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Topic R1.2 · HL only

Energy cycles in reactions: notes and practice questions

5 min read Jump to 5 practice questions
Summary
  • This topic covers the application of Hess's law using standard enthalpy changes of formation and combustion, and the interpretation of Born-Haber cycles for ionic compounds.
  • The standard enthalpy change of a reaction can be calculated using standard enthalpies of formation: ΔH⊖=Σ(ΔHf⊖ products)−Σ(ΔHf⊖ reactants)\Delta H^\ominus = \Sigma(\Delta H_\text{f}^\ominus \text{ products}) - \Sigma(\Delta H_\text{f}^\ominus \text{ reactants}).
  • The standard enthalpy change of a reaction can be calculated using standard enthalpies of combustion: ΔH⊖=Σ(ΔHc⊖ reactants)−Σ(ΔHc⊖ products)\Delta H^\ominus = \Sigma(\Delta H_\text{c}^\ominus \text{ reactants}) - \Sigma(\Delta H_\text{c}^\ominus \text{ products}).
  • Born-Haber cycles illustrate energy changes in the formation of ionic compounds, incorporating ionization energies, enthalpy of atomization, electron affinities, lattice enthalpy, and enthalpy of formation.
  • Interpret Born-Haber cycles for univalent and divalent ionic compounds to determine unknown energy values.

How it is examined

Bond enthalpy calculations are 2 to 3 marks: bonds broken, bonds formed, difference. Only average bond enthalpies work for gaseous species, which is a standard "explain why the calculated value differs" follow-up worth 1 mark. May 2025 HL Paper 2 TZ1 gave a partly filled Born-Haber-style cycle and asked candidates to complete boxes with names of processes and formulas of species including state symbols for [3], with a Notes line refusing "ionization energy" singular where "ionization energies" was needed. It also ran a paired part asking for the enthalpy of a reaction from bond enthalpies [3] and then from ΔHf⦵ data [2], then why the two answers differ [1].

Given in the booklet

Average bond enthalpies. HL: standard enthalpies of formation and combustion, the two Hess's law summation equations, ionization energies, electron affinities and lattice enthalpies. What is recall: the direction of the two summation equations, which is the part students reverse. Products minus reactants for formation, reactants minus products for combustion.

Key ideas
  • 1.2.1 Bond-breaking absorbs and bond-forming releases energy. Students calculate the enthalpy change of a reaction from given average bond enthalpy data.
  • 1.2.2 Hess's law states that the enthalpy change for a reaction is independent of the pathway between the initial and final states. Students apply Hess's law to calculate enthalpy changes in multistep reactions.
Not assessed

HL: the construction of a complete Born-Haber cycle will not be assessed. Students interpret a given cycle or fill in parts of one. Do not generate a question that asks a student to build the whole cycle from nothing.

Guiding questions

  • How does application of the law of conservation of energy help us to predict energy changes during reactions?

Linking questions

  • Structure 2.2 How would you expect bond enthalpy data to relate to bond length and polarity? (HL) Would you expect allotropes of an element, such as diamond and graphite, to have different ΔHf⦵ values?
  • Reactivity 3.4 How does the strength of a carbon-halogen bond affect the rate of a nucleophilic substitution reaction?
  • Structure 2.1 (HL) What are the factors that influence the strength of lattice enthalpy in an ionic compound?

Practice questions

5 questions · 4 medium · 1 hard
Showing 5 of 5

Question 1

MediumPaper 1A · calculator1 mark

Which equation represents the standard enthalpy of atomization, ΔHat⊖\Delta H_{at}^{\ominus}, of bromine?

A. Br2(l)→2Br(g)Br_2(l) \rightarrow 2Br(g)

B. Br2(g)→2Br(g)Br_2(g) \rightarrow 2Br(g)

C. 12Br2(g)→Br(g)\frac{1}{2}Br_2(g) \rightarrow Br(g)

D. 12Br2(l)→Br(g)\frac{1}{2}Br_2(l) \rightarrow Br(g)

Question 2

HardPaper 2 · calculator23 marks
(a)

A sample of chlorine consists of two isotopes, 35Cl^{35}Cl and 37Cl^{37}Cl.

(a) Contrast the sub-atomic structure of these two isotopes.

[1]
(b)(i)

(b) (i) The sample of chlorine is analysed in a mass spectrometer, producing a spectrum for the Cl2+Cl_2^+ ion. The spectrum shows three peaks at m/z values of 70, 72 and 74. Explain the origin and relative heights of these three peaks, given that the abundance of 35Cl^{35}Cl is approximately three times that of 37Cl^{37}Cl.

[3]
(b)(ii)

(ii) A more precise measurement finds the composition by mass to be: 35Cl^{35}Cl: 75.76%, 37Cl^{37}Cl: 24.24%. Calculate the relative atomic mass of chlorine from this sample, giving your answer to two decimal places. (Use isotopic masses of 35.0 and 37.0 for this calculation).

[2]
(c)(i)

Magnesium chloride, MgCl2MgCl_2, and manganese(II) chloride, MnCl2MnCl_2, are two ionic compounds.

(c) (i) Deduce the type of bonding in magnesium chloride, MgCl2MgCl_2, using electronegativity values from section 9 of the data booklet.

[2]
(c)(ii)

(ii) Determine the lattice enthalpy of magnesium chloride, assuming the bonding is purely ionic. Use sections 9, 10 and 12 of the data booklet and the following data:

Enthalpy of formation of magnesium chloride = −641 kJ mol−1-641 \text{ kJ mol}^{-1}

[3]
(c)(iii)

(iii) Explain, with reference to electron configurations, why the ionic radii of Mg2+Mg^{2+}, Mn2+Mn^{2+} and Cl−Cl^- are different. Use section 10 of the data booklet.

[3]
(c)(iv)

(iv) Predict, with a reason, which has the stronger ionic bonding, manganese(II) chloride, MnCl2MnCl_2, or magnesium chloride.

[2]
(d)(i)

Magnesium chloride is white, but manganese(II) chloride is pale pink.

(d) (i) State the condensed electron configuration of a manganese atom.

[1]
(d)(ii)

(ii) State the reason, in terms of electron configuration, why manganese(II) chloride is coloured.

[1]
(d)(iii)

(iii) Manganese(II) chloride absorbs light with a wavelength of approximately 530 nm. Describe why this is consistent with the observed colour of the compound. Use sections 2 and 15 of the data booklet.

[2]
(e)(i)

A copper key is to be electroplated with manganese using an aqueous solution of manganese(II) chloride as the electrolyte.

(e) (i) Deduce the half-equations for the reactions occurring at the anode (made of pure manganese) and the cathode (the copper key).

[2]
(e)(ii)

(ii) Deduce a balanced chemical equation for the reaction of fluorine gas with the aqueous chloride ions in the electrolyte.

[1]

Question 3

MediumPaper 1A · calculator1 mark

Ethanol is considered a renewable fuel source. Its complete combustion can be represented by the following equation:

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)

Use the following standard enthalpy of formation values to determine the enthalpy change for the reaction in kJ mol−1\text{kJ mol}^{-1}.

ΔHf⊖(C2H5OH(l))=−277.6 kJ mol−1\Delta H_f^{\ominus} (C_2H_5OH(l) ) = -277.6 \text{ kJ mol}^{-1}

ΔHf⊖(CO2(g))=−393.5 kJ mol−1\Delta H_f^{\ominus} (CO_2(g) ) = -393.5 \text{ kJ mol}^{-1}

ΔHf⊖(H2O(l))=−285.8 kJ mol−1\Delta H_f^{\ominus} (H_2O(l) ) = -285.8 \text{ kJ mol}^{-1}

A. 2(−393.5)+3(−285.8)−(−277.6)2(-393.5) + 3(-285.8) - (-277.6)

B. −277.6−(2(−393.5)+3(−285.8))-277.6 - (2(-393.5) + 3(-285.8) )

C. 2(−393.5)−3(−285.8)+(−277.6)2(-393.5) - 3(-285.8) + (-277.6)

D. 2(−393.5)+3(−285.8)+(−277.6)2(-393.5) + 3(-285.8) + (-277.6)

Question 4

MediumPaper 1A · calculator1 mark

The hydrogenation of ethene is an important industrial process to produce ethane. Use the provided standard enthalpy of combustion data to calculate the enthalpy change for the hydrogenation of ethene in kJ mol−1kJ\ mol^{-1}.

C2H4(g)+H2(g)→C2H6(g)C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)

Given standard enthalpy of combustion values:

I. C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l) ΔH⊖=−1410 kJ mol−1\Delta H^{\ominus} = -1410\ kJ\ mol^{-1}

II. C2H6(g)+3.5O2(g)→2CO2(g)+3H2O(l)C_2H_6(g) + 3.5O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) ΔH⊖=−1560 kJ mol−1\Delta H^{\ominus} = -1560\ kJ\ mol^{-1}

III. H2(g)+0.5O2(g)→H2O(l)H_2(g) + 0.5O_2(g) \rightarrow H_2O(l) ΔH⊖=−285 kJ mol−1\Delta H^{\ominus} = -285\ kJ\ mol^{-1}

A. −3255-3255

B. −135-135

C. +135+135

D. +2685+2685

Question 5

MediumPaper 1A · calculator1 mark

Which of the following enthalpy changes associated with the Born-Haber cycle for magnesium oxide are exothermic?

I. Mg(g)→Mg2+(g)+2e−Mg(g) \rightarrow Mg^{2+}(g) + 2e^-

II. O(g)+2e−→O2−(g)O(g) + 2e^- \rightarrow O^{2-}(g)

III. Mg2+(g)+O2−(g)→MgO(s)Mg^{2+}(g) + O^{2-}(g) \rightarrow MgO(s)

A. I and II only

B. I and III only

C. III only

D. I, II and III

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What does Energy cycles in reactions cover in IB Chemistry?

This topic covers the application of Hess's law using standard enthalpy changes of formation and combustion, and the interpretation of Born-Haber cycles for ionic compounds. The standard enthalpy change of a reaction can be calculated using standard enthalpies of formation: Δ H^ominus = Σ(Δ H_f^ominus products) - Σ(Δ H_f^ominus reactants). The standard enthalpy change of a reaction can be calculated using standard enthalpies of combustion: Δ H^ominus = Σ(Δ H_c^ominus reactants) - Σ(Δ H_c^ominus products).

Is Energy cycles in reactions SL or HL?

Energy cycles in reactions is HL only. SL students are not examined on it.

How do I revise Energy cycles in reactions for IB Chemistry?

Start from the core idea: this topic covers the application of Hess's law using standard enthalpy changes of formation and combustion, and the interpretation of Born-Haber cycles for ionic compounds. In the exam: bond enthalpy calculations are 2 to 3 marks: bonds broken, bonds formed, difference. Only average bond enthalpies work for gaseous species, which is a standard "explain why the calculated value differs" follow-up worth 1 mark. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Energy cycles in reactions?

FourtyFive has 5 Energy cycles in reactions questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

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