4.5 Energy changes — revision question pack

5 specification points · notes, questions, answers and worked methods

Checked against AQA 8462 section 4.5. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

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4.5.1.1 · Energy transfer during exothermic and endothermic reactions

Explanation

An exothermic reaction transfers energy to the surroundings, so the temperature of the surroundings increases; combustion, many oxidation reactions and neutralisation are examples. An endothermic reaction takes in energy from the surroundings, so the temperature of the surroundings decreases; thermal decomposition and some instant cold packs are examples.
Measure the initial temperature, mix the reactants, stir and record the highest or lowest temperature reached; compare temperature changes while keeping quantities and apparatus controlled.
Energy is conserved: a temperature rise does not mean energy was created.
A common error is to describe the reacting chemicals, rather than the surroundings, as getting hotter or colder.
A fair comparison needs insulated, identical apparatus and consistent reactant amounts so unwanted heat transfer does not dominate the result.

Worked example

A student compares two neutralisation reactions. For each test, the student uses the same cup and the same total volume of solution. Give three other features of the method that should be kept the same or carried out consistently so the temperature changes can be compared fairly.

1.Choose controls that could otherwise alter the measured temperature change: the initial temperature, solution concentrations, mixing or stirring, and the measuring instrument or rule for selecting the maximum temperature. Any three valid consistent features score.

Answer: Use solutions with the same starting temperature and stated concentrations, stir in the same way, and record the maximum temperature using the same thermometer or temperature probe.

Common mistakes

Don't fall into the trap of describing the reacting chemicals, rather than the surroundings, as getting hotter or colder.
Don't fall into the trap of using the final temperature alone instead of calculating and comparing the temperature change.

Exam tip

For required practical 4, state how temperature change is measured and identify the quantities and apparatus kept constant.

Tier 1 · Easy

A reaction mixture starts at $19.6\,^\circ\text{C}$ and reaches $27.1\,^\circ\text{C}$ . State whether the reaction is exothermic or endothermic.
[1 mark]
Total for this question: 1
During a demonstration, the liquid warms from $17.8\,^\circ\text{C}$ to $24.6\,^\circ\text{C}$ . Calculate the temperature change and state the direction of energy transfer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An instant cold pack falls from $21.0\,^{\circ}\text{C}$ to $8.5\,^{\circ}\text{C}$ when its chemicals mix. Classify the process and describe the direction of energy transfer.
[2 marks]
Total for this question: 2
Two reactions use equal masses of similar aqueous solutions in identical insulated cups. Test R changes from $18.4\,^\circ\text{C}$ to $29.7\,^\circ\text{C}$ ; test S changes from $21.3\,^\circ\text{C}$ to $30.1\,^\circ\text{C}$ . Calculate both temperature changes and determine which test transfers more energy to the surroundings.
[3 marks]
Total for this question: 3
A student says that a temperature rise during neutralisation shows that the reaction created energy. Explain why this statement is incorrect. Refer to the reacting chemicals and the surroundings.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Two reusable hand warmers are tested from the same room temperature. Warmer P raises the temperature by $18\,^\circ\text{C}$ for $11$ minutes and costs $\pounds1.70$ per use. Warmer Q raises it by $12\,^\circ\text{C}$ for $34$ minutes and costs $\pounds0.95$ per use. Evaluate which warmer is more suitable for a walker who needs gentle heating for a $30$ -minute journey.
[4 marks]
Total for this question: 4
A group claims reaction U transfers more energy than reaction V because U reaches $34.2\,^\circ\text{C}$ whereas V reaches $31.8\,^\circ\text{C}$ . U starts at $22.7\,^\circ\text{C}$ and uses $75\,\text{cm}^3$ of solution. V starts at $18.6\,^\circ\text{C}$ and uses $25\,\text{cm}^3$ . Evaluate the claim and give two changes needed for a fair comparison.
[6 marks]
Total for this question: 6
Student P measures a temperature change of $7\,^\circ\text{C}$ using a thermometer marked in $1\,^\circ\text{C}$ divisions. Student Q measures $7.4\,^\circ\text{C}$ using a temperature probe that reads to $0.1\,^\circ\text{C}$ . Compare the precision of the measurements, explain why the two results can agree, and suggest an apparatus change that would make the measured change closer to the true change.
[5 marks]
Total for this question: 5
Equal amounts of reactants are tested in identical insulated cups. Every test starts at $20.0\,^\circ\text{C}$ . Reaction A reaches $27.4$ , $27.6$ and $27.5\,^\circ\text{C}$ in three repeats. Reaction B reaches $14.1$ , $14.3$ and $18.8\,^\circ\text{C}$ . Identify the anomalous result, calculate a mean temperature change for each reaction using the consistent results, and compare the two reactions.
[6 marks]
Total for this question: 6
A reaction starts at $18.6\,^\circ\text{C}$ and gives a measured temperature rise of $8.4\,^\circ\text{C}$ . A calibration test shows that this apparatus records only $70\%$ of the true temperature change because of unwanted energy transfer. Calculate the estimated true temperature change and maximum temperature, identify the reaction type, and suggest one change that would reduce the correction needed.
[5 marks]
Total for this question: 5

4.5.1.2 · Reaction profiles

Explanation

A reaction profile plots energy against progress of reaction; the curved line rises to a maximum before falling or rising to the products' energy level. Activation energy is the minimum energy that colliding particles must have for a reaction to occur, shown from the reactants' energy level to the peak.
Products below reactants indicate an exothermic reaction and a negative overall energy change; products above reactants indicate an endothermic reaction and a positive change.
A common error is to draw activation energy from the vertical axis or from the products.
For the forward reaction, measure it from the reactants' level to the peak.
The sign of the overall energy change follows products minus reactants, whereas the activation-energy arrow always rises to the peak.

An exothermic reaction profile with products below reactants and activation energy measured to the peak.

Worked example

A reaction profile places the reactants at $74\,\text{kJ mol}^{-1}$ , the peak at $139\,\text{kJ mol}^{-1}$ and the products at $102\,\text{kJ mol}^{-1}$ . Calculate the activation energy and the overall energy change, then identify whether the reaction is exothermic or endothermic.

1.The forward activation energy is peak minus reactants: $139-74=65\,\text{kJ mol}^{-1}$ . The overall change is products minus reactants: $102-74=+28\,\text{kJ mol}^{-1}$ . The positive change, with products above reactants, identifies an endothermic reaction.

Answer: Activation energy $=65\,\text{kJ mol}^{-1}$ ; overall energy change $=+28\,\text{kJ mol}^{-1}$ ; the reaction is endothermic.

Common mistakes

Don't fall into the trap of drawing activation energy from the vertical axis or from the products instead of from the reactants' level to the peak.
Don't fall into the trap of reversing exothermic and endothermic profiles by ignoring the relative product and reactant energy levels.

Exam tip

Label reactants, products, activation energy and overall energy change; activation energy starts at the reactant level.

Tier 1 · Easy

On a reaction profile, the products are at a lower energy level than the reactants. Identify the type of reaction.
[1 mark]
Total for this question: 1
State what activation energy means and identify the two points joined by its arrow on a forward reaction profile.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Profile X has products below its reactants. Profile Y has products above its reactants. Identify the reaction type shown by each profile.
[2 marks]
Total for this question: 2
A blank graph has energy on its vertical axis and reaction progress on its horizontal axis. Describe how to complete it as an exothermic reaction profile, including the activation energy.
[4 marks]
Total for this question: 4
A reaction profile places the reactants at $90\,\text{kJ mol}^{-1}$ , the peak at $210\,\text{kJ mol}^{-1}$ and the products at $60\,\text{kJ mol}^{-1}$ . A student calls the reaction endothermic because the peak is above the reactants. Explain why the student is incorrect.
[3 marks]
Total for this question: 3

Tier 3 · Hard

The reactants, peak and products on a reaction profile have relative energies of $181$ , $337$ and $126\,\text{kJ mol}^{-1}$ respectively. Determine the forward activation energy, the reverse activation energy and the overall energy change for the forward reaction.
[5 marks]
Total for this question: 5
A student intends to draw an endothermic profile. The products are placed below the reactants, the activation-energy arrow starts on the vertical axis, and the overall energy-change arrow points down from reactants to products. Identify each error and give the correct version.
[4 marks]
Total for this question: 4
A student says every collision between the reactant particles produces products because the products lie below the reactants on the reaction profile. Explain why this statement is incorrect and relate the condition for reaction to the profile.
[4 marks]
Total for this question: 4
A reaction profile has its reactants at $72$ relative energy units. Its forward activation energy is $138$ units and its overall energy change is $+46$ units. Calculate the energy levels of the peak and products, calculate the reverse activation energy, and identify the reaction type.
[5 marks]
Total for this question: 5
Profile P has reactants, peak and products at $80$ , $164$ and $52$ relative energy units. Profile Q has them at $65$ , $121$ and $97$ units. Experiment X makes the surroundings warmer; experiment Y makes them cooler. Match each experiment to a profile, calculate both forward activation energies and both overall energy changes, then compare their magnitudes.
[6 marks]
Total for this question: 6

4.5.1.3 · The energy change of reactions (HT only)

Explanation

Higher tier: breaking bonds in reactants requires energy, whereas forming bonds in products releases energy. Calculate the overall energy change using $\Delta E=\sum E(\text{bonds broken})-\sum E(\text{bonds formed})$ and include every bond shown by the balanced equation.
If bond formation releases more energy than bond breaking requires, $\Delta E$ is negative and the reaction is exothermic; the reverse balance gives a positive, endothermic change.
A common error is to reverse the subtraction or count molecules instead of bonds.
Multiply each bond energy by the number of that bond broken or formed.
Higher tier: a negative result means more energy is released during bond formation than is supplied for bond breaking.

Worked example

Higher tier: use bond energies H-H $=436$ , Cl-Cl $=243$ and H-Cl $=431\,\text{kJ mol}^{-1}$ to calculate the energy change for $\mathrm{H_2+Cl_2\rightarrow2HCl}$ .

1.Bonds broken: one H-H and one Cl-Cl, so $436+243=679\,\text{kJ mol}^{-1}$ .
2.Bonds formed: two H-Cl, so $2\times431=862\,\text{kJ mol}^{-1}$ .
3.Calculate $\Delta E=679-862=-183\,\text{kJ mol}^{-1}$ .

Answer: $-183\,\text{kJ mol}^{-1}$ ; the reaction is exothermic.

Common mistakes

Don't fall into the trap of reversing the subtraction or counting molecules instead of every bond broken and formed.
Don't fall into the trap of using the number of molecules as the bond count without multiplying by every bond within each molecule.

Exam tip

Higher tier: tabulate every bond broken and formed, total each column, then calculate broken minus formed with the sign.

Tier 1 · Easy

State whether energy is taken in or released when a bond is broken.
[1 mark]
Total for this question: 1
State whether bond formation takes in or releases energy. Then compare the energy needed for bond breaking with the energy released by bond formation in an endothermic reaction.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Use the equation $\mathrm{CH_4+2O_2\rightarrow CO_2+2H_2O}$ and these bond energies in $\text{kJ mol}^{-1}$ : $\mathrm{C-H}=413$ , $\mathrm{O=O}=498$ , $\mathrm{C=O}$ in carbon dioxide $=805$ , and $\mathrm{O-H}=464$ . Calculate the overall energy change.
[4 marks]
Total for this question: 4
Use the displayed change $\mathrm{H_3C-CH_3\rightarrow H_2C=CH_2+H-H}$ and these bond energies in $\text{kJ mol}^{-1}$ : $\mathrm{C-C}=347$ , $\mathrm{C=C}=612$ , $\mathrm{C-H}=413$ and $\mathrm{H-H}=436$ . Calculate the overall energy change.
[4 marks]
Total for this question: 4
Explain how a reaction can be exothermic even though energy must be supplied to break every bond in the reactants.
[3 marks]
Total for this question: 3

Tier 3 · Hard

For $\mathrm{N_2+3H_2\rightarrow2NH_3}$ , the overall energy change is $-92\,\text{kJ mol}^{-1}$ . The bond energies are $\mathrm{N\equiv N}=945\,\text{kJ mol}^{-1}$ and $\mathrm{H-H}=436\,\text{kJ mol}^{-1}$ . Calculate the mean $\mathrm{N-H}$ bond energy.
[5 marks]
Total for this question: 5
Hydrogen peroxide decomposes as $\mathrm{2H-O-O-H\rightarrow2H-O-H+O=O}$ . The overall energy change is $-196\,\text{kJ mol}^{-1}$ and the $\mathrm{O=O}$ bond energy is $498\,\text{kJ mol}^{-1}$ . Use the unchanged $\mathrm{O-H}$ bonds to calculate the mean $\mathrm{O-O}$ bond energy.
[5 marks]
Total for this question: 5
For $\mathrm{N_2+O_2\rightarrow2NO}$ , use bond energies $\mathrm{N\equiv N}=945$ , $\mathrm{O=O}=498$ and $\mathrm{N=O}=631\,\text{kJ mol}^{-1}$ . A student calculates $945+498-631=+812\,\text{kJ mol}^{-1}$ and concludes that the reaction is exothermic. Identify the bond-counting error, calculate the correct overall energy change and determine the reaction type.
[4 marks]
Total for this question: 4
Higher Tier: methane reacts by $\mathrm{CH_4+X_2\rightarrow CH_3X+HX}$ . The unchanged bonds may be cancelled. For chlorine, use $\mathrm{C-H}=413$ , $\mathrm{Cl-Cl}=243$ , $\mathrm{C-Cl}=328$ and $\mathrm{H-Cl}=431\,\text{kJ mol}^{-1}$ . For iodine, use $\mathrm{I-I}=151$ , $\mathrm{C-I}=240$ and $\mathrm{H-I}=299\,\text{kJ mol}^{-1}$ . Calculate the energy change for each reaction and choose which is more exothermic.
[6 marks]
Total for this question: 6
Higher Tier: for $\mathrm{2CO+O_2\rightarrow2CO_2}$ , an experiment gives an overall energy change of $-566\,\text{kJ mol}^{-1}$ . Use $\mathrm{O=O}=498$ and the $\mathrm{C=O}$ bond in carbon dioxide $=805\,\text{kJ mol}^{-1}$ to calculate the mean $\mathrm{C\equiv O}$ bond energy in carbon monoxide. A second table lists $\mathrm{C\equiv O}=1072\,\text{kJ mol}^{-1}$ ; calculate its predicted energy change and evaluate which table agrees with the experiment.
[6 marks]
Total for this question: 6

4.5.2.1 · Cells and batteries (chemistry only)

Explanation

A simple cell uses two different metals in contact with an electrolyte; chemical reactions transfer energy electrically and produce a potential difference. The voltage depends on the electrode materials and the electrolyte.
Data about relative metal reactivity can be used to compare or predict cell voltages. Cells connected in series have their voltages added; a battery contains two or more cells connected together in series to provide a greater voltage.
Non-rechargeable cells stop when a reactant is used up.
Rechargeable cells use an external current to reverse the reactions; a common error is to claim that recharging creates new reactants from nothing.
A larger reactivity difference between suitable electrodes generally produces a larger potential difference, but the electrolyte also matters.

Worked example

A cell made from zinc and copper produces $1.08\,\text{V}$ . Three identical cells are connected in series. Calculate the battery voltage and explain why copper-copper electrodes would not make the same cell.

1.Series cell voltages add, so $V=3\times1.08=3.24\,\text{V}$ . A simple cell requires different electrode materials; using copper for both removes that difference and would not reproduce the zinc-copper potential difference.

Answer: $3.24\,\text{V}$ ; identical copper electrodes do not provide the required difference between electrode materials.

Common mistakes

Don't fall into the trap of claiming that recharging creates new reactants from nothing instead of reversing the chemical reactions.
Don't fall into the trap of assuming the more reactive metal must always be labelled positive without using the stated cell arrangement or data.

Exam tip

For a series battery, add cell voltages with their directions; for evaluation, link lifetime and recharging to the chemical reactions.

Tier 1 · Easy

State the two essential electrode features needed to make a simple chemical cell with an electrolyte.
[2 marks]
Total for this question: 2
Three $0.85\,\text{V}$ cells are connected in series in the same direction. Calculate the battery voltage.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Simple cells using the same electrolyte give these voltages: zinc-copper $1.10\,\text{V}$ , zinc-iron $0.60\,\text{V}$ and iron-copper $0.50\,\text{V}$ . Choose the electrode pair for the largest voltage and explain what the data suggest about electrode reactivity.
[3 marks]
Total for this question: 3
A student investigates how electrolyte type affects cell voltage. The first cell uses zinc and copper in sodium chloride solution; the second uses magnesium and copper in dilute acid. Explain why this comparison is invalid and describe how to correct it.
[3 marks]
Total for this question: 3
Explain what happens chemically when a non-rechargeable alkaline cell runs down, and explain how supplying an external electrical current allows a rechargeable cell to be used again.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A torch needs at least $3.6\,\text{V}$ . Cell P is non-rechargeable, gives $1.5\,\text{V}$ and costs $\pounds0.60$ . Cell Q is rechargeable, gives $1.2\,\text{V}$ , costs $\pounds3.50$ and can provide $400$ evening-use cycles. Evaluate which type is more suitable for a torch used every evening. Assume each new P cell lasts one evening.
[5 marks]
Total for this question: 5
A holder contains four $1.35\,\text{V}$ cells. Three cells point in one direction and the fourth is reversed. Calculate the terminal voltage, explain whether it can operate a $5.0\,\text{V}$ sensor, and calculate the voltage after correcting the reversed cell.
[4 marks]
Total for this question: 4
Four cells are tested under identical conditions. A zinc-copper cell gives $1.10\,\text{V}$ in sodium chloride solution and $1.35\,\text{V}$ in dilute acid. A magnesium-copper cell gives $2.60\,\text{V}$ in sodium chloride solution and $2.85\,\text{V}$ in dilute acid. Calculate the effect of changing the electrolyte for each electrode pair and the effect of replacing zinc with magnesium in each electrolyte. Evaluate the claim that electrode choice has the greater effect on voltage, using only these data.
[6 marks]
Total for this question: 6
Metals J, K, L and M are listed in decreasing reactivity. In one electrolyte, J-K gives $0.72\,\text{V}$ , K-L gives $0.48\,\text{V}$ and L-M gives $0.36\,\text{V}$ . Assume the voltage gaps add along this reactivity order. Predict the voltages of J-L and J-M cells, then determine whether three identical J-M cells in series can operate a $4.5\,\text{V}$ device.
[6 marks]
Total for this question: 6
A student tests whether moving fixed zinc and copper electrodes farther apart changes cell voltage. At $1.0\,\text{cm}$ separation the repeat readings are $1.08$ , $1.10$ and $1.09\,\text{V}$ ; at $4.0\,\text{cm}$ they are $1.07$ , $1.11$ and $1.09\,\text{V}$ . Calculate the mean and range for each separation, evaluate the claim that greater separation increases voltage, and state one variable that must be controlled.
[6 marks]
Total for this question: 6

4.5.2.2 · Fuel cells (chemistry only)

Explanation

A fuel cell receives a continuous external supply of fuel and oxygen or air; the fuel is oxidised electrochemically to produce a potential difference. In a hydrogen fuel cell, hydrogen is oxidised and the overall reaction forms water: $\mathrm{2H_2+O_2\rightarrow2H_2O}$ .
Hydrogen fuel cells can operate while reactants are supplied, whereas rechargeable cells store reactants and must be recharged; comparisons should include storage, refuelling, lifetime and environmental effects.
Higher tier: alkaline-cell half-equations may be written $\mathrm{2H_2+4OH^-\rightarrow4H_2O+4e^-}$ and $\mathrm{O_2+2H_2O+4e^-\rightarrow4OH^-}$ ; atoms and charge must balance.
Do not call hydrogen automatically pollution-free without considering its production.
Higher tier: the electrode half-equations must cancel electrons and hydroxide ions to give the overall formation of water.

Worked example

Give two differences between a hydrogen fuel cell and a rechargeable cell when each is used to power a vehicle.

1.Make paired comparisons rather than listing isolated facts. Contrast external fuel supply and refuelling with stored reactants and electrical recharging, then give a valid consequence such as operating emissions, infrastructure, mass or recharge time.

Answer: A fuel cell needs continuing supplies of hydrogen and oxygen and can be refuelled, whereas a rechargeable cell stores its reactants and needs an external current to reverse its reactions. The fuel cell produces water during use, but the overall environmental impact depends on hydrogen production and storage.

Common mistakes

Don't fall into the trap of calling hydrogen automatically pollution-free without considering the source of hydrogen and energy used in its production.
Don't fall into the trap of treating a fuel cell as a rechargeable battery that stores a fixed quantity of reactants inside.

Exam tip

Higher tier: balance atoms and total charge in each alkaline hydrogen-fuel-cell half-equation before combining them.

Tier 1 · Easy

Hydrogen and oxygen are supplied to a fuel cell. Name the substance formed.
[1 mark]
Total for this question: 1
State which substance is oxidised inside a hydrogen fuel cell.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A hydrogen fuel-cell vehicle produces only water at the point of use. Explain why this does not prove that using the vehicle has no environmental impact.
[3 marks]
Total for this question: 3
A fuel cell powers a remote monitoring station. Explain why electrical recharging is unnecessary while its supplies remain connected, and why the cell stops if the hydrogen tank becomes empty.
[3 marks]
Total for this question: 3
During a test, a hydrogen fuel cell completely reacts $3.50\,\text{g}$ of hydrogen with $28.0\,\text{g}$ of oxygen. Water is the only substance produced. Calculate the mass of water and explain why it is greater than the initial mass of hydrogen.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A hydrogen system stores $118\,\text{MJ}$ per kilogram of hydrogen and delivers $52\%$ of this as useful electrical energy. A rechargeable battery stores $0.90\,\text{MJ}$ per kilogram and delivers $84\%$ usefully. Evaluate the two systems for a long-distance vehicle. Include calculated useful energies per kilogram and one environmental limitation of hydrogen.
[6 marks]
Total for this question: 6
A mountain rescue station needs emergency power during outages that may last $10$ hours. A hydrogen fuel-cell unit can operate while hydrogen and oxygen are supplied. A rechargeable lithium-ion battery stores its reactants and must be recharged using an external electrical current. Evaluate which system is more suitable for the station by comparing operating time, storage or recharging, and environmental impact.
[5 marks]
Total for this question: 5
A storage system receives $80.0\,\text{MJ}$ of surplus renewable electricity. Making hydrogen stores $68\%$ of this energy, and a fuel cell later converts $57\%$ of the stored energy back to electricity. A rechargeable battery returns $83\%$ of the input energy. Calculate the electrical energy returned by each route, using unrounded values in the hydrogen chain. Evaluate which route is more suitable when the priority is to return as much electrical energy as possible, and give one possible advantage of the hydrogen route.
[6 marks]
Total for this question: 6
A hydrogen fuel cell completely reacts $4.80\,\text{g}$ of hydrogen with $38.4\,\text{g}$ of oxygen, forming only water. A student collects $40.6\,\text{g}$ of water and claims that all the product was collected. Calculate the expected water mass, the uncollected mass and the percentage collected, then evaluate the claim.
[5 marks]
Total for this question: 5
A fuel-cell car uses $0.80\,\text{kg}$ of hydrogen for a journey. Producing $1.0\,\text{kg}$ of hydrogen uses $52\,\text{kWh}$ of electricity. A battery car needs $28\,\text{kWh}$ of charging electricity for the same journey. The electricity source releases $0.15\,\text{kg}$ of carbon dioxide per $\text{kWh}$ . Calculate the electricity used and carbon dioxide released for each route, then evaluate the claim that the fuel-cell journey has no carbon dioxide impact because its only product is water.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.5.1.1 · Energy transfer during exothermic and endothermic reactions

Tier 1 · Easy

Mark scheme for 4.5.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Exothermic.	The surroundings have warmed by $27.1-19.6=7.5\,^\circ\text{C}$ . A reaction that transfers energy to the surroundings is exothermic.	1
Total Question 1			1
02.1	$6.8\,^\circ\text{C}$ ; energy is transferred from the reacting chemicals to the surroundings.	The temperature change is $24.6-17.8=6.8\,^\circ\text{C}$ . The surroundings warm, so the reaction transfers energy to the surroundings.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The process is endothermic. Energy is transferred from the surroundings to the reacting chemicals, so the surroundings cool.	A fall in the measured temperature means the surroundings have lost thermal energy. The reaction takes in that energy, so it is endothermic.	2
Total Question 1			2
02.1	R changes by $11.3\,^\circ\text{C}$ and S changes by $8.8\,^\circ\text{C}$ ; R transfers more energy to the surroundings.	For R, $\Delta T=29.7-18.4=11.3\,^\circ\text{C}$ . For S, $\Delta T=30.1-21.3=8.8\,^\circ\text{C}$ . The controlled masses, solutions and apparatus make the larger rise in R evidence of greater energy transfer to the surroundings.	3
Total Question 2			3
03.1	Energy is conserved, so the reaction does not create energy. Energy is transferred from the reacting chemicals to the surroundings. The surroundings gain energy and become warmer, so the reaction is exothermic.	Track an energy transfer rather than describing energy as being made. The reacting chemicals lose energy and the surroundings gain the same amount. The measured temperature rise belongs to the surroundings and identifies an exothermic reaction.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Warmer Q is more suitable because its heating lasts beyond $30$ minutes and it costs less per use, although P gives the larger temperature rise.	Use every relevant comparison. P heats more strongly because $18>12$ , but its $11$ -minute duration is too short. Q lasts $34$ minutes, which covers the journey, and saves $\pounds1.70-\pounds0.95=\pounds0.75$ per use. Therefore Q best matches the stated need, while the lower temperature rise is its disadvantage.	4
Total Question 1			4
02.1	U rises by $11.5\,^\circ\text{C}$ and V rises by $13.2\,^\circ\text{C}$ , so U does not have the larger temperature change. However, assuming similar densities and specific heat capacities, U transfers about $\frac{75\times11.5}{25\times13.2}=2.61$ times as much energy to its larger volume of solution. The final temperatures alone do not justify the claim, and the unequal volumes prevent a fair comparison. Use equal volumes and concentrations of solution and give both tests the same starting temperature.	Calculate changes rather than comparing final temperatures: U gives $34.2-22.7=11.5\,^\circ\text{C}$ and V gives $31.8-18.6=13.2\,^\circ\text{C}$ . V has the larger rise. For similar solutions, the energy transferred to the solution is proportional to volume multiplied by temperature change, so U's value is $75(11.5)=862.5$ compared with V's $25(13.2)=330$ ; U's is about $2.61$ times greater because much more solution is heated. This does not make the comparison fair. Repeat with equal reactant amounts or equal solution volumes and concentrations, the same initial temperature and identical insulated apparatus.	6
Total Question 2			6
03.1	Student Q's measurement is more precise because the probe has the finer $0.1\,^\circ\text{C}$ resolution. $7.4\,^\circ\text{C}$ rounds to $7\,^\circ\text{C}$ to the nearest whole degree, so P's result does not conflict with Q's. Use an insulated cup and a lid to reduce energy transfer to the apparatus and the air, bringing the measured change closer to the true change.	A smaller scale division gives finer resolution, so Q reports more precise detail. Rounding Q's $7.4\,^\circ\text{C}$ result to the resolution used by P gives $7\,^\circ\text{C}$ , making the readings compatible. Precision does not remove systematic energy transfer; insulation and a lid reduce transfer to the cup and air.	5
Total Question 3			5
04.1	The $18.8\,^\circ\text{C}$ result for B is anomalous. A has rises of $7.4$ , $7.6$ and $7.5\,^\circ\text{C}$ , giving a mean rise of $7.5\,^\circ\text{C}$ . B's consistent results are falls of $5.9$ and $5.7\,^\circ\text{C}$ , giving a mean fall of $5.8\,^\circ\text{C}$ . A is exothermic because it warms the surroundings. B is endothermic because it takes in energy from the surroundings, which cool. Under the controlled conditions, A produces the larger temperature-change magnitude because $7.5>5.8$ .	Subtract the common starting temperature from each final temperature and compare the repeats before averaging. A changes by $+7.4$ , $+7.6$ and $+7.5\,^\circ\text{C}$ , so its mean rise is $(7.4+7.6+7.5)\div3=7.5\,^\circ\text{C}$ . For B, the $1.2\,^\circ\text{C}$ fall is far from the other two and is excluded; the mean of the $5.9$ and $5.7\,^\circ\text{C}$ falls is $5.8\,^\circ\text{C}$ . A transfers energy to the surroundings and is exothermic; B takes energy from them and is endothermic. The controlled comparison gives A the larger change in temperature.	6
Total Question 4			6
05.1	The estimated true temperature rise is $8.4\div0.70=12.0\,^\circ\text{C}$ . The estimated maximum temperature is $18.6+12.0=30.6\,^\circ\text{C}$ . The reaction is exothermic. Energy is transferred from the reacting chemicals to the surroundings, causing the rise. Use better insulation or add a lid so less energy is transferred to the cup and air.	The recorded rise is $70\%$ of the true rise, so divide rather than multiply: $8.4\div0.70=12.0\,^\circ\text{C}$ . Adding this correction to the initial temperature gives $30.6\,^\circ\text{C}$ . A rise in the surroundings identifies an exothermic reaction and an outward energy transfer. Better insulation or a lid reduces the unwanted transfer that made the measured rise too small.	5
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4.5.1.2 · Reaction profiles

Tier 1 · Easy

Mark scheme for 4.5.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Exothermic.	Products with less energy than the reactants mean that energy has been transferred to the surroundings, so the reaction is exothermic.	1
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02.1	Activation energy is the minimum energy reacting particles need for a reaction to occur; its arrow runs from the reactants' energy level to the peak.	The forward-reaction particles begin at the reactant level. The required minimum energy is the rise from that level to the maximum of the curved pathway.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	X is exothermic. Y is endothermic.	Compare product energy with reactant energy. Products lower means energy has been transferred to the surroundings, so X is exothermic. Products higher means energy has been taken in from the surroundings, so Y is endothermic.	2
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02.1	Draw a reactant energy level above the product energy level. Join them with a curved line that rises to a peak before falling. Show activation energy with an arrow from the reactant level to the peak.	An exothermic profile ends below its starting level because energy is transferred to the surroundings. The reaction pathway must curve over a maximum, and the forward activation energy is the vertical rise from the reactants to that maximum.	4
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03.1	Reaction type is determined by comparing the product and reactant energy levels, not by the height of the peak. The products are below the reactants, so the reaction is exothermic. The peak shows the activation-energy barrier that must be overcome; it does not show that the overall reaction is endothermic.	Separate the two vertical intervals. The rise from the reactant level to the peak is the activation energy. The products finish $30\,\text{kJ mol}^{-1}$ below the reactants, so the overall change is negative and the reaction is exothermic.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$E_{\text{a, forward}}=156\,\text{kJ mol}^{-1}$ $E_{\text{a, reverse}}=211\,\text{kJ mol}^{-1}$ Overall energy change $=-55\,\text{kJ mol}^{-1}$	For the forward reaction, $E_{\text{a}}=337-181=156\,\text{kJ mol}^{-1}$ . For the reverse reaction, start from the products: $E_{\text{a}}=337-126=211\,\text{kJ mol}^{-1}$ . The forward overall change is $126-181=-55\,\text{kJ mol}^{-1}$ , so the forward reaction is exothermic.	5
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02.1	Place the products above the reactants. Draw the activation-energy arrow from the reactant level to the peak. Draw the overall energy-change arrow upwards from the reactant level to the product level.	An endothermic reaction stores more energy in its products than in its reactants, so the product level and the overall-change arrow must be higher. Activation energy is measured from the reactants' level to the top of the curved pathway, not from an axis.	4
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03.1	Reactant particles must collide for a reaction to occur. The colliding particles must have at least the activation energy. On the profile, the activation energy is the vertical rise from the reactant level to the peak. Collisions with less energy than this do not produce a reaction, even if the products are at a lower energy level.	A lower product level describes the overall energy change, not whether an individual collision succeeds. Particles still have to collide with enough energy to reach the peak of the pathway. Any collision below that minimum energy cannot cross the activation-energy barrier.	4
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04.1	The peak is at $72+138=210$ units. The products are at $72+46=118$ units. The reverse activation energy is $210-118=92$ units. The reaction is endothermic. The positive overall change places the products above the reactants.	Reconstruct the profile from its vertical intervals. The forward barrier raises the pathway from $72$ to $210$ units. The positive overall change raises the products from $72$ to $118$ units. For the reverse reaction, the particles start at the product level, so the rise to the same peak is $210-118=92$ units. Products above reactants show an endothermic forward reaction.	5
Total Question 4			5
05.1	Experiment X matches P because P's products are below its reactants, so P is exothermic. Experiment Y matches Q because Q's products are above its reactants, so Q is endothermic. P has a forward activation energy of $164-80=84$ units. Q has a forward activation energy of $121-65=56$ units. P's overall change is $52-80=-28$ units and Q's is $97-65=+32$ units. Q has the smaller activation energy but the larger overall-change magnitude because $56<84$ and $32>28$ .	First use the product levels to connect the profiles to the observations: P transfers energy to the surroundings and X warms them, whereas Q takes energy from the surroundings and Y cools them. Forward activation energy is peak minus reactants, giving $84$ units for P and $56$ units for Q. Overall change is products minus reactants, giving $-28$ and $+32$ units. Compare absolute overall changes, so Q's magnitude is $32$ units rather than treating its positive sign as part of the size.	6
Total Question 5			6

4.5.1.3 · The energy change of reactions (HT only)

Tier 1 · Easy

Mark scheme for 4.5.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Energy is taken in when a bond is broken.	Bond breaking requires an input of energy. Bond formation, not bond breaking, releases energy.	1
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02.1	Bond formation releases energy. In an endothermic reaction, breaking the reactant bonds needs more energy than forming the product bonds releases.	Bond breaking requires energy and bond formation releases it. A reaction is endothermic when the required breaking total is greater than the released formation total.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$-818\,\text{kJ mol}^{-1}$	Breaking four $\mathrm{C-H}$ bonds and two $\mathrm{O=O}$ bonds requires $4(413)+2(498)=2648\,\text{kJ mol}^{-1}$ . Forming two $\mathrm{C=O}$ bonds and four $\mathrm{O-H}$ bonds releases $2(805)+4(464)=3466\,\text{kJ mol}^{-1}$ . Therefore $\Delta E=2648-3466=-818\,\text{kJ mol}^{-1}$ .	4
Total Question 1			4
02.1	$+125\,\text{kJ mol}^{-1}$	The reactant bonds require $347+6(413)=2825\,\text{kJ mol}^{-1}$ to break. Forming the product bonds releases $612+4(413)+436=2700\,\text{kJ mol}^{-1}$ . Therefore $\Delta E=2825-2700=+125\,\text{kJ mol}^{-1}$ .	4
Total Question 2			4
03.1	Breaking bonds in the reactants requires energy. Forming bonds in the products releases energy. The reaction is exothermic when bond formation releases more energy than bond breaking requires.	Consider both stages of the energy balance. Bond breaking is an energy input, but product-bond formation is an energy output. If the formation total is larger, the excess is transferred to the surroundings and the overall change is exothermic.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$391\,\text{kJ mol}^{-1}$ to three significant figures	Breaking one $\mathrm{N\equiv N}$ bond and three $\mathrm{H-H}$ bonds requires $945+3(436)=2253\,\text{kJ mol}^{-1}$ . Six $\mathrm{N-H}$ bonds form; let their mean energy be $x$ . Then $-92=2253-6x$ , so $6x=2345$ and $x=390.833\ldots\,\text{kJ mol}^{-1}$ . To three significant figures, the mean bond energy is $391\,\text{kJ mol}^{-1}$ .	5
Total Question 1			5
02.1	$151\,\text{kJ mol}^{-1}$	Four $\mathrm{O-H}$ bonds occur on each side, so their energies cancel. Two $\mathrm{O-O}$ bonds are broken and one $\mathrm{O=O}$ bond forms. If the mean $\mathrm{O-O}$ energy is $x$ , then $-196=2x-498$ . Hence $2x=302$ and $x=151\,\text{kJ mol}^{-1}$ .	5
Total Question 2			5
03.1	Two $\mathrm{N=O}$ bonds form, so the student should use $2\times631\,\text{kJ mol}^{-1}$ . The correct overall energy change is $+181\,\text{kJ mol}^{-1}$ . The reaction is endothermic because the overall energy change is positive.	Breaking one $\mathrm{N\equiv N}$ bond and one $\mathrm{O=O}$ bond requires $945+498=1443\,\text{kJ mol}^{-1}$ . Forming two $\mathrm{N=O}$ bonds releases $2(631)=1262\,\text{kJ mol}^{-1}$ . Therefore $\Delta E=1443-1262=+181\,\text{kJ mol}^{-1}$ . The positive result means more energy is taken in for bond breaking than is released by bond formation, so the reaction is endothermic.	4
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04.1	For chlorine, the bonds broken require $413+243=656\,\text{kJ mol}^{-1}$ . The bonds formed release $328+431=759\,\text{kJ mol}^{-1}$ . The chlorination energy change is $656-759=-103\,\text{kJ mol}^{-1}$ . For iodine, the broken total is $413+151=564$ and the formed total is $240+299=539\,\text{kJ mol}^{-1}$ . The iodination energy change is $564-539=+25\,\text{kJ mol}^{-1}$ , so it is endothermic. Chlorination is more exothermic: its energy change is $128\,\text{kJ mol}^{-1}$ more negative than the iodination value.	Only one $\mathrm{C-H}$ bond and the halogen bond are replaced, so the other three $\mathrm{C-H}$ bonds cancel. For chlorine, broken minus formed is $(413+243)-(328+431)=-103\,\text{kJ mol}^{-1}$ . For iodine it is $(413+151)-(240+299)=+25\,\text{kJ mol}^{-1}$ . Chlorination releases energy whereas iodination takes it in; the chlorination value is $25-(-103)=128\,\text{kJ mol}^{-1}$ more negative.	6
Total Question 4			6
05.1	Let the $\mathrm{C\equiv O}$ bond energy be $x$ ; breaking the bonds requires $2x+498\,\text{kJ mol}^{-1}$ . Four $\mathrm{C=O}$ bonds form, releasing $4\times805=3220\,\text{kJ mol}^{-1}$ . $-566=(2x+498)-3220$ , so $x=1078\,\text{kJ mol}^{-1}$ . Using $1072\,\text{kJ mol}^{-1}$ predicts $2(1072)+498-3220=-578\,\text{kJ mol}^{-1}$ . The derived $1078\,\text{kJ mol}^{-1}$ value agrees exactly with the measured $-566\,\text{kJ mol}^{-1}$ . The second table predicts an energy change $12\,\text{kJ mol}^{-1}$ more negative than the experiment.	Two carbon-monoxide bonds and one oxygen bond are broken, while the two carbon-dioxide molecules contain four $\mathrm{C=O}$ bonds in total. Therefore $-566=2x+498-4(805)$ . Rearranging gives $2x=2156$ and $x=1078\,\text{kJ mol}^{-1}$ . Substituting the alternative value gives $-578\,\text{kJ mol}^{-1}$ , so it does not reproduce the experimental estimate and differs by $12\,\text{kJ mol}^{-1}$ .	6
Total Question 5			6

4.5.2.1 · Cells and batteries (chemistry only)

Tier 1 · Easy

Mark scheme for 4.5.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Two electrodes made from different metals, both in contact with the electrolyte.	Award one idea for using two metal electrodes and one for the metals being different. The electrolyte provides the ionic contact needed by the cell.	2
Total Question 1			2
02.1	$2.55\,\text{V}$	Series voltages add when the cells face the same direction: $3\times0.85=2.55\,\text{V}$ .	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.5.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Use zinc and copper. They give the largest voltage, suggesting that they have the greatest difference in reactivity of the three pairs.	Compare the measured voltages directly. With the electrolyte controlled, a larger separation in electrode reactivity generally gives a larger potential difference.	3
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02.1	The electrolyte and one electrode metal both change, so either could cause a voltage difference. Use the same two metals in both cells and change only the electrolyte, while keeping conditions such as volume and temperature constant.	A valid comparison changes only the independent variable. Keep the electrode pair fixed, use equal electrolyte volumes at the same temperature and compare the chosen electrolyte types.	3
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03.1	In the non-rechargeable cell, one of the reactants is eventually used up. The chemical reaction then stops, so the cell no longer produces a potential difference. In a rechargeable cell, an external electrical current reverses the chemical reactions.	A cell produces a potential difference while its chemical reaction can continue. Using up a reactant stops that reaction in a non-rechargeable cell. A rechargeable system is different because supplied electrical energy drives its reactions in the reverse direction.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Three P cells give $4.5\,\text{V}$ and cost $\pounds1.80$ per replacement; three Q cells give exactly $3.6\,\text{V}$ and cost $\pounds10.50$ initially but can be reused for up to $400$ evenings. Q is more suitable for frequent long-term use, provided charging is available.	P needs three cells because $3(1.5)=4.5\,\text{V}$ , costing $3(0.60)=\pounds1.80$ each evening. Q also needs three because $3(1.2)=3.6\,\text{V}$ , with an initial cost of $3(3.50)=\pounds10.50$ . Repeated use makes Q much cheaper per evening and creates less discarded-cell waste, but P has the advantage of no charger and a larger voltage margin. For daily use, Q is the justified choice.	5
Total Question 1			5
02.1	The terminal voltage is $2.70\,\text{V}$ , so it cannot operate the $5.0\,\text{V}$ sensor. Turning the reversed cell around gives $5.40\,\text{V}$ .	The reversed cell opposes one of the others, giving $3(1.35)-1.35=2.70\,\text{V}$ , which is below $5.0\,\text{V}$ . With all four aligned, the voltages add to $4(1.35)=5.40\,\text{V}$ .	4
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03.1	Changing from sodium chloride solution to dilute acid increases the zinc-copper voltage by $0.25\,\text{V}$ and the magnesium-copper voltage by $0.25\,\text{V}$ . Replacing zinc with magnesium increases the sodium-chloride voltage by $1.50\,\text{V}$ and the dilute-acid voltage by $1.50\,\text{V}$ . The data support the claim because $1.50\,\text{V}$ is greater than $0.25\,\text{V}$ , but the conclusion is limited to the metals, electrolytes and conditions tested.	For the electrolyte change, calculate $1.35-1.10=0.25\,\text{V}$ and $2.85-2.60=0.25\,\text{V}$ . For the electrode change, calculate $2.60-1.10=1.50\,\text{V}$ and $2.85-1.35=1.50\,\text{V}$ . The electrode substitution has the larger measured effect in both controlled comparisons. Other metals, electrolytes or conditions could give a different pattern, so do not generalise beyond the supplied results.	6
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04.1	J-L spans the J-K and K-L gaps. Its predicted voltage is $0.72+0.48=1.20\,\text{V}$ . J-M spans all three stated gaps. Its predicted voltage is $0.72+0.48+0.36=1.56\,\text{V}$ . Three J-M cells give $3\times1.56=4.68\,\text{V}$ in series. The battery can operate the device because $4.68\,\text{V}$ exceeds $4.5\,\text{V}$ by $0.18\,\text{V}$ .	Treat each measured voltage as the gap between neighbouring metals in the stated order. Adding the first two gives $1.20\,\text{V}$ for J-L, while adding all three gives $1.56\,\text{V}$ for J-M. Series voltages then add, so three J-M cells provide $4.68\,\text{V}$ . Comparing this with the operating threshold leaves a $0.18\,\text{V}$ margin.	6
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05.1	The mean at $1.0\,\text{cm}$ is $1.09\,\text{V}$ . Its range is $1.10-1.08=0.02\,\text{V}$ . The mean at $4.0\,\text{cm}$ is also $1.09\,\text{V}$ . Its range is $1.11-1.07=0.04\,\text{V}$ . These data do not support the claim because increasing the separation leaves the mean unchanged; the spread at $4.0\,\text{cm}$ is larger, not its mean voltage. Keep a factor such as electrolyte type and concentration, temperature, electrode surface area or immersed depth constant.	Average each set of three readings: both totals are $3.27\,\text{V}$ , so both means are $1.09\,\text{V}$ . Subtract the lowest repeat from the highest to obtain ranges of $0.02$ and $0.04\,\text{V}$ . Identical means give no evidence for the claimed increase. Only electrode separation should change, so the electrolyte and the exposed electrodes need controlled conditions.	6
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4.5.2.2 · Fuel cells (chemistry only)

Tier 1 · Easy

Mark scheme for 4.5.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Water.	Hydrogen is oxidised by oxygen in the fuel cell, giving water as the overall reaction product.	1
Total Question 1			1
02.1	Hydrogen.	The supplied fuel is hydrogen, and the fuel is the substance oxidised electrochemically in the cell.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.5.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Hydrogen must first be produced and transported, and those processes require energy. If that energy comes from fossil fuels, greenhouse gases or other pollutants may be released even though the fuel cell itself forms water.	Separate emissions at the vehicle from impacts over the whole fuel cycle. The source of the energy used to manufacture and supply hydrogen determines whether indirect emissions occur.	3
Total Question 1			3
02.1	Hydrogen and oxygen or air are supplied from outside the cell, so their reaction keeps producing a potential difference without electrical recharging. If the hydrogen tank is empty, a reactant is missing and the reaction stops.	Unlike a rechargeable cell with a fixed store of reactants, a fuel cell operates from continuing external reactant supplies. Removing its hydrogen supply prevents the oxidation reaction that produces the potential difference.	3
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03.1	The mass of water is $31.5\,\text{g}$ . The water contains atoms from the oxygen supply as well as from the hydrogen supply. Mass is conserved, so the product mass equals the total mass of hydrogen and oxygen that reacted.	Both reactants contribute to the only product. By conservation of mass, $m(\mathrm{H_2O})=3.50+28.0=31.5\,\text{g}$ . Comparing the water only with the hydrogen ignores the oxygen atoms incorporated into the water.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Hydrogen delivers about $61\,\text{MJ kg}^{-1}$ and the battery about $0.76\,\text{MJ kg}^{-1}$ . Hydrogen offers much greater useful energy per kilogram, but storage is difficult and its production may cause emissions or require substantial energy. A justified choice depends on range, storage and the hydrogen source.	For hydrogen, useful energy is $118\times0.52=61.36\,\text{MJ kg}^{-1}$ . For the battery it is $0.90\times0.84=0.756\,\text{MJ kg}^{-1}$ . Hydrogen therefore has the large mass-specific energy advantage needed for long range. Balance this against bulky or high-pressure storage, refuelling infrastructure and the fact that producing hydrogen can use fossil fuels or considerable electrical energy. State a conclusion linked to those data rather than claiming either system is always best.	6
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02.1	The fuel cell can keep producing a potential difference throughout a long outage if enough hydrogen and oxygen are available, and its fuel supply can be replaced without electrical recharging. However, hydrogen must be stored and delivered to the remote station, and producing it may use fossil fuels or substantial electrical energy. The rechargeable battery needs no hydrogen supply, but its operating time is limited by the energy stored and it cannot be recharged from the mains during an outage. The fuel cell is more suitable if a safe, reliable hydrogen supply can cover the $10$ hours; otherwise a sufficiently large charged battery may be more practical.	Build linked comparison chains. Continuous external reactant supplies can extend fuel-cell operation, but they create hydrogen storage and delivery requirements. A rechargeable cell carries a fixed store of reactants, so it must have enough capacity before the outage and needs an external current afterwards. Environmental impact depends on how the hydrogen or charging electricity is produced. Finish with a conditional judgement tied to the station's $10$ -hour requirement.	5
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03.1	The hydrogen route returns $31.0\,\text{MJ}$ to three significant figures. The battery returns $66.4\,\text{MJ}$ . The battery is more suitable for maximising returned electrical energy because it returns $35.4\,\text{MJ}$ more from the same input. Hydrogen can be stored outside the cell and supplied continuously, so the available operating time can be extended by storing more fuel.	The hydrogen first stores $80.0\times0.68=54.4\,\text{MJ}$ . The fuel cell returns $54.4\times0.57=31.008\,\text{MJ}$ , which is $31.0\,\text{MJ}$ to three significant figures. The battery returns $80.0\times0.83=66.4\,\text{MJ}$ . Its advantage for the stated priority is $66.4-31.008=35.392\,\text{MJ}$ , or $35.4\,\text{MJ}$ to three significant figures. Hydrogen returns less of the original energy but can be kept as an external fuel supply rather than as a fixed store of reactants inside the cell.	6
Total Question 3			6
04.1	Conservation of mass gives an expected water mass of $4.80+38.4=43.2\,\text{g}$ . The uncollected mass is $43.2-40.6=2.6\,\text{g}$ . The percentage collected is $40.6\div43.2\times100=94.0\%$ to three significant figures. The claim is not supported because the collected mass is below the expected mass. Some water may have remained in the apparatus or escaped as water vapour.	Both supplied reactants contribute to the sole product, so conservation of mass gives an expected mass of $43.2\,\text{g}$ . The shortfall is $2.6\,\text{g}$ , and the collection percentage is $93.981\ldots\%$ , which rounds to $94.0\%$ . Since this is below $100\%$ , the result contradicts the student's claim and suggests incomplete collection rather than loss of mass from the reaction.	5
Total Question 4			5
05.1	Hydrogen production uses $0.80\times52=41.6\,\text{kWh}$ . This releases $41.6\times0.15=6.24\,\text{kg}$ of carbon dioxide. Battery charging releases $28\times0.15=4.20\,\text{kg}$ of carbon dioxide. For these data, the hydrogen route releases $6.24-4.20=2.04\,\text{kg}$ more carbon dioxide. The claim is incorrect for this electricity source: water is the only product at the vehicle, but making the hydrogen causes indirect emissions. If low-carbon electricity produced the hydrogen, its indirect carbon dioxide impact would be lower, so the conclusion depends on the energy source.	Account for hydrogen production before comparing the vehicles. The fuel route requires $41.6\,\text{kWh}$ and therefore releases $6.24\,\text{kg}$ of carbon dioxide with the stated electricity. The battery route releases $4.20\,\text{kg}$ , making it lower by $2.04\,\text{kg}$ for this journey. Water at the point of use does not include upstream emissions, and changing the electricity source could change the comparison.	6
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