Calculating molar enthalpy change

Every "calculate the molar enthalpy change" question follows the same five steps. Learn the route once and it works for combustion, neutralisation, dissolving and displacement.

Step 1 — Find the heat energy, Q.

$Q = m \times c \times \Delta T$

• Q = heat energy transferred, in joules (J).
• m = mass of the WATER (or solution) being heated, in grams (g) — not the mass of fuel.
• c = specific heat capacity of water = 4.18 J/g/°C (given on the data sheet — but learn it).
• ΔT = temperature change = final − start (the size of 1 °C and 1 K is the same, so the number is the same either way).

Step 2 — Convert Q from joules to kilojoules.

Molar enthalpy is quoted in kJ/mol, so divide your Q by 1000:

$Q\,(\text{kJ}) = \dfrac{Q\,(\text{J})}{1000}$

Step 3 — Find the moles of the substance that reacted, n.

• For a solid (a fuel, or a solid dissolving): $n = \dfrac{\text{mass}}{M_r}$
• For a solution (acid or alkali in neutralisation): $n = \dfrac{c \times V}{1000}$ (c in mol/dm³, V in cm³).

Step 4 — Divide to get ΔH per mole.

$\Delta H = \dfrac{Q\,(\text{kJ})}{n}$

This gives the energy change per mole — the units must be kJ/mol.

Step 5 — Put in the correct sign.

• Temperature ROSE → reaction gave out heat → exothermic → ΔH is NEGATIVE.
• Temperature FELL → reaction took in heat → endothermic → ΔH is POSITIVE.

Watch out: the most common slip is forgetting Step 2 (leaving Q in joules) or forgetting Step 5 (no sign). Both lose marks even when the maths is right.

The number you calculate from Q ÷ n is only half the answer — the sign carries just as many marks, and it tells you whether the reaction is exothermic or endothermic.

Why there is a sign at all.

The overall enthalpy change depends on the balance between bonds broken and bonds made:

• Breaking bonds takes energy IN (endothermic step).
• Making bonds gives energy OUT (exothermic step).
• If more energy is released making new bonds than is taken in breaking old bonds → the reaction releases heat overall → exothermic → ΔH negative.
• If more energy is taken in breaking bonds than is released making them → the reaction absorbs heat overall → endothermic → ΔH positive.

The shortcut from the experiment.

You almost never count bonds in these calculations — you read the sign straight off the thermometer:

What you observe	Reaction is…	Sign of ΔH
Temperature RISES (mixture warms)	EXOthermic	NEGATIVE
Temperature FALLS (mixture cools)	ENDOthermic	POSITIVE

The trap. Q from Q = mcΔT always comes out positive (you use the size of ΔT). The sign of ΔH is not decided by the maths — you must add it from whether the mixture warmed up or cooled down. Most combustion and neutralisation answers are negative; dissolving ammonium nitrate is one of the few positive ones.

ΔH = Q ÷ n is only as good as your value of n. Picking the wrong moles formula — or using the wrong substance — is where most marks are thrown away.

Combustion or dissolving a SOLID → use mass and Mr.

$n = \dfrac{\text{mass of substance (g)}}{M_r}$

• For a fuel, n is the moles of fuel burned (weigh the burner before and after; the difference is the mass burned).
• For a dissolving experiment, n is the moles of the solid added.
• Mr is the relative formula mass (add up the relative atomic masses) — not the mass of water.

Neutralisation or a reaction in SOLUTION → use concentration and volume.

$n = \dfrac{c \times V}{1000}$

• c = concentration in mol/dm³.
• V = volume in cm³ (dividing by 1000 converts cm³ to dm³).
• In a 1 : 1 neutralisation (e.g. HCl + NaOH), the acid and the alkali have the same moles, so use either one.

Worked check — which n do I divide by?

Experiment	n is the moles of…	Formula
Burning a fuel	the fuel	mass ÷ Mr
Dissolving a salt	the solid salt	mass ÷ Mr
Neutralising acid + alkali	the acid (or alkali)	cV ÷ 1000
Displacement (e.g. Zn + CuSO₄)	the limiting reactant	cV ÷ 1000 (for the solution)

Golden rule: the m in Q = mcΔT is the mass of water/solution heated, but the n in ΔH = Q ÷ n is the moles of the reacting chemical. They are two different quantities — never use the same number for both.

The final answer is only fully correct if it has three features: the right number, the right units (kJ/mol), and the right sign.

Unit checklist.

• Q from Q = mcΔT is in joules — divide by 1000 before you divide by n.
• ΔH must be quoted in kJ/mol (not J, not kJ).
• Give a sign: − for exothermic, + for endothermic.

Sensible-value check.

• Strong acid + strong alkali neutralisation is always about −57 kJ/mol — if you get −57 000 or −0.057, you forgot the ÷1000 or used the wrong moles.
• Combustion of a fuel from a school experiment is usually in the hundreds to low thousands of kJ/mol (and negative).
• If your ΔH comes out tiny (e.g. −2 kJ/mol) you probably divided by the mass of water instead of the moles of reactant.

Why school values are "too small".

Experimental combustion values are usually less negative than the data-book value because of heat loss to the surroundings (air, the calorimeter, draughts) and incomplete combustion. The calculation is still correct — the apparatus just doesn't capture all the heat.

Rounding. Keep full figures through the working and only round the final ΔH — usually to 3 significant figures unless the question says otherwise.