OxfordAQA A-Level Chemistry: Energetics (9620) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions are aligned to the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams from May/June 2020 onwards, section 3.1.4 Energetics (3.1.4.1 to 3.1.4.4). Revise first with the study guide or the revision notes. The course page is OxfordAQA A-level Chemistry and the printable list is the checklist. Take the specific heat capacity of water and of every solution as 4.18 J g⁻¹ K⁻¹, with solution density 1.00 g cm⁻³. Ar: Mg 24.3. The enthalpy data are invented for practice.

Questions

1. Give the meaning of the term standard enthalpy of combustion. [3]

2. (a) Write the equation, with state symbols, for the reaction whose enthalpy change is the standard enthalpy of formation of ethanoic acid, CH₃COOH(l). [2] (b) Explain why the standard enthalpy of formation of I₂(s) is zero. [1]

3. 60.0 cm³ of 0.800 mol dm⁻³ silver nitrate solution is poured into a polystyrene cup. Excess copper powder is added and the temperature rises by 14.0 K. Calculate the enthalpy change, in kJ mol⁻¹, for the reaction of one mole of silver ions. [4]

4. A student burns 0.74 g of butan-1-ol (Mr = 74.0) in a spirit burner. The flame warms 250 g of water in a metal can and the temperature goes up by 20.0 K. (a) Work out the enthalpy of combustion of butan-1-ol, in kJ mol⁻¹, from these results. [4] (b) A data table gives −2676 kJ mol⁻¹. Give two reasons why the student’s result is different. [2] (c) Propose a modification to the equipment that would bring the result nearer to the table value, and say why. [2]

5. 50.0 cm³ of 1.20 mol dm⁻³ hydrochloric acid at 19.6 °C is added to 50.0 cm³ of 1.20 mol dm⁻³ sodium hydroxide at the same temperature. The enthalpy change of neutralisation is −57.0 kJ mol⁻¹. Calculate the highest temperature the mixture should reach, assuming no heat is lost. [4]

6. Describe how you would carry out an experiment to measure the enthalpy change when a solid dissolves in water, so that the temperature change can be found reliably. Include how you would treat the temperature readings. [4]

7. Use these enthalpies of combustion to calculate the standard enthalpy of formation of propan-2-ol, C₃H₇OH(l): carbon −394, hydrogen −286, propan-2-ol −2006 kJ mol⁻¹. [3]

8. (a) Calculate ΔH for 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(l) + CO₂(g) from these standard enthalpies of formation in kJ mol⁻¹: NaHCO₃(s) −948, Na₂CO₃(s) −1131, H₂O(l) −286, CO₂(g) −394. [3] (b) The reaction 2NO(g) + O₂(g) → 2NO₂(g) has ΔH = −114 kJ mol⁻¹. ΔfH° of NO(g) is +90 kJ mol⁻¹. Calculate ΔfH° of NO₂(g). [3]

9. Calculate ΔH for C₆H₁₂(l) → C₆H₆(l) + 3H₂(g) from these enthalpies of combustion in kJ mol⁻¹: C₆H₁₂(l) −3920, C₆H₆(l) −3267, H₂(g) −286. State whether the reaction is endothermic or exothermic. [4]

10. (a) Define mean bond enthalpy. [2] (b) Using the mean bond enthalpies below, in kJ mol⁻¹ (C–H 413, C–C 347, O=O 498, C=O 805, O–H 464) to calculate ΔH for C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g). [3] (c) Use ΔfH° values of C₃H₈(g) = −105, CO₂(g) = −394 and H₂O(g) = −242 kJ mol⁻¹ to calculate ΔH for the same reaction. [2] (d) Explain why your answers to (b) and (c) are different. [2]

11. (a) ΔH for H₂(g) + Br₂(g) → 2HBr(g) is −103 kJ mol⁻¹. Mean bond enthalpies: H–H 436 and Br–Br 193 kJ mol⁻¹. Calculate the H–Br bond enthalpy. [3] (b) Explain why bond enthalpies cannot be used directly to find ΔH for the combustion of ethanol, C₂H₅OH(l), forming H₂O(l). [2]

12. The enthalpy of formation of magnesium oxide cannot be measured directly with a calorimeter. A student measures two enthalpy changes using 100 cm³ of excess hydrochloric acid each time. Experiment A: 0.486 g of magnesium ribbon, temperature rise 22.3 K. Experiment B: 0.806 g of magnesium oxide (Mr = 40.3), temperature rise 6.9 K. (a) Calculate the enthalpy change per mole of magnesium in Experiment A. [3] (b) Calculate the enthalpy change per mole of magnesium oxide in Experiment B. [2] (c) Use these results and ΔfH°(H₂O(l)) = −286 kJ mol⁻¹ to calculate the standard enthalpy of formation of magnesium oxide. Show your cycle. [3]

Answers

1. The heat change at constant pressure when exactly one mole of a substance [1] reacts fully with oxygen [1] at standard conditions (100 kPa, stated temperature), all species in standard states [1]. [3] Examiner insight: A version missing “one mole” or “completely” is likely to lose credit.

2. (a) 2C(s) + 2H₂(g) + O₂(g) → CH₃COOH(l): correct species, balanced, one mole of product [1] and correct state symbols [1] [2] (b) Solid iodine is itself an element in its standard state, so nothing has to change to make it [1] [1] Examiner insight: For formation equations, check that you have one mole of product and elements as reactants. Fractions on the left are allowed in a formation equation.

3. q = 60.0 × 4.18 × 14.0 [1] = 3511 J = 3.51 kJ [1] moles of Ag⁺ = 0.800 × 0.0600 = 0.0480 mol [1] ΔH = −3.511 ÷ 0.0480 = −73.2 kJ mol⁻¹ [1] [4] Examiner insight: The sign is a separate decision from the number. State that the temperature rose and so the reaction is exothermic.

4. (a) moles = 0.74 ÷ 74.0 = 0.0100 mol [1] q = 250 × 4.18 × 20.0 = 20 900 J [1] = 20.9 kJ [1] ΔcH = −20.9 ÷ 0.0100 = −2090 kJ mol⁻¹ [1] (b) Any two from: heat lost to the surroundings; incomplete combustion; heat absorbed by the can; evaporation of fuel. [2] [2] (c) Add a draught shield or lid (or lag the can) [1] so that less heat is lost to the surroundings [1]. [2] Examiner insight: “Human error” is vague. Naming a specific process, such as heat loss or incomplete combustion, is how you earn the reason marks.

5. moles = 1.20 × 0.0500 = 0.0600 mol [1] q = 57.0 × 1000 × 0.0600 = 3420 J [1] ΔT = 3420 ÷ (100 × 4.18) = 8.18 K [1] final temperature = 19.6 + 8.18 = 27.8 °C [1] [4] Examiner insight: This reverses the usual calculation, so write the formula as q = mcΔT first and then rearrange it, rather than working from memory.

6. Use a polystyrene cup with a lid, in a beaker for stability, and measure a known volume of water [1]. Record the water temperature at fixed time intervals for a few minutes before adding the solid [1]. Add a weighed mass of solid, stir, and keep recording at the same intervals [1]. Plot temperature against time and extrapolate the line back to the time of addition to find ΔT, which allows for heat exchange with the surroundings [1]. [4] Examiner insight: Mention both what you measure (mass, volume, temperature) and what you do with the readings. Your own school’s method may differ in detail.

7. Cycle: 3C(s) + 4H₂(g) + ½O₂(g) → C₃H₇OH(l), with all species burned to CO₂ and H₂O. ΔfH = 3ΔcH(C) + 4ΔcH(H₂) − ΔcH(C₃H₇OH) [1] = 3(−394) + 4(−286) − (−2006) [1] = −1182 − 1144 + 2006 = −320 kJ mol⁻¹ [1] [3] Examiner insight: Show the multiplying numbers (3 and 4) explicitly. A cycle drawn first makes the sign choices easy to check.

8. (a) ΔH = [ΔfH(Na₂CO₃) + ΔfH(H₂O) + ΔfH(CO₂)] − [2 × ΔfH(NaHCO₃)] [1] = (−1131 − 286 − 394) − (2 × −948) = −1811 + 1896 [1] = +85 kJ mol⁻¹ [1] (b) −114 = [2x] − [2(+90)] [1] −114 = 2x − 180, so 2x = 66 [1] x = +33 kJ mol⁻¹ [1] Examiner insight: When the unknown is inside the formula, write the equation with x in it before rearranging, so the method is visible even if the algebra slips.

9. ΔH = ΔcH(C₆H₁₂) − [ΔcH(C₆H₆) + 3ΔcH(H₂)] [1] = −3920 − [−3267 + 3(−286)] [1] = −3920 − (−4125) = +205 kJ mol⁻¹ [1] Endothermic, because ΔH is positive [1] [4] Examiner insight: Combustion data reverse the order of subtraction compared with formation data. Writing the formula down first helps you notice this.

10. (a) The enthalpy change when one mole of a given type of bond is broken in the gaseous phase [1], averaged over a range of different molecules containing that bond [1]. [2] (b) broken: 8 × 413 + 2 × 347 + 5 × 498 = 6488 kJ [1] formed: 6 × 805 + 8 × 464 = 8542 kJ [1] ΔH = 6488 − 8542 = −2054 kJ mol⁻¹ [1] [3] (c) ΔH = [3(−394) + 4(−242)] − [−105] [1] = −2150 + 105 = −2045 kJ mol⁻¹ [1] [2] (d) Mean bond enthalpies are averages over many molecules, not values for these specific molecules [1], whereas formation values are measured for these actual compounds [1]. [2] Examiner insight: Part (d) asks for the cause. A statement such as “bond enthalpies are less accurate” repeats the question without explaining why.

11. (a) 436 + 193 − 2x = −103 [1] 629 + 103 = 2x [1] x = 366 kJ mol⁻¹ [1] [3] (b) Ethanol and the water formed are liquids, and bond enthalpies are for gaseous species [1]. The energy needed to change between liquid and gas states is not included in the bond enthalpy calculation [1]. [2] Examiner insight: In (a) the final answer is a bond enthalpy, so it is positive. Write the equation with the unknown before rearranging.

12. (a) q = 100 × 4.18 × 22.3 = 9321 J [1]; moles Mg = 0.486 ÷ 24.3 = 0.0200 mol [1] ΔH = −9.321 ÷ 0.0200 = −466 kJ mol⁻¹ [1] (b) q = 100 × 4.18 × 6.9 = 2884 J, moles MgO = 0.806 ÷ 40.3 = 0.0200 mol [1] ΔH = −2.884 ÷ 0.0200 = −144 kJ mol⁻¹ [1] (c) Equations: A is Mg + 2HCl → MgCl₂ + H₂; B is MgO + 2HCl → MgCl₂ + H₂O. The target is Mg + ½O₂ → MgO. Route: A, then reverse B, then form H₂O from H₂ and ½O₂ [1] ΔfH(MgO) = ΔH(A) − ΔH(B) + ΔfH(H₂O) = −466 − (−144) + (−286) [1] = −608 kJ mol⁻¹ [1] Examiner insight: Reversing B changes its sign. State clearly which equation you are reversing, so the sign change is easy to follow.

Where marks are usually lost

• Omitting “one mole” or “standard conditions” from a definition.
• Using the mass of the solid or fuel in q = mcΔT instead of the mass of the liquid that heated up.
• Dividing joules by moles without first converting to kilojoules.
• Forgetting the negative sign on an exothermic molar enthalpy change.
• Writing wrong state symbols in formation equations.
• Not multiplying each enthalpy by the number of moles in the balanced equation.
• Subtracting in the wrong order in Hess’s law, especially when switching between combustion and formation data.
• In a bond enthalpy calculation, missing a bond because the structure was not drawn out in full.
• Giving “experimental error” as the reason for a difference rather than naming the heat loss or incomplete combustion.

Next steps

• Revise from the revision notes and the study guide.
• Find the topic on the course page and tick it off on the checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Frequently asked questions

How do you avoid sign errors in calorimetry calculations? Treat the sign as a separate decision from the number. Work out the size of the enthalpy change, then state whether the temperature rose or fell and give an exothermic change a negative sign.

Which mass do you use in q = mcΔT? Use the mass of the water or solution that changed temperature, not the mass of the fuel or solid that reacted. For solutions this is usually taken from the volume using a density of 1.00 g cm⁻³.

Why should you draw a cycle in Hess’s law questions? A cycle drawn first makes the sign choices easy to check, and showing the multiplying numbers from the balanced equation makes the method visible. Remember that combustion data reverse the order of subtraction compared with formation data.

What unit mistakes are common in enthalpy calculations? A frequent slip is dividing joules by moles without first converting to kilojoules, so the answer is out by a factor of 1000. Convert q to kJ before dividing to give a value in kJ mol⁻¹.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.4 Energetics.