Secondary 6 Chemistry Practice Paper: Equilibrium And Kc, Electrochemistry And The Nernst Equation, Enthalpy And Hess Law, Reaction Kinetics, Organic Synthesis And Polymers, Solubility And Titration

Secondary 6 Chemistry Practice Paper: Equilibrium and Kc, Electrochemistry and the Nernst Equation, Enthalpy and Hess Law, Reaction Kinetics, Organic Synthesis and Polymers, Solubility and Titration

2026-11-13

Chemistry – Secondary 6 – Practice Paper

Secondary 6 chemistry practice paper with answers: Kc and Kp, cell emf, Nernst equation, enthalpy by Hess law, Arrhenius kinetics, esterification, Ksp and titration.

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Paper details　Secondary 6 chemistry practice paper with answers: equilibrium constant from equilibrium amounts, Kp from Kc, cell emf from standard electrode potentials, Gibbs energy of a cell reaction, Nernst equation with non-standard concentration, enthalpy from bond enthalpies and from standard enthalpies of formation, mass of fuel for heating water, rate equation and rate constant, Arrhenius activation energy, esterification equilibrium yield, solubility product and titration volume.

Part 1: Chemical Equilibrium and the Equilibrium Constant

Nitrogen and hydrogen form ammonia in a 2.0 dm3 vessel. At equilibrium:

#	Question	Answer and marking notes
1	At equilibrium the vessel holds 0.40 mol N2, 0.60 mol H2 and 0.80 mol NH3. Write the expression for Kc in mol dm-3 units and calculate Kc to 2 significant figures.	Answer: Kc = [NH3]^2 / ([N2][H2]^3). Concentrations are 0.20, 0.30 and 0.40 mol dm-3, so Kc = 0.160 / (0.20 x 0.0270) = 29.6 mol-2 dm6.
2	Assuming only N2 and H2 were present initially, find the initial amounts of both reactants.	Answer: Forming 0.80 mol NH3 consumes 0.40 mol N2 and 1.20 mol H2, so initial N2 = 0.80 mol and initial H2 = 1.80 mol.
3	At 700 K with Kc = 29.6 mol-2 dm6, calculate Kp in atm units using R = 0.0821 dm3 atm K-1 mol-1.	Answer: delta n = 2 – 4 = -2, so Kp = 29.6 x (0.0821 x 700)^-2 = 29.6 / 3302.8 = 8.97 x 10^-3 atm-2.
4	If the vessel is compressed to 1.0 dm3 at constant temperature, state and explain how the amount of ammonia changes.	Answer: The amount of ammonia increases. Halving the volume raises all concentrations, the reaction quotient falls below Kc, and the system shifts towards the side with fewer gas moles, which is the products.

Part 2: Electrochemistry and Cell Emf

Referred to the standard hydrogen electrode, E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V.

#	Question	Answer and marking notes
5	Write the cell reaction of a Zn/Cu cell and state which electrode is the negative terminal.	Answer: Zn(s) + Cu2+(aq) gives Zn2+(aq) + Cu(s). Zinc has the lower electrode potential, so zinc is the negative terminal and is oxidised.
6	Calculate the standard cell emf.	Answer: E cell = 0.34 – (-0.76) = 1.10 V.
7	Two moles of electrons are transferred per mole of reaction and the Faraday constant is 96500 C mol-1. Find the standard Gibbs energy change in kJ mol-1.	Answer: delta G = -2 x 96500 x 1.10 = -212300 J mol-1, that is -212.3 kJ mol-1; the negative sign shows the reaction is spontaneous.
8	If the copper ion concentration falls to 0.010 mol dm-3 with all other conditions standard, find the new cell emf to 2 decimal places using the 0.0592/n form of the Nernst equation.	Answer: Q = 1/0.010 = 100, so E = 1.10 – (0.0592/2) x 2 = 1.04 V.

Part 3: Enthalpy Change, Bond Enthalpy and Hess Law

Combustion of methane: CH4(g) + 2O2(g) gives CO2(g) + 2H2O(l).

#	Question	Answer and marking notes
9	Using bond enthalpies C-H 413, O=O 498, C=O 805 and O-H 464 kJ mol-1, calculate the enthalpy change of the reaction.	Answer: Bonds broken need 4 x 413 + 2 x 498 = 2648 kJ mol-1; bonds formed release 2 x 805 + 4 x 464 = 3466 kJ mol-1; delta H = -818 kJ mol-1.
10	Using standard enthalpies of formation of -74.8 for methane, -393.5 for carbon dioxide and -285.8 kJ mol-1 for liquid water, calculate the enthalpy change.	Answer: delta H = (-393.5 – 571.6) + 74.8 = -890.3 kJ mol-1.
11	Explain why the two values differ.	Answer: Bond enthalpies are average values for a type of bond rather than the actual bond energy in methane, and the formation route refers to liquid water which releases condensation enthalpy, so the two cannot agree.
12	Methane heats 2.0 kg of water from 25 to 75 C at an efficiency of 0.60, with water specific heat capacity 4.18 J g-1 K-1. Find the mass of methane needed using -890.3 kJ mol-1 and molar mass 16.0.	Answer: Water needs 2000 x 4.18 x 50 = 418 kJ; combustion must supply 696.7 kJ; that is 0.7825 mol, so the mass is 12.5 g.

Part 4: Reaction Kinetics and Activation Energy

For A + B giving C, three experiments at constant temperature (concentrations in mol dm-3, initial rates in mol dm-3 s-1).

#	Question	Answer and marking notes
13	Experiment 1 gives rate 2.0e-3 at 0.10 and 0.10; experiment 2 gives 8.0e-3 at 0.20 and 0.10; experiment 3 gives 4.0e-3 at 0.10 and 0.20. Find the order with respect to each reactant.	Answer: From experiments 1 and 2, doubling A quadruples the rate, so the order in A is 2; from 1 and 3, doubling B doubles the rate, so the order in B is 1.
14	Write the rate equation and give the value and unit of the rate constant.	Answer: rate = k[A]^2[B]; k = 2.0e-3 / 1.0e-3 = 2.0 dm6 mol-2 s-1.
15	Find the initial rate when A is 0.30 and B is 0.050 mol dm-3.	Answer: rate = 2.0 x 0.090 x 0.050 = 9.0e-3 mol dm-3 s-1.
16	At 300 K the rate constant is 2.0 and the activation energy is 50 kJ mol-1. With R = 8.314 J K-1 mol-1, find the rate constant at 310 K to 3 significant figures.	Answer: ln(k2/k1) = 0.647, so k2 = 2.0 x 1.909 = 3.82.

Part 5: Organic Synthesis and Polymers

An organic compound contains only carbon, hydrogen and oxygen, with mass fractions 0.400 carbon, 0.067 hydrogen and 0.533 oxygen.

#	Question	Answer and marking notes
17	Find the empirical formula of the compound.	Answer: The mole ratio is 3.33 : 6.7 : 3.33, that is 1 : 2 : 1, so the empirical formula is CH2O.
18	If the molar mass is 180 g mol-1, find the molecular formula.	Answer: The empirical formula mass is 30 and 180/30 = 6, so the molecular formula is C6H12O6.
19	The esterification of ethanol with ethanoic acid has Kc = 4.0. Starting from 0.10 mol of each in 1.0 dm3, find the amount of ester at equilibrium.	Answer: Let the ester be x mol, then x^2/(0.10 – x)^2 = 4.0, giving x/(0.10 – x) = 2 and x = 0.0667 mol.
20	Ethyl ethanoate has molar mass 88.0 g mol-1. Find the mass of ester obtained and state the type of polymerisation its monomers can undergo.	Answer: The mass is 0.0667 x 88.0 = 5.87 g. A carboxylic acid and an alcohol undergo condensation polymerisation to give a polyester.

Part 6: Solubility Product and Acid-Base Titration

At 298 K, Ksp(AgCl) = 1.8e-10 and the molar mass of AgCl is 143.4 g mol-1.

#	Question	Answer and marking notes
21	Find the solubility of AgCl in pure water in mol dm-3.	Answer: With solubility s, Ksp = s^2 = 1.8e-10, so s = 1.34e-5 mol dm-3.
22	Convert that solubility into g dm-3.	Answer: 1.34e-5 x 143.4 = 1.92e-3 g dm-3.
23	Find the solubility of AgCl in 0.010 mol dm-3 sodium chloride solution and compare it with pure water.	Answer: The common ion fixes the chloride concentration at 0.010, so s = 1.8e-8 mol dm-3, about three orders of magnitude lower than in pure water.
24	Sulfuric acid of 0.0500 mol dm-3 titrates 25.0 cm3 of 0.100 mol dm-3 sodium hydroxide. Find the acid volume at the end point and the pH of the alkali (log 2 = 0.301).	Answer: The alkali contains 2.50e-3 mol, so 1.25e-3 mol of acid is needed and the volume is 25.0 cm3. Then pOH = 1.30 and pH = 12.70.

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