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General Chemistry I · Thermochemistry

Standard Enthalpies of Formation

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Study tools
  7. Sources & references

In 30 seconds

The standard enthalpy of formation (ΔH°f) of a substance is the enthalpy change when 1 mole of that substance is formed from its elements in their standard states. The "standard state" is the most stable form of a substance at 1 atm and a specified temperature (usually 25 °C = 298 K). Because formation reactions always build compounds from elements, they provide a universal set of "building blocks." Hess's law then gives a powerful shortcut: the enthalpy of any reaction can be computed from tabulated ΔH°f values with

ΔH°rxn = Σ n·ΔH°f(products) − Σ n·ΔH°f(reactants)

Why this matters

ΔH°f tables are the chemist's energy ledger. Once a table exists, the enthalpy of millions of possible reactions can be computed in seconds without doing a single experiment — just balance the equation and apply products-minus-reactants. This underpins fuel selection, explosive design, metabolic calculations, and industrial process design. It also gives a clean criterion for stability: the more negative ΔH°f, the more thermodynamically stable the compound is relative to its elements.

The college version

Key Ideas

  • Standard state examples: O₂(g), H₂(g), N₂(g), F₂(g), Cl₂(g), Br₂(l), I₂(s), C(s, graphite), and solid metals (Fe(s), Na(s), Hg(l), etc.).
  • ΔH°f = 0 for any element in its standard state. (They are already "formed" from themselves.)
  • ΔH°f is defined per mole of the compound formed.
  • The formation reaction for a compound shows 1 mol of product made from elements, e.g., C(s) + O₂(g) → CO₂(g), ΔH°f = −393.5 kJ/mol.
  • Negative ΔH°f means the compound is more stable (lower energy) than its elements; most stable compounds have negative ΔH°f. (Some, like NO and C₂H₂, are positive — they are unstable relative to their elements.)
  • The products-minus-reactants formula works for any balanced reaction, not just formation reactions.

Equations and Variables

SymbolMeaningUnits
ΔH°fStandard enthalpy of formationkJ/mol
ΔH°rxnStandard enthalpy of reactionkJ
n (or ν)Stoichiometric coefficient—
°Superscript denoting standard state (1 atm)—
  • ΔH°rxn = Σ n·ΔH°f(products) − Σ n·ΔH°f(reactants)
  • ΔH°f(element, standard state) = 0

How It Works

Because enthalpy is a state function, you can imagine any reaction as first decomposing all reactants back into their elements (costing −ΔH°f of each reactant) and then reforming the products from those elements (gaining ΔH°f of each product). The net heat change is the difference — hence "products minus reactants."

Steps:

  1. Balance the chemical equation.
  2. Look up ΔH°f for every compound in a table (per mole).
  3. Set ΔH°f = 0 for every element in its standard state.
  4. Multiply each ΔH°f by its stoichiometric coefficient.
  5. Sum the products' contributions, sum the reactants' contributions, and subtract.

Worked Example

Calculate ΔH°rxn for the combustion of propane: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l).

Given ΔH°f (kJ/mol): C₃H₈(g) = −103.85; CO₂(g) = −393.5; H₂O(l) = −285.8; O₂(g) = 0 (element in standard state).

Step 1 — Products: 3 mol CO₂ and 4 mol H₂O.

Σ n·ΔH°f(products) = 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ

Step 2 — Reactants: 1 mol C₃H₈ and 5 mol O₂.

Σ n·ΔH°f(reactants) = 1(−103.85) + 5(0) = −103.85 kJ

Step 3 — Subtract:

ΔH°rxn = (−2323.7) − (−103.85) = −2323.7 + 103.85 = −2219.9 kJ

The combustion of 1 mole of propane releases about 2220 kJ of heat (exothermic). This is the energy content that makes propane a useful fuel.

Common Confusions

  • Which state is "standard"? Br₂ is a liquid and I₂ is a solid at standard conditions; C is graphite, not diamond. Using the wrong form breaks the "ΔH°f = 0" rule.
  • Sign of the formula. It is products MINUS reactants — reversing it flips every answer.
  • Forgetting coefficients. Each ΔH°f must be multiplied by its stoichiometric coefficient (e.g., 4 × ΔH°f for 4H₂O).
  • ΔH°f of O₂ = 0, but ΔH°f of O₃ ≠ 0. Ozone is not the standard state of oxygen.
  • Units. ΔH°f is kJ/mol; ΔH°rxn is kJ (for the reaction as written). Don't drop the "per mole" distinction.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a LEGO set. The "elements" are the loose bricks — carbon, hydrogen, oxygen. The "compounds" are finished models built from those bricks. ΔH°f tells you how much heat is released or absorbed when you snap together exactly one model (one mole) from loose bricks. Loose bricks cost nothing to "form" because they're already just bricks — that's why an element's ΔH°f is zero. Now, if you want to know how much heat comes from transforming one finished model into another (a reaction), you just add up the heat of taking apart the starting models and building the new ones. That's the products-minus-reactants rule: products' formation heat minus reactants' formation heat. Build the new stuff, knock down the old stuff, and the difference is your answer.

Key takeaways

  • ΔH°f = enthalpy to form 1 mol of compound from elements in standard states.
  • ΔH°f of an element in its standard state = 0 (by definition).
  • ΔH°rxn = Σ n·ΔH°f(products) − Σ n·ΔH°f(reactants).
  • Standard state: 1 atm, 25 °C, most stable physical form.
  • More negative ΔH°f → more stable compound.
  • The formula is Hess's law applied to formation reactions.
  • What is the standard state of iodine, and what is its ΔH°f?
  • Write the formation reaction and ΔH°f for water vapor, H₂O(g).
  • For 2 H₂(g) + O₂(g) → 2 H₂O(l), calculate ΔH°rxn if ΔH°f(H₂O(l)) = −285.8 kJ/mol.
  • Why is the formula "products minus reactants" and not the reverse?
  • Answers: (1) I₂(s), ΔH°f = 0; (2) H₂(g) + ½O₂(g) → H₂O(g), ΔH°f ≈ −241.8 kJ/mol; (3) 2(−285.8) − [0 + 0] = −571.6 kJ; (4) because the reaction is conceptually products formed from elements minus reactants decomposed to elements.

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Practice General Chemistry I

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Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

  • Define the standard state and the standard enthalpy of formation (ΔH°f).
  • State and apply the rule that ΔH°f of an element in its standard state is zero.
  • Calculate ΔH°rxn from tabulated ΔH°f values using the products-minus-reactants formula.

Sources & references

  1. OpenStax, *Chemistry 2e*, §5.3 Enthalpy (standard enthalpy of formation).
  2. NIST Chemistry WebBook — standard enthalpies of formation.
  3. IUPAC "Gold Book" — standard state and standard enthalpy of formation.

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

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