How to calculate heat of reaction

Heat of reaction calculations look intimidating but they are actually two short formulas used in the same order every time. Once you can find the heat released with Q = mcθ and then turn it into a value per mole, you can handle heat of neutralisation, displacement, precipitation and combustion with one method. This guide sets out that method using the thermochemistry syllabus, with a fully worked example.

The two formulas you need

Formula 1, heat change:

Q = mcθ

• Q is the heat change, in joules (J).
• m is the mass of the solution being heated, in grams (g).
• c is the specific heat capacity, taken as 4.2 J g⁻¹ °C⁻¹ for water and dilute solutions.
• θ is the temperature change, in °C (the final temperature minus the initial temperature).

Formula 2, heat of reaction per mole:

ΔH = Q ÷ n

where n is the number of moles of the reactant that determines the reaction, and ΔH is usually quoted in kJ mol⁻¹. The full method is on our heat of reaction per mole page.

The method in five steps

1. Find the mass, m. Assume the solution’s density is 1 g cm⁻³, so a volume in cm³ equals a mass in g. If you mix 50 cm³ + 50 cm³, the mass is 100 g.
2. Find θ, the temperature change: final minus initial. A rise for exothermic, a fall for endothermic.
3. Calculate Q = mcθ in joules, then divide by 1000 to get kilojoules.
4. Find the number of moles, n, of the key reactant, using moles = molarity × volume (in dm³).
5. Calculate ΔH = Q ÷ n and attach the sign: negative for exothermic, positive for endothermic.

The step-by-step for the heat part alone is on heat change of reaction if you want to drill just that.

Getting the sign right

This is where marks are lost. If the temperature rises, heat is given out, the reaction is exothermic, and ΔH is negative. If the temperature falls, heat is absorbed, the reaction is endothermic, and ΔH is positive. Always write the sign, a heat of reaction without it is not a complete answer.

Worked example: heat of neutralisation

50 cm³ of 1.0 mol dm⁻³ hydrochloric acid is mixed with 50 cm³ of 1.0 mol dm⁻³ sodium hydroxide. The temperature rises from 28.0 °C to 34.8 °C. Find the heat of neutralisation.

Step 1, mass. Total volume = 50 + 50 = 100 cm³, so m = 100 g.

Step 2, temperature change. θ = 34.8 − 28.0 = 6.8 °C.

Step 3, heat change.

Q = mcθ = 100 × 4.2 × 6.8 = 2856 J = 2.856 kJ

Step 4, moles. Moles of HCl = 1.0 × (50 ÷ 1000) = 0.05 mol. NaOH is also 0.05 mol, and they react 1 : 1, so 0.05 mol of water is formed.

Step 5, per mole and sign. The temperature rose, so the reaction is exothermic and ΔH is negative:

ΔH = −Q ÷ n = −2.856 ÷ 0.05 = −57.1 kJ mol⁻¹

That is the classic value for a strong acid neutralised by a strong alkali, which is a good sanity check. The reaction itself is on our heat of neutralisation page.

The same method for other heats of reaction

• Heat of displacement, for example excess zinc powder added to copper(II) sulfate. Use the moles of the limiting reactant (the salt in solution). The mass m is the mass of the solution.
• Heat of precipitation, mix two solutions that form a precipitate; m is the total volume of the mixture.
• Heat of combustion, here the water in a copper can is heated by a burning fuel, so m is the mass of water, and n is the moles of fuel burned (from the loss in mass of the burner). This one is measured differently, but Q = mcθ then ΔH = Q ÷ n still applies.

Common mistakes to avoid

• Using the wrong mass. In neutralisation and displacement, m is the mass of the solution, not just the water added and not the mass of the solid. In combustion, m is the mass of water heated.
• Forgetting to convert to kilojoules. ΔH is quoted in kJ mol⁻¹, so divide Q by 1000.
• Dropping the sign. Exothermic is negative, endothermic is positive, always shown.
• Using the wrong moles. Use the moles of the reactant the definition is based on (for neutralisation, the acid or alkali that gives the water formed).

Practise until the order is automatic

The sequence never changes: mass, temperature change, Q = mcθ, moles, then ΔH = Q ÷ n with a sign. Work through one neutralisation, one displacement and one combustion question and the pattern locks in. If choosing the right mass or the right moles keeps tripping you, that is a quick fix with a teacher, our online one-to-one lessons run in English from RM50 an hour, with a paid one-hour trial. See how it works if you would like these drilled against real Paper 2 questions.