The enthalpy change formula

The enthalpy change formula used in calorimetry is ΔH = −q ÷ n, where q = mcΔT is the heat taken in or given out by the water or solution, and n is the number of moles that reacted. Put together, ΔH = −mcΔT ÷ n, usually in kJ mol⁻¹. The minus sign is there because when a reaction warms its surroundings (ΔT positive), the reaction itself has lost energy, so ΔH is negative: it is exothermic. If the temperature falls, ΔH is positive and the reaction is endothermic.

Step 1: the heat change, q = mcΔT

AQA's A-level specification gives q = mcΔT, where m is the mass of the substance that has the temperature change ΔT, and says students will not be expected to recall the value of c: the question gives it. Chemists usually work in grams with c = 4.18 J g⁻¹ K⁻¹, and take 1 cm³ of a dilute solution as 1 g.

The most common mistake is the mass. m is the mass of the water or solution that warms up, not the mass of the solid or fuel that reacts.

Step 2: the molar enthalpy change, ΔH = −q ÷ n

Here n is the number of moles of the substance you want the enthalpy change for: the fuel burned, the solid dissolved, or the water formed in a neutralisation. Divide q in J by n, then divide by 1000 to get kJ mol⁻¹. AQA's specification asks students to use q = mcΔT to calculate the molar enthalpy change for a reaction, and its required practical is the measurement of an enthalpy change.

Enthalpy change, ΔH, is the heat energy change at constant pressure. Standard enthalpy changes, ΔH⦵, refer to standard conditions, which AQA gives as 100 kPa and a stated temperature, such as 298 K.

Enthalpy change worked examples

Example 1: neutralisation

30.0 cm³ of 1.50 mol dm⁻³ hydrochloric acid is mixed with 30.0 cm³ of 1.50 mol dm⁻³ sodium hydroxide. The temperature rises by 9.8 °C. Find the enthalpy change of neutralisation. Take c = 4.18 J g⁻¹ K⁻¹ and the solution's density as 1.00 g cm⁻³.

The solution's mass is 30.0 + 30.0 = 60.0 g.

Moles of water formed = moles of acid = 1.50 × 30.0 ÷ 1000 = 0.0450 mol.

Example 2: combustion of an alcohol

Burning 0.46 g of ethanol (M_r = 46) under a copper can heats 150 g of water by 12.0 °C. Find the enthalpy change of combustion of ethanol.

Moles of ethanol = 0.46 ÷ 46 = 0.0100 mol.

Spirit burner experiments give values much less negative than data book values, because a lot of the heat warms the air and the can instead of the water, and some ethanol burns incompletely. This is the classic evaluation question.

Example 3: an endothermic dissolving

5.0 g of potassium chloride (M_r = 74.6) dissolves in 50.0 g of water, and the temperature falls by 6.5 °C. Find the enthalpy change of solution.

Moles of KCl = 5.0 ÷ 74.6 = 0.0670 mol. The temperature fell, so ΔT is negative, and the minus signs cancel:

Always write the sign. +20.3 kJ mol⁻¹ and −20.3 kJ mol⁻¹ mean opposite things.

Example 4: a displacement reaction

Excess zinc powder is added to 50.0 cm³ of 0.200 mol dm⁻³ copper(II) sulfate solution. The temperature rises by 10.2 °C. Find ΔH for Zn + CuSO₄ → ZnSO₄ + Cu.

q = 50.0 × 4.18 × 10.2 = 2132 J. The zinc is in excess, so the copper sulfate controls the amount: n = 0.200 × 50.0 ÷ 1000 = 0.0100 mol.

Finding ΔT from a cooling curve

In a real experiment, the solution starts cooling before all the heat has been released, so the highest reading is too low. The fix is to plot temperature against time, then extrapolate the cooling line back to the moment of mixing.

Here x is the time in minutes and y is the temperature in °C. The solution was at 20.0 °C before mixing at 3 minutes. The highest reading is 32.8 °C, but extending the straight cooling line back to 3 minutes gives 33.0 °C, so the corrected ΔT is 33.0 − 20.0 = 13.0 °C, not 12.8 °C.

Other enthalpy change formulas

Calorimetry measures ΔH directly. Two other methods calculate it from data:

• Hess's law: the enthalpy change of a reaction is the same whichever route you take. From enthalpies of formation, ΔH = ΣΔ_fH(products) − ΣΔ_fH(reactants). From enthalpies of combustion, ΔH = ΣΔ_cH(reactants) − ΣΔ_cH(products).
• Mean bond enthalpies: ΔH = Σ(bonds broken) − Σ(bonds formed). This gives an approximate value for reactions in the gas phase.

Our enthalpy calculator guide works through both methods, and the Enthalpy Calculator builds the Hess cycle or adds up the bond enthalpies for you.

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Enthalpy Calculator

Use Hess's law and bond enthalpies to find energy changes

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Try a bond enthalpy question in the course, then press Mark to check it there.

Common mistakes

• Using the mass of the reactant instead of the water or solution in q = mcΔT.
• Adding the masses of the solids in a solution reaction. Use the volume of solution as its mass in grams.
• Forgetting the sign. Temperature rise means ΔH is negative.
• Forgetting to convert J to kJ. q = mcΔT gives joules; ΔH is usually quoted in kJ mol⁻¹.
• Using moles of the reagent in excess. Use the moles of the substance that is used up, the limiting reagent.
• Too many significant figures. If ΔT is measured to 2 significant figures, the answer cannot honestly have more.

Practise energetics in a short timed paper

A-level Chemistry: Year 1 and GCSE Chemistry Higher each give you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so energy changes turn up regularly but not in every paper. The specific heat capacity side of q = mcΔT is covered in our specific heat capacity guide.

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Chemistry · EnergeticsEnthalpy calculator: ΔH by Hess's law or bond enthalpiesFind the enthalpy change of a reaction from enthalpies of formation or combustion, combined equations or bond enthalpies, with working and worked examples.