200 cm^3 of water rises by 5.0 \,^C. Calculate the energy released. [2 marks]

Burning 0.50 g of propanol (GFM = 60.0 g) heats 200 cm^3 of water by 9.0 \,^C. Calculate the enthalpy of combustion. [3 marks]

SQA SQA Higher 2018-style practice: In a calorimetry experiment, burning 0.450 g of ethanol raised the temperature of 200 cm^3 of water by 18.5 \,^C. Using E_h = cm T, calculate the enthalpy of combustion of ethanol in kJ mol^-1. (Take c = 4.18 kJ kg^-1\,^C^-1, density of water = 1 g cm^-3, GFM of ethanol = 46.0 g.)

Markers reward the heat gained by the water, the moles of ethanol, division to get per mole, and a negative sign for the exothermic enthalpy. Heat gained by the water (mass = 0.200 kg): $E_h = cm T = 4.18 × 0.200 × 18.5 = 15.5 kJ$ Moles of ethanol burned: $n = (m)/(GFM) = (0.450)/(46.0) = 9.78 × 10^-3 mol$ Enthalpy of combustion per mole: $H = -15.59.78 × 10^-3 = -1580 kJ mol^-1$ The sign must be negative because combustion is exothermic; markers deduct for a missing sign or for leaving the mass in grams.

SQA SQA Higher 2021-style practice: Using Hess's law and the enthalpies of combustion of carbon (-394 kJ mol^-1), hydrogen (-286 kJ mol^-1) and methane (-891 kJ mol^-1), calculate the enthalpy of formation of methane, C(s) + 2H_2(g) → CH_4(g).

A Hess cycle links formation to the three combustion reactions. Forming methane from its elements can be reached by burning the elements and then "un-burning" methane: $H_f = H_c(C) + 2 H_c(H_2) - H_c(CH_4)$ Substituting: $H_f = (-394) + 2(-286) - (-891)$ $H_f = -394 - 572 + 891 = -75 kJ mol^-1$ Markers reward doubling the hydrogen term (two moles) and reversing the sign of the methane combustion because that step is run backwards in the cycle.

ScotlandChemistrySyllabus dot point

How do chemists measure and calculate the energy released by reactions?

Enthalpy changes of combustion, formation and neutralisation, the calculation of energy released using the formula relating energy, mass, specific heat capacity and temperature change, and the use of Hess's law to find an unknown enthalpy change.

An SQA Higher Chemistry answer on chemical energy, covering the enthalpies of combustion, formation and neutralisation, calculating energy released using the formula linking energy, mass, specific heat capacity and temperature change, and using Hess's law to find an unknown enthalpy change.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this key area is asking
Enthalpy changes
Calculating energy released
Worked example: enthalpy of combustion by calorimetry
Hess's law
Examples in context
Try this

What this key area is asking

The SQA wants you to define the enthalpies of combustion, formation and neutralisation, calculate the energy released or absorbed using the calorimetry formula, and use Hess's law to find an enthalpy change that cannot be measured directly. The calorimetry calculation and the Hess's law cycle are both staple paper-2 questions, so the unit-handling here is worth securing.

Enthalpy changes

Calculating energy released

The energy taken in or given out by the water (or solution) in an experiment is:

To find the enthalpy change per mole, divide the energy by the number of moles of the substance that reacted, and attach a negative sign for an exothermic reaction.

Worked example: enthalpy of combustion by calorimetry

Worked example

Burning $0.32 \text{ g}$ of methanol ( $GFM = 32.0 \text{ g}$ ) raised the temperature of $150 \text{ cm}^3$ of water by $12.0 \,^{\circ}\text{C}$ . Find the enthalpy of combustion of methanol.

Step 1: Heat gained by the water

The mass of water is $150 \text{ g} = 0.150 \text{ kg}$ :

E_h = cm\Delta T = 4.18 \times 0.150 \times 12.0 = 7.52 \text{ kJ}

Step 2: Moles of methanol burned

n = \frac{m}{GFM} = \frac{0.32}{32.0} = 0.0100 \text{ mol}

Step 3: Enthalpy per mole

\Delta H = -\frac{7.52}{0.0100} = -752 \text{ kJ mol}^{-1}

The experimental value is less exothermic than the data-book figure because heat is lost to the surroundings and combustion is incomplete, a standard evaluation point.

Hess's law

To use Hess's law, construct an energy cycle linking the unknown reaction to known reactions (often combustion or formation data), then add the steps, reversing the sign of any reaction you run backwards.

Examples in context

Calorimetry and Hess's law are exactly how the energy content printed on a food packet is determined: a measured mass of food is burned in a bomb calorimeter and the heat released raises the temperature of a known mass of water, giving the kilojoules per gram. Hess's law lets industrial chemists work out the enthalpy of reactions that are too slow, too dangerous or too incomplete to measure directly, such as the formation of a hydrocarbon from carbon and hydrogen, by combining easily measured combustion enthalpies. Rocket engineers use the same combustion enthalpies to compare fuels: the very exothermic combustion of hydrogen with oxygen, releasing about $286 \text{ kJ mol}^{-1}$ , is why liquid hydrogen and oxygen power many launch vehicles.

Try this

Q1. $200 \text{ cm}^3$ of water rises by $5.0 \,^{\circ}\text{C}$ . Calculate the energy released. [2 marks]

Cue. $E_h = cm\Delta T = 4.18 \times 0.200 \times 5.0 = 4.18 \text{ kJ}$ .

Q2. State Hess's law. [1 mark]

Cue. The enthalpy change for a reaction is independent of the route taken between the same reactants and products.

Q3. Burning $0.50 \text{ g}$ of propanol ( $GFM = 60.0 \text{ g}$ ) heats $200 \text{ cm}^3$ of water by $9.0 \,^{\circ}\text{C}$ . Calculate the enthalpy of combustion. [3 marks]

Cue. $E_h = 4.18 \times 0.200 \times 9.0 = 7.52 \text{ kJ}$ ; $n = 0.50 / 60.0 = 8.33 \times 10^{-3}$ mol; $\Delta H = -7.52 / 8.33 \times 10^{-3} = -903 \text{ kJ mol}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20184 marksIn a calorimetry experiment, burning

0.450 \text{ g}

of ethanol raised the temperature of

200 \text{ cm}^3

of water by

18.5 \,^{\circ}\text{C}

. Using

E_h = cm\Delta T

, calculate the enthalpy of combustion of ethanol in

\text{kJ mol}^{-1}

. (Take

c = 4.18 \text{ kJ kg}^{-1}\,^{\circ}\text{C}^{-1}

, density of water

= 1 \text{ g cm}^{-3}

GFM

of ethanol

= 46.0 \text{ g}

Show worked answer →

Markers reward the heat gained by the water, the moles of ethanol, division to get per mole, and a negative sign for the exothermic enthalpy.

Heat gained by the water (mass $= 0.200 \text{ kg}$ ):

E_h = cm\Delta T = 4.18 \times 0.200 \times 18.5 = 15.5 \text{ kJ}

Moles of ethanol burned:

n = \frac{m}{GFM} = \frac{0.450}{46.0} = 9.78 \times 10^{-3} \text{ mol}

Enthalpy of combustion per mole:

\Delta H = -\frac{15.5}{9.78 \times 10^{-3}} = -1580 \text{ kJ mol}^{-1}

The sign must be negative because combustion is exothermic; markers deduct for a missing sign or for leaving the mass in grams.

SQA Higher 20213 marksUsing Hess's law and the enthalpies of combustion of carbon (

-394 \text{ kJ mol}^{-1}

), hydrogen (

-286 \text{ kJ mol}^{-1}

) and methane (

-891 \text{ kJ mol}^{-1}

), calculate the enthalpy of formation of methane,

C(s) + 2H_2(g) \rightarrow CH_4(g)

Show worked answer →

A Hess cycle links formation to the three combustion reactions.

Forming methane from its elements can be reached by burning the elements and then "un-burning" methane:

\Delta H_f = \Delta H_c(C) + 2\Delta H_c(H_2) - \Delta H_c(CH_4)

Substituting:

\Delta H_f = (-394) + 2(-286) - (-891)

\Delta H_f = -394 - 572 + 891 = -75 \text{ kJ mol}^{-1}

Markers reward doubling the hydrogen term (two moles) and reversing the sign of the methane combustion because that step is run backwards in the cycle.

Related dot points

Sources & how we know this

SQA Higher Chemistry Course Specification — SQA (2018)