SACE Board SACE 2022-style practice: When 1.50\ g of ammonium nitrate dissolved in 50.0\ g of water, the temperature fell from 21.0\ ^C to 16.2\ ^C. Calculate the molar enthalpy of solution of ammonium nitrate. State whether the process is exothermic or endothermic. (c(water) = 4.18\ J g^-1\ ^C^-1; M(NH_4NO_3) = 80.04\ g mol^-1.)

Step 1: q = mc T = 50.0 × 4.18 × (16.2 - 21.0) = 50.0 × 4.18 × (-4.8) = -1003\ J. The water lost 1003\ J, so the dissolving absorbed +1003\ J = +1.003\ kJ. (2 marks) Step 2: n(NH_4NO_3) = 1.5080.04 = 1.874 × 10^-2\ mol. (1 mark) Step 3: H = +1.0031.874 × 10^-2 = +53.5\ kJ mol^-1. (1 mark) Step 4: the temperature fell, so heat was absorbed from the surroundings; the process is **endothermic** (H positive). (1 mark)

§-Syllabus dot point

SAChemistrySyllabus dot point

How is the heat released or absorbed by a reaction measured and calculated?

Calculate enthalpy changes from calorimetry data using q = mcΔT, and interpret exothermic and endothermic reactions.

Using q = mcΔT to find heat transferred, converting to molar enthalpy change, the sign convention for exothermic and endothermic reactions, sources of heat loss, and fully worked SACE-style calorimetry calculations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Lead worked calculation
The calorimetry relationship
Sign convention
Energy profiles and bond changes
Heat loss and accuracy
Common procedural steps
Why it matters for managing processes

What this dot point is asking

SACE expects you to calculate $q$ , convert to $\Delta H$ per mole with the correct sign, interpret exothermic versus endothermic, and identify and account for heat losses.

Lead worked calculation

Enthalpy of neutralisation

$50.0\ \text{mL}$ of $1.00\ \text{mol L}^{-1}$ $\text{HCl}$ is mixed with $50.0\ \text{mL}$ of $1.00\ \text{mol L}^{-1}$ $\text{NaOH}$ . The temperature rises from $20.0\ ^\circ\text{C}$ to $26.7\ ^\circ\text{C}$ . Find the molar enthalpy of neutralisation. (Assume the mixture has the density and specific heat of water; $c = 4.18\ \text{J g}^{-1}\ ^\circ\text{C}^{-1}$ .)

Step 1: Identify the mass of solution and the temperature change

The calorimetry equation $q = mc\Delta T$ uses the mass of the solution that absorbs the heat, not the mass of any individual reactant. When the two solutions are combined, the total volume is $50.0 + 50.0 = 100.0\ \text{mL}$ . With the stated assumption that the mixture has the same density as water ( $1.00\ \text{g mL}^{-1}$ ), this gives a mass of $100.0\ \text{g}$ . The temperature change is measured from initial to final.

\Delta T = 26.7 - 20.0 = 6.7\ ^\circ\text{C}

Step 2: Calculate the heat released using $q = mc\Delta T$

This equation converts the observed temperature rise into a quantity of heat energy. The specific heat capacity $c = 4.18\ \text{J g}^{-1}\ ^\circ\text{C}^{-1}$ tells you how much energy is needed to raise one gram of water by one degree. Multiplying by mass and temperature change gives the total heat transferred to the solution. Because the temperature rose, the heat flowed from the reaction into the solution, meaning the reaction is exothermic.

q = mc\Delta T = 100.0 \times 4.18 \times 6.7 = 2801\ \text{J} = 2.80\ \text{kJ}

Step 3: Find the moles of limiting reactant

Molar enthalpy is heat per mole, so you need to know how many moles actually reacted. Both solutions have equal concentration and equal volume, so they contain identical mole amounts and neither is in excess. The neutralisation proceeds $1:1$ , so all of each reactant is consumed.

n(\text{HCl}) = cV = 1.00 \times 0.0500 = 0.0500\ \text{mol}

The $1:1$ stoichiometry means $0.0500\ \text{mol}$ of water is produced and $0.0500\ \text{mol}$ of each reactant is consumed.

Step 4: Divide to get the molar enthalpy and assign the sign

Dividing the total heat released by the number of moles reacted gives the heat released per mole. The sign must be negative for an exothermic process: the system (reaction) lost energy, so $\Delta H$ is negative. Always assign the sign from reasoning about whether heat was released or absorbed, not mechanically from the sign of $q$ .

\Delta H = -\frac{q}{n} = -\frac{2.80}{0.0500} = -56.0\ \text{kJ mol}^{-1}

Final answer: $\Delta H = -56.0\ \text{kJ mol}^{-1}$ . The negative sign confirms the reaction is exothermic, releasing $56.0\ \text{kJ}$ per mole of water formed. This is close to the accepted standard value for neutralisation of a strong acid by a strong base.

The calorimetry relationship

To get molar enthalpy, divide the heat by the moles of the substance reacting:

\Delta H = \pm\frac{q}{n}

Sign convention

The key logic: when a reaction warms the water, the heat came from the reaction, so the reaction is exothermic and $\Delta H$ is negative, even though $q$ for the water is positive. Always assign the sign to $\Delta H$ from the direction of temperature change, not blindly from the sign of $q$ .

Energy profiles and bond changes

Enthalpy change reflects the balance between energy absorbed to break bonds (always endothermic) and energy released when new bonds form (always exothermic). If more energy is released forming bonds than is absorbed breaking them, the reaction is exothermic overall. On an energy profile, exothermic reactions have products below reactants; endothermic reactions have products above.

Heat loss and accuracy

Real calorimetry under-measures the heat because some escapes to the surroundings and the apparatus.

Common procedural steps

Record the mass of water (or solution) and the initial temperature.
Carry out the reaction and record the maximum or minimum temperature.
Calculate $\Delta T$ , then $q = mc\Delta T$ .
Calculate moles of the reacting substance.
Divide and assign the sign to find $\Delta H$ in $\text{kJ mol}^{-1}$ .

Why it matters for managing processes

Calorimetry quantifies the energy of reactions, essential for comparing fuels, designing safe exothermic industrial processes, and predicting how temperature changes will affect equilibrium-controlled reactions through Le Chatelier's principle.

Exam-style practice questions

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

SACE 20225 marksWhen

1.50\ \text{g}

of ammonium nitrate dissolved in

50.0\ \text{g}

of water, the temperature fell from

21.0\ ^\circ\text{C}

16.2\ ^\circ\text{C}

. Calculate the molar enthalpy of solution of ammonium nitrate. State whether the process is exothermic or endothermic. (

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

;

M(\text{NH}_4\text{NO}_3) = 80.04\ \text{g mol}^{-1}

Show worked answer →

Step 1: $q = mc\Delta T = 50.0 \times 4.18 \times (16.2 - 21.0) = 50.0 \times 4.18 \times (-4.8) = -1003\ \text{J}$ . The water lost $1003\ \text{J}$ , so the dissolving absorbed $+1003\ \text{J} = +1.003\ \text{kJ}$ . (2 marks)

Step 2: $n(\text{NH}_4\text{NO}_3) = \dfrac{1.50}{80.04} = 1.874 \times 10^{-2}\ \text{mol}$ . (1 mark)

Step 3: $\Delta H = \dfrac{+1.003}{1.874 \times 10^{-2}} = +53.5\ \text{kJ mol}^{-1}$ . (1 mark)

Step 4: the temperature fell, so heat was absorbed from the surroundings; the process is endothermic ( $\Delta H$ positive). (1 mark)

SACE 20204 marksIn a calorimetry experiment, burning

0.450\ \text{g}

of ethanol raised the temperature of

200.0\ \text{g}

of water by

12.5\ ^\circ\text{C}

. Calculate the experimental molar enthalpy of combustion of ethanol, and suggest one reason the result is less exothermic than the data-book value. (

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

;

M(\text{C}_2\text{H}_5\text{OH}) = 46.07\ \text{g mol}^{-1}

Show worked answer →

Step 1: $q = mc\Delta T = 200.0 \times 4.18 \times 12.5 = 1.045 \times 10^{4}\ \text{J} = 10.45\ \text{kJ}$ gained by the water. (1 mark)

Step 2: $n(\text{ethanol}) = \dfrac{0.450}{46.07} = 9.77 \times 10^{-3}\ \text{mol}$ . (1 mark)

Step 3: $\Delta H_c = -\dfrac{10.45}{9.77 \times 10^{-3}} = -1.07 \times 10^{3}\ \text{kJ mol}^{-1}$ (negative, exothermic). (1 mark)

Step 4: heat is lost to the surroundings and the apparatus rather than all going to the water, and combustion may be incomplete, so the measured value is less exothermic than the true value. (1 mark)