Define enthalpy change, classify reactions as exothermic or endothermic, and use calorimetry data with energy profile diagrams.

Exothermic and endothermic reactions, enthalpy change, energy profile diagrams, and calorimetry calculations using q equals mcT.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-30

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What this dot point is asking

You must define enthalpy change, classify and represent reactions energetically, and calculate energy changes from calorimetry data.

Exothermic and endothermic reactions

Every reaction involves breaking bonds (which absorbs energy) and making bonds (which releases energy). The net result determines the energy change.

In an exothermic reaction more energy is released forming new bonds than is absorbed breaking old ones, so heat flows out to the surroundings and the temperature rises. $\Delta H$ is negative.
In an endothermic reaction more energy is absorbed breaking bonds than is released, so heat flows in from the surroundings and the temperature falls. $\Delta H$ is positive.

Energy profile diagrams

An energy profile plots enthalpy against reaction progress. The reactants and products sit at different enthalpy levels, and the activation energy is the height of the peak above the reactants.

For an exothermic reaction the products lie below the reactants, so $\Delta H$ is negative. For an endothermic reaction the products lie above the reactants, so $\Delta H$ is positive. The difference in height between reactants and products is $\Delta H$ .

Calorimetry

We measure enthalpy changes by capturing the heat in a known mass of water and recording the temperature change. The heat transferred is found from:

To convert to a molar enthalpy change, divide the heat by the number of moles reacting and attach the correct sign: $\Delta H = -q / n$ for an exothermic process (negative because heat left the reaction).

example

Enthalpy of combustion from calorimetry

Burning $0.50\ \text{g}$ of ethanol raises the temperature of $200\ \text{g}$ of water by $19.0\ ^{\circ}\text{C}$ . Find the molar enthalpy of combustion. ( $M(\text{ethanol}) = 46\ \text{g mol}^{-1}$ .)

Step 1: $q = mc\Delta T = 200 \times 4.18 \times 19.0 = 15\,884\ \text{J} = 15.88\ \text{kJ}$ .

Step 2: $n(\text{ethanol}) = 0.50 / 46 = 0.01087\ \text{mol}$ .

Step 3: $\Delta H = -q / n = -15.88 / 0.01087 = -1461\ \text{kJ mol}^{-1}$ .

The reaction releases heat, so $\Delta H$ is negative, confirming combustion is exothermic.

In the exam, always show $q = mc\Delta T$ with the correct mass and specific heat, divide by moles for a molar value, and attach the sign that matches an exothermic or endothermic process.

Exam-style practice questions

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

2022 TASC4 marksA class determined the enthalpy of combustion of butan-1-ol by calorimetry. Mass of water used: 100.0 g; initial temperature 25.0 degrees C; final temperature 31.5 degrees C; mass of butan-1-ol burned 0.152 g; M(butan-1-ol) = 74.12 g mol-1. Calculate the heat of combustion of butan-1-ol in kJ mol-1.

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Step 1 - heat absorbed by the water. q = m c delta T = 100.0 x 4.18 x (31.5 - 25.0) = 100.0 x 4.18 x 6.5 = 2717 J = 2.717 kJ. (2 marks)

Step 2 - moles of butan-1-ol burned. n = 0.152 / 74.12 = 2.051 x 10^-3 mol. (1 mark)

Step 3 - molar heat of combustion. delta H = -q / n = -2.717 / 2.051 x 10^-3 = -1325 kJ mol-1. The negative sign shows combustion is exothermic. (1 mark) Note the experimental value is well below the data-book value (about -2676 kJ mol-1) because of heat loss to the surroundings and incomplete combustion.

2021 TASC3 marksA Year 12 student determined the experimental heat of combustion of ethanol using a spirit burner. Volume of water heated = 200.0 mL; mass of ethanol burned = 1.75 g; initial water temperature = 20.2 degrees C; final water temperature = 73.5 degrees C. Use these values to calculate the experimental heat of reaction. (Mr of ethanol = 46.07.)

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Step 1 - heat gained by the water. Take the density of water as 1.00 g mL-1, so m = 200.0 g. q = m c delta T = 200.0 x 4.18 x (73.5 - 20.2) = 200.0 x 4.18 x 53.3 = 44 559 J = 44.56 kJ. (1 mark)

Step 2 - moles of ethanol burned. n = 1.75 / 46.07 = 0.03799 mol. (1 mark)

Step 3 - experimental molar heat of combustion. delta H = -q / n = -44.56 / 0.03799 = -1173 kJ mol-1. The value is less negative than the theoretical -1367 kJ mol-1 because of heat lost to the surroundings and to the apparatus. (1 mark)

2023 TASC5 marksCaCl2(s) -> Ca2+(aq) + 2Cl-(aq), delta H = +82.8 kJ mol-1, and KOH(s) -> K+(aq) + OH-(aq), delta H = -56.6 kJ mol-1. One mole of each is added to separate beakers of the same volume of water. Qualitatively compare the temperature changes of the water in the two beakers, and explain the difference in terms of enthalpy and bond strength.

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Temperature comparison (1 mark). Dissolving CaCl2 is endothermic (delta H positive), so the water in that beaker cools. Dissolving KOH is exothermic (delta H negative), so that water warms up.

Explanation (4 marks). When an ionic solid dissolves, energy is absorbed to break the ionic lattice and to separate water molecules, and energy is released when the ions are hydrated (ion-dipole attractions form). The sign of delta H is the net of these. For CaCl2 the energy needed to break the lattice is greater than the energy released on hydration, so the process is endothermic overall and absorbs heat from the water. For KOH the hydration energy released exceeds the lattice energy needed, so the process is exothermic overall and releases heat to the water.

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