Chemistry: Atoms First 2e · Thermochemistry

Enthalpy

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On this page 9 sections

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

In 30 seconds

When a reaction happens in an open beaker, the pressure is constant — the atmosphere pushes on the system the whole time. Under those conditions, the heat the reaction absorbs or releases is the enthalpy change, ΔH. Enthalpy (H) is defined as internal energy plus a pressure–volume term, H = E + PV, but in practice think of it as "the heat of a process at constant pressure." It is a : its value depends only on where the system starts and ends, not on the path taken.

This topic explains the definition of enthalpy, the / distinction, how to read and use thermochemical equations, how to scale ΔH with reactant amounts, and the two big calculation tools — and standard enthalpies of formation — that let you find ΔH without running the reaction in the lab.

Why this matters

Every is a prediction of heat. Knowing whether a reaction releases heat (exothermic) or absorbs it (endothermic), and how much, matters for reactor design, safety, cooking, and biology.
Hess's law is a lab shortcut. Some reactions are too fast, too slow, too dangerous, or too incomplete to measure directly; Hess's law computes their ΔH from reactions that can be measured.
Standard enthalpies of formation make thermochemistry routine. Given a table of ΔH_f^° values, you can calculate the heat of any reaction — the most-used calculation in this chapter and a guaranteed exam item.
Food and fuel. The enthalpy of combustion of fuels and of metabolism of food are the numbers behind energy content claims; nutrition labels trace back to ΔH_c^° measurements.
Real systems run at constant pressure. Open beakers, industrial vats, and your own body operate at ~constant atmospheric pressure, so ΔH — not ΔE — matches everyday experience.

The college version

Core Concepts

Defining enthalpy

Enthalpy is defined as:

$H = E + PV$

where E is internal energy, P pressure, and V volume. For a process at constant pressure:

$Δ H = Δ E + P Δ V$

Substituting the first law (ΔE = q + w) and pressure–volume work (w = -PΔV) gives the result that matters:

$Δ H = q P$

At constant pressure, the enthalpy change equals the heat exchanged with the surroundings. That is why the coffee-cup calorimeter — a constant-pressure device — measures ΔH directly. Because E, P, and V are all state functions, enthalpy is too: ΔH depends only on initial and final states, never on the path. This property is the foundation of Hess's law.

Exothermic and endothermic processes

Exothermic: heat flows out of the system into the surroundings. ΔH < 0, surroundings warm up. Combustion, neutralization of strong acids and bases, and freezing are exothermic.
Endothermic: heat flows into the system from the surroundings. ΔH > 0, surroundings cool down. Melting ice, dissolving ammonium nitrate in water (the "instant cold pack"), and photosynthesis are endothermic.

A useful reading of thermochemical equations: treat ΔH like a product (exothermic) or a reactant (endothermic):

$CH 4 (g) + 2 O 2 (g) \to CO 2 (g) + 2 H 2 O (l) Δ H = -890.4 kJ$

This one equation states: when 1 mol of methane reacts with 2 mol of oxygen at constant pressure, 890.4 kJ of heat is released to the surroundings.

Thermochemical equations: scaling with amounts

A thermochemical equation is a recipe with a built-in amount: the ΔH value belongs to the reaction as written, with its specific coefficients. To scale:

Multiply the whole equation by a factor → multiply ΔH by the same factor (2 mol of CH₄ releases 2 × 890.4 = 1780.8 kJ).
Reverse the equation → change the sign of ΔH (making CO₂ and H₂O react to form CH₄ would absorb 890.4 kJ).
Physical states matter. ΔH depends on state: ΔH for reactions producing H₂O(l) differs from those producing H₂O(g). Always note (s), (l), (g), (aq).

Hess's law: adding reactions adds their enthalpy changes

Hess's law states that because enthalpy is a state function, the ΔH of a reaction is the sum of the ΔH values of any sequence of reactions that adds up to the target:

$Δ H target = Δ H 1 + Δ H 2 + Δ H 3 + \dots$

Strategy: (1) write the target equation; (2) find known reactions containing its reactants and products; (3) reverse or scale them as needed; (4) add the equations, canceling species on both sides; (5) add the ΔH values. If the final equation matches the target, the arithmetic is correct.

Standard enthalpies of formation

The standard enthalpy of formation, ΔH_f^°, of a compound is the enthalpy change when 1 mole forms from its elements in their standard states at 1 atm (and typically 25 °C). By definition:

$Δ H f ° (element in standard state) = 0$

So O₂(g), C(s, graphite), H₂(g), and Fe(s) all have ΔH_f^°= 0, while most compounds have nonzero (usually negative) values. Tables of ΔH_f^° give ΔH for any reaction through the summation formula:

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

where n_p and n_r are the stoichiometric coefficients. Read it as "products minus reactants, each multiplied by its coefficient."

Common Confusions

Do Not Confuse	With	Difference
ΔH (constant pressure)	ΔE (constant volume)	ΔH = q_P from coffee-cup calorimetry; ΔE = q_V from bomb calorimetry; they differ by the PΔV work term
Exothermic	Endothermic	ΔH < 0 heat out (surroundings warm); ΔH > 0 heat in (surroundings cool)
Enthalpy H	Enthalpy change ΔH	H is a property of a state; ΔH is the difference between two states — only differences are measurable
ΔH per mole	ΔH for the equation as written	ΔH_f^° is per mole formed; a thermochemical equation's ΔH belongs to its coefficients
Products minus reactants	Reactants minus products	The formula is products first; reversing the order flips the sign and gives a wrong answer
ΔH_f^°= 0 for elements	ΔH_f^°= 0 for all pure substances	Only elements in their standard states have zero formation enthalpy
Standard state (1 atm)	Standard temperature (25 °C)	"Standard state" refers to pressure (1 atm); 25 °C is a reporting convention, not part of the definition

Eli explains

The same idea, in plain words

Explain it like I’m 10

Enthalpy is a way of counting how much heat a reaction gives away or takes in while it runs at normal air pressure. Think of a campfire: burning wood gives off heat — that's exothermic, a negative number, because the wood's "heat account" goes down. A cold pack, though, soaks up heat from your skin when you crack it — that's endothermic. The cool trick of Hess's law is that you can add up smaller, easy reactions to figure out the heat of one big reaction you can't easily run — like adding up the prices of ingredients to know the price of the whole cake.

Worked example

Example 1: Hess's law for the formation of carbon monoxide

Carbon monoxide forms from carbon and oxygen, but the reaction always "overshoots" to CO₂, so its ΔH can't be measured directly. Use Hess's law with:

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

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

Step 1 — Target equation:

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

Step 2 — Reverse reaction 2 so CO appears as a product, flipping its sign:

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

Step 3 — Add the two reactions, canceling CO₂ and the O₂ halves:

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

After cancellation: C(s) + 12O₂(g) → CO(g) — matches the target.

Step 4 — Add the ΔH values:

$Δ H = (-393.5 kJ) + (+283.0 kJ) = -110.5 kJ$

So ΔH_f^°[CO(g)] = -110.5 kJ/mol, the standard table value. The pathway didn't matter — only the starting and ending states did.

Example 2: ΔH from standard enthalpies of formation

Calculate the standard enthalpy of combustion of methane:

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

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

Step 1 — Write the summation formula:

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

Step 2 — Sum the products:

$\sum n p Δ H f ° = (1)(-393.5) + (2)(-285.8) = -393.5 - 571.6 = -965.1 kJ$

Step 3 — Sum the reactants (O₂ is an element in standard state, so ΔH_f^°= 0):

$\sum n r Δ H f ° = (1)(-74.6) + (2)(0) = -74.6 kJ$

Step 4 — Subtract:

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

The reaction releases 890.5 kJ per mole of methane — the well-known heat of combustion of methane. (Dimensional check: kJ/mol × mol = kJ.)

Example 3: Scaling a thermochemical equation

Using the methane combustion from Example 2, how much heat is released when 4.00 g of CH₄ (molar mass 16.04 g/mol) is burned completely?

Step 1 — Convert mass to moles:

$n = 4.00 g 16.04 g mol -1 = 0.249 mol CH 4$

Step 2 — Scale ΔH by the mole ratio (1 mol CH₄ : −890.5 kJ):

$q = 0.249 mol CH 4 \times -890.5 kJ 1 mol CH 4 = -222 kJ$

About 222 kJ of heat is released. The mole ratio comes straight from the thermochemical equation's coefficients — the ΔH value belongs to the reaction as written, so scaling it by actual moles is the only legitimate use.

Key takeaways

Definition: H = E + PV; at constant pressure, ΔH = q_P — enthalpy change equals heat.
Enthalpy is a state function — path-independent; this is why Hess's law works.
Exothermic: ΔH < 0, heat released, surroundings warm; endothermic: ΔH > 0, heat absorbed, surroundings cool.
Scaling rules: multiply coefficients → multiply ΔH; reverse equation → flip sign; physical states matter.
Hess's law: ΔH_target = ∑ΔH_steps for any valid pathway.
Standard formation enthalpy: ΔH_f^° of an element in its standard state = 0; ΔH_rxn^°= ∑n_pΔH_f^°(products) - ∑n_rΔH_f^°(reactants).
Standard state = 1 atm pressure; tables usually report 25 °C (298 K).

Check yourself

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

Define enthalpy and state why ΔH = q_P at constant pressure.
Show answer
H = E + PV. At constant pressure, ΔH = ΔE + PΔV = (q + w) + PΔV = q - PΔV + PΔV = q_P. The PΔV work terms cancel, leaving heat.
Classify as exothermic or endothermic: (a) ice melting; (b) burning gasoline; (c) an instant cold pack dissolving; (d) freezing water.
Show answer
(a) endothermic (absorbs heat); (b) exothermic; (c) endothermic; (d) exothermic.
2H₂(g) + O₂(g) → 2H₂O(l) has ΔH = -571.6 kJ. What is ΔH when the equation is reversed, and what is ΔH per mole of H₂O formed?
Show answer
Reversed: ΔH = +571.6 kJ. Per mole of H₂O: -571.6/2 = -285.8 kJ/mol.
Why can Hess's law add ΔH values from different pathways?
Show answer
Because enthalpy is a state function — its change depends only on initial and final states, so any valid pathway gives the same ΔH.
Why is O₂(g)'s contribution always zero in the formation-enthalpy summation?
Show answer
Elements in their standard states are the reference zero of the formation-enthalpy scale: ΔH_f^°= 0 by definition, so forming O₂(g) from O₂(g) involves no enthalpy change.
A reaction's ΔH changes if a product's physical state changes (H₂O(l) vs H₂O(g)). Why?
Show answer
Different physical states have different energies (condensation releases heat), so the products' total enthalpy differs. That's why thermochemical equations must specify (s), (l), (g), (aq).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Enthalpy (H): Internal energy plus pressure–volume term, H = E + PV
Enthalpy change (Δ H): Heat absorbed or released by a process at constant pressure
Exothermic: Process with ΔH < 0, releasing heat to surroundings
Endothermic: Process with ΔH > 0, absorbing heat from surroundings
Thermochemical equation: Balanced equation with its ΔH for the reaction as written
State function: Property depending only on current state, not the path
Hess's law: ΔH of a reaction equals the sum of ΔH of any valid pathway
Standard enthalpy of formation (Δ Hf°): Heat to form 1 mol of compound from elements in standard states
Standard state: Physical form of a substance at 1 atm (often 25 °C)

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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