Enthalpy and thermochemistry: TCE Chemistry (Tasmania)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

Enthalpy change $\Delta H$ is the heat energy absorbed or released by a reaction at constant pressure. Exothermic reactions release heat ( $\Delta H$ negative, surroundings warm) and endothermic reactions absorb it ( $\Delta H$ positive, surroundings cool). Calorimetry measures these changes with $q = mc\Delta T$ , then $\Delta H = -q/n$ gives the molar value.

What this dot point is asking

TASC expects you to define enthalpy change, classify and represent reactions on energy profiles, and calculate molar enthalpy changes from calorimetry data.

Exothermic and endothermic reactions

Every reaction breaks bonds (absorbing energy) and makes bonds (releasing energy); the net result sets the energy change.

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

Enthalpy change

The heat energy absorbed or released by a reaction at constant pressure, symbol $\Delta H$ , usually quoted in $\text{kJ mol}^{-1}$ . Negative for exothermic, positive for endothermic.

Energy profile diagrams

An energy profile plots enthalpy against reaction progress. Reactants and products sit at different enthalpy levels, and the activation energy $E_a$ is the height of the peak above the reactants. For an exothermic reaction the products lie below the reactants ( $\Delta H < 0$ ); for an endothermic reaction the products lie above ( $\Delta H > 0$ ). The vertical gap 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:

Heat transferred

q = mc\Delta T

q

is heat (J),

m

is mass of water (g),

c

is the specific heat capacity of water (

4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}

), and

\Delta T

is the temperature change. To get a molar value,

\Delta H = -q/n

, where

n

is the moles reacting; the minus sign makes

\Delta H

negative for an exothermic process.

Enthalpy of combustion from calorimetry

Find the heat gained by the water

Burning $0.50\ \text{g}$ of ethanol raises the temperature of $200\ \text{g}$ of water by $19.0\ ^\circ\text{C}$ . ( $M(\text{ethanol}) = 46.07\ \text{g mol}^{-1}$ .)

q = mc\Delta T = 200 \times 4.18 \times 19.0 = 1.59 \times 10^{4}\ \text{J} = 15.9\ \text{kJ}.

Find the moles of fuel burned

n(\text{ethanol}) = \frac{0.50}{46.07} = 1.085 \times 10^{-2}\ \text{mol}.

Convert to a molar enthalpy and assign the sign

\Delta H = -\frac{q}{n} = -\frac{15.9}{1.085 \times 10^{-2}} = -1.46 \times 10^{3}\ \text{kJ mol}^{-1}.

The reaction releases heat, so

\Delta H

is negative, confirming combustion is exothermic.

Sign convention

A negative $\Delta H$ means energy is released (exothermic, surroundings warm). A positive $\Delta H$ means energy is absorbed (endothermic, surroundings cool). Always state the sign explicitly.

Standard enthalpy changes

Chemists define standard enthalpy changes so values can be compared fairly. The standard enthalpy of combustion ( $\Delta H_c^\circ$ ) is the heat released when one mole of a substance burns completely in oxygen under standard conditions. The standard enthalpy of formation ( $\Delta H_f^\circ$ ) is the heat change when one mole of a compound forms from its elements in their standard states; elements in their standard state are defined as zero. The standard enthalpy of neutralisation is the heat released when an acid and base react to form one mole of water, and for strong acid and strong base it is almost constant (about $-57\ \text{kJ mol}^{-1}$ ) because the reaction is essentially $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ in every case.

Sources of experimental error in calorimetry

A school calorimetry result is almost always less exothermic than the data-book value, for predictable reasons: heat is lost to the surroundings and to the apparatus rather than all going into the water; combustion may be incomplete, leaving soot and producing less energy; and some fuel evaporates without burning. Recognising these systematic errors, and stating that they make the measured $\Delta H$ less negative, is frequently worth a mark in TASC questions.

Forgetting that the water absorbs the heat

In a combustion calorimetry calculation, the mass in $q = mc\Delta T$ is the mass of water being heated, not the mass of fuel burned; using the fuel mass is a common error. The fuel mass is used only to find the moles for the molar enthalpy. A second slip is dropping the negative sign, so an exothermic reaction is wrongly reported as positive.

In the exam, show $q = mc\Delta T$ with the correct mass and specific heat, divide by moles for a molar value, attach the sign that matches an exothermic or endothermic process, and explain experimental losses when comparing with a data-book value.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

A class determined the enthalpy of combustion of butan-1-ol by calorimetry. Mass of water heated

= 100.0\ \text{g}

; initial temperature

25.0\ ^\circ\text{C}

; final temperature

31.5\ ^\circ\text{C}

; mass of butan-1-ol burned

= 0.152\ \text{g}

;

M(\text{butan-1-ol}) = 74.12\ \text{g mol}^{-1}

;

c(\text{water}) = 4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}

. Calculate the experimental molar heat of combustion in

\text{kJ mol}^{-1}

Show worked answer →

Heat absorbed by the water: $q = mc\Delta T = 100.0 \times 4.18 \times (31.5 - 25.0) = 2717\ \text{J} = 2.717\ \text{kJ}$ . (2 marks)

Moles of butan-1-ol burned: $n = 0.152 / 74.12 = 2.05 \times 10^{-3}\ \text{mol}$ . (1 mark)

Molar heat of combustion: $\Delta H = -q/n = -2.717 / (2.05 \times 10^{-3}) = -1.33 \times 10^{3}\ \text{kJ mol}^{-1}$ . The negative sign confirms combustion is exothermic; the value is well below the data-book figure (about $-2676\ \text{kJ mol}^{-1}$ ) because of heat loss to the surroundings and incomplete combustion. (1 mark)

Practice3 marks

A student measured the heat of combustion of ethanol with a spirit burner. Mass of water heated

= 200.0\ \text{g}

; mass of ethanol burned

= 1.75\ \text{g}

; initial water temperature

= 20.2\ ^\circ\text{C}

; final

= 73.5\ ^\circ\text{C}

;

M(\text{ethanol}) = 46.07\ \text{g mol}^{-1}

. Calculate the experimental molar heat of combustion and explain why it differs from the data-book value of

-1367\ \text{kJ mol}^{-1}

Show worked answer →

Heat gained by the water: $q = mc\Delta T = 200.0 \times 4.18 \times (73.5 - 20.2) = 4.456 \times 10^{4}\ \text{J} = 44.56\ \text{kJ}$ . (1 mark)

Moles of ethanol: $n = 1.75 / 46.07 = 0.03799\ \text{mol}$ . (1 mark)

$\Delta H = -q/n = -44.56 / 0.03799 = -1173\ \text{kJ mol}^{-1}$ . It is less negative than $-1367\ \text{kJ mol}^{-1}$ because heat is lost to the surroundings and apparatus and combustion may be incomplete. (1 mark)

Practice5 marks

\text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq)

\Delta H = +82.8\ \text{kJ mol}^{-1}

, and

\text{KOH}(s) \rightarrow \text{K}^+(aq) + \text{OH}^-(aq)

\Delta H = -56.6\ \text{kJ mol}^{-1}

. One mole of each is dissolved in separate, equal volumes of water. (a) Compare the temperature change of the water in each beaker. (b) Explain the difference in terms of lattice and hydration energy.

Show worked answer →

(a) Dissolving $\text{CaCl}_2$ is endothermic ( $\Delta H > 0$ ), so that water cools. Dissolving $\text{KOH}$ is exothermic ( $\Delta H < 0$ ), so that water warms. (1 mark)

(b) When an ionic solid dissolves, energy is absorbed to break the lattice and is released when the ions are hydrated (ion-dipole attractions form). The sign of $\Delta H$ is the net of these. For $\text{CaCl}_2$ the lattice energy needed exceeds the hydration energy released, so the process is endothermic and absorbs heat from the water. For $\text{KOH}$ the hydration energy released exceeds the lattice energy needed, so it is exothermic and releases heat to the water. (4 marks)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC