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D. Energy Changes and Rates of Reaction · Energy Changes and Rates of Reaction

Standard Enthalpies of Formation

D2.7 D2.5 D3.4 A1.12About 70 minutes
Learning goals
  • I can define standard enthalpy of formation and explain why it is zero for elements in their standard states.
  • I can write formation equations that make exactly one mole of a compound.
  • I can calculate ΔH° for a reaction from a table of standard enthalpies of formation.
  • I can work backward from a measured ΔH° to an unknown enthalpy of formation.

Words to know

standard molar enthalpy of formation (ΔH°f)
The enthalpy change when 1 mol of a compound forms from its elements in their standard states at 25 °C and 100 kPa.
standard state
The most stable form of a substance at SATP, such as O₂(g), Br₂(l) or C(graphite).
formation equation
A balanced equation that makes exactly 1 mol of a compound from its elements in their standard states.
Σ (sigma)
A symbol meaning "the sum of".

One table to rule them all

Hess's law is powerful, but searching for exactly the right set of equations for every reaction would be slow. Chemists found a shortcut: list one number for each compound, the enthalpy change for making it from its elements. With a table of these standard enthalpies of formation, you can calculate ΔH for thousands of reactions with one simple formula. Engineers use exactly this method to compare fuels, design furnaces and predict whether an industrial process will need heating or cooling.

Standard enthalpy of formation

The standard molar enthalpy of formation, ΔH°f, is the enthalpy change when one mole of a compound is formed from its elements in their standard states at SATP (25 °C and 100 kPa).

The standard state of an element is its most stable form at SATP: O₂(g), H₂(g), N₂(g), Cl₂(g), Br₂(l), I₂(s), Hg(l), C(s, graphite), Na(s), Fe(s), and so on.

By definition, ΔH°f of an element in its standard state is zero. (Forming O₂(g) from O₂(g) is no change at all.) But O₃(g), C(diamond) and Br₂(g) are not standard states, so their ΔH°f values are not zero.

A formation equation always makes exactly 1 mol of product, so fractions are common:

  • ½N₂(g) + 3/2H₂(g) → NH₃(g) ΔH°f = −46.1 kJ/mol
  • 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) ΔH°f = −277.7 kJ/mol
  • Na(s) + ½Cl₂(g) → NaCl(s) ΔH°f = −411.2 kJ/mol
SubstanceΔH°f (kJ/mol)SubstanceΔH°f (kJ/mol)
CO₂(g)−393.5NH₃(g)−46.1
CO(g)−110.5NO(g)+90.3
H₂O(l)−285.8NO₂(g)+33.2
H₂O(g)−241.8SO₂(g)−296.8
CH₄(g)−74.8SO₃(g)−395.7
C₂H₆(g)−84.7HCl(g)−92.3
C₃H₈(g)−103.8NaCl(s)−411.2
C₈H₁₈(l)−250.1CaCO₃(s)−1206.9
C₂H₄(g)+52.4CaO(s)−635.1
C₂H₂(g)+226.7MgO(s)−601.6
CH₃OH(l)−239.1Fe₂O₃(s)−824.2
C₂H₅OH(l)−277.7Al₂O₃(s)−1675.7
C₆H₁₂O₆(s)−1273.3
Standard molar enthalpies of formation at 25 °C and 100 kPa. Elements in their standard states are 0. Values in different textbooks and data tables can differ by a few tenths of a kilojoule; use the table you are given.
+2000−200−400−600−800ΔH°f (kJ/mol)C₂H₂(g): +226.7NO(g): +90.3C₂H₄(g): +52.4O₂, H₂, C(s), N₂ ...: 0NH₃(g): −46.1CH₄(g): −74.8CO(g): −110.5H₂O(g): −241.8H₂O(l): −285.8CO₂(g): −393.5MgO(s): −601.6Fe₂O₃(s): −824.2above 0: less stable thantheir elementsfar below 0: very stablecompounds
Formation enthalpies on one scale. Elements in their standard states sit at zero. Compounds with large negative values, such as CO₂ and Fe₂O₃, are much lower in enthalpy (more stable) than their elements.

The formula

ΔH°r = Σ n ΔH°f(products) − Σ n ΔH°f(reactants)

(Σ means "the sum of", and n is each substance's coefficient in the balanced equation.)

Why it works (Hess's law): imagine the reaction happening in two steps. First, break the reactants apart into their elements: that is the reverse of their formation, so it contributes −ΣnΔH°f(reactants). Second, build the products from those same elements: +ΣnΔH°f(products). Because enthalpy is a state function, the sum of the two steps equals ΔH for the reaction.

Helements: C(s) + 2H₂(g) + 2O₂(g) (H set to 0)CH₄(g) + 2O₂(g)CO₂(g) + 2H₂O(l)step 1 (up): un-form CH₄, +74.8 kJstep 2: form products−393.5 + 2(−285.8)= −965.1 kJoverall:ΔH = −890.3 kJ
Burning methane, imagined through its elements. Step 1 reverses the formation of CH₄ (+74.8 kJ); step 2 forms the products (−965.1 kJ). The sum, −890.3 kJ, is the enthalpy of combustion. (Drawn to scale.)
Worked example

Example 1: Combustion of propane

Calculate ΔH°comb for C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).

  1. Products: 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ.
  2. Reactants: 1(−103.8) + 5(0) = −103.8 kJ. (O₂ is an element in its standard state.)
  3. ΔH° = −2323.7 − (−103.8) = −2219.9 kJ.

ΔH°comb = −2219.9 kJ per mole of propane (the −2220 kJ/mol you have been using).

Worked example

Example 2: Making lime from limestone

Calculate ΔH° for CaCO₃(s) → CaO(s) + CO₂(g). Is it exothermic or endothermic?

  1. Products: (−635.1) + (−393.5) = −1028.6 kJ.
  2. Reactants: −1206.9 kJ.
  3. ΔH° = −1028.6 − (−1206.9) = +178.3 kJ.

ΔH° = +178.3 kJ, endothermic. Lime kilns must heat limestone strongly (to roughly 900 °C), which is one reason cement and lime production uses so much fuel; the CO₂ released by the reaction itself adds to the fuel's emissions.

Worked example

Example 3: Working backward to a ΔH°f

The standard enthalpy of combustion of liquid ethanol is −1366.7 kJ/mol: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). Find ΔH°f of ethanol.

  1. Write the formula with the unknown x: −1366.7 = [2(−393.5) + 3(−285.8)] − [x + 3(0)].
  2. Products: −787.0 + (−857.4) = −1644.4 kJ.
  3. −1366.7 = −1644.4 − x, so x = −1644.4 + 1366.7 = −277.7 kJ/mol.

ΔH°f(C₂H₅OH, l) = −277.7 kJ/mol, matching the table.

Worked example

Example 4: The thermite reaction

Calculate ΔH° for 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s), and the heat released per gram of reactant mixture (M: Al 26.98, Fe₂O₃ 159.70 g/mol).

  1. ΔH° = [(−1675.7) + 2(0)] − [2(0) + (−824.2)] = −851.5 kJ.
  2. Mass of mixture in the equation: 2(26.98) + 159.70 = 213.66 g.
  3. Per gram: 851.5 kJ ÷ 213.66 g = 3.985 kJ/g.

ΔH° = −851.5 kJ, about 3.99 kJ per gram of mixture. That is far less than a fuel like propane (about 50 kJ per gram of fuel), but thermite releases its heat in a small space, producing molten iron hot enough to weld railway rails.

Four checks before you calculate

  1. Is the equation balanced? 2. Did you use the right state (H₂O(l) or H₂O(g))? 3. Did you multiply each ΔH°f by its coefficient? 4. Did you subtract the reactants from the products (not the other way round)? Elements in standard states count as zero, but write them in so you can see you have not forgotten anything.
Hands-on activity

Formation equations from the kitchen

You need: a pencil, paper, and a few labelled containers from the kitchen: table salt (NaCl), sugar (sucrose, C₁₂H₂₂O₁₁), baking soda (NaHCO₃), and water (H₂O). Do not taste or mix anything; you are only reading labels.

  1. For each substance, write a balanced formation equation that makes exactly 1 mol of the compound from its elements in their standard states. Use fractions where needed.
  2. Use the table to look up ΔH°f where available (NaCl, H₂O). Write the thermochemical formation equations.
  3. For sucrose, use ΔH°f = −2226 kJ/mol to calculate its enthalpy of combustion, C₁₂H₂₂O₁₁(s) + 12O₂(g) → 12CO₂(g) + 11H₂O(l), and compare with the bomb calorimeter value of −5645 kJ/mol from the combustion calorimetry lesson.

Answers: Na(s) + ½Cl₂(g) → NaCl(s), −411.2 kJ; H₂(g) + ½O₂(g) → H₂O(l), −285.8 kJ; 12C(s) + 11H₂(g) + 11/2O₂(g) → C₁₂H₂₂O₁₁(s); Na(s) + ½H₂(g) + C(s) + 3/2O₂(g) → NaHCO₃(s). Sucrose combustion: [12(−393.5) + 11(−285.8)] − (−2226) = −4722.0 − 3143.8 + 2226 = −5639.8 kJ, within about 0.1% of the calorimeter value.

Just for fun

Formation enthalpies make chemists sound like accountants: every compound's "balance" is measured relative to its elements, which start with a value of exactly zero. Acetylene, at +226.7 kJ/mol, is the friend who is always in debt, which is why it pays out so much when it burns.

Your turn

1

Write the formation equation for CH₃OH(l), including ΔH°f.

2

Why is ΔH°f for Br₂(l) zero but ΔH°f for Br₂(g) not zero?

3

Calculate ΔH° for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).

4

Calculate ΔH° for 2SO₂(g) + O₂(g) → 2SO₃(g).

5

Calculate ΔH° for C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l) (cellular respiration overall).

6

Calculate ΔH° for 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) (a step in making nitric acid).

7

Which substance in the table is formed most exothermically per mole?

More practice: warm-up

1

What is ΔH°f for O₂(g)?

2

How many moles of product does a formation equation make?

3

Write the formula for ΔH°r using formation enthalpies.

4

What is the standard state of carbon?

5

Is the formation of NO(g) exothermic or endothermic?

More practice: on your own

1

Calculate ΔH°comb of octane per mole, 2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(l), and the energy released per gram (M = 114.26 g/mol).

2

Calculate ΔH° for Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) (a blast-furnace reaction).

3

Calculate ΔH° for C₂H₄(g) + H₂O(l) → C₂H₅OH(l) and compare with your Hess's law answer.

4

The enthalpy of combustion of benzene, C₆H₆(l), is −3267.5 kJ/mol (water as a liquid). Find ΔH°f of benzene.

5

Calculate ΔH° for CaO(s) + H₂O(l) → Ca(OH)₂(s), given ΔH°f(Ca(OH)₂, s) = −986.1 kJ/mol.

Challenge

1

Which fuel releases more energy per gram when burned to CO₂(g) and H₂O(l): methane or methanol? Show calculations. (M: CH₄ 16.05, CH₃OH 32.05 g/mol)

2

For propane burning to CO₂(g) and H₂O(g), find ΔH° and explain why it differs from −2219.9 kJ.

3

A reaction has ΔH° = −92.2 kJ for N₂(g) + 3H₂(g) → 2NH₃(g). Without the table, find ΔH°f of NH₃ and explain your reasoning.

Check your understanding

Quiz

1.

Which substance has a standard enthalpy of formation of exactly zero?

2.

Which equation is a correct standard formation equation for NH₃(g)?

3.

Using ΔH°f values, ΔH° for 2CO(g) + O₂(g) → 2CO₂(g) is

4.

In ΔH°r = ΣnΔH°f(products) − ΣnΔH°f(reactants), the subtraction of the reactants represents

5.

ΔH°f(CO₂) = −393.5 kJ/mol and ΔH°f(H₂O, l) = −285.8 kJ/mol. A fuel C₂H₆(g) has ΔH°f = −84.7 kJ/mol. What is its molar enthalpy of combustion (water as liquid)?

6.

A compound has a large positive ΔH°f. This means

7.

Using ΔH°f values, ΔH° for CaCO₃(s) → CaO(s) + CO₂(g) is

For parents and teachers

Big ideas (D2.7, D2.5, D3.4): the standard enthalpy of formation, ΔH°f, is the enthalpy change 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° can be calculated as ΣnΔH°f(products) − ΣnΔH°f(reactants), which is Hess's law applied through the elements.

How to help: Have your teen explain, using the methane diagram, why the reactants are subtracted. Ask them to check each calculation with the four checks in the tip box.

Common difficulties:

  • Forgetting to multiply by coefficients.
  • Subtracting products from reactants (backward), which flips the sign.
  • Using H₂O(g) instead of H₂O(l) (or the reverse).
  • Writing formation equations that make 2 mol, or starting from atoms like O(g) instead of O₂(g).
  • Thinking O₂ "has no energy" because its ΔH°f is zero. Zero is just the reference point, like sea level.

Activity safety: reading labels only; nothing is mixed or tasted.