(I once asked a class, "Why does a reaction release energy?" Most students said, "Because it gets hot." But that is a description, not an explanation. The real answer is that the products are more stable than the reactants — they have lower enthalpy. The energy released is the difference. In this chapter, we are going to go deeper than ever before into the energy changes that drive every chemical reaction. We will learn how to calculate enthalpy changes that are impossible to measure directly. We will see how the same reaction can release energy in one step or in many. And we will meet two new ideas — entropy and free energy — that tell us whether a reaction will happen at all. Once you understand these ideas, you understand why anything happens in chemistry. Let us begin.) In your earlier studies, you learned about exothermic and endothermic reactions, energy level diagrams, and bond energies. That is a solid foundation — but it is only the beginning. At the advanced level, we ask deeper questions. How can we calculate the enthalpy change for a reaction that is too dangerous or too slow to carry out directly? What does the energy difference between reactants and products tell us about whether a reaction is spontaneous? Why do some endothermic reactions happen spontaneously while some exothermic reactions do not? These are the questions that this chapter answers. We will explore Hess's law in detail, learn about standard enthalpies of formation, combustion, and neutralisation. We will see how to use bond energies and Born-Haber cycles. We will meet the concept of entropy — the measure of disorder — and understand how it combines with enthalpy to determine whether a reaction is spontaneous. By the end, you will have a deep understanding of the energy that drives every chemical reaction. Let us begin.
Enthalpy change (ΔH) is the heat energy transferred during a reaction at constant pressure. To compare enthalpy changes fairly, chemists define standard conditions: 298 K (25°C) and 1 atmosphere pressure.
| Type | Symbol | Definition |
|---|---|---|
| Standard enthalpy of reaction | ΔH°r | Enthalpy change when molar quantities of reactants react completely under standard conditions. |
| Standard enthalpy of formation | ΔH°f | Enthalpy change when one mole of a compound is formed from its elements in their standard states. |
| Standard enthalpy of combustion | ΔH°c | Enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. |
| Standard enthalpy of neutralisation | ΔH°n | Enthalpy change when one mole of water is formed from the reaction of an acid and a base under standard conditions. |
| Standard enthalpy of solution | ΔH°sol | Enthalpy change when one mole of a substance is dissolved in a large excess of water. |
| Standard enthalpy of atomisation | ΔH°at | Enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. |
| Electron affinity | ΔH°ea | Enthalpy change when one mole of gaseous atoms gains one mole of electrons. |
| Ionisation energy | ΔH°ie | Enthalpy change when one mole of gaseous atoms loses one mole of electrons. |
Important: The standard enthalpy of formation of an element in its standard state is zero by definition. For example, ΔH°f for O₂(g), H₂(g), and C(graphite) are all zero.
Hess's law states that the total enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same. This is because enthalpy is a state function — it depends only on the current state, not on how that state was reached.
Why it is useful: Hess's law allows us to calculate enthalpy changes that are difficult or impossible to measure directly. For example, we can calculate the enthalpy of formation of carbon monoxide (CO) even though it cannot be synthesised directly from carbon and oxygen without also forming CO₂.
Example: Calculate the enthalpy of formation of methane (CH₄) given:
Solution: We want: C(s) + 2H₂(g) → CH₄(g).
Reverse reaction 3: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890 kJ/mol.
Add reaction 1 and 2 × reaction 2:
C(s) + O₂(g) → CO₂(g) ΔH = −393
2H₂(g) + O₂(g) → 2H₂O(l) ΔH = −572
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890
Sum: C(s) + 2H₂(g) → CH₄(g) ΔH = −393 − 572 + 890 = −75 kJ/mol.
So the standard enthalpy of formation of methane is −75 kJ/mol.
Hess's law can be applied systematically using standard enthalpies of formation:
ΔH°reaction = Σ ΔH°f(products) − Σ ΔH°f(reactants)
Example: Calculate the enthalpy of combustion of methane using standard enthalpies of formation:
Solution: Reaction: CH₄ + 2O₂ → CO₂ + 2H₂O
ΔH°reaction = [ΔH°f(CO₂) + 2 × ΔH°f(H₂O)] − [ΔH°f(CH₄) + 2 × ΔH°f(O₂)]
= [−393 + 2(−286)] − [−75 + 0]
= [−393 − 572] − [−75]
= −965 + 75 = −890 kJ/mol.
Every chemical reaction involves breaking bonds in the reactants and forming new bonds in the products.
The overall enthalpy change is the difference between the energy absorbed to break bonds and the energy released when new bonds form:
ΔH = Σ(bond energies of bonds broken) − Σ(bond energies of bonds formed)
Example: Calculate the enthalpy change for the reaction H₂ + Cl₂ → 2HCl. Bond energies (kJ/mol): H–H = 436, Cl–Cl = 243, H–Cl = 431.
Solution:
Bonds broken: H–H + Cl–Cl = 436 + 243 = 679 kJ/mol.
Bonds formed: 2 × H–Cl = 2 × 431 = 862 kJ/mol.
ΔH = 679 − 862 = −183 kJ/mol. The reaction is exothermic.
Note: Average bond energies are used because the exact energy of a bond can vary depending on the molecule it is in.
A Born-Haber cycle is a special application of Hess's law used to calculate the lattice enthalpy of an ionic compound. The lattice enthalpy is the energy required to break one mole of an ionic compound into its gaseous ions.
Steps in a Born-Haber cycle for sodium chloride (NaCl):
Lattice enthalpy calculation:
ΔH°f(NaCl) = ΔH°sub(Na) + ΔH°ie(Na) + ½ΔH°diss(Cl₂) + ΔH°ea(Cl) − ΔH°lattice
−411 = 108 + 496 + 122 − 349 − ΔH°lattice
−411 = 377 − ΔH°lattice
ΔH°lattice = 377 + 411 = 788 kJ/mol
So the lattice enthalpy of NaCl is +788 kJ/mol.
Why this matters: Born-Haber cycles allow us to calculate lattice enthalpies that cannot be measured directly. They also help us understand the stability of ionic compounds.
Entropy (S) is a measure of the disorder or randomness of a system. The greater the disorder, the higher the entropy.
Second law of thermodynamics: The total entropy of the universe increases in any spontaneous process.
Change in entropy (ΔS):
Example: When ice melts, the ordered structure of the solid becomes the more disordered structure of the liquid. ΔS is positive.
| Process | Entropy Change |
|---|---|
| Solid → Liquid | Increases (+) |
| Liquid → Gas | Increases (+) |
| Gas → Liquid | Decreases (−) |
| Dissolving a solid in water | Usually increases (+) |
| Increasing temperature | Increases (+) |
| More gas moles on product side | Increases (+) |
| Fewer gas moles on product side | Decreases (−) |
The Gibbs free energy change (ΔG) determines whether a reaction is spontaneous. It combines enthalpy and entropy:
ΔG = ΔH − TΔS
Where:
| ΔG | Meaning |
|---|---|
| Negative | Reaction is spontaneous (thermodynamically favourable). |
| Zero | System is at equilibrium. |
| Positive | Reaction is non-spontaneous (not thermodynamically favourable). |
Important: A negative ΔG means the reaction can happen, but it does not mean it will happen quickly. Kinetics (activation energy) determines the rate.
| ΔH | ΔS | Spontaneity |
|---|---|---|
| Negative (exothermic) | Positive (more disorder) | Always spontaneous |
| Positive (endothermic) | Negative (less disorder) | Never spontaneous |
| Negative (exothermic) | Negative (less disorder) | Spontaneous at low temperatures |
| Positive (endothermic) | Positive (more disorder) | Spontaneous at high temperatures |
(This table is one of the most important in all of thermodynamics. It tells you that whether a reaction happens depends on a tug-of-war between enthalpy and entropy. Sometimes enthalpy wins, sometimes entropy wins, and sometimes temperature decides the winner. Once you understand this, you understand why some reactions happen and some do not.)
Example: For a reaction, ΔH = −50 kJ/mol and ΔS = +100 J/K·mol. Calculate ΔG at 298 K and determine whether the reaction is spontaneous.
Solution: ΔG = ΔH − TΔS. Convert ΔS to kJ: 100 J/K·mol = 0.1 kJ/K·mol.
ΔG = −50 − (298 × 0.1) = −50 − 29.8 = −79.8 kJ/mol.
Since ΔG is negative, the reaction is spontaneous.
Example 2: For a reaction, ΔH = +20 kJ/mol and ΔS = −50 J/K·mol. Is the reaction spontaneous at any temperature?
Solution: ΔG = ΔH − TΔS = +20 − T(−0.05) = +20 + 0.05T. Since both terms are positive, ΔG is always positive for all temperatures. The reaction is never spontaneous.
| Field | Application |
|---|---|
| Industry | Designing efficient chemical processes; predicting whether reactions will occur; optimising temperature and pressure. |
| Biology | Understanding metabolism, protein folding, and the spontaneity of biochemical reactions. |
| Environmental science | Predicting the fate of pollutants; understanding climate change. |
| Materials science | Designing new materials with desired stability and properties. |
| Energy | Developing efficient fuels and energy storage systems. |
Question: Calculate the enthalpy of combustion of ethanol (C₂H₅OH) given:
Solution: Reaction: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
ΔH°reaction = [2 × ΔH°f(CO₂) + 3 × ΔH°f(H₂O)] − [ΔH°f(C₂H₅OH) + 3 × ΔH°f(O₂)]
= [2(−393) + 3(−286)] − [−278 + 0]
= [−786 − 858] − [−278]
= −1644 + 278 = −1366 kJ/mol.
Question: Calculate the enthalpy change for the reaction: CH₄ + 2O₂ → CO₂ + 2H₂O. Bond energies (kJ/mol): C–H = 413, O=O = 498, C=O = 745, O–H = 464.
Solution:
Bonds broken: 4 × C–H + 2 × O=O = 4(413) + 2(498) = 1652 + 996 = 2648 kJ/mol.
Bonds formed: 2 × C=O + 4 × O–H = 2(745) + 4(464) = 1490 + 1856 = 3346 kJ/mol.
ΔH = 2648 − 3346 = −698 kJ/mol.
Question: Calculate the lattice enthalpy of magnesium oxide (MgO) given:
Solution: ΔH°f(MgO) = ΔH°sub + ΔH°ie1 + ΔH°ie2 + ½ΔH°diss + ΔH°ea1 + ΔH°ea2 − ΔH°lattice
−602 = 148 + 738 + 1451 + 249 − 141 + 798 − ΔH°lattice
−602 = 3243 − ΔH°lattice
ΔH°lattice = 3243 + 602 = 3845 kJ/mol.
Question: For a reaction, ΔH = −100 kJ/mol and ΔS = −200 J/K·mol. Calculate ΔG at 298 K and determine whether the reaction is spontaneous.
Solution: ΔS = −200 J/K·mol = −0.2 kJ/K·mol.
ΔG = ΔH − TΔS = −100 − (298 × −0.2) = −100 + 59.6 = −40.4 kJ/mol.
Since ΔG is negative, the reaction is spontaneous at 298 K.
1. State Hess's law.
Hess's law states that the total enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same.
2. Define standard enthalpy of formation.
The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.
3. What is a Born-Haber cycle used for?
A Born-Haber cycle is used to calculate the lattice enthalpy of an ionic compound using Hess's law.
4. Define entropy.
Entropy is a measure of the disorder or randomness of a system. The greater the disorder, the higher the entropy.
5. State the equation for Gibbs free energy change.
ΔG = ΔH − TΔS.
6. What does a negative ΔG indicate?
A negative ΔG indicates that the reaction is spontaneous (thermodynamically favourable).
7. Predict the sign of ΔS for the following processes: (a) melting ice, (b) condensing steam, (c) dissolving salt in water.
(a) Melting ice: ΔS positive (solid → liquid, more disorder). (b) Condensing steam: ΔS negative (gas → liquid, less disorder). (c) Dissolving salt in water: ΔS positive (solid → dissolved ions, more disorder).
Question 1 [10 marks]
(a) State Hess's law and explain why it is useful. [4]
(b) Calculate the enthalpy of formation of methane (CH₄) given:
C(s) + O₂(g) → CO₂(g) ΔH = −393 kJ/mol
H₂(g) + ½O₂(g) → H₂O(l) ΔH = −286 kJ/mol
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890 kJ/mol [6]
(a) Hess's law states that the total enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same. It is useful because it allows us to calculate enthalpy changes that are difficult or impossible to measure directly. (4 marks: 2 for law, 2 for usefulness.)
(b) We want: C(s) + 2H₂(g) → CH₄(g). Reverse reaction 3: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890. Add reaction 1 and 2 × reaction 2: C(s) + O₂(g) → CO₂(g) ΔH = −393; 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = −572. Sum: C(s) + 2H₂(g) → CH₄(g) ΔH = −393 − 572 + 890 = −75 kJ/mol. (6 marks: 2 for reversal, 2 for addition, 2 for answer.)
Question 2 [10 marks]
(a) Define lattice enthalpy. [2]
(b) Describe the steps in a Born-Haber cycle for sodium chloride and explain how it is used to calculate the lattice enthalpy. [5]
(c) Explain why the lattice enthalpy of MgO is much higher than that of NaCl. [3]
(a) Lattice enthalpy is the energy required to break one mole of an ionic compound into its gaseous ions. (2 marks: 1 for "break into gaseous ions", 1 for "one mole".)
(b) Steps: sublimation of Na, ionisation of Na, dissociation of Cl₂, electron affinity of Cl, formation of NaCl. Using Hess's law, ΔH°f(NaCl) = ΔH°sub + ΔH°ie + ½ΔH°diss + ΔH°ea − ΔH°lattice. Rearranging gives ΔH°lattice = ΔH°sub + ΔH°ie + ½ΔH°diss + ΔH°ea − ΔH°f. (5 marks: 2 for steps, 2 for equation, 1 for rearrangement.)
(c) MgO has a much higher lattice enthalpy than NaCl because the ions in MgO carry higher charges (Mg²⁺ and O²⁻ vs Na⁺ and Cl⁻). The electrostatic force of attraction is proportional to the product of the charges, so the attraction in MgO is much stronger. Also, the ions in MgO are smaller, bringing them closer together and further strengthening the attraction. (3 marks: 1 for higher charges, 1 for smaller ions, 1 for stronger attraction.)
Question 3 [10 marks]
(a) Define entropy and state the second law of thermodynamics. [3]
(b) Predict the sign of ΔS for: (i) 2H₂(g) + O₂(g) → 2H₂O(l), (ii) CaCO₃(s) → CaO(s) + CO₂(g). [4]
(c) For a reaction, ΔH = +50 kJ/mol and ΔS = +150 J/K·mol. Calculate ΔG at 298 K and determine whether the reaction is spontaneous. [3]
(a) Entropy is a measure of the disorder or randomness of a system. The second law of thermodynamics states that the total entropy of the universe increases in any spontaneous process. (3 marks: 1 for entropy, 1 for second law, 1 for clarity.)
(b) (i) ΔS is negative (3 moles of gas → 2 moles of liquid, less disorder). (ii) ΔS is positive (solid → solid + gas, more disorder). (4 marks: 2 for each prediction + reason.)
(c) ΔS = 150 J/K·mol = 0.15 kJ/K·mol. ΔG = ΔH − TΔS = +50 − (298 × 0.15) = 50 − 44.7 = +5.3 kJ/mol. ΔG is positive, so the reaction is not spontaneous at 298 K. (3 marks: 1 for conversion, 1 for calculation, 1 for conclusion.)
Challenge 1: Explain why a reaction with a positive ΔH (endothermic) can still be spontaneous.
A reaction with a positive ΔH (endothermic) can still be spontaneous if the entropy change (ΔS) is positive and large enough to make ΔG negative. According to the equation ΔG = ΔH − TΔS, a positive ΔH can be overcome by a large positive TΔS term. This means that at high temperatures, the entropy term dominates, and the reaction becomes spontaneous. A classic example is the melting of ice: ΔH is positive (heat is absorbed), but the reaction is spontaneous above 0°C because the entropy increase (solid → liquid) is large enough. Another example is the dissolution of ammonium nitrate in water, which is endothermic but spontaneous because the entropy of the solution is much higher than that of the solid.
Challenge 2: Explain why the lattice enthalpy of an ionic compound cannot be measured directly, and how Born-Haber cycles solve this problem.
The lattice enthalpy of an ionic compound cannot be measured directly because it would require separating the ions in a crystal lattice into gaseous ions — a process that cannot be carried out in a single step in the laboratory. You would need to break the ionic bonds without also breaking other bonds or causing other reactions. Born-Haber cycles solve this problem by using Hess's law to calculate the lattice enthalpy indirectly. The cycle relates the lattice enthalpy to other enthalpy changes that can be measured experimentally, such as the enthalpy of formation, sublimation, ionisation energy, dissociation energy, and electron affinity. By constructing a cycle and applying Hess's law, the lattice enthalpy can be calculated from these measurable quantities. This is one of the most powerful applications of Hess's law.
Challenge 3: Explain the difference between thermodynamic spontaneity and kinetic rate, using an example.
Thermodynamic spontaneity refers to whether a reaction is energetically favourable — whether ΔG is negative. A spontaneous reaction can occur without external input of energy. Kinetic rate refers to how fast a reaction actually happens, which depends on the activation energy. A reaction can be thermodynamically spontaneous but kinetically very slow. A classic example is the reaction between hydrogen and oxygen to form water: 2H₂ + O₂ → 2H₂O. This reaction is highly spontaneous (ΔG is very negative), but at room temperature, it does not happen at any noticeable rate because the activation energy is very high. A spark or catalyst is needed to provide the activation energy and start the reaction. Once started, it proceeds rapidly and explosively. So thermodynamic spontaneity tells you if a reaction can happen, but kinetic rate tells you if it will happen quickly. Both are important in understanding chemical reactions.
"Energy and entropy — the two forces that decide whether a reaction happens. Now you understand the tug-of-war that drives all of chemistry."
