Energetics / Thermochemistry — IB Chemistry HL HL Study Guide

For: IB Chemistry HL candidates sitting IB Chemistry HL.

Covers: Enthalpy of reaction, Hess's law and bond enthalpies, HL-only Born-Haber cycles, entropy, and Gibbs free energy, with worked examples and exam-specific tips.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Chemistry HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Energetics / Thermochemistry?

Energetics (also called thermochemistry) is the study of energy transfers that occur during chemical reactions and physical state changes, and it forms Topic 5 of the IB Chemistry HL syllabus. All reactions exchange heat with their surroundings, and we measure this exchange using enthalpy ($H$), the total heat content of a system at constant pressure (the standard condition for most lab and industrial reactions). We almost exclusively measure changes in enthalpy ($\Delta H$) rather than absolute enthalpy values, as only relative differences between reactants and products are practically useful for predicting reaction behavior.

2. Enthalpy of reaction

The standard enthalpy of reaction ($\Delta H_r^{\circ }$) is the heat energy transferred at constant pressure when molar quantities of reactants react exactly as written in the balanced chemical equation, under standard conditions: 298 K temperature, 100 kPa pressure, 1 mol dm⁻³ concentration for solutions, and all elements in their most stable standard state.

Key sign convention

• If $\Delta H<0$, the reaction is exothermic: heat is released to the surroundings (e.g., combustion, neutralization of strong acids and bases).
• If $\Delta H>0$, the reaction is endothermic: heat is absorbed from the surroundings (e.g., photosynthesis, thermal decomposition of calcium carbonate).

Examiners frequently ask you to sketch enthalpy level diagrams: for exothermic reactions, draw reactants on a higher horizontal line than products, with a downward arrow between them labeled $\Delta H$ (negative value). For endothermic reactions, products are higher than reactants, with an upward arrow labeled $\Delta H$ (positive value).

Common standard enthalpy change definitions

• $\Delta H_f^{\circ }$ (standard enthalpy of formation): Enthalpy change when 1 mole of a compound is formed from its elements in standard state. The $\Delta H_f^{\circ }$ of any element in its standard state is 0 by definition.
• $\Delta H_c^{\circ }$ (standard enthalpy of combustion): Enthalpy change when 1 mole of a substance burns completely in excess oxygen.
• $\Delta H_{neut}^{\circ }$ (standard enthalpy of neutralization): Enthalpy change when 1 mole of water is formed from the reaction of a strong acid and strong base, always ~-57 kJ mol⁻¹ for strong acid-strong base pairs.

The formula to calculate $\Delta H_r^{\circ }$ from enthalpy of formation values is: $$\Delta H_r^{\circ }=\sum \Delta H_f^{\circ }(\text{products})-\sum \Delta H_f^{\circ }(\text{reactants})$$

Worked example

Calculate $\Delta H_r^{\circ }$ for the reaction $2H_2(g)+O_2(g)\to 2H_{2}O(l)$ given $\Delta H_f^{\circ }(H_{2}O(l))=-286$ kJ mol⁻¹.

1. Sum of product enthalpies: $2\times (-286)=-572$ kJ mol⁻¹
2. Sum of reactant enthalpies: $\Delta H_f^{\circ }(H_2(g))+\Delta H_f^{\circ }(O_2(g))=0+0=0$
3. $\Delta H_r^{\circ }=-572-0=-572$ kJ mol⁻¹ (highly exothermic, as expected for hydrogen combustion)

3. Hess's law and bond enthalpies

Hess's law states that the total enthalpy change for a reaction is independent of the pathway taken between the initial reactants and final products. It is a direct consequence of the first law of thermodynamics, which says energy cannot be created or destroyed, only transferred. You can use Hess's law to calculate enthalpy changes for reactions that cannot be measured directly, via enthalpy cycles or bond enthalpy values.

Bond enthalpies

Mean bond enthalpy ($E$) is the average energy required to break 1 mole of a specific covalent bond in gaseous molecules, averaged across a wide range of different compounds. Breaking bonds requires energy (endothermic, positive value), while forming bonds releases energy (exothermic, negative value). The formula for reaction enthalpy from bond enthalpies is: $$\Delta H_r^{\circ }=\sum E(\text{bonds broken})-\sum E(\text{bonds formed})$$

Important note: Bond enthalpies only apply to gaseous species. If your reactants or products are solids or liquids, you must add the relevant enthalpy of vaporization or fusion to convert them to gas before using this formula.

Worked example

Calculate $\Delta H_r^{\circ }$ for the gas-phase reaction $C{H}_4(g)+2O_2(g)\to C{O}_2(g)+2H_{2}O(g)$ using the following bond enthalpies: $E(C-H)=413$, $E(O=O)=498$, $E(C=O)=746$, $E(O-H)=464$ kJ mol⁻¹.

1. Energy to break all reactant bonds: $4(C-H)+2(O=O)=(4\times 413)+(2\times 498)=1652+996=2648$ kJ mol⁻¹
2. Energy released forming all product bonds: $2(C=O)+4(O-H)=(2\times 746)+(4\times 464)=1492+1856=3348$ kJ mol⁻¹
3. $\Delta H_r^{\circ }=2648-3348=-700$ kJ mol⁻¹

4. Born-Haber cycle (HL)

The Born-Haber cycle is a specialized Hess's law cycle used to calculate the lattice enthalpy of ionic compounds, a value that cannot be measured directly in the lab. Lattice enthalpy ($\Delta H_{lat}^{\circ }$) is defined as the enthalpy change when 1 mole of solid ionic compound is formed from its isolated gaseous ions, and it is always negative (exothermic, as strong ionic bonds are formed).

A standard Born-Haber cycle for an ionic compound $M_a{X}_b$ includes the following steps:

1. Atomization of the metal: $M(s)\to M(g)$ (endothermic, $\Delta H_{at}^{\circ }$)
2. Ionization of the gaseous metal: $M(g)\to M^{n+}(g)+n{e}^{-}$ (sum of relevant ionization energies, endothermic)
3. Atomization of the non-metal: $\frac{1}{2}{X}_2(g)\to X(g)$ (endothermic, $\Delta H_{at}^{\circ }$, multiplied by stoichiometric coefficient $b$)
4. Electron affinity of the non-metal: $X(g)+e^{-}\to X^{m-}(g)$ (sum of relevant electron affinities; first electron affinity is exothermic, second and higher are endothermic)
5. Lattice formation: $a{M}^{n+}(g)+b{X}^{m-}(g)\to M_a{X}_b(s)$ (exothermic, $\Delta H_{lat}^{\circ }$)
6. Direct enthalpy of formation of the ionic compound (the reference path)

By Hess's law: $$\Delta H_f^{\circ }(M_a{X}_b)=a\Delta H_{at}^{\circ }(M)+a\sum IE(M)+b\Delta H_{at}^{\circ }(X)+b\sum EA(X)+\Delta H_{lat}^{\circ }$$

When drawing a Born-Haber cycle diagram, endothermic steps are drawn as upward arrows (increasing enthalpy) and exothermic steps as downward arrows, which helps you avoid sign errors when summing the steps.

Worked example

Calculate the lattice enthalpy of NaCl given $\Delta H_f^{\circ }(NaCl)=-411$ kJ mol⁻¹, $\Delta H_{at}^{\circ }(Na)=+107$ kJ mol⁻¹, $I{E}_1(Na)=+496$ kJ mol⁻¹, $\Delta H_{at}^{\circ }(Cl)=+122$ kJ mol⁻¹, $E{A}_1(Cl)=-349$ kJ mol⁻¹.

1. Rearrange the Born-Haber formula to solve for $\Delta H_{lat}^{\circ }$: $$\Delta H_{lat}^{\circ }=\Delta H_f^{\circ }-[\Delta H_{at}^{\circ }(Na)+I{E}_1(Na)+\Delta H_{at}^{\circ }(Cl)+E{A}_1(Cl)]$$
2. Plug in values: $\Delta H_{lat}^{\circ }=-411-(107+496+122-349)=-411-376=-787$ kJ mol⁻¹

5. Entropy and Gibbs free energy (HL)

Enthalpy change alone cannot predict if a reaction will occur spontaneously, as disorder (entropy) also contributes to reaction feasibility. Entropy ($S$) is a measure of the randomness or disorder of a system, with units J K⁻¹ mol⁻¹. Higher disorder = higher entropy: gases > liquids > solids, solutions > pure solute + solvent, and reactions that produce more moles of gas have a positive entropy change.

The standard entropy change of reaction is calculated as: $$\Delta S_r^{\circ }=\sum S^{\circ }(\text{products})-\sum S^{\circ }(\text{reactants})$$

Gibbs free energy ($G$) combines enthalpy and entropy into a single value to predict reaction spontaneity at a given temperature: $$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$$ where $T$ is temperature in Kelvin. A reaction is spontaneous if $\Delta G^{\circ }<0$, non-spontaneous if $\Delta G^{\circ }>0$, and at equilibrium if $\Delta G^{\circ }=0$.

Spontaneity rules

You can calculate the minimum temperature for a reaction to become spontaneous by setting $\Delta G^{\circ }=0$: $T=\frac{\Delta H^{\circ }}{\Delta S^{\circ }}$. This only works if $\Delta H^{\circ }$ and $\Delta S^{\circ }$ have the same sign.

Worked example

Calculate the minimum temperature (in °C) at which the decomposition of calcium carbonate becomes spontaneous: $CaC{O}_3(s)\to CaO(s)+C{O}_2(g)$, given $\Delta H^{\circ }=+178$ kJ mol⁻¹, $\Delta S^{\circ }=+160$ J K⁻¹ mol⁻¹.

1. Convert $\Delta S^{\circ }$ to kJ to match units of $\Delta H^{\circ }$: $160$ J K⁻¹ mol⁻¹ = $0.16$ kJ K⁻¹ mol⁻¹
2. Calculate minimum $T$: $T=\frac{178}{0.16}=1112.5$ K
3. Convert to °C: $1112.5-273=839.5$ °C

6. Common Pitfalls (and how to avoid them)

• Wrong move: Mixing up the sign of lattice enthalpy or electron affinity. Why: Students confuse the definition of lattice enthalpy (formation from ions vs dissociation into ions) and forget that second electron affinities are endothermic. Correct move: Always start with the definition: bond/ionic bond formation is exothermic (-ve), breaking is endothermic (+ve); if a question asks for lattice enthalpy of dissociation, it is the positive inverse of the formation lattice enthalpy.
• Wrong move: Using bond enthalpies for solid/liquid species without adding phase change enthalpies. Why: Students forget mean bond enthalpies are only defined for gaseous molecules. Correct move: Always add enthalpy of vaporization/fusion values to convert non-gaseous species to gas before using the bond enthalpy formula.
• Wrong move: Forgetting stoichiometric multipliers in Born-Haber cycles (e.g., using 1x EA of Cl for MgCl₂ instead of 2x). Why: Students rush through cycle drawing without matching the ionic formula stoichiometry. Correct move: Write the balanced formation equation first, count how many of each ion you need, and multiply atomization, ionization, and electron affinity values accordingly.
• Wrong move: Using entropy in J K⁻¹ mol⁻¹ directly with enthalpy in kJ mol⁻¹ in the Gibbs free energy equation. Why: Students skip unit consistency checks. Correct move: Always convert $\Delta S^{\circ }$ to kJ K⁻¹ mol⁻¹ by dividing by 1000 before plugging into the $\Delta G$ formula.
• Wrong move: Assuming a negative $\Delta G$ means a reaction will occur quickly. Why: Students confuse thermodynamic feasibility with kinetic rate. Correct move: $\Delta G$ only tells you if a reaction is possible, not how fast it will happen: e.g., diamond turning to graphite has $\Delta G<0$ but is kinetically inert at room temperature, a distinction examiners test frequently.

7. Practice Questions (IB Chemistry HL Style)

Question 1

Calculate the standard enthalpy of combustion of ethanol: $C_2{H}_{5}OH(l)+3O_2(g)\to 2C{O}_2(g)+3H_{2}O(l)$, using the following $\Delta H_f^{\circ }$ values: $\Delta H_f^{\circ }(C_2{H}_{5}OH(l))=-278$ kJ mol⁻¹, $\Delta H_f^{\circ }(C{O}_2(g))=-394$ kJ mol⁻¹, $\Delta H_f^{\circ }(H_{2}O(l))=-286$ kJ mol⁻¹.

Worked solution

$$\Delta H_r^{\circ }=\sum \Delta H_f^{\circ }(\text{products})-\sum \Delta H_f^{\circ }(\text{reactants})$$

1. Sum of product enthalpies: $(2\times -394)+(3\times -286)=-788-858=-1646$ kJ mol⁻¹
2. Sum of reactant enthalpies: $\Delta H_f^{\circ }(C_2{H}_{5}OH(l))+3\Delta H_f^{\circ }(O_2(g))=-278+0=-278$ kJ mol⁻¹
3. $\Delta H_c^{\circ }=-1646-(-278)=-1368$ kJ mol⁻¹

Question 2

Use the Born-Haber cycle to calculate the second electron affinity of oxygen for MgO, given: $\Delta H_f^{\circ }(MgO)=-602$ kJ mol⁻¹, $\Delta H_{at}^{\circ }(Mg)=+148$ kJ mol⁻¹, $I{E}_1(Mg)=+738$ kJ mol⁻¹, $I{E}_2(Mg)=+1451$ kJ mol⁻¹, $\Delta H_{at}^{\circ }(O)=+249$ kJ mol⁻¹, $E{A}_1(O)=-141$ kJ mol⁻¹, $\Delta H_{lat}^{\circ }(MgO)=-3795$ kJ mol⁻¹.

Worked solution

Rearrange the Born-Haber formula to solve for $E{A}_2(O)$: $$E{A}_2(O)=\Delta H_f^{\circ }-[\Delta H_{at}^{\circ }(Mg)+I{E}_1(Mg)+I{E}_2(Mg)+\Delta H_{at}^{\circ }(O)+E{A}_1(O)+\Delta H_{lat}^{\circ }]$$

1. Plug in values: $$E{A}_2(O)=-602-[148+738+1451+249-141-3795]$$ $$E{A}_2(O)=-602-(-1350)=+748$$ kJ mol⁻¹ The positive value makes sense, as adding a second electron to the negative O⁻ ion requires energy to overcome electrostatic repulsion.

Question 3

A reaction has $\Delta H^{\circ }=+92$ kJ mol⁻¹ and $\Delta S^{\circ }=+125$ J K⁻¹ mol⁻¹. (a) Calculate the minimum temperature in °C at which the reaction becomes spontaneous. (b) Explain why the reaction is non-spontaneous at 298 K.

Worked solution

(a) Set $\Delta G^{\circ }=0$ to find minimum spontaneous $T$: $$T=\frac{\Delta H^{\circ }}{\Delta S^{\circ }}=\frac{92}{0.125}=736\text{ K}=736-273=463\text{ °C}$$ (b) At 298 K: $$\Delta G^{\circ }=92-(298\times 0.125)=92-37.25=+54.75{\text{ kJ mol}}^{-1}$$ $\Delta G^{\circ }$ is positive, so the reaction is non-spontaneous. The positive (endothermic) enthalpy term outweighs the positive entropy change at low temperatures.

8. Quick Reference Cheatsheet

9. What's Next

The energetics concepts you learned in this guide form the foundation for multiple later IB Chemistry HL topics. For example, Gibbs free energy directly relates to equilibrium constants in Topic 7 (Equilibrium) via the equation $\Delta G^{\circ }=-RT\ln K$, which lets you predict how temperature changes will shift equilibrium position. Energetics also underpins Topic 9 (Redox Processes), where cell potential is directly proportional to Gibbs free energy change, and Topic 16 (Chemical Kinetics), where you will distinguish between thermodynamic feasibility (ΔG) and kinetic reaction rate. Understanding lattice enthalpy also helps explain trends in solubility of ionic compounds in Topic 4 (Chemical Bonding and Structure).

If you’re stuck on any part of energetics, from drawing Born-Haber cycles to calculating Gibbs free energy for spontaneous reactions, you can ask Ollie, our AI tutor, for personalized walkthroughs, extra practice questions, or clarification of any confusing concepts. You can also find more topic-specific study guides, past paper walkthroughs, and flashcards on the homepage to help you prepare for your IB Chemistry HL exams.