O2(aq) -> 2H2O(l) + O2(g). In one experiment the concentration of H2O2 is 0.800 mol dm-3 at 0 s and falls to 0.320 mol dm-3 at 200 s, with the temperature kept constant.

#	Question	Answer and marking notes
14	(a) Calculate the average rate of decomposition of H2O2 over the first 200 s.	Answer: Rate = (0.800 – 0.320) / 200 = 0.480 / 200 = 2.40 x 10-3 mol dm-3 s-1. Marking: concentration change 2, divided by time 1, unit 1.
15	(b) Find the average rate of formation of oxygen over the same period.	Answer: From the stoichiometry, 2 mol of H2O2 give 1 mol of O2, so the rate of formation of oxygen = 2.40 x 10-3 / 2 = 1.20 x 10-3 mol dm-3 s-1. Marking: 1 to 2 relationship 2, value with unit 1.
16	(c) Adding manganese(IV) oxide makes the reaction finish in a shorter time. Explain using collision theory.	Answer: Manganese(IV) oxide acts as a catalyst and provides an alternative pathway of lower activation energy. At the same temperature a larger fraction of molecules has enough energy to pass the barrier, so the frequency of effective collisions rises and the reaction is faster; the enthalpy change is unchanged. Marking: lower activation energy 2, more effective collisions 1, enthalpy unchanged 1.
17	(d) Raising the temperature by 10 K roughly doubles the rate. Explain, and state how this differs from the action of a catalyst.	Answer: Heating raises the average kinetic energy of the molecules, so the energy distribution shifts to higher energies and flattens; the fraction of molecules with energy above the activation energy rises sharply and effective collisions roughly double. Unlike a catalyst, heating raises the energy of the molecules rather than lowering the barrier, yet both increase the rate. Marking: kinetic energy 2, distribution 2, comparison 1.

Question 5 – Enthalpy Change and Calorimetry (16 marks)

Given these enthalpies of combustion: C(s) + O2(g) -> CO2(g) with dH = -394 kJ mol-1, and CO(g) + half O2(g) -> CO2(g) with dH = -283 kJ mol-1. In another experiment, burning 5.00 g of a fuel raises the temperature of 200 g of water by 20.0 degrees Celsius; the specific heat capacity of water is 4.2 J g-1 K-1.

#	Question	Answer and marking notes
18	(a) Use Hess law to find the standard enthalpy change of C(s) + half O2(g) -> CO(g).	Answer: Subtract the second equation from the first: C(s) + O2 -> CO2 minus (CO + half O2 -> CO2) gives C(s) + half O2 -> CO. Hence dH = (-394) – (-283) = -111 kJ mol-1. Marking: combination 2, sign handling 2, value with unit 1.
19	(b) Use the calorimetry data to find the heat released per gram of the fuel.	Answer: Heat absorbed by water = 200 x 4.2 x 20.0 = 16800 J = 16.8 kJ. Heat released per gram = 16.8 / 5.00 = 3.36 kJ g-1. Marking: q = mc dT 2, conversion to kJ 2, divided by mass 2.
20	(c) Explain why experimental values are usually lower than literature values.	Answer: Some heat is lost to the surroundings and absorbed by the container, combustion may be incomplete, and some fuel may vaporise without burning, so the measured heat is too small and the calculated value is below the literature value. Marking: heat loss 1, incomplete combustion 1, other factor 1.
21	(d) State the definition of the standard enthalpy of formation of CO(g).	Answer: The enthalpy change when 1 mol of CO(g) is formed from its elements in their standard states at a specified temperature, that is the enthalpy change of C(s) + half O2(g) -> CO(g). Marking: 1 mol of product 1, elements in standard states 1.

Question 6 – Volumetric Analysis (16 marks)

A 25.0 cm3 portion of sodium carbonate solution is titrated with standard 0.100 mol dm-3 hydrochloric acid using methyl orange as indicator; the mean titre is 22.40 cm3. The equation is Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g).

#	Question	Answer and marking notes
22	(a) Calculate the number of moles of hydrochloric acid used.	Answer: Moles = 0.100 x 22.40 / 1000 = 2.24 x 10-3 mol. Marking: conversion to dm3 2, multiply by concentration 1.
23	(b) Calculate the number of moles of sodium carbonate in the portion.	Answer: From the equation 1 mol of Na2CO3 reacts with 2 mol of HCl, so moles of sodium carbonate = 2.24 x 10-3 / 2 = 1.12 x 10-3 mol. Marking: 1 to 2 ratio 3, value 1.
24	(c) Find the concentration of the sodium carbonate solution.	Answer: Concentration = 1.12 x 10-3 / (25.0 / 1000) = 1.12 x 10-3 / 0.0250 = 0.0448 mol dm-3. Marking: volume conversion 2, division 1, unit 1.
25	(d) Explain why methyl orange is suitable here but phenolphthalein is not.	Answer: At the equivalence point the solution contains sodium chloride and dissolved carbon dioxide, so it is weakly acidic with pH about 4 to 5, which lies inside the methyl orange range. Phenolphthalein changes at pH 8 to 10, before the equivalence point, so the end point would come early. Marking: acidic equivalence point 2, range comparison 1.
26	(e) If the burette were rinsed with distilled water only and not with the acid, how would the calculated concentration be affected?	Answer: The residual water dilutes the acid, so a larger titre is needed, the calculated moles of acid are too high and the concentration of sodium carbonate comes out too high. Marking: acid diluted 1, concentration too high 1.

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