Enthalpy changes depend on conditions, so a value measured in a hot lab in Karachi and one measured in a cold lab in Oslo are not comparable. Chemists fixed this by agreeing on one set of conditions. The definitions on this page look like dull bookwork — they are also some of the most reliably examined marks in the whole course.
📘 What you need to know
Standard conditions: a pressure of 100 kPa, solutions at 1 mol dm−3, and every substance in its standard state.
Temperature is not part of the definition of standard state, but 298 K is normally the specified temperature.
The superscript symbol (a plimsoll line, printed as a small circle with a bar through it) means “measured under standard conditions”. On this page it is written as ΔH°.
ΔH°r reaction — for the equation exactly as written. Can be positive or negative.
ΔH°f formation — 1 mol of a compound from its elements. Can be positive or negative.
ΔH°c combustion — 1 mol of a substance burnt completely in excess oxygen. Always negative.
ΔH°neut neutralisation — 1 mol of water formed from an acid and an alkali. Always negative.
The ΔH°f of an element in its standard state is zero.
Why we need a standard at all
Enthalpy changes are sensitive to conditions. Change the pressure and a gas-phase reaction gives a slightly different value. Change the concentration of a solution and the dissolving contributes differently. Worst of all, change the state of a product and the value moves a lot: condensing water vapour to liquid water releases a large amount of extra energy, so any combustion value depends on whether you counted the water as steam or as liquid.
So the agreement is simple. Quote everything under one fixed set of conditions, mark it with a symbol, and then any two values in the world can be compared or added together.
The amber box is the odd one out. If an exam question asks you to list standard conditions, mention 298 K as the specified temperature rather than as part of the definition of standard state.
Standard state means the physical state a substance is naturally in at 100 kPa and the stated temperature. So at 298 K water is a liquid, oxygen is a gas, sodium chloride is a solid, and bromine is a liquid. Writing H2O(g) in a standard enthalpy of combustion equation is a real mistake, not a technicality.
The four standard enthalpy changes
All four are enthalpy changes, all four are quoted in kJ mol−1, and all four are measured under the same standard conditions. What separates them is what one mole refers to.
Two of the four are locked to a negative sign, and knowing which two lets you check an answer instantly. Burning something and neutralising an acid both give out energy, always.
Standard enthalpy of reaction
The enthalpy change when the amounts in the balanced equation as written react under standard conditions. Change the coefficients and you change the value, which is why the equation must always be quoted alongside it.
Standard enthalpy of formation
The enthalpy change when one mole of a compound is formed from its elements in their standard states. Two conditions, both strict: exactly one mole of the product, and elements on the left.
This gives us a very useful shortcut. If a substance already is an element in its standard state, then forming it from itself involves no change at all, so ΔH°f = 0 for O2(g), Na(s), C(graphite), Br2(l) and every other element. Zero is a real value here, not a missing one.
Watch the state of carbon. The standard state is graphite, so ΔH°f of graphite is zero but ΔH°f of diamond is not. Diamond is a different form, and forming it from graphite costs energy.
Standard enthalpy of combustion
The enthalpy change when one mole of a substance is burnt completely in excess oxygen under standard conditions. “Completely” and “excess” matter: carbon must end up as CO2, not CO or soot, and hydrogen must end up as H2O(l).
Combustion always releases energy, so this value is always negative. If you calculate a positive enthalpy of combustion, you have made an arithmetic or sign error.
Standard enthalpy of neutralisation
The enthalpy change when an acid and an alkali react to form one mole of water under standard conditions. Note carefully that the “per mole” refers to the water, not to the acid. Neutralisation is always exothermic, so the value is always negative.
The trap: one mole of what?
This is where marks disappear, so it is worth slowing down. Consider these three equations:
Equation
Is it a formation equation?
Why
Na(s) + ½Cl2(g) → NaCl(s)
Yes
Exactly 1 mol of the compound, elements on the left. Half-equations are allowed and are often needed.
2Na(s) + Cl2(g) → 2NaCl(s)
No
2 mol of product, so this is ΔH°r and equals 2 × ΔH°f.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
No
Not made from elements. This one is ΔH°neut, because 1 mol of water forms.
Fractional coefficients are your friend. Formation equations often need ½O2 or ½Cl2 so that exactly one mole of product appears. Students often “tidy up” the halves and accidentally turn a formation enthalpy into a reaction enthalpy.
WORKED EXAMPLE
The standard enthalpy of formation of ammonia is −46 kJ mol−1. Calculate ΔH°r for N2(g) + 3H2(g) → 2NH3(g).
Step 1: Write the formation equation½N₂(g) + 1½H₂(g) → NH₃(g), giving 1 mol of productStep 2: Compare with the equation asked forthe question makes 2 mol of NH₃, so everything doublesStep 3: Scale the valueΔH°r = 2 × (−46) = −92 kJ mol⁻¹ΔH°r = −92 kJ mol⁻¹the formation value is per mole of ammonia, the reaction value is per mole of equation
WORKED EXAMPLE
Identify each enthalpy change as ΔH°r, ΔH°f, ΔH°c or ΔH°neut. More than one label may apply.
(a) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
(b) CaCO3(s) → CaO(s) + CO2(g)
(c) H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
(d) Mg(s) + ½O2(g) → MgO(s)
(a) One mole of ethanol burnt in excess oxygencomplete combustion, water as liquidΔH°c (and also ΔH°r)(b) Thermal decompositionnot made from elements, nothing burnt, no water formedΔH°r only(c) Acid plus alkali, but look at the water2 mol of water forms, so this is 2 × ΔH°neutΔH°r, not ΔH°neut(d) One mole of MgO from its elementsΔH°f (also ΔH°c for magnesium, and ΔH°r)(a) c and r (b) r (c) r (d) f, c and r
WORKED EXAMPLE
ΔH°c for ethanol is −1367 kJ mol−1. Calculate the energy released when 4.60 g of ethanol burns completely. (M = 46.08 g mol−1)
Step 1: Convert mass to molesn = 4.60 ÷ 46.08 = 0.0998 molStep 2: Multiply by the energy per mole0.0998 × 1367 = 136.5 kJStep 3: Answer with the right unit and signthis is for a fixed amount, so the unit is kJ, not kJ mol⁻¹137 kJ released, ΔH = −137 kJ“energy released” is a positive quantity; ΔH for it is negative
🧩 Naming any enthalpy change in four questions
Is oxygen in excess on the left and are the products fully oxidised? If exactly one mole of fuel is burnt, it is combustion.
Are only elements on the left and exactly one mole of one compound on the right? It is formation.
Is it an acid plus an alkali making exactly one mole of water? It is neutralisation.
If none of those fit, or the amounts do not match, call it enthalpy of reaction — which is always technically correct.
💡 Exam tip
Learn all four definitions word for word. They are short, they come up constantly, and paraphrasing usually loses the mark.
Every definition needs three parts: the amount (one mole of what), the process, and the phrase under standard conditions.
Always include state symbols. A combustion equation with H2O(g) instead of H2O(l) is not a standard equation.
Quote the balanced equation whenever you quote a ΔH°r value; without it the number is meaningless.
If an answer for combustion or neutralisation comes out positive, stop and find the sign error.
Remember ΔH°f = 0 for elements. Questions rely on you knowing this rather than telling you.
⚠ Common mix-up
Saying standard conditions include 298 K by definition. Temperature must be specified separately; 298 K is conventional, not definitional.
Writing 1 atm instead of 100 kPa. The IB uses 100 kPa.
Treating 2Na + Cl2 → 2NaCl as a formation equation. Two moles of product means it is a reaction enthalpy.
Dividing a neutralisation value by the moles of acid. It is per mole of water formed.
Assuming ΔH°f is always negative. Plenty of compounds have positive formation enthalpies.
Thinking ΔH°f = 0 for an element means the element has no energy. It means we chose it as the zero point on the scale.
Avoiding fractions in equations. Half a mole of O2 is perfectly legitimate when you need one mole of product.
Up next: Calorimetry — time to stop defining enthalpy changes and start measuring them, with nothing more than a thermometer, a balance and a polystyrene cup.
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