IB DP Chemistry Reactivity 1.2: Energy cycles in reactions – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions test Reactivity 1.2, “Energy cycles in reactions”, from the IB Diploma Programme Chemistry guide (first assessment 2025, published February 2023). Understandings 1.2.1 to 1.2.5 each appear at least once. Questions 1 to 4 and question 10 cover SL and HL content. Questions labelled “HL only” use the additional higher level understandings 1.2.3, 1.2.4 and 1.2.5. The set is for the course examined from 2025 onwards.

Study the unit beforehand with the Reactivity 1.2 study guide and the condensed notes. Other units are listed on the course hub; track progress with the unit checklist; a free diagnostic helps you find weak spots.

Use these average bond enthalpies where needed (kJ mol⁻¹): H–H 436, C–H 414, C–C 346, C=C 614, C–O 358, C=O 804, O–H 463, O=O 498, N–H 391, N≡N 945, C–N 305. All other data are given in the question. Give answers in kJ mol⁻¹ to the precision of the data.

Questions

1. (SL and HL)

(a) Explain why breaking a covalent bond is an endothermic process. [1] (b) Explain why enthalpy changes calculated from bond enthalpy tables are described as estimates, even for reactions in which every species is a gas. [2]

2. (SL and HL) Methylamine, CH₃NH₂, is a gas at room temperature. It burns as follows: CH₃NH₂(g) + 2¼O₂(g) → CO₂(g) + 2½H₂O(g) + ½N₂(g).

(a) Use the bond enthalpies above to estimate ΔH for this reaction. [3] (b) Measured thermochemical data give −975.0 kJ mol⁻¹ for the same equation. Suggest why the two values differ. [1]

3. (SL and HL) Copper forms two oxides.

• 2Cu(s) + ½O₂(g) → Cu₂O(s) ΔH = −168.6 kJ mol⁻¹
• Cu(s) + ½O₂(g) → CuO(s) ΔH = −157.3 kJ mol⁻¹

Calculate ΔH for Cu₂O(s) + ½O₂(g) → 2CuO(s). [3]

4. (SL and HL) At very high temperatures, iron(III) oxide releases oxygen: 6Fe₂O₃(s) → 4Fe₃O₄(s) + O₂(g).

• (1) 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = −1648.4 kJ mol⁻¹
• (2) 3Fe(s) + 2O₂(g) → Fe₃O₄(s) ΔH = −1118.4 kJ mol⁻¹

(a) Use Hess’s law to calculate ΔH for the decomposition as written. [3] (b) Calculate the enthalpy change per mole of Fe₃O₄ formed. [1]

5. (HL only)

(a) Give the equation, including state symbols, that represents ΔHf⦵ of urea, CO(NH₂)₂(s). [1] (b) Give the equation, including state symbols, that represents ΔHc⦵ of butan-1-ol, C₄H₉OH(l). [1] (c) State the value of ΔHf⦵ for N₂(g) and explain your answer. [1]

6. (HL only) Gently heated ammonium nitrate gives dinitrogen monoxide: NH₄NO₃(s) → N₂O(g) + 2H₂O(l). Calculate ΔH⦵ using these ΔHf⦵ values (kJ mol⁻¹): NH₄NO₃(s) −365.6; N₂O(g) +82.1; H₂O(l) −285.8. [3]

7. (HL only) Consider the conversion of methane into liquid methanol: 2CH₄(g) + O₂(g) → 2CH₃OH(l). Calculate ΔH⦵ using these ΔHc⦵ values (kJ mol⁻¹): CH₄(g) −890.3; CH₃OH(l) −726.0. [3]

8. (HL only) Data for potassium oxide, K₂O (kJ mol⁻¹): atomization of K +89; IE₁(K) +419; ½O₂(g) → O(g) +249; first electron affinity of O −141; ΔHf⦵(K₂O) −361; lattice enthalpy of K₂O +2283.

(a) Use the data to determine the enthalpy change when a second electron is added to oxygen, O⁻(g) + e⁻ → O²⁻(g). [3] (b) Explain the sign of your answer. [1]

9. (HL only) Data for calcium fluoride, CaF₂ (kJ mol⁻¹): atomization of Ca +178; IE₁(Ca) +590; IE₂(Ca) +1145; F–F bond enthalpy +158; first electron affinity of F −328; ΔHf⦵(CaF₂) −1228.

(a) Determine the lattice enthalpy of CaF₂. [3] (b) Explain why the lattice enthalpy of CaF₂ is much larger than that of potassium fluoride. [2]

10. (SL and HL) Methoxymethane, CH₃OCH₃, is a gaseous fuel.

(a) Use bond enthalpies to estimate ΔH for CH₃OCH₃(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(g). [3] (b) The enthalpy change of vaporization of water is +44.0 kJ mol⁻¹. Use it to estimate ΔH when the water is formed as a liquid. [2] (c) The measured enthalpy of combustion of methoxymethane is −1460 kJ mol⁻¹. Calculate the percentage difference between your answer to (b) and this value, and give one reason for the difference. [2]

11. (HL only) Cyclopropane, C₃H₆, rearranges to its isomer propene when heated. Use these ΔHc⦵ values (kJ mol⁻¹): C(graphite) −393.5; H₂ −285.8; cyclopropane(g) −2091.3; propene(g) −2058.0.

(a) Write the formation equation for cyclopropane and calculate its ΔHf⦵. [3] (b) Calculate ΔH⦵ for cyclopropane(g) → propene(g). [2] (c) Use bond enthalpies to estimate ΔH for the same rearrangement. [2] (d) Suggest why the estimate in (c) is so different from the answer to (b). [1]

Answers

1. (a) Energy must be supplied to overcome the attraction between the shared electron pair and the two nuclei [1]. (b) Each value is an average over many compounds [1], but the actual strength of a bond depends on its environment in the particular molecule, so the true values in these molecules differ from the mean [1]. [3] Examiner insight: In (b), saying “the values are inaccurate” without explaining that they are averages taken from different molecules is unlikely to gain credit.

2. (a) Broken: 3 C–H + C–N + 2 N–H + 2¼ O=O = 1242 + 305 + 782 + 1120.5 = 3449.5 kJ [1]. Formed: 2 C=O + 5 O–H + ½ N≡N = 1608 + 2315 + 472.5 = 4395.5 kJ [1]. ΔH = 3449.5 − 4395.5 = −946 kJ mol⁻¹ [1] (b) Each tabulated value is a mean taken over many compounds, so it does not match the actual strength of that bond in methylamine, oxygen, carbon dioxide or water [1]. [4] Examiner insight: All species are gases here, so “water is a liquid” is not a valid reason; pick the reason that fits the states in the equation. Fractional coefficients are fine: multiply each bond enthalpy by them just as you would by whole numbers.

3. Reverse the Cu₂O equation: +168.6 [1]. Double the CuO equation: 2(−157.3) = −314.6 [1]. ΔH = 168.6 − 314.6 = −146.0 kJ mol⁻¹ [1] [3] Examiner insight: Show which equations you reversed and scaled; a bare final number gives the reader no way to see where any slip happened.

4. (a) Reverse (1) and multiply by 3: 6Fe₂O₃ → 12Fe + 9O₂, ΔH = +4945.2 [1]. Multiply (2) by 4: 12Fe + 8O₂ → 4Fe₃O₄, ΔH = −4473.6 [1]. ΔH = 4945.2 − 4473.6 = +471.6 kJ mol⁻¹ [1] (b) 471.6 ÷ 4 = +117.9 kJ mol⁻¹ [1] [4] Examiner insight: Check the sum of your adjusted equations: 9O₂ released and 8O₂ used leaves one O₂, matching the target, which confirms the factors.

5. (a) C(s, graphite) + ½O₂(g) + N₂(g) + 2H₂(g) → CO(NH₂)₂(s) [1] (b) C₄H₉OH(l) + 6O₂(g) → 4CO₂(g) + 5H₂O(l) [1] (c) Zero, because N₂(g) is the element in its standard state, so no change occurs in forming it from itself [1]. [3] Examiner insight: One mole of the named substance is required; doubling (a) to clear the fraction changes it into a different enthalpy change.

6. Σ ΔHf⦵(products) = 82.1 + 2(−285.8) = −489.5 [1]. Σ ΔHf⦵(reactants) = −365.6 [1]. ΔH⦵ = −489.5 − (−365.6) = −123.9 kJ mol⁻¹ [1] [3] Examiner insight: The coefficient 2 for water is the step most often skipped; write each product term separately before adding.

7. Σ ΔHc⦵(reactants) = 2(−890.3) + 0 = −1780.6 [1]. Σ ΔHc⦵(products) = 2(−726.0) = −1452.0 [1]. ΔH⦵ = −1780.6 − (−1452.0) = −328.6 kJ mol⁻¹ [1] [3] Examiner insight: With combustion data it is reactants minus products; using the formation order here gives +328.6, the right size with the wrong sign.

8. (a) Steps from elements to ions, excluding the second electron affinity: 2(89) + 2(419) + 249 + (−141) = +1124 [1]. ΔHf⦵ = 1124 + EA₂ − 2283, so −361 = 1124 + EA₂ − 2283 [1]. EA₂ = +798 kJ mol⁻¹ [1] (b) The incoming electron is repelled by the negative charge of O⁻, so energy must be supplied [1]. [4] Examiner insight: K₂O has two potassium ions, so both the atomization and IE₁ of potassium appear twice; missing this gives an answer 508 kJ mol⁻¹ too large.

9. (a) Steps: 178 + 590 + 1145 + 158 + 2(−328) = +1415 [1]. −1228 = 1415 − lattice enthalpy [1]. Lattice enthalpy = +2643 kJ mol⁻¹ [1] (b) The calcium ion carries a 2+ charge, double that of the potassium ion [1], and it has a smaller ionic radius, so the electrostatic attraction to F⁻ is stronger [1]. [5] Examiner insight: The F–F bond enthalpy is used once, because one F₂ molecule supplies both fluorine atoms; using 2 × 158 is a common slip.

10. (a) Broken: 6 C–H + 2 C–O + 3 O=O = 2484 + 716 + 1494 = 4694 kJ [1]. Formed: 4 C=O + 6 O–H = 3216 + 2778 = 5994 kJ [1]. ΔH = 4694 − 5994 = −1300 kJ mol⁻¹ [1] (b) Condensing 3 mol of water releases 3 × 44.0 = 132 kJ [1]. ΔH = −1300 − 132 = −1432 kJ mol⁻¹ [1] (c) (1460 − 1432) ÷ 1460 × 100 = 1.9% [1]. Bond enthalpies are averages, not the actual values in methoxymethane, carbon dioxide and water [1]. [7] Examiner insight: In (b), condensation is exothermic, so ΔH becomes more negative; adding 132 instead of subtracting it is a sign error that carries through to (c).

11. (a) 3C(s, graphite) + 3H₂(g) → C₃H₆(g) [1]. Σ ΔHc⦵(elements) = 3(−393.5) + 3(−285.8) = −2037.9 [1]. ΔHf⦵ = −2037.9 − (−2091.3) = +53.4 kJ mol⁻¹ [1] (b) Combustion data, so reactants minus products: ΔH⦵ = −2091.3 − (−2058.0) [1] = −33.3 kJ mol⁻¹ [1] (c) Broken: 3 C–C in the ring = 1038 kJ; formed: C=C + C–C = 960 kJ [1]. The six C–H bonds are unchanged, so ΔH ≈ 1038 − 960 = +78 kJ mol⁻¹ [1] (d) The C–C bonds in the strained three-membered ring are weaker than the average C–C value, so using the average overestimates the energy needed to break them [1]. [8] Examiner insight: A positive estimate against a negative measured value is not a calculation error to hide; what (d) asks for is a reason rooted in the use of average values.

Where marks are usually lost

• Writing Σ(formed) − Σ(broken) for bond enthalpies, giving the right size with the wrong sign.
• Missing bonds in a condensed formula: CH₃OCH₃ has two C–O bonds and six C–H bonds.
• Using bond enthalpies for a reaction that forms liquid water without the vaporization correction.
• Reversing a Hess equation but keeping the original sign of ΔH.
• Not multiplying ΔHf⦵ or ΔHc⦵ values by the coefficients in the balanced equation.
• Putting a ΔHc⦵ term in for O₂, or a ΔHf⦵ term in for N₂, H₂ or Mg.
• Forgetting IE₂ for Ca²⁺ or Mg²⁺, or doubling the wrong terms in an MX₂ or M₂X cycle.
• Treating the lattice enthalpy as negative when the question defines it in the separating direction.

Next steps

• Recap with the short-form notes.
• For any question you missed, go back to the matching section of the full study guide.
• See every DP Chemistry unit on the course hub.
• Use the DP Chemistry checklist and all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme Chemistry guide, first assessment 2025 (published February 2023): Reactivity 1.2, “Energy cycles in reactions”.