Enthalpy Change: Definitions Revision Notes | A Level AQA Chemistry

1. Why Enthalpy Definitions Matter

Every time a fuel burns or a compound forms, energy is exchanged with the surroundings. To compare these energy changes meaningfully — across different labs, different countries, different years — chemists need a shared reference point. That reference point is built into the definitions of standard enthalpy changes.

At A Level, two definitions dominate the energetics topic: standard enthalpy of combustion and standard enthalpy of formation. They look similar on the surface, but they describe fundamentally different chemical processes. Getting them confused in an exam costs marks — and it happens more often than you might expect.

This lesson pins down both definitions with exam-standard precision, shows you how to spot the difference, and walks you through the kind of equation-checking questions that regularly appear on AQA papers.

2. Standard Conditions: The Shared Reference Point

Before defining either enthalpy change, you need to understand what standard conditions actually means. Standard conditions are a temperature of 298 K (25 °C) and a pressure of 100 kPa. These are the conditions under which all standard enthalpy values are measured and tabulated.

Alongside standard conditions, every definition also requires substances to be in their standard state — the most stable physical form of a substance at 298 K and 100 kPa. Carbon's standard state is graphite (not diamond). Bromine's standard state is liquid $Br_2(l)$. Oxygen's standard state is $O_2(g)$. These details matter when you write equations for $\Delta_f H^\ominus$.

The superscript $\ominus$ (the Plimsoll symbol) on any enthalpy term signals that the value was measured under these standard conditions. You will see it written as $\Delta H^\ominus$, and it is non-negotiable in AQA definitions — omitting it in an exam answer can cost you a mark.

3. Standard Enthalpy of Combustion ($\Delta_c H^\ominus$)

Think about burning methane in a Bunsen burner. The fuel reacts completely with oxygen and releases heat. The standard enthalpy of combustion captures exactly this idea — but with strict conditions attached to make it a reproducible, comparable quantity.

Three features define a valid $\Delta_c H^\ominus$ equation. First, exactly one mole of the substance being combusted appears on the left. Second, combustion is complete — carbon becomes $CO_2(g)$, hydrogen becomes $H_2O(l)$, sulphur becomes $SO_2(g)$. Third, water is produced as a liquid, not steam, because at 298 K water is in its liquid standard state.

This means the coefficients in a combustion equation are often non-integer fractions. For example, the combustion of propane requires $5O_2$, but the combustion of ethane requires $\frac{7}{2}O_2(g)$. Fractional coefficients are perfectly acceptable in thermochemical equations because you are defining the enthalpy change per mole of the substance combusted — not writing a balanced equation for a practical reaction.

4. Standard Enthalpy of Formation ($\Delta_f H^\ominus$)

Now consider a different question: how much energy is released or absorbed when a compound is built from scratch? That is what the standard enthalpy of formation measures. Instead of breaking a compound down by burning it, you are assembling it from its elements.

Three features define a valid $\Delta_f H^\ominus$ equation. First, exactly one mole of the compound being formed appears on the right. Second, all reactants are elements in their standard states — $C(s)$ as graphite, $H_2(g)$, $O_2(g)$, $N_2(g)$, and so on. Third, fractional coefficients on the left are not only acceptable but often necessary to ensure exactly one mole of product forms.

A direct consequence of this definition is that the $\Delta_f H^\ominus$ of any element in its standard state is zero. There is no enthalpy change when, say, $O_2(g)$ is formed from $O_2(g)$ — the element is already in its reference form. This zero-value convention underpins Hess's Law calculations you will meet later in the course.

5. Comparing the Two Definitions

Both definitions share the requirement for standard conditions and standard states, and both specify exactly one mole of a key substance. The critical difference lies in the direction of the process and the identity of the reactants and products.

For organic compounds, $\Delta_c H^\ominus$ values are always negative — combustion of organic substances always releases energy to the surroundings, making it exothermic. By contrast, $\Delta_f H^\ominus$ values can be either negative (exothermic formation, e.g. $CO_2$) or positive (endothermic formation, e.g. $NO_2$). This distinction is worth remembering for data-interpretation questions.

6. Writing and Checking Thermochemical Equations

Exam questions frequently ask you to write or identify the equation that represents a specific standard enthalpy change. A reliable method is to work through a short checklist for each equation you write or evaluate.

Worked example: Writing the equation for $\Delta_f H^\ominus$ of ethanol, $C_2H_5OH(l)$

Write the thermochemical equation that represents the standard enthalpy of formation of ethanol, $C_2H_5OH(l)$.

1. Product: $C_2H_5OH(l)$ — exactly one mole on the right-hand side. — The definition requires one mole of the compound as the sole product. State symbol (l) is correct because ethanol is a liquid at 298 K.
2. Ethanol contains C, H, and O. Standard states: $C(s)$ (graphite), $H_2(g)$, $O_2(g)$. — All reactants must be elements in their standard states. Carbon must be graphite, not diamond.
3. Ethanol formula: $C_2H_5OH$ contains 2 C, 6 H, 1 O. Carbon: need 2 mol $C(s)$ Hydrogen: need 3 mol $H_2(g)$ (since $3 \times 2 = 6$ H atoms) Oxygen: need $\frac{1}{2}$ mol $O_2(g)$ (since $\frac{1}{2} \times 2 = 1$ O atom) — Fractional coefficients are acceptable in thermochemical equations. The priority is one mole of product.
4. $2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)$ — Check: 2 C on each side, 6 H on each side, 1 O on each side. Balanced. One mole of ethanol on the right. All reactants are elements in standard states.
5. One mole of compound on right? Yes. Reactants are elements in standard states? Yes. State symbols included? Yes. — This equation satisfies all conditions for $\Delta_f H^\ominus$. It is ready to use in a Hess's Law cycle or Born-Haber-style calculation.

Answer: $2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)$

Worked example: Writing the equation for $\Delta_c H^\ominus$ of propane, $C_3H_8(g)$

Write the thermochemical equation that represents the standard enthalpy of combustion of propane, $C_3H_8(g)$.

1. Reactant: $C_3H_8(g)$ — exactly one mole on the left-hand side. — The definition requires one mole of the substance being combusted. Propane is a gas at 298 K.
2. Complete combustion: C $\rightarrow$ $CO_2(g)$; H $\rightarrow$ $H_2O(l)$. — Complete combustion means no CO or unburnt carbon. Water must be liquid at 298 K.
3. 3 C atoms in propane $\rightarrow$ 3 $CO_2(g)$ — One carbon atom produces one molecule of $CO_2$.
4. 8 H atoms in propane $\rightarrow$ 4 $H_2O(l)$ (since $4 \times 2 = 8$ H atoms) — Two hydrogen atoms produce one water molecule.
5. Oxygen needed: $3 \times 2 + 4 \times 1 = 6 + 4 = 10$ O atoms $\rightarrow$ 5 $O_2(g)$ — Count oxygen in products: 3 $CO_2$ gives 6 O atoms; 4 $H_2O$ gives 4 O atoms; total = 10 O atoms = 5 $O_2$ molecules.
6. $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$ — Check: 3 C each side, 8 H each side, 10 O each side. Balanced. One mole of propane on the left. Water is liquid.

Answer: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$

7. Exam Technique and Common Mistakes

AQA mark schemes for definition questions are precise. A definition that is almost right often scores zero. The following guidance reflects the exact language and conditions that examiners look for.

When an exam question asks you to 'state' a definition, give the full AQA-standard wording in one sentence. When it asks you to 'write an equation', include state symbols, balance the equation, and ensure the key substance appears as exactly one mole. These two tasks are assessed separately and each carries its own marks.

These two definitions — $\Delta_c H^\ominus$ and $\Delta_f H^\ominus$ — are the foundation for every Hess's Law calculation and Born-Haber cycle you will encounter in the rest of the energetics topic. Precision here pays dividends throughout the course.