Unit 6 · Topic 6.8 Beta

Enthalpy of Formation

A standard enthalpy of formation is ΔH for making 1 mol of a compound from its elements in their standard states; elements in their standard states have ΔH°f = 0.

Practice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

Chemists have measured the enthalpy of maybe a few thousand reactions directly, yet you can find ΔH for millions. The trick: measure one number per compound, the energy of making it from its elements, and let every reaction pass through the elements on paper.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Reversing a thermochemical equation

  1. changes the sign of ΔH
  2. doubles ΔH
  3. leaves ΔH unchanged
  4. makes ΔH zero
Show the answer

The energy flows the other way, so ΔH has the same size and opposite sign.

  • Correct: changes the sign of ΔH:
  • doubles ΔH:
  • leaves ΔH unchanged:
  • makes ΔH zero:

2. Which is an element?

  1. O₂
  2. H₂O
  3. CO₂
  4. NH₃
Show the answer

O₂ contains only oxygen atoms.

  • Correct: O₂:
  • H₂O:
  • CO₂:
  • NH₃:

Part 4 · See it

See it first

Enthalpy ladder for CH₄ + 2 O₂ → CO₂ + 2 H₂O(l). A dashed reference line marks the elements in their standard states (ΔH°f = 0). The reactants lie 74.8 kJ below it and the products 965.1 kJ below it. Step 1 takes the reactants apart into elements (+74.8 kJ); step 2 builds the products from elements (−965.1 kJ); overall ΔH° = −890.3 kJ.
Every reaction can be routed through the elements: take the reactants apart, then build the products. Only ΔH°f values are needed. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. ΔH°f is the enthalpy change to make 1 mol of a compound from its elements in their standard stateseach compound gets one tabulated number; elements in their standard states get zero
  2. A reaction can be imagined as reactants → elements → productstaking reactants apart costs −Σ ΔH°f(reactants); building products costs +Σ ΔH°f(products)
  3. The route does not change the overall enthalpy changeΔH°rxn = Σ n ΔH°f(products) − Σ n ΔH°f(reactants)
  4. ΔH°f values are measured for each substance and statethe result is more exact than a bond-enthalpy estimate

Part 6 · Key ideas

Key ideas

  • The standard enthalpy of formation, ΔH°f, is ΔH for making 1 mol of a compound from its elements in their standard states (most stable form at 1 atm, usually 25 °C).
  • ΔH°f = 0 for an element in its standard state: O₂(g), H₂(g), C(graphite), Br₂(l), Fe(s).
  • ΔH°rxn = Σ n ΔH°f(products) − Σ n ΔH°f(reactants): products minus reactants, each times its coefficient.
  • States matter: H₂O(l) −285.8 and H₂O(g) −241.8 kJ/mol.
  • Bond enthalpies: broken − formed. Formation: products − reactants. Do not mix them up.

Part 7 · Misconception

A common mistake

The wrong idea: Every pure substance made of one element, like O₃ or O atoms, has ΔH°f = 0.

What actually happens: Only the standard state, the most stable form at 1 atm and 25 °C, has ΔH°f = 0. O₂(g) is zero; O₃(g) and O(g) are positive.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Standard enthalpies of formation at 25 °C

Use these values for every item in this set.

ΔH°f values (kJ/mol)
SubstanceΔH°f (kJ/mol)SubstanceΔH°f (kJ/mol)
C₂H₅OH(l)−277.7NH₃(g)−46.1
CO₂(g)−393.5NO(g)+90.3
H₂O(l)−285.8NO₂(g)+33.2
H₂O(g)−241.8Fe₂O₃(s)−824.2
CO(g)−110.5Al₂O₃(s)−1675.7

1. Calculate ΔH° for the combustion of ethanol: C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l).

Type a number and its unit.

Show the answer

ΔH° = Σ ΔH°f(products) − Σ ΔH°f(reactants) = [2(−393.5) + 3(−285.8)] − [(−277.7) + 3(0)] = (−787.0 − 857.4) + 277.7 = −1366.7 kJ/mol. O₂ is an element in its standard state, so its ΔH°f is 0.

  • Answer: -1366.7 kJ/mol

2. Calculate ΔH°, in kJ/mol, for 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g), the first step of making nitric acid.

Type a number and its unit.

Show the answer

ΔH° = [4(+90.3) + 6(−241.8)] − [4(−46.1) + 5(0)] = (361.2 − 1450.8) − (−184.4) = −905.2 kJ/mol.

  • Answer: -905.2 kJ/mol

3. The thermite reaction is 2 Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2 Fe(s). Calculate ΔH°, in kJ/mol.

Type a number and its unit.

Show the answer

Al(s) and Fe(s) are elements in their standard states (ΔH°f = 0). ΔH° = [(−1675.7) + 2(0)] − [2(0) + (−824.2)] = −851.5 kJ/mol.

  • Answer: -851.5 kJ/mol

4. Which equation has ΔH° equal to the standard enthalpy of formation of NO₂(g), +33.2 kJ/mol?

  1. N₂(g) + 2 O₂(g) → 2 NO₂(g)
  2. NO(g) + ½ O₂(g) → NO₂(g)
  3. N(g) + 2 O(g) → NO₂(g)
  4. ½ N₂(g) + O₂(g) → NO₂(g)
Show the answer

A formation reaction makes exactly 1 mol of the compound from its elements in their standard states, using fractions if needed.

  • N₂(g) + 2 O₂(g) → 2 NO₂(g): This makes 2 mol of NO₂, so its ΔH° is 2 × 33.2 = 66.4 kJ/mol.
  • NO(g) + ½ O₂(g) → NO₂(g): NO is a compound, not an element; a formation reaction starts from elements only.
  • N(g) + 2 O(g) → NO₂(g): Nitrogen and oxygen atoms are not the standard states; the elements are N₂(g) and O₂(g).
  • Correct: ½ N₂(g) + O₂(g) → NO₂(g): Right: one mole of NO₂ made from its elements, each in its standard state.

Particle view

Forms of two elements

Box 1Box 2Box 3Box 4

Key: red circle, oxygen atom; dark circle, carbon atom; touching circles or a line, bonded atoms. All four boxes are at 25 °C and 1 atm.

5. Which box shows an element in its standard state, so its ΔH°f is zero?

  1. Box 1, separate oxygen atoms
  2. Box 2, O₂ molecules
  3. Box 4, O₃ molecules
  4. Box 3 and Box 4 together
Show the answer

The standard state of an element is its most stable form at 1 atm and the stated temperature (usually 25 °C). For oxygen that is O₂(g); for carbon, graphite.

  • Box 1, separate oxygen atoms: Single O atoms are far less stable than O₂; making them from O₂ absorbs energy, so their ΔH°f is positive.
  • Correct: Box 2, O₂ molecules: Right: oxygen's most stable form at 25 °C and 1 atm is O₂ gas.
  • Box 4, O₃ molecules: Ozone is a less stable form of oxygen; its ΔH°f is positive (about +143 kJ/mol).
  • Box 3 and Box 4 together: Box 3 is a standard state (graphite), but Box 4, ozone, is not.

6. Which boxes show substances with ΔH°f = 0? Select all that apply.

  1. Box 1
  2. Box 2
  3. Box 3
  4. Box 4
Show the answer

ΔH°f = 0 belongs to elements in their standard states: O₂(g) and C(graphite) here.

  • Box 1: Separate O atoms are not oxygen's standard state.
  • Correct: Box 2: O₂(g) is oxygen's standard state.
  • Correct: Box 3: Graphite is carbon's standard state, so its ΔH°f is 0.
  • Box 4: O₃(g) is not the most stable form of oxygen; its ΔH°f is positive.

7. Use ΔH°f values (C₆H₁₂O₆(s) −1273.3, CO₂(g) −393.5, H₂O(l) −285.8 kJ/mol) to calculate ΔH° for C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(l).

Type a number and its unit.

Show the answer

ΔH° = [6(−393.5) + 6(−285.8)] − [(−1273.3) + 6(0)] = (−2361.0 − 1714.8) + 1273.3 = −2802.5 kJ/mol.

  • Answer: -2802.5 kJ/mol

Part 9 · Summary

Summary

A standard enthalpy of formation is ΔH for making 1 mol of a compound from its elements in their standard states; elements in their standard states have ΔH°f = 0. Any reaction's ΔH° is the sum of ΔH°f of the products minus the sum for the reactants, each multiplied by its coefficient.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections