- What is enthalpy
- Moles from mass, volume and concentration
- Acids, alkalis and neutralisation · coming soon
One reaction, two answers, both correct
Look up the energy released when hydrogen burns, and you may find two different numbers for the same reaction. One table says . Another says . Neither table is wrong.
The difference is the water. In the first number, the water produced ends up as a liquid. In the second, it stays as steam. Steam holds more enthalpy than liquid water, so less energy has been released when the water is still a gas. The gap, 44 kJ/mol, is exactly the energy given out when one mole of steam condenses.
So a ΔH value on its own is not enough. You also need to know the exact equation, the physical state of every substance, and the conditions. The last lecture insisted on "per mole of the reaction as written". This lecture adds the second half of the agreement: a set of standard conditions, so that every chemist's numbers can be compared and collected in one table.
Standard conditions and the ⦵ symbol
A ΔH depends on more than the equation. It depends on the physical states, as the hook shows. For solutions, it depends on the concentration. It also changes a little with temperature and pressure. So chemists agree on one set of conditions.
A pressure of 100 kPa, a stated temperature (almost always 298 K, which is 25 °C), and solutions at a concentration of 1 mol/dm³.
Every substance is in its standard state: its most stable form under those conditions.
An enthalpy change measured, or worked out, for standard conditions is written with the symbol ⦵, as ΔH⦵, and read "delta H standard". Some books print a small circle instead, ΔH°. They mean the same thing. Some older books also use 101 kPa (1 atmosphere) as the standard pressure. The international rule changed to 100 kPa in 1982, and the difference to ΔH values is tiny.
Standard states need care. At 298 K and 100 kPa, oxygen is O₂(g), bromine is Br₂(l), mercury is Hg(l) and sodium is Na(s). Carbon is graphite, C(s, graphite), not diamond, because graphite is the more stable form. Water is H₂O(l). That is why the standard value for burning hydrogen is the −286 kJ/mol one.
Each kind of reaction then gets its own standard definition. Three of them are used again and again: formation, combustion and neutralisation.
Formation: building a compound from its elements
Written ΔHf⦵. The enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions.
Three words in the definition decide every equation. "One mole" means the compound has a coefficient of 1, even if the elements need fractions. "Elements" means only elements on the left, nothing else. "Standard states" means each element in its most stable form. For example:
C(s, graphite) + O₂(g) → CO₂(g) ΔHf⦵ = −393.5 kJ/mol
Na(s) + ½Cl₂(g) → NaCl(s) ΔHf⦵ = −411.2 kJ/mol
The ½ in front of O₂ and Cl₂ does not mean half a molecule. It means half a mole of molecules, which is a perfectly real amount.
Now a strange-looking rule follows from the definition. The standard enthalpy of formation of an element in its standard state is zero. Forming O₂(g) "from O₂(g)" means no change at all, so ΔH must be 0.
Why is this rule so useful? Recall from the last lecture that no one can measure the total enthalpy of a substance. Only changes can be measured. So chemists choose a reference level, exactly as map-makers measure every height from sea level. Elements in their standard states are chemistry's sea level. Every compound's ΔHf⦵ is its height above or below that level.
Look at NO(g) in the diagram. Its ΔHf⦵ is +90.3 kJ/mol, so forming it from nitrogen and oxygen takes in energy. A standard enthalpy of formation can be positive. Most compounds have negative values, which means they are lower in enthalpy than their elements. But "most" is not "all". Diamond is another example: forming it from graphite takes in about 1.9 kJ/mol, and that is why its value is not zero.
Combustion: burning one mole completely
Written ΔHc⦵. The enthalpy change when one mole of a substance is completely burned in oxygen, under standard conditions.
Here the "one mole" belongs to the substance being burned, not to a product. "Completely" means there is enough oxygen for every carbon atom to become CO₂ and every hydrogen atom to become water. Under standard conditions, that water is H₂O(l). Combustion always releases energy, so ΔHc⦵ is always negative.
Notice something about the combustion of graphite: C(s, graphite) + O₂(g) → CO₂(g). This is the same equation as the formation of CO₂. So ΔHc⦵ of graphite and ΔHf⦵ of CO₂ are the same number, −393.5 kJ/mol. The name depends on which substance you are describing. The value depends only on the equation.
The equation H₂(g) + ½O₂(g) → H₂O(l) has ΔH = −285.8 kJ/mol. Is this a standard enthalpy of formation or a standard enthalpy of combustion?
Show the answer
Both. It forms one mole of H₂O(l) from its elements in their standard states, so it is ΔHf⦵ of water. It also burns one mole of H₂(g) completely in oxygen, so it is ΔHc⦵ of hydrogen. One equation can fit two definitions. Always check an equation against each definition separately, word by word.
Combustion values are how fuels are compared. Octane, a main part of petrol, has ΔHc⦵ ≈ −5470 kJ/mol, and ethanol has −1367 kJ/mol (Engineering ToolBox, standard heats of combustion). As the last lecture showed, a fair comparison of fuels uses energy per gram. Octane gives about 48 kJ/g, ethanol about 30 kJ/g, and methane about 56 kJ/g.
Neutralisation: one mole of water
Written ΔHneut⦵. The enthalpy change when an acid and a base react to form one mole of water, under standard conditions.
This time the one mole belongs to the water. That matters for an acid like sulfuric acid, H₂SO₄, which gives two moles of water for every mole of acid.
Something remarkable happens with strong acids and strong bases. Hydrochloric acid with sodium hydroxide, nitric acid with potassium hydroxide, and many other pairs all give almost the same value. Measured values lie between about −57 and −58 kJ/mol (Chemguide). Why should different chemicals give the same number?
Write the ionic equation. A strong acid and a strong base are completely split into ions in solution. In HCl(aq) + NaOH(aq), the Na⁺ and Cl⁻ ions are there before the reaction and still there after it. They are spectator ions. The only change is:
That is it. That is the whole idea. Every strong acid with every strong base is really the same reaction, so it has the same ΔH. Books quote values such as −57.1 or −57.3 kJ/mol from experiments. Data tables for exactly standard conditions give about −55.8 kJ/mol. In an exam, use the value your question gives.
With a weak acid, the value is usually a little less negative. A weak acid such as ethanoic acid is mostly not split into ions. As it reacts, it must ionise, and for ethanoic acid this takes in a small amount of energy. One source gives −56.1 kJ/mol for ethanoic acid, against −57.9 kJ/mol for hydrochloric acid, both with sodium hydroxide (Chemguide). This is not a law, though. Hydrofluoric acid is a weak acid, yet its value is more negative, about −68 kJ/mol in commonly quoted data, because other energy changes in the solution matter too.
Sort the equations below by the definition each one matches.
Put each equation in the box it matches exactly, as written. Check three things: what is exactly one mole, what the reaction starts from, and the states.
Nine equations, sorted by the definition each one matches exactly.
| Equation | Matches | Reason |
|---|---|---|
| Na(s) + ½Cl₂(g) → NaCl(s) | Formation | One mole of NaCl from elements in standard states. |
| 2Na(s) + Cl₂(g) → 2NaCl(s) | None as written | Two moles of NaCl are formed. |
| C(s, graphite) + 2H₂(g) → CH₄(g) | Formation | One mole of CH₄ from its elements. |
| CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) | Combustion | One mole of CH₄ burned completely. |
| CO(g) + ½O₂(g) → CO₂(g) | Combustion | One mole of CO burned. CO is not an element, so this is not formation. |
| C(s, diamond) + O₂(g) → CO₂(g) | Combustion | One mole of diamond burned. Diamond is not carbon's standard state, so this is not the formation of CO₂. |
| C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g) | None as written | The water is a gas, not its standard state. |
| HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l) | Neutralisation | One mole of water from an acid and a base. |
| 2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2H₂O(l) | None as written | Two moles of water are formed. |
What should you have seen? Most mistakes come from the "one mole" and from the states. A correct-looking equation can still fail the definition because of a 2 in the wrong place, or a (g) where there should be an (l).
Worked examples
Write the equation for the standard enthalpy of formation of ethanol, C₂H₅OH(l).
Start from the product: exactly one mole of C₂H₅OH(l). The elements in it are carbon, hydrogen and oxygen, so the left side has C(s, graphite), H₂(g) and O₂(g).
Balance the atoms. 2 C atoms need 2C. 6 H atoms need 3H₂. 1 O atom needs ½O₂.
Check: C 2 = 2, H 6 = 6, O 1 = 1. ✓ The coefficient of ethanol is 1, and every element is in its standard state. ✓ No laboratory can actually make ethanol this way. The definition does not need the reaction to be possible. It only needs the number, which can be worked out from other reactions, as the lecture on Hess's law will show.
Write the equation for the standard enthalpy of combustion of ethanol.
Now the one mole is the ethanol, on the left. Complete combustion gives CO₂(g) and H₂O(l).
Balance: 2 C atoms give 2CO₂. 6 H atoms give 3H₂O. The oxygen atoms on the right are 4 + 3 = 7. Ethanol supplies 1, so 6 must come from O₂, which is 3O₂.
Check: O on the left 1 + 6 = 7, on the right 4 + 3 = 7. ✓ Compare with Example 1. Formation starts from elements and ends with one mole of ethanol. Combustion starts with one mole of ethanol and ends with oxides.
Using ΔHc⦵(octane) = −5470 kJ/mol and ΔHc⦵(ethanol) = −1367 kJ/mol, find the energy released by burning 1.00 kg of each. Use M(C₈H₁₈) = 114 g/mol and M(C₂H₅OH) = 46 g/mol.
Octane: n = 1000 g ÷ 114 g/mol ≈ 8.77 mol. Energy ≈ 8.77 × 5470 ≈ 48 000 kJ, which is about 48 MJ.
Ethanol: n = 1000 ÷ 46 ≈ 21.7 mol. Energy ≈ 21.7 × 1367 ≈ 29 700 kJ, which is about 30 MJ.
Check: 48 MJ ÷ 1 kg = 48 kJ/g, as in the text. ✓ Ethanol already contains an oxygen atom, so part of each molecule is "already burned". That is one reason it releases less energy per gram.
50.0 cm³ of 2.00 mol/dm³ HCl(aq) is mixed with 50.0 cm³ of 2.00 mol/dm³ NaOH(aq). Using ΔHneut⦵ = −57.1 kJ/mol, find the heat released.
Moles of each: n = cV = 2.00 × 50.0 ÷ 1000 = 0.100 mol.
HCl + NaOH → NaCl + H₂O is 1 : 1 : 1 : 1, so 0.100 mol of water forms.
Heat released = 0.100 × 57.1 = 5.71 kJ.
Check: the amount that matters is the moles of water, not of acid. Here they happen to be equal. The next interactive shows a case where they are not.
Here is a calculation where that difference matters. Find the mistake.
A student finds the enthalpy change of neutralisation from an experiment in which 2.85 kJ of heat was released. Tap the line with the mistake.
The working divides by the moles of acid instead of the moles of water. The mistake is in line 4.
- H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
- Moles of H₂SO₄ = 1.00 × 25.0 ÷ 1000 = 0.0250 mol
- Moles of NaOH = 1.00 × 50.0 ÷ 1000 = 0.0500 mol
- ΔH = −2.85 kJ ÷ 0.0250 mol = −114 kJ/mol (The error: the definition is per mole of water. 0.0500 mol of water formed, so ΔH = −2.85 ÷ 0.0500 = −57.0 kJ/mol.)
- So sulfuric acid gives twice the usual enthalpy of neutralisation.
Three classic mistakes
Most are, but not all. NO(g) is +90.3 kJ/mol, and diamond is +1.9 kJ/mol. Only combustion and neutralisation are always exothermic. A formation value can have either sign.
Formation: one mole of the compound formed. Combustion: one mole of the substance burned. Neutralisation: one mole of water. Writing 2H₂ + O₂ → 2H₂O gives twice the standard value, and dividing a neutralisation by the moles of acid can give twice the value too.
Only an element in its standard state has zero. Br₂(g), O(g) and C(s, diamond) are not standard states, so their values are not zero. Always check the state symbol and the form.
Work each one on paper first. Open a hint only when you are stuck.
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Write the equation for the standard enthalpy of formation of glucose, C₆H₁₂O₆(s).
Hint 1
Exactly one mole of glucose on the right; only elements in their standard states on the left.
Hint 2
Hydrogen and oxygen are diatomic gases: H₂(g) and O₂(g). Carbon is C(s, graphite).
Full answer
6C(s, graphite) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s). Check: C 6 = 6, H 12 = 12, O 6 = 6. ✓ The coefficient of glucose is 1, and no fractions are needed this time.
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Write the equation for the standard enthalpy of combustion of propane, C₃H₈(g). Using ΔHc⦵ = −2220 kJ/mol, find the heat released when 11.0 g of propane burns completely. (M = 44 g/mol.)
Hint 1
One mole of propane on the left; CO₂(g) and H₂O(l) on the right.
Hint 2
3CO₂ + 4H₂O need 10 O atoms, so 5O₂. Then n = 11.0 ÷ 44.
Full answer
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). Check: O 10 = 6 + 4. ✓ Moles of propane = 11.0 ÷ 44 = 0.250 mol. Heat released = 0.250 × 2220 = 555 kJ. Check: a quarter of a mole releases a quarter of 2220 kJ. ✓
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100 cm³ of 0.500 mol/dm³ HNO₃(aq) is mixed with 40.0 cm³ of 1.00 mol/dm³ KOH(aq). Using ΔHneut⦵ = −57.1 kJ/mol, find the heat released.
Hint 1
Find the moles of each. They react 1 : 1. Which one runs out first?
Hint 2
HNO₃: 0.0500 mol. KOH: 0.0400 mol. The water formed is set by the limiting reactant.
Full answer
Moles of HNO₃ = 0.500 × 100 ÷ 1000 = 0.0500 mol. Moles of KOH = 1.00 × 40.0 ÷ 1000 = 0.0400 mol. KOH is the limiting reactant, so 0.0400 mol of water forms, and 0.0100 mol of acid is left over. Heat released = 0.0400 × 57.1 ≈ 2.28 kJ. Check: using the acid's 0.0500 mol would give 2.86 kJ, more heat than the reaction can release. ✓
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In an engine, the water from burning a fuel leaves as steam, not as liquid. Use ΔHf⦵[H₂O(l)] = −285.8 kJ/mol and ΔHf⦵[H₂O(g)] = −241.8 kJ/mol. (a) Find the energy released per gram of hydrogen when the water stays as steam. (b) Methane releases 890 kJ/mol when its water is liquid. Find what it releases per gram when its 2 mol of water stays as steam. (c) Which fuel releases more energy per gram, and by roughly how many times?
Hint 1
Each mole of water that stays as steam keeps 285.8 − 241.8 = 44.0 kJ that would otherwise be released.
Hint 2
H₂: 241.8 kJ per 2 g. CH₄: 890 − 2 × 44.0 kJ per 16 g.
Full answer
(a) H₂(g) + ½O₂(g) → H₂O(g) releases 241.8 kJ per mole of hydrogen, which is 2 g. So 241.8 ÷ 2 ≈ 121 kJ/g. (b) Steam keeps 44.0 kJ per mole of water, so methane releases 890 − 2 × 44.0 = 802 kJ/mol, and 802 ÷ 16 ≈ 50 kJ/g. (c) Hydrogen releases about 121 ÷ 50 ≈ 2.4 times as much per gram. Check: the standard (liquid-water) values, 143 kJ/g and about 56 kJ/g, give a similar ratio, about 2.6. ✓ This is the hook's 44 kJ/mol at work: engineers compare fuels using the steam values, because that is what an engine really delivers.
Standard values turn single experiments into shared knowledge. Because every chemist uses the same conditions and the same definitions, a value measured in one laboratory can be used in another, decades later. Tables of ΔHf⦵ now cover thousands of substances.
And the sea-level picture has a powerful consequence. If every compound has a height measured from the same level, then the enthalpy change for any reaction is just the difference in height between its products and its reactants. You could find ΔH for a reaction that no one has ever carried out, using only a table. The next lecture, on Hess's law, shows exactly how.
- Standard conditions are 100 kPa, a stated temperature (usually 298 K), 1 mol/dm³ for solutions, and every substance in its standard state, shown by the symbol ⦵.
- The standard enthalpy of formation is for one mole of a compound formed from its elements in their standard states, and it is zero for an element in its standard state.
- The standard enthalpy of combustion is for one mole of a substance burned completely in oxygen, with water as a liquid, and it is always negative.
- The standard enthalpy of neutralisation is per mole of water, and strong acids with strong bases all give almost the same value because each is really H⁺ + OH⁻ → H₂O.
- Formation values can be positive or negative, so the sign must be read, never assumed.