VCE Chemistry energetics calculations walkthrough: The 2026 guide

What this guide is for

VCE Chemistry energetics calculations appear in Section B of every Unit 3-4 exam. This guide walks through the three main calculation methods (calorimetry, Hess's law, enthalpy of formation), works through specific examples, and notes the marking conventions VCAA expects.

Calorimetry

A reaction profile makes the calorimetry calculation legible: $Q$ is the magnitude of energy released to the surroundings, and the activation energy $E_a$ explains why a catalyst speeds the reaction without changing $\Delta H$.

where:

• $Q$ = energy absorbed/released by the surroundings (J).
• $m$ = mass of the substance absorbing heat (kg).
• $c$ = specific heat capacity (J kg$^{-1}$ K$^{-1}$).
• $\Delta T$ = temperature change (K or degrees C).

For water: $c = 4186$ J kg$^{-1}$ K$^{-1}$ = 4.186 J g$^{-1}$ K$^{-1}$.

Procedure for calorimetry problems

1. Identify the surroundings (usually water in the calorimeter).
2. Calculate $Q$ for the surroundings using $Q = mc\Delta T$.
3. Apply conservation of energy. Heat released by the reaction = heat absorbed by the surroundings (for exothermic).
4. Divide by moles of reactant. Get the molar enthalpy change.
5. Apply sign convention. Exothermic $\Delta H$ is negative; endothermic is positive.

Worked example

50.0 mL of 1.00 M HCl is mixed with 50.0 mL of 1.00 M NaOH in a calorimeter. Initial temperature 22.0 degrees C. Final temperature 28.5 degrees C. Find $\Delta H$ per mole of water produced.

Step 1: Heat absorbed by the solution.
Total volume = 100 mL, mass ~100 g.
$\Delta T = 28.5 - 22.0 = 6.5$ degrees C.
$Q = 100 \times 4.186 \times 6.5 = 2721$ J.

Step 2: Moles of water produced.
Moles HCl = 0.050 L × 1.00 M = 0.050 mol.
Moles NaOH = 0.050 L × 1.00 M = 0.050 mol.
Reaction: HCl + NaOH → NaCl + H2O.
Moles H2O = 0.050 mol.

Step 3: Molar enthalpy.
$\Delta H = -Q / \text{moles} = -2721 / 0.050 = -54420$ J/mol = -54.4 kJ/mol.

The negative sign indicates exothermic. The accepted value is around -55 to -57 kJ/mol; the small discrepancy is due to heat loss.

Enthalpy of formation

where $n$ is the stoichiometric coefficient.

Elements in their standard state have $\Delta H_f = 0$ (e.g., O2 (g), C (s, graphite), H2 (g)).

Worked example

Calculate $\Delta H$ for the combustion of methane: CH4 + 2 O2 → CO2 + 2 H2O (l).

$\Delta H_f$ values (kJ/mol):

• CH4 (g) = -74.8
• O2 (g) = 0
• CO2 (g) = -393.5
• H2O (l) = -285.8

$\Delta H_{rxn} = [1(-393.5) + 2(-285.8)] - [1(-74.8) + 2(0)]$
$= [-393.5 - 571.6] - [-74.8]$
$= -965.1 + 74.8 = -890.3$ kJ/mol.

The negative value indicates exothermic. Matches the accepted value for methane combustion.

Hess's law

For a target reaction, find a series of reactions whose sum gives the target. The target $\Delta H$ is the sum of the constituent $\Delta H$'s.

Procedure

1. Write the target reaction.
2. Identify constituent reactions with known $\Delta H$.
3. Manipulate each (reverse if needed, change sign of $\Delta H$; multiply if needed, multiply $\Delta H$).
4. Sum the manipulated reactions; the result should equal the target.
5. Sum the $\Delta H$ values with their adjusted signs.

Worked example

Target: 2 C (s) + O2 (g) → 2 CO (g). $\Delta H = ?$

Given:

• C (s) + O2 (g) → CO2 (g). $\Delta H_1 = -393.5$ kJ/mol.
• 2 CO (g) + O2 (g) → 2 CO2 (g). $\Delta H_2 = -566.0$ kJ/mol.

Manipulate. Multiply reaction 1 by 2: 2 C (s) + 2 O2 (g) → 2 CO2 (g). $\Delta H = -787.0$.

Reverse reaction 2: 2 CO2 (g) → 2 CO (g) + O2 (g). $\Delta H = +566.0$.

Sum: 2 C + 2 O2 + 2 CO2 → 2 CO2 + 2 CO + O2.
Cancel: 2 C + O2 → 2 CO.

$\Delta H = -787.0 + 566.0 = -221.0$ kJ/mol.

A Hess cycle makes the bookkeeping visible. For the combustion of ethanol, the target reaction $\text{C}_2\text{H}_5\text{OH(l)} + 3\,\text{O}_2 \rightarrow 2\,\text{CO}_2 + 3\,\text{H}_2\text{O(l)}$ can be reached either directly or by way of the elements (formation enthalpies).

Bond energies

Bond energies are tabulated average values. The calculation gives an estimate; actual values differ.

Worked example

H2 + Cl2 → 2 HCl.

Bonds broken: 1 H-H (436 kJ/mol) + 1 Cl-Cl (243 kJ/mol) = 679 kJ/mol.

Bonds formed: 2 H-Cl (2 × 431 = 862 kJ/mol).

$\Delta H = 679 - 862 = -183$ kJ/mol.

Negative, consistent with exothermic.

Sign conventions and units

Sign of $\Delta H$

Exothermic: $\Delta H < 0$. Endothermic: $\Delta H > 0$.

Units

kJ/mol (per mole of the specified reactant or product).

Per mole of what

Always specify. "$\Delta H$ of combustion of methane = -890 kJ/mol of methane" is precise.

Standard conditions

$\Delta H^\circ$ refers to standard state (1 atm, 25 degrees C, 1 M for solutions).

Check your knowledge

A mix of definitional, calorimetry, Hess and bond-energy questions in the VCAA Unit 3/4 style. Aim to attempt under exam conditions (about 1.5 minutes per mark) before checking the solutions block.

1. Define the term standard enthalpy of formation and explain why $\Delta H_f^\circ$ for $O_2(g)$ is zero by convention. (2 marks)
2. A bomb calorimeter is calibrated by passing a current of 2.50 A at 12.0 V through the heater for 90.0 s. The temperature rises by 0.842 degrees C. Calculate the calibration factor in $\text{J K}^{-1}$. (2 marks)
3. (a) A 0.785 g sample of ethanol ($C_2H_5OH$, $M = 46.07$ g mol$^{-1}$) is combusted in a calorimeter with calibration factor $1.92 \times 10^{3} \ \text{J K}^{-1}$. The temperature rises by 11.20 degrees C. Calculate the molar enthalpy of combustion of ethanol. (b) The data-book value is $-1364 \ \text{kJ mol}^{-1}$. Calculate the percent error and identify the most likely source of any discrepancy. (5 marks)
4. Brown coal from Loy Yang in the Latrobe Valley contains roughly 60 percent water by mass. Treat the dry coal as carbon and the combustion of carbon as $C(s) + O_2(g) \rightarrow CO_2(g)$, $\Delta H = -394 \ \text{kJ mol}^{-1}$. Treat vaporising the water as $H_2O(l) \rightarrow H_2O(g)$, $\Delta H = +44.0 \ \text{kJ mol}^{-1}$. (a) Calculate the net energy released per kilogram of wet brown coal burned. (b) Compare with the $-890 \ \text{kJ mol}^{-1}$ for methane and comment on the relative greenhouse output per joule of useful heat. (6 marks)
5. Using bond energies (H-H 436, O=O 498, O-H 463, all $\text{kJ mol}^{-1}$), estimate $\Delta H$ for the reaction $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$. Compare with the standard value of $-484 \ \text{kJ mol}^{-1}$ and account for any difference. (4 marks)
6. Use Hess's law and the data $\Delta H_f^\circ[CO_2(g)] = -394 \ \text{kJ mol}^{-1}$, $\Delta H_f^\circ[H_2O(l)] = -286 \ \text{kJ mol}^{-1}$, $\Delta H_f^\circ[C_3H_8(g)] = -104 \ \text{kJ mol}^{-1}$ to calculate the standard enthalpy of combustion of propane. (4 marks)
7. (a, 2) State Le Chatelier's principle as it applies to a temperature change for an exothermic reaction. (b, 3) The galvanic cell $Zn | Zn^{2+} \| Cu^{2+} | Cu$ has $E^\circ_{cell} = +1.10 \ \text{V}$. Calculate the maximum electrical work obtainable per mole of zinc reacted ($F = 96\,500 \ \text{C mol}^{-1}$). (c, 2) Comment on why the measured cell voltage under load is always less than $E^\circ_{cell}$. (7 marks)
8. A student investigates the dissolution of $NH_4NO_3$ in a polystyrene-cup calorimeter. Adding 5.00 g of solid to 100.0 mL of water at 22.4 degrees C drops the temperature to 18.7 degrees C. ($M(NH_4NO_3) = 80.05 \ \text{g mol}^{-1}$, treat the solution as water with $c = 4.18 \ \text{J g}^{-1} \text{K}^{-1}$.) (a) Calculate $\Delta H_{sol}$ for ammonium nitrate. (b) Explain why the dissolution is endothermic in terms of lattice and hydration energies. (5 marks)