Chemistry 2e · Thermochemistry

Enthalpy

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Enthalpy (H) is a thermodynamic property defined as the internal energy of a system plus the pressure–volume work needed to make room for it:

$H = E + PV$

Chemists use enthalpy because of a remarkable practical fact: for any process at constant pressure — which describes most laboratory and biological chemistry, including reactions open to the atmosphere — the change in enthalpy equals the heat transferred:

$Δ H = q p$

So measuring heat at constant pressure directly gives the enthalpy change, ΔH, without ever knowing the absolute values of E, P, or V. This topic explains what enthalpy change means, how to write thermochemical equations, and two tools for predicting it: formation enthalpies and .

Why this matters

Enthalpy changes answer a question people ask about every reaction: does it give off or require heat, and how much? That information drives safety and design, and food energy comes from metabolic enthalpy changes. Hess's law lets chemists calculate heats of reactions that are difficult or impossible to measure directly, such as forming carbon monoxide from its elements.

The college version

Core Concepts

Enthalpy change: endothermic vs. exothermic

The enthalpy change of a reaction is the difference between products and reactants:

$Δ H = H products - H reactants$

If ΔH < 0, the reaction is : it releases heat, and the surroundings warm up.
If ΔH > 0, the reaction is : it absorbs heat, and the surroundings cool down.

The sign convention is easy to invert: heat released by the system is negative for the system, even though it feels positive to your hand.

Thermochemical equations

A is a balanced equation that includes physical states and the enthalpy change:

$CH 4 (g) + 2 O 2 (g) \to CO 2 (g) + 2 H 2 O (l) Δ H = -890.5 kJ$

Three rules govern them:

Physical states matter: different states have different enthalpies (H₂O(l) vs. H₂O(g)).
The ΔH value applies to the reaction exactly as written.
If you multiply the equation by a factor, multiply ΔH by the same factor; if you reverse the equation, change the sign of ΔH.

Standard enthalpy of formation

The (ΔH_f^°) of a compound is the enthalpy change when one mole forms from its elements in their standard states (most stable form at 1 bar and a specified temperature, usually 25 °C). By definition, ΔH_f^°= 0 for elements in their standard states — O₂(g), graphite, and so on. Tabulated formation values make these otherwise unmeasurable reactions known numbers.

Hess's law: enthalpy is a state function

Hess's law states that a reaction's enthalpy change is the same whether it happens in one step or many, because enthalpy is a — it depends only on starting and ending states, not the path. The payoff: add, subtract, reverse, and scale known reactions to build a target reaction, doing the same to their ΔH values. This is how heats of reactions that never run cleanly are determined.

Computing ΔH from formation values

The most common exam calculation uses tabulated values directly:

$Δ H rxn ° = \sum Δ H f ° (products) - \sum Δ H f ° (reactants)$

Each term is multiplied by its coefficient from the balanced equation; forgetting this is the single most common error.

Common Confusions

Do Not Confuse	With	Difference
ΔH	ΔE	ΔH is heat at constant pressure; ΔE is heat at constant volume. They differ by the PΔV work term.
Exothermic sign	Endothermic sign	Exothermic: ΔH < 0 (heat out); endothermic: ΔH > 0 (heat in).
ΔH_f^° value	Per-reaction heat	Formation values are per mole of the compound formed; multiply by the coefficient in the reaction.
Reversing an equation	Scaling an equation	Reversing flips the sign of ΔH; scaling by a factor multiplies ΔH by that factor.
Standard state	STP (0 °C, 1 atm)	Standard state for thermochemistry is usually 1 bar and 25 °C; STP is a gas-law reference condition.
Hess's law arithmetic	Simple averaging	You add, reverse, and scale equations — never average heats unless the reactions actually combine that way.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Enthalpy change is the "heat receipt" of a reaction: how much heat the reaction gives off or takes in when it runs. If a reaction gives off heat, it's exothermic, like a campfire warming your hands. If it takes in heat, it's endothermic, like an ice pack getting cold as it absorbs heat. Hess's law says you can figure out the heat of a big reaction by adding up the heats of smaller steps — like knowing the total price of a shopping trip by adding up each item's receipt.

Worked example

Example 1: Combustion of methane from formation values

Calculate ΔH_rxn^° for:

$CH 4 (g) + 2 O 2 (g) \to CO 2 (g) + 2 H 2 O (l)$

Using ΔH_f^°: CH₄(g) = −74.6 kJ/mol, CO₂(g) = −393.5 kJ/mol, H₂O(l) = −285.8 kJ/mol, O₂(g) = 0.

Write the equation before substituting:

$Δ H rxn ° = [Δ H f ° (CO 2) + 2 Δ H f ° (H 2 O)] - [Δ H f ° (CH 4) + 2 Δ H f ° (O 2)]$

Substitute values (note the coefficient 2 on water):

$Δ H rxn ° = [(-393.5) + 2(-285.8)] - [(-74.6) + 2(0)]$

$Δ H rxn ° = (-965.1) - (-74.6) = -890.5 kJ$

The negative value confirms combustion is exothermic, releasing 890.5 kJ per mole of methane — matching the accepted −890 kJ/mol.

Example 2: Hess's law for the formation of carbon monoxide

CO forms directly from carbon and oxygen but also produces CO₂, so it is hard to measure cleanly. Use Hess's law with these known reactions:

$C (s) + O 2 (g) \to CO 2 (g) Δ H 1 = -393.5 kJ$

$CO (g) + 12 O 2 (g) \to CO 2 (g) Δ H 2 = -283.0 kJ$

Target: C(s) + 12O₂(g) → CO(g).

Reverse the second equation (CO₂ becomes a reactant) and flip its sign:

$CO 2 (g) \to CO (g) + 12 O 2 (g) Δ H = +283.0 kJ$

Add the first equation and the reversed second:

$C (s) + O 2 (g) + CO 2 (g) \to CO 2 (g) + CO (g) + 12 O 2 (g)$

Cancel CO₂ and one O₂ on each side to recover the target reaction:

$Δ H = -393.5 + 283.0 = -110.5 kJ$

So ΔH_f^°(CO) = -110.5 kJ/mol, the tabulated value — correctly canceling species confirms the manipulations.

Example 3: The thermite reaction

The thermite reaction reduces iron oxide with aluminum:

$2 Al (s) + Fe 2 O 3 (s) \to Al 2 O 3 (s) + 2 Fe (s)$

Using ΔH_f^°: Fe₂O₃(s) = −824.2 kJ/mol, Al₂O₃(s) = −1675.7 kJ/mol; Al and Fe are elements, so both are 0.

$Δ H rxn ° = [Δ H f ° (Al 2 O 3) + 2 Δ H f ° (Fe)] - [2 Δ H f ° (Al) + Δ H f ° (Fe 2 O 3)]$

$Δ H rxn ° = [(-1675.7) + 0] - [0 + (-824.2)] = -851.5 kJ$

The large negative value explains why thermite is violently exothermic — used in railway welding because it delivers intense heat in a compact package.

Key takeaways

H = E + PV; at constant pressure, ΔH = q_p — enthalpy change is heat at constant pressure.
Exothermic: ΔH < 0, heat released; endothermic: ΔH > 0, heat absorbed.
Thermochemical equations must include physical states; ΔH scales with coefficients and flips sign when reversed.
ΔH_f^°= 0 for elements in their standard states; formation values are per mole of compound.
Hess's law works because enthalpy is a state function — path independent.
ΔH_rxn^°= ∑ΔH_f^°(products) - ∑ΔH_f^°(reactants), multiplied by coefficients.
The accepted heat of methane combustion is about −890 kJ/mol — a benchmark exothermic value.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Write the defining relationship between enthalpy, internal energy, and pressure–volume work.
Show answer
H = E + PV.
Under what condition does ΔH equal the heat transferred?
Show answer
At constant pressure — the usual condition for reactions open to the atmosphere — ΔH = q_p.
A reaction is reversed. What happens to its ΔH?
Show answer
The sign reverses: a reaction that released heat absorbs heat when run backward.
Why is ΔH_f^°= 0 for O₂(g)?
Show answer
ΔH_f^° is defined as the heat of forming a substance from its elements in their standard states; forming an element from itself involves no change, so the value is zero by definition.
Using the formation values from Example 1, would burning 2 mol of CH₄ release more or less than 890.5 kJ? How much?
Show answer
More: enthalpy change scales with the amount of reaction. Doubling the coefficients doubles the heat, so 2 mol of CH₄ release 2 × 890.5 = 1781 kJ.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Enthalpy (H): Internal energy plus pressure–volume work: H = E + PV.
Enthalpy change (Δ H): Heat absorbed or released by a process at constant pressure.
Exothermic: Releases heat; ΔH < 0.
Endothermic: Absorbs heat; ΔH > 0.
Thermochemical equation: Balanced equation with states and an enthalpy change.
Standard enthalpy of formation: Heat to form 1 mol of a compound from elements in standard states.
Standard state: Most stable form of a substance at 1 bar and set temperature.
Hess's law: Total ΔH is path independent; add step heats to get the total.
State function: Property depending only on current state, not path.

Sources & references

openstax.org — Chemistry 2e

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