How Chemists Measure the Heat Released When Fuel Burns
Define enthalpy and relate it to internal energy, understand enthalpies of combustion, formation and neutralisation, and calculate enthalpy changes using Hess's law and bond enthalpies.
Why do chemists measure heat as enthalpy?
Most reactions in a laboratory, a kitchen or a factory take place in open vessels at constant atmospheric pressure. Under those conditions, the heat taken in or given out equals the change in a quantity called enthalpy — which makes enthalpy the most practical way to compare fuels, foods and reactions.
This lesson covers enthalpy and its main types, and how to calculate enthalpy changes using Hess's law and bond enthalpies.
This lesson covers enthalpy and its main types, and how to calculate enthalpy changes using Hess's law and bond enthalpies.
What is enthalpy, and what are enthalpies of combustion, formation and neutralisation?
**Enthalpy, , is a state function whose change equals the heat exchanged at constant pressure, and standard enthalpies of combustion, formation and neutralisation describe the heat changes of particular kinds of reaction under standard conditions.
Enthalpy and internal energy:**
where is the change in the number of moles of gas.
Worked example 1. For , kJ at 298 K and :
Exothermic and endothermic. A negative means heat is released; a positive means heat is absorbed.
**Standard enthalpy of combustion, .** The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions — for methane, , it is kJ mol.
**Standard enthalpy of formation, .** The enthalpy change when one mole of a compound forms from its elements in their most stable states:
- Water: , kJ mol
- The standard enthalpy of formation of an element in its most stable state, such as oxygen gas or graphite, is zero
Enthalpy of neutralisation. The enthalpy change when one mole of water forms from the reaction of an acid with a base in dilute solution:
- For any strong acid with any strong base it is about kJ mol, because the reaction is always
- For a weak acid or base it is less negative, because some energy is used to ionise the weak electrolyte
Other enthalpy changes. Enthalpies of fusion, vaporisation, atomisation and solution apply the same idea to other physical and chemical changes.
An everyday example. Comparing LPG with kerosene as a cooking fuel is partly a comparison of enthalpies of combustion — how much heat each releases per kilogram burned.
The substance. Formation enthalpies are measured against elements, not against an absolute zero — assigning zero to elements in their standard states gives every compound a common reference point.
Enthalpy and internal energy:**
where is the change in the number of moles of gas.
Worked example 1. For , kJ at 298 K and :
Exothermic and endothermic. A negative means heat is released; a positive means heat is absorbed.
**Standard enthalpy of combustion, .** The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions — for methane, , it is kJ mol.
**Standard enthalpy of formation, .** The enthalpy change when one mole of a compound forms from its elements in their most stable states:
- Water: , kJ mol
- The standard enthalpy of formation of an element in its most stable state, such as oxygen gas or graphite, is zero
Enthalpy of neutralisation. The enthalpy change when one mole of water forms from the reaction of an acid with a base in dilute solution:
- For any strong acid with any strong base it is about kJ mol, because the reaction is always
- For a weak acid or base it is less negative, because some energy is used to ionise the weak electrolyte
Other enthalpy changes. Enthalpies of fusion, vaporisation, atomisation and solution apply the same idea to other physical and chemical changes.
An everyday example. Comparing LPG with kerosene as a cooking fuel is partly a comparison of enthalpies of combustion — how much heat each releases per kilogram burned.
The substance. Formation enthalpies are measured against elements, not against an absolute zero — assigning zero to elements in their standard states gives every compound a common reference point.
Formula
How do you calculate enthalpy changes using Hess's law and bond enthalpies?
**Hess's law states that the total enthalpy change of a reaction is the same whether it happens in one step or several, so , while for gaseous reactions the enthalpy change can be estimated as the bond enthalpies of reactants minus those of products.
Hess's law of constant heat summation:**
Worked example 1 — two routes to carbon dioxide.
- , kJ
- , kJ
- Direct route: kJ, the enthalpy of formation of carbon dioxide
Worked example 2 — combustion of methane from formation enthalpies. With values of kJ mol for methane, for carbon dioxide and for liquid water:
Bond enthalpies, for gaseous reactions:
Worked example 3. For , with bond enthalpies of 436 kJ mol for H-H, 242 for Cl-Cl and 431 for H-Cl:
More energy is released in forming the new bonds than is used in breaking the old ones, so the reaction is exothermic.
An everyday example. Working out the heat released by the methane in biogas uses exactly these formation enthalpies, without anyone measuring each step separately.
The substance. Bond enthalpy calculations are only approximate — they use average bond enthalpies and strictly apply to gases, so formation enthalpies give more reliable results.
Hess's law of constant heat summation:**
Worked example 1 — two routes to carbon dioxide.
- , kJ
- , kJ
- Direct route: kJ, the enthalpy of formation of carbon dioxide
Worked example 2 — combustion of methane from formation enthalpies. With values of kJ mol for methane, for carbon dioxide and for liquid water:
Bond enthalpies, for gaseous reactions:
Worked example 3. For , with bond enthalpies of 436 kJ mol for H-H, 242 for Cl-Cl and 431 for H-Cl:
More energy is released in forming the new bonds than is used in breaking the old ones, so the reaction is exothermic.
An everyday example. Working out the heat released by the methane in biogas uses exactly these formation enthalpies, without anyone measuring each step separately.
The substance. Bond enthalpy calculations are only approximate — they use average bond enthalpies and strictly apply to gases, so formation enthalpies give more reliable results.
Exam tip
What earns full marks on enthalpy calculations?
Write every equation with state symbols and multiply each enthalpy by its coefficient before adding — a missed coefficient is the easiest slip in Hess's law problems.
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- Reaction enthalpy from formation data: products minus reactants
- Reaction enthalpy from bond enthalpies: bonds broken minus bonds formed
- Strong acid with strong base: about kJ mol
- Elements in their standard states have zero enthalpy of formation
The trap. Using the same order for both methods. Formation enthalpies use products minus reactants, but bond enthalpies use reactants minus products.
-
- Reaction enthalpy from formation data: products minus reactants
- Reaction enthalpy from bond enthalpies: bonds broken minus bonds formed
- Strong acid with strong base: about kJ mol
- Elements in their standard states have zero enthalpy of formation
The trap. Using the same order for both methods. Formation enthalpies use products minus reactants, but bond enthalpies use reactants minus products.
Did you know
How do instant cold packs and hot packs work?
First-aid kits for sports often carry instant cold packs. Squeezing the pack breaks an inner pouch of water, which dissolves a salt such as ammonium nitrate. That dissolution is strongly endothermic, so the pack turns icy cold within seconds.
Instant hot packs use the opposite effect: dissolving salts such as calcium chloride, or letting a supersaturated solution crystallise, releases heat.
Both are everyday uses of the enthalpy of solution — the heat change when a substance dissolves.
Instant hot packs use the opposite effect: dissolving salts such as calcium chloride, or letting a supersaturated solution crystallise, releases heat.
Both are everyday uses of the enthalpy of solution — the heat change when a substance dissolves.
Exam relevance
How do JEE Main and NEET test enthalpy, Hess's law and bond enthalpies?
Thermodynamics is a recurring chapter in both JEE Main and NEET, and its enthalpy questions usually appear as numericals.
What gets asked. **Converting between and , reaction enthalpies from formation or combustion data, Hess's law cycles, bond enthalpy calculations, and why the enthalpy of neutralisation is less negative for weak acids.
Question types. Mostly numerical questions, with some statement-based questions on definitions and sign conventions.
Why it matters later. Enthalpy combines with entropy to give Gibbs energy in the next part of this chapter, and enthalpy changes explain temperature effects in Equilibrium.
The trap that costs marks. Counting liquids and solids in ** — only gaseous moles count when converting between and .
What gets asked. **Converting between and , reaction enthalpies from formation or combustion data, Hess's law cycles, bond enthalpy calculations, and why the enthalpy of neutralisation is less negative for weak acids.
Question types. Mostly numerical questions, with some statement-based questions on definitions and sign conventions.
Why it matters later. Enthalpy combines with entropy to give Gibbs energy in the next part of this chapter, and enthalpy changes explain temperature effects in Equilibrium.
The trap that costs marks. Counting liquids and solids in ** — only gaseous moles count when converting between and .
Key takeaways
What must you be able to do from this lesson?
- Enthalpy: , the heat change at constant pressure, with
- Types: enthalpies of combustion, formation and neutralisation, and why elements in their standard states have zero formation enthalpy
- Calculations: Hess's law from formation enthalpies, and bond enthalpies as bonds broken minus bonds formed
Using bond enthalpies of 436 kJ mol for H-H, 498 for O=O and 463 for O-H, can you estimate for ?
- Types: enthalpies of combustion, formation and neutralisation, and why elements in their standard states have zero formation enthalpy
- Calculations: Hess's law from formation enthalpies, and bond enthalpies as bonds broken minus bonds formed
Using bond enthalpies of 436 kJ mol for H-H, 498 for O=O and 463 for O-H, can you estimate for ?