What does VCE Chemistry energetics cover?

VCE Chemistry energetics (Unit 3 AoS 1, and aspects of Unit 4 AoS 1) covers enthalpy change in chemical reactions, exothermic and endothermic reactions, calorimetry calculations using $Q = mc\Delta T$, enthalpy of formation, enthalpy of combustion, Hess's law for indirect calculation, and applications including fuel choice and bond energies.

How do calorimetry calculations work?

$Q = mc\Delta T$ where $Q$ is energy (J), $m$ is mass of the surroundings (often water in kg), $c$ is specific heat capacity (J kg K), and $\Delta T$ is temperature change. Calorimetry experiments measure the heat absorbed or released by surrounding water; this equals the negative of the enthalpy change of the reaction (assuming a perfectly insulated calorimeter). Real systems lose heat to the environment, requiring calibration.

Hess's law states that the enthalpy change of a reaction is independent of the path taken. You can calculate the enthalpy change of a target reaction by adding the enthalpy changes of any series of reactions that combine to give the target reaction. Useful when the direct enthalpy change is difficult or impossible to measure (e.g., for combustion of solid carbon directly to CO without producing CO2 as an intermediate).

How is enthalpy of formation used?

$\Delta H_{rxn} = \sum \Delta H_f \text{ (products)} - \sum \Delta H_f \text{ (reactants)}$. The enthalpy of formation is the enthalpy change when 1 mol of a compound is formed from its constituent elements in their standard states. Elements in standard state have $\Delta H_f = 0$. Useful when you have tabulated $\Delta H_f$ values for all species.

How are bond energies used?

$\Delta H_{rxn} = \sum \text{(bond energies of bonds broken)} - \sum \text{(bond energies of bonds formed)}$. Bond energies are tabulated as average values; the bond-energy calculation gives an estimate. Less accurate than $\Delta H_f$ method but useful when $\Delta H_f$ is unavailable.

What are the most common Section B energetics questions?

Three main types. (1) Calorimetry: given data from a calorimeter experiment, calculate the enthalpy change of the reaction. (2) Hess's law: combine given reactions to get a target reaction's enthalpy. (3) Standard enthalpy of formation: use tabulated $\Delta H_f$ to find $\Delta H_{rxn}$. All three appear in VCE Section B every year or two.

VICChemistry

VCE Chemistry energetics calculations walkthrough: the 2026 guide

A complete walkthrough of VCE Chemistry energetics calculations. Enthalpy change calculation methods (calorimetry $Q = mc\Delta T$ , enthalpy of formation, Hess's law), worked examples, and the moves that secure top marks in Section B.

Generated by Claude Opus 4.816 min readVCAA-CHEM-ENERGETICSUpdated 2026-06-21

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

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What this guide is for
Calorimetry
Enthalpy of formation
Hess's law
Bond energies
Sign conventions and units
Check your knowledge

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 $\Delta H$ .

Exothermic reaction profile: the catalyst trims the activation energy from

E_a

E_a

(cat.) while leaving

\Delta H

unchanged.

Q = m c \Delta T

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

Identify the surroundings (usually water in the calorimeter).
Calculate $Q$ for the surroundings using $Q = mc\Delta T$ .
Apply conservation of energy. Heat released by the reaction = heat absorbed by the surroundings (for exothermic).
Divide by moles of reactant. Get the molar enthalpy change.
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

\Delta H_{rxn} = \sum n \Delta H_f \text{ (products)} - \sum n \Delta H_f \text{ (reactants)}

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

Write the target reaction.
Identify constituent reactions with known $\Delta H$ .
Manipulate each (reverse if needed, change sign of $\Delta H$ ; multiply if needed, multiply $\Delta H$ ).
Sum the manipulated reactions; the result should equal the target.
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).

Closing the Hess triangle for ethanol combustion gives

\Delta H_c = -1366.7

kJ mol

^{-1}

, the same value VCAA tables list for the bioethanol fuel used in Australian E10 petrol blends.

Bond energies

\Delta H_{rxn} = \sum \text{BE(bonds broken)} - \sum \text{BE(bonds formed)}

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.

Bonds broken minus bonds formed for

\text{H}_2 + \text{Cl}_2 \rightarrow 2\,\text{HCl}

: the 183 kJ mol

^{-1}

surplus from bond formation is released as the negative

\Delta H

of reaction.

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.

Define the term standard enthalpy of formation and explain why $\Delta H_f^\circ$ for $O_2(g)$ is zero by convention. (2 marks)
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)
(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)
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)
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)
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)
(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)
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)

Solutions

Q1: The standard enthalpy of formation $\Delta H_f^\circ$ is the enthalpy change when 1 mole of a substance is formed from its constituent elements in their standard states at 1 atm (100 kPa) and 25 degrees C. Because $O_2(g)$ is the standard state of the element oxygen, forming it from itself involves no chemical change, so $\Delta H_f^\circ[O_2(g)] = 0$ by definition.
Q2: Electrical energy input $Q = VIt = 12.0 \times 2.50 \times 90.0 = 2700 \ \text{J}$ . Calibration factor $= Q / \Delta T = 2700 / 0.842 = 3.21 \times 10^{3} \ \text{J K}^{-1}$ . The calorimeter calibration factor is $3.21 \times 10^{3} \ \text{J K}^{-1}$ .
Q3: (a) Heat released by combustion $= 1.92 \times 10^{3} \times 11.20 = 2.150 \times 10^{4} \ \text{J} = 21.50 \ \text{kJ}$ . Moles of ethanol $= 0.785 / 46.07 = 1.704 \times 10^{-2} \ \text{mol}$ . Molar $\Delta H_c = -21.50 / 1.704 \times 10^{-2} = -1.262 \times 10^{3} \ \text{kJ mol}^{-1}$ . (b) Percent error $= |-1262 - (-1364)| / 1364 \times 100 = 7.5$ percent. The experimental value is less negative than the literature value, consistent with incomplete combustion (some soot) and unrecovered heat carried off in exhaust gases.
Q4: (a) 1 kg of wet coal contains 0.40 kg of dry carbon and 0.60 kg of liquid water. Moles of $C = 400 / 12.01 = 33.31 \ \text{mol}$ , releasing $33.31 \times 394 = 1.312 \times 10^{4} \ \text{kJ}$ . Moles of water vaporised $= 600 / 18.02 = 33.30 \ \text{mol}$ , absorbing $33.30 \times 44.0 = 1.465 \times 10^{3} \ \text{kJ}$ . Net energy $= 1.312 \times 10^{4} - 1.465 \times 10^{3} = 1.165 \times 10^{4} \ \text{kJ kg}^{-1}$ . (b) Methane releases $890 / 16.04 = 55.5 \ \text{kJ g}^{-1} = 5.55 \times 10^{4} \ \text{kJ kg}^{-1}$ , almost five times the useful energy per kilogram of wet brown coal. Per unit of useful heat, brown coal emits roughly $33.3 / 11\,650 \approx 2.86 \ \text{mmol CO}_2 \text{ per kJ}$ versus methane at $1 / (890) \approx 1.12 \ \text{mmol CO}_2 \text{ per kJ}$ , so Loy Yang brown coal emits about 2.5 times the $CO_2$ per joule of useful heat.
Q5: Bonds broken: $2 \times H{-}H + 1 \times O{=}O = 2(436) + 498 = 1370 \ \text{kJ mol}^{-1}$ . Bonds formed: $4 \times O{-}H = 4(463) = 1852 \ \text{kJ mol}^{-1}$ . $\Delta H = 1370 - 1852 = -482 \ \text{kJ mol}^{-1}$ , in close agreement with the data-book value of $-484 \ \text{kJ mol}^{-1}$ for gaseous water. The small difference reflects that tabulated bond energies are averages across many molecules, while the standard value is measured for this specific reaction.
Q6: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$ . $\Delta H_c = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) = [3(-394) + 4(-286)] - [(-104) + 5(0)] = [-1182 - 1144] - [-104] = -2326 + 104 = -2222 \ \text{kJ mol}^{-1}$ . The enthalpy of combustion of propane is $-2222 \ \text{kJ mol}^{-1}$ .
Q7: (a) Le Chatelier: if a system at equilibrium is disturbed, the position of equilibrium shifts in the direction that opposes the disturbance. For an exothermic reaction, increasing temperature shifts equilibrium toward the reactants (the endothermic direction). (b) $n(e^-) = 2$ per Zn. Work $= nFE^\circ_{cell} = 2 \times 96\,500 \times 1.10 = 2.12 \times 10^{5} \ \text{J mol}^{-1} = 212 \ \text{kJ mol}^{-1}$ . (c) Under load, current flows and the cell exhibits internal resistance ( $IR$ drop) plus polarisation overpotentials at each electrode. These dissipate energy as heat, so the terminal voltage is lower than $E^\circ_{cell}$ , and the realised electrical work per mole of Zn is less than 212 kJ.
Q8: (a) Mass of solution $\approx 105.0 \ \text{g}$ . Heat absorbed by solution $Q = mc\Delta T = 105.0 \times 4.18 \times (18.7 - 22.4) = 105.0 \times 4.18 \times (-3.7) = -1624 \ \text{J}$ . The solution lost 1624 J to the dissolution process, so the reaction absorbed $+1624 \ \text{J}$ . Moles of $NH_4NO_3 = 5.00 / 80.05 = 6.246 \times 10^{-2} \ \text{mol}$ . $\Delta H_{sol} = 1624 / 6.246 \times 10^{-2} = 2.60 \times 10^{4} \ \text{J mol}^{-1} = +26.0 \ \text{kJ mol}^{-1}$ . (b) The lattice enthalpy (energy required to separate the $NH_4^+$ and $NO_3^-$ ions in the solid) is larger than the hydration enthalpy (energy released when the gaseous ions become hydrated). The net energy required is supplied by the surroundings, cooling the solution and giving an overall endothermic $\Delta H_{sol} = +26.0 \ \text{kJ mol}^{-1}$ .