Chemical Energetics (Thermodynamics) — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: Enthalpy change classification, Hess's law and enthalpy cycle calculations, average bond enthalpy applications, Born-Haber cycles for ionic compounds, and entropy + Gibbs free energy for reaction feasibility analysis.

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 A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Chemical Energetics?

Chemical energetics (also called chemical thermodynamics) is the branch of physical chemistry that quantifies energy transfers during chemical reactions and physical state changes, and predicts whether a reaction will occur spontaneously under given conditions. It uses standardised notation and experimental data to avoid direct measurement of unobservable properties like absolute enthalpy, and accounts for 15-20% of total marks across A-Level Chemistry Papers 1, 2 and 4, with AS content focusing on enthalpy calculations and A2 content extending to feasibility predictions. All energy change values are reported under standard conditions: 298 K temperature, 100 kPa pressure, 1 mol dm⁻³ concentration for solutions, and elements in their most stable standard state.

2. Enthalpy change $\Delta H$ — exothermic vs endothermic

Enthalpy ($H$) is defined as the total heat content of a system at constant pressure. Absolute enthalpy cannot be measured directly, so we only calculate the change in enthalpy for a reaction: $$\Delta H=H_{\text{products}}-H_{\text{reactants}}$$ Units of $\Delta H$ are kJ mol⁻¹, and the superscript $\ominus$ denotes values measured under standard conditions ($\Delta H^{\ominus }$).

Reactions are classified based on the sign of $\Delta H$:

• Exothermic: $\Delta H<0$. Heat is released from the system to the surroundings, so products have lower enthalpy than reactants. Common examples include combustion, neutralization, and metal oxidation.
• Endothermic: $\Delta H>0$. Heat is absorbed by the system from the surroundings, so products have higher enthalpy than reactants. Common examples include thermal decomposition of carbonates and dissolving ammonium nitrate in water.

Worked Example

The complete combustion of 1 mole of ethanol releases 1367 kJ of heat under standard conditions. Classify the reaction and write its standard enthalpy change of combustion. Solution: Combustion releases heat, so it is exothermic. $\Delta H_c^{\ominus }=-1367$ kJ mol⁻¹. Examiners deduct 1 mark for missing the negative sign, so always include it for exothermic reactions.

3. Hess's law and enthalpy cycles

Hess's law is derived from the first law of thermodynamics (energy cannot be created or destroyed), and states that the total enthalpy change for a reaction is independent of the route taken from reactants to products, as long as initial and final conditions are identical. This allows you to calculate $\Delta H$ for reactions that cannot be measured directly, using two methods: algebraic manipulation of given reaction enthalpies, or enthalpy cycle (triangle) diagrams.

Key rules for Hess's law calculations:

1. If you reverse a reaction equation, flip the sign of its $\Delta H$ value
2. If you scale a reaction by a factor, scale its $\Delta H$ value by the same factor

Worked Example

Calculate the standard enthalpy of formation of ethene (${\text{C}}_2{\text{H}}_4$) using the given data:

1. $\text{C(s)}+{\text{O}}_2\text{(g)}\to {\text{CO}}_2\text{(g)}$ $\Delta H_1=-394$ kJ mol⁻¹
2. ${\text{H}}_2\text{(g)}+\frac{1}{2}{\text{O}}_2\text{(g)}\to {\text{H}}_2\text{O(l)}$ $\Delta H_2=-286$ kJ mol⁻¹
3. ${\text{C}}_2{\text{H}}_4\text{(g)}+3{\text{O}}_2\text{(g)}\to 2{\text{CO}}_2\text{(g)}+2{\text{H}}_2\text{O(l)}$ $\Delta H_3=-1411$ kJ mol⁻¹

Solution: Target reaction: $2\text{C(s)}+2{\text{H}}_2\text{(g)}\to {\text{C}}_2{\text{H}}_4\text{(g)}$

• Reverse reaction 3: $2{\text{CO}}_2\text{(g)}+2{\text{H}}_2\text{O(l)}\to {\text{C}}_2{\text{H}}_4\text{(g)}+3{\text{O}}_2\text{(g)}$ $\Delta H_3^{\prime }=+1411$ kJ mol⁻¹
• Multiply reaction 1 by 2: $2\text{C(s)}+2{\text{O}}_2\text{(g)}\to 2{\text{CO}}_2\text{(g)}$ $\Delta H_1^{\prime }=2(-394)=-788$ kJ mol⁻¹
• Multiply reaction 2 by 2: $2{\text{H}}_2\text{(g)}+{\text{O}}_2\text{(g)}\to 2{\text{H}}_2\text{O(l)}$ $\Delta H_2^{\prime }=2(-286)=-572$ kJ mol⁻¹
• Sum the modified reactions: $\Delta H_f=\Delta H_1^{\prime }+\Delta H_2^{\prime }+\Delta H_3^{\prime }=-788-572+1411=+51$ kJ mol⁻¹

4. Bond enthalpies and average values

Bond dissociation enthalpy ($E$) is the energy required to break 1 mole of a specific covalent bond in the gaseous state under standard conditions. Bond breaking is always endothermic ($E>0$), while bond formation is always exothermic ($E<0$ for the reverse process).

Since bond strength varies slightly depending on the surrounding atoms in a molecule, we use average bond enthalpies: the mean value of bond dissociation enthalpies for a given bond across a range of gaseous compounds. This allows us to estimate reaction enthalpy using the formula: $$\Delta H=\sum E(\text{bonds broken})-\sum E(\text{bonds formed})$$

Worked Example

Calculate the enthalpy change for the reaction of ethene with hydrogen to form ethane: ${\text{C}}_2{\text{H}}_4\text{(g)}+{\text{H}}_2\text{(g)}\to {\text{C}}_2{\text{H}}_6\text{(g)}$ Given average bond enthalpies: $E(\text{C=C})=612$ kJ mol⁻¹, $E(\text{C-C})=347$ kJ mol⁻¹, $E(\text{C-H})=413$ kJ mol⁻¹, $E(\text{H-H})=436$ kJ mol⁻¹

Solution:

• Bonds broken: 1 C=C, 4 C-H, 1 H-H = $612+(4\times 413)+436=2700$ kJ mol⁻¹
• Bonds formed: 1 C-C, 6 C-H = $347+(6\times 413)=2825$ kJ mol⁻¹
• $\Delta H=2700-2825=-125$ kJ mol⁻¹

Note that this value is approximate, as average bond enthalpies are not specific to the molecules in this reaction. Examiners frequently ask you to state this limitation as a 1-mark question.

5. Born-Haber cycles for ionic compounds

The Born-Haber cycle is a specialized enthalpy cycle used to calculate lattice enthalpy of ionic compounds, a value that cannot be measured directly. The A-Level Chemistry syllabus defines lattice enthalpy ($\Delta H_{\text{latt}}^{\ominus }$) as the enthalpy change when 1 mole of an ionic solid is formed from its gaseous ions under standard conditions, which is always negative (exothermic, due to electrostatic attraction between oppositely charged ions).

Components of a Born-Haber cycle (all values are standard enthalpy changes):

1. Atomization enthalpy ($\Delta H_{\text{at}}^{\ominus }$): Energy to form 1 mole of gaseous atoms from an element in its standard state (always positive)
2. Ionization energy ($IE$): Energy to remove electrons from gaseous metal atoms to form positive ions (always positive)
3. Electron affinity ($EA$): Energy change when gaseous non-metal atoms gain electrons to form negative ions (first EA is usually negative, second EA is always positive due to repulsion between negative ions and incoming electrons)
4. Enthalpy of formation ($\Delta H_f^{\ominus }$): Measured experimentally, enthalpy change when 1 mole of ionic solid forms from its elements in standard state

Worked Example

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

Solution: By Hess's law: $$\Delta H_f^{\ominus }=\Delta H_{\text{at}}(\text{Na})+\Delta H_{\text{at}}(\text{Cl})+I{E}_1(\text{Na})+E{A}_1(\text{Cl})+\Delta H_{\text{latt}}^{\ominus }$$ Rearrange to solve for lattice enthalpy: $$\Delta H_{\text{latt}}^{\ominus }=-411-(107+122+496-349)=-787{\text{ kJ mol}}^{-1}$$

6. Entropy and Gibbs free energy (where in syllabus)

This is an A2 topic tested primarily in Paper 4, and is used to predict reaction feasibility.

• Entropy ($S$): A measure of the disorder of a system, with units J K⁻¹ mol⁻¹. Gases have higher entropy than liquids, which have higher entropy than solids. Dissolving solids or increasing temperature increases entropy. The entropy change of a reaction is calculated as: $$\Delta S^{\ominus }=\sum S^{\ominus }(\text{products})-\sum S^{\ominus }(\text{reactants})$$
• Gibbs free energy ($\Delta G$): The quantity that determines if a reaction is spontaneous (feasible) at a given temperature, calculated as: $$\Delta G^{\ominus }=\Delta H^{\ominus }-T\Delta S^{\ominus }$$ Where $T$ is temperature in Kelvin. If $\Delta G<0$, the reaction is feasible; if $\Delta G>0$, it is not feasible; if $\Delta G=0$, the reaction is at equilibrium.

Worked Example

For the thermal decomposition of calcium carbonate: ${\text{CaCO}}_3\text{(s)}\to \text{CaO(s)}+{\text{CO}}_2\text{(g)}$, $\Delta H^{\ominus }=+178$ kJ mol⁻¹, $\Delta S^{\ominus }=+161$ J K⁻¹ mol⁻¹. Calculate the minimum temperature at which the reaction is feasible.

Solution: First convert $\Delta S$ to kJ K⁻¹ mol⁻¹ to match units of $\Delta H$: $161$ J K⁻¹ mol⁻¹ = $0.161$ kJ K⁻¹ mol⁻¹. Set $\Delta G=0$ (threshold for feasibility): $$0=178-T(0.161)$$ $$T=\frac{178}{0.161}=1106\text{ K (or }833^{\circ }\text{C)}$$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to flip the sign of $\Delta H$ when reversing a reaction in Hess's law calculations. Why: Students focus on numerical values before accounting for reaction direction. Correct move: Flip and write the new $\Delta H$ sign immediately when reversing a reaction, before doing any arithmetic.
• Wrong move: Using Joule units for $\Delta S$ directly in the Gibbs free energy formula without conversion. Why: $\Delta H$ is given in kJ, leading to unit mismatch. Correct move: Divide $\Delta S$ by 1000 to convert to kJ K⁻¹ mol⁻¹ before substituting into the $\Delta G$ formula.
• Wrong move: Confusing lattice enthalpy formation and dissociation definitions. Why: Some textbooks define lattice enthalpy as the energy to break the ionic lattice (positive value), but A-Level uses the formation definition. Correct move: Memorize the A-Level definition explicitly: lattice enthalpy for formation of ionic solid from gaseous ions is negative.
• Wrong move: Stating that average bond enthalpy values are exact. Why: Students assume bond strengths are identical across all molecules. Correct move: Always note that average bond enthalpies are approximate, as they are mean values across multiple compounds, and only apply to gaseous reactions.
• Wrong move: Using negative values for second electron affinity in Born-Haber cycles. Why: First electron affinity is negative, so students assume all EA values are negative. Correct move: Second electron affinity is always positive, as you have to overcome repulsion between a negative ion and an incoming electron.

8. Practice Questions (A-Level Chemistry Style)

Question 1 (AS Level, 3 marks)

Classify each of the following reactions as exothermic or endothermic, and state the sign of $\Delta H$: a) Reaction of magnesium with dilute hydrochloric acid b) Melting of ice c) Neutralization of sulfuric acid with potassium hydroxide

Solution: a) Metal-acid reaction releases heat: exothermic, $\Delta H<0$ (1 mark) b) Melting requires heat input: endothermic, $\Delta H>0$ (1 mark) c) Neutralization releases heat: exothermic, $\Delta H<0$ (1 mark)

Question 2 (AS Level, 5 marks)

Use average bond enthalpies to calculate the enthalpy change of combustion of methanol (${\text{CH}}_3\text{OH}$): ${\text{CH}}_3\text{OH(g)}+1\frac{1}{2}{\text{O}}_2\text{(g)}\to {\text{CO}}_2\text{(g)}+2{\text{H}}_2\text{O(g)}$ Given: $E(\text{C-O})=358$ kJ mol⁻¹, $E(\text{O-H})=464$ kJ mol⁻¹, $E(\text{C-H})=413$ kJ mol⁻¹, $E(\text{O=O})=498$ kJ mol⁻¹, $E(\text{C=O})=746$ kJ mol⁻¹

Solution:

1. Bonds broken: 3 C-H, 1 C-O, 1 O-H, 1.5 O=O = $(3\times 413)+358+464+(1.5\times 498)=2808$ kJ mol⁻¹ (2 marks)
2. Bonds formed: 2 C=O, 4 O-H = $(2\times 746)+(4\times 464)=3348$ kJ mol⁻¹ (2 marks)
3. $\Delta H=2808-3348=-540$ kJ mol⁻¹ (1 mark, accept ±20 kJ mol⁻¹ for rounding differences)

Question 3 (A2 Level, 6 marks)

For the reaction ${\text{N}}_2\text{(g)}+3{\text{H}}_2\text{(g)}\rightleftharpoons 2{\text{NH}}_3\text{(g)}$, $\Delta H^{\ominus }=-92$ kJ mol⁻¹, $\Delta S^{\ominus }=-198$ J K⁻¹ mol⁻¹. a) Calculate $\Delta G^{\ominus }$ at 298 K b) State if the reaction is feasible at 298 K c) Calculate the temperature above which the reaction is no longer feasible

Solution: a) Convert $\Delta S$ to kJ K⁻¹ mol⁻¹: $-198/1000=-0.198$ kJ K⁻¹ mol⁻¹. $\Delta G=-92-(298\times -0.198)=-33.0$ kJ mol⁻¹ (2 marks) b) $\Delta G$ is negative, so reaction is feasible at 298 K (1 mark) c) Set $\Delta G=0$: $0=-92-(T\times -0.198)\to T=92/0.198=465$ K (or 192 °C) (2 marks). Above this temperature, $\Delta G$ becomes positive so reaction is not feasible (1 mark)

9. Quick Reference Cheatsheet

10. What's Next

Chemical energetics is a foundational topic that connects directly to multiple high-weight sections of the A-Level Chemistry syllabus. Enthalpy change calculations are used to explain activation energy in reaction kinetics, while Gibbs free energy directly determines the position of chemical equilibrium and the value of equilibrium constants $K_c$ and $K_p$. Lattice enthalpy values from Born-Haber cycles are also used to explain solubility trends of ionic compounds and their melting point variations. Examiners frequently combine energetics with these topics in 8-10 mark long answer questions in Paper 4, so mastering the calculations and sign conventions covered here will give you a significant advantage in later topics.

If you struggle with any step of cycle drawing, sign convention, or calculation, you can get personalized support from Ollie, our AI tutor, at any time by visiting the homepage. Ollie can generate extra practice problems tailored to your weak points, walk you through tricky Born-Haber or Gibbs free energy questions, and check your working against Cambridge mark scheme conventions to help you maximize your score on exam day.

Aligned with the Cambridge International AS & A Level Chemistry 9701 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.