Q: What does H2 Chemistry Notes: Topic 7 - Chemical Energetics cover? A: Integrate enthalpy changes, bond energies, Born-Haber cycles, entropy, and Gibbs free energy for Core Idea 3 (Chemical Energetics) in the 2026 H2 Chemistry syllabus.
Core Idea
CORE IDEA 3 - TRANSFORMATION, 7 CHEMICAL ENERGETICS: THERMOCHEMISTRY AND THERMODYNAMICS (GIBBS FREE ENERGY AND ENTROPY)
Energetics connects molecular behaviour to spontaneous change. Mastering enthalpy cycles, entropy arguments, and ΔG calculations is essential for structured questions across Papers 2 and 3. Use this note alongside worksheets stored at https://eclatinstitute.sg/blog/h2-chemistry-notes. For the full topic map and paper weightings, see our H2 Chemistry Syllabus 2026-27 overview.
Status: SEAB's current H2 Chemistry (9476) syllabus PDF is labelled for 2026, and the current Chemistry Data Booklet is labelled 8873/9476/9813 for use from 2026 in non-practical papers. Core Idea 3 Topic 7 is assessed across Papers 1-3.
The core idea is simple: Energetics is sign discipline: know what is being formed, broken, released, or absorbed.
Use it as a working check: Draw the cycle before doing arithmetic. Reverse a reaction, reverse the sign; multiply a reaction, multiply the enthalpy change.
Then go one layer deeper: Example: for bond energies, add energy to break reactant bonds, subtract energy released when product bonds form, then state whether the final ΔH is exothermic or endothermic.
Quick revision box
What this topic tests: Enthalpy cycles, entropy, Gibbs free energy, and feasibility arguments.
Sign errors in ΔH/ΔS/ΔG; weak Hess-cycle setup; qualitative feasibility without thermodynamic linkage.
20-minute sprint plan: 5 min sign convention recap; 10 min Hess/ΔG calculations; 5 min spontaneity explanation drill.
Route map: choose the energetics method first
If the question asks about...
Start with...
Then connect to...
Trap to avoid
Standard enthalpy definition
The exact process per mole
Balanced equation with standard states
Do not define combustion, formation, atomisation, and bond dissociation with the same wording.
Missing reaction enthalpy from data
Hess' law cycle
Reverse, scale, and add the given equations
Do not change an equation without changing the sign or coefficient of its enthalpy value.
Approximate ΔH from bonds
Bond counts in gaseous species
Energy to break reactant bonds minus energy released forming product bonds
Do not use mean bond energies as if they are exact for liquid or solid species.
Ionic solid lattice energy
Born-Haber cycle
Atomisation, ionisation, electron affinity, and lattice formation
Do not omit second ionisation energy for Group 2 compounds.
Feasibility at a temperature
ΔG=ΔH−TΔS
Sign of ΔG after unit conversion
Do not combine kJ⋅mol−1
Why high temperature changes feasibility
Signs of ΔH and ΔS
Whether the TΔS term becomes large enough
Do not say "endothermic means impossible"; entropy can dominate at high temperature.
Use this map before substituting numbers. Most energetics errors happen because the arithmetic is attached to the wrong cycle, definition, or sign convention.
1 Enthalpy Changes
Enthalpy change
Definition
Example equation
Standard enthalpy of formation ΔHf∘
Enthalpy change when one mole of compound forms from elements in standard states.
HX2(g)+21OX2(g)HX2O(l)
Standard enthalpy of combustion ΔHc∘
Enthalpy change when one mole of substance burns completely in excess oxygen.
CHX4(g)+2OX2(g)COX2(g)+2HX2O(l)
Enthalpy of atomisation ΔHatom
Enthalpy required to form one mole of gaseous atoms from the element.
21ClX2(g)Cl(g)
Bond dissociation enthalpy
Enthalpy to break one mole of a specified gaseous bond.
HX2(g)2H(g)
State conditions: 298K, 1bar, substances in standard states. Use bond energies (and a few constants like c for water) from the SEAB Chemistry Data Booklet where relevant; ΔH and S∘ data tables are typically provided in the question when needed.
2 Hess' Law and Enthalpy Cycles
Hess' law uses the state function property-route independent.
2.1 Formation/Combustion Cycle
Draw arrows from elements to compound (formation) or compounds to oxides (combustion).
Combine given ΔH∘ values; reverse sign if reversing a reaction; multiply when scaling.
Hess equation-combination checkpoint
Before adding enthalpy values, make the given equations behave like algebra lines. Every species that is not in the target equation must cancel.
If the given equation...
Do this to the equation
Do this to its enthalpy value
Common trap
Points in the opposite direction from the target step
Reverse the whole equation.
Change the sign of ΔH.
Reversing reactants and products but keeping the old sign.
Has half or double the needed amount
Multiply every coefficient by the same factor.
Multiply ΔH by the same factor.
Scaling the equation but leaving ΔH unchanged.
Contains a species absent from the final equation
Pair it with another adjusted equation so it cancels.
Keep the adjusted value until all lines are added.
Cancelling a species that appears on the same side in both equations.
Already matches one part of the target
Leave it unchanged.
Leave ΔH unchanged.
Changing a matching line just because other lines were adjusted.
Worked check: if a combustion equation gives C(s)+OX2(g)COX2(g) and the target needs COX2(g)C(s)+OX2(g), reverse the equation and change ΔH from ΔHc∘ to −ΔHc∘. If the target needs two moles of COX2, reverse and then double both the equation and the enthalpy value.
Misconception check: Hess' law does not let you add any convenient numbers. The adjusted equations must add up to the target equation after cancellation.
2.2 Bond Energy Approach
Approximate reaction enthalpy:
ΔH≈∑(Ebroken)−∑(Eformed)
Ensure all species are gaseous; adjust for physical states if necessary (include enthalpy of vaporisation or atomisation).
When using mean bond energies, take the data from the SEAB Chemistry Data Booklet unless the question provides specific values.
Methane: standard bond-counting reference for combustion enthalpy calculations.
A fast bond-energy check uses methane combustion: count four C−H and two O=O bonds broken, then four C=O and four O−H bonds formed.
Bond-counting checkpoint
For mean bond energy questions, build a two-column bond ledger before substituting. The sign comes from the ledger: breaking reactant bonds absorbs energy, forming product bonds releases energy.
Ledger side
What to count
Energy sign in ΔH
Common trap
Reactants
Every bond that must be broken after applying the balanced equation coefficients.
Add these bond energies.
Counting only one molecule when the equation has a coefficient.
Products
Every bond formed in the product molecules after applying coefficients.
Subtract these bond energies.
Forgetting that two HX2O molecules contain four O−H bonds.
Physical states
Check whether the species in the question are gaseous.
Add any required state-change correction if data is provided.
Using mean bond energies directly for liquid water without noticing the state.
Worked check: for CHX4(g)+2OX2(g)COX2(g)+2HX2O(g), the broken ledger is four C−H bonds plus two O=O bonds. The formed ledger is two C=O bonds in COX2 plus four O−H bonds in 2HX2O. Therefore
ΔH≈4E(C−H)+2E(O=O)−2E(C=O)−4E(O−H).
Misconception check: products formed are not "positive because they are made". Bond formation releases energy, so product bond energies are subtracted in this approximation.
3 Lattice Energy and Born-Haber Cycles
Born-Haber cycle for ionic compounds involves:
Atomisation of metal and non-metal.
First (and second) ionisation energies of metal.
Electron affinity of non-metal.
Lattice energy (formation of solid from gaseous ions).
Construct cycles carefully, labelling each step with values and sign convention (exothermic negative). Remember standard enthalpy of formation sits at bottom of cycle.
Born-Haber step-direction checkpoint
Before adding numbers, write the particle change for each step. The sign follows the direction of that particle change.
Step in the cycle
What is changing
Usual sign
Common trap
Atomisation of the metal
Solid metal becomes gaseous atoms.
Positive
Treating atomisation as bond formation because the final compound is forming later.
Atomisation or bond dissociation of the non-metal
Molecules become gaseous atoms.
Positive
Forgetting to halve the bond dissociation enthalpy when only half a mole of diatomic molecules is needed.
Ionisation energy
A gaseous atom loses electron(s).
Positive
Omitting the second ionisation energy for a Group 2 metal ion.
Electron affinity
A gaseous atom gains electron(s).
Usually negative for the first electron affinity.
Making every electron-gain step negative; second electron affinity can be endothermic.
Lattice formation
Gaseous ions form the ionic solid.
Negative
Using the opposite sign for lattice dissociation data without reversing it.
Worked check: for magnesium chloride, magnesium must become MgX2+(g), so the cycle needs both the first and second ionisation energies of magnesium. Chlorine gas also needs bond dissociation to form chlorine atoms before electron affinity can be applied.
Misconception check: a Born-Haber cycle is not a list of memorised energies. It is a route from elements in their standard states to gaseous ions, then to the ionic solid, with each arrow carrying its own sign.
Use the standard molar entropy values provided in the question data; convert to kJ⋅mol−1⋅K−1 before combining with enthalpy terms.
4.2 Gibbs Free Energy
ΔG=ΔH−TΔS
ΔG<0: spontaneous under given conditions.
Temperature dependence: if both ΔH and ΔS are positive, spontaneity increases with temperature.
4.3 Equilibrium Connection
At the threshold where ΔG=0, T≈ΔSΔH. This approximation is useful for estimating the temperature where spontaneity flips (e.g., some decomposition reactions) when ΔH and ΔS are treated as roughly constant over the range.
4.4 Gibbs sign checkpoint
Before substituting into ΔG=ΔH−TΔS, decide what the two signs are trying to do.
If ΔH<0 and ΔS>0, both terms favour ΔG<0. The reaction is feasible at all temperatures under the simplified constant-data assumption.
If ΔH>0 and ΔS<0, both terms oppose ΔG<0. The reaction is not feasible at any temperature under the simplified constant-data assumption.
If ΔH>0 and ΔS>0, high temperature can make TΔS large enough for ΔG<0
If ΔH<0 and ΔS<0, low temperature helps because the unfavourable −TΔS term stays smaller.
Worked check: for ΔH=178kJ⋅mol−1 and ΔS=161J⋅mol−1⋅K−1, first convert ΔS to 0.161kJ⋅mol−1⋅K−1. The threshold is T=178/0.161=1.11⋅103K. Above this temperature, TΔS is larger than ΔH, so ΔG becomes negative.
Common trap: do not memorise "high temperature helps" as a universal rule. High temperature helps only when ΔS is positive; when ΔS is negative, increasing T makes −TΔS more positive.
5 Worked Example
Question:
Given the data below, calculate the lattice energy of NaCl.
ΔHf∘(NaCl)=−411kJmol−1
Enthalpy of atomisation of sodium: +108kJmol−1
Bond dissociation enthalpy of ClX2: +242kJmol−1
First ionisation energy of sodium: +496kJmol−1
Electron affinity of chlorine: −349kJmol−1
Solution:
Construct the cycle:
Na(s)Na(g): +108kJmol−1.
Na(g)NaX+(g)+eX−
21ClX2(g)Cl(g)
Cl(g)+eX−ClX−(g)
NaX+(g)+ClX−(g)NaCl(s)
Apply Hess' law:
ΔHf∘=108+496+121−349+U
Solve for U:
U=ΔHf∘−(108+496+121−349)=−787kJmol−1
State final answer: lattice energy of NaCl is −787kJmol−1.
6 Experimental Considerations
Calorimetry (Paper 4):
Use polystyrene cup to minimise heat loss.
Record initial and maximum temperatures; extrapolate to correction for heat loss if required.
Compute q=mcΔT with appropriate units c=4.18Jg−1K−1 for water (value from the SEAB Chemistry Data Booklet).
Error sources: heat exchange with environment, incomplete reaction, inaccurate mass/volume measurements. Suggest improvements (insulation, repeating and averaging, using lid).
Calorimetry calculation checkpoint
When a temperature change is measured, separate the heat gained or lost by the solution from the enthalpy change of the reaction.
Step
What to write first
Common trap
Temperature change
ΔT=Tfinal−Tinitial for the solution.
Using a negative temperature change when the thermometer reading rises.
Heat change of solution
q=mcΔT, with mass in grams and c=4.18Jg−1K−1
Amount reacting
Find the limiting amount or the moles of product specified by the enthalpy definition.
Dividing by the volume instead of the amount in moles.
Reaction enthalpy sign
If the solution warms, the reaction released heat, so ΔH for the reaction is negative.
Reporting a positive ΔH just because ΔT is positive.
Worked check: 50.0cm3 acid is mixed with 50.0cm3 alkali and the temperature rises by 6.5K. Approximating the solution mass as 100g,
q=100×4.18×6.5=2717J=2.72kJ
If 0.0500mol of water forms, then
ΔH=−0.05002.72=−54.4kJmol−1
Misconception check: a temperature rise belongs to the solution. The reaction enthalpy has the opposite sign because the reaction released the heat that the solution gained.
In spirit-burner practicals, ethanol is a common fuel for comparing enthalpy change setups across repeat trials.
Ethanol: common liquid fuel reference in calorimetry planning and error analysis.
7 Common Misconceptions
Forgetting to convert ΔS to kJ⋅mol−1⋅K−1 when combining with ΔH (divide by 1000).
Reversing sign for electron affinity (exothermic values negative).
Using bond energies from data booklet without adjusting for stoichiometric coefficients.
Assuming ΔH<0 guarantees spontaneity; emphasise the TΔS term.
8 Quick Drills
Determine ΔH for 2CX2HX6(g)+7OX2(g)4COX2(g)+6HX2O(l) using bond energies (Data Booklet) and compare with a tabulated ΔHc∘ value when it is provided.
For CaCOX3(s)CaO(s)+COX2(g)
Sketch a Born-Haber cycle for MgClX2 including the second ionisation energy of magnesium and the second electron affinity of chlorine (endothermic). Annotate each energy change.
Common exam mistakes
Unit mismatch when calculating ΔG: ΔH is typically in kJ⋅mol−1 while ΔS is in J⋅mol−1⋅K−1; forgetting to divide ΔS by 1000 before substituting into ΔG=ΔH−TΔS is the most common numerical error in this topic.
Reversing the sign of electron affinity: Electron affinity is exothermic (negative) for most elements; treating it as positive in a Born-Haber cycle gives the wrong lattice energy and is penalised heavily.
Applying bond energies to non-gaseous species: Mean bond energy calculations assume all species are in the gaseous state; failing to account for enthalpy of vaporisation or atomisation when a liquid or solid is involved gives an inaccurate answer.
Assuming ΔH < 0 guarantees spontaneity: A reaction is only guaranteed spontaneous when ΔG<0, which depends on both ΔH and TΔS; ignoring the entropy term is a recurring conceptual error.
Mishandling stoichiometric coefficients with bond energies: Bond counts must match the balanced equation exactly; students often forget to multiply bond energies by the correct coefficient (e.g. for four O−H bonds in two HX2O molecules).
Wrong arrow direction in Born-Haber cycles: Each step must be drawn in the correct direction with the correct sign; a common slip is drawing atomisation or ionisation steps as exothermic, when they are endothermic.
Ignoring second ionisation energy for Group 2 compounds: MgClX2 Born-Haber cycles require both the first and second ionisation energies of magnesium; omitting one gives an incorrect lattice energy.
Frequently asked questions
How do I decide whether a reaction is spontaneous at a given temperature? Calculate ΔG=ΔH−TΔS at the stated temperature (using consistent units). If ΔG<0, the reaction is spontaneous under those conditions. If ΔG>0, it is non-spontaneous. At the temperature where ΔG=0, the system is at equilibrium.
What is the difference between bond dissociation enthalpy and mean bond energy? Bond dissociation enthalpy is the energy to break a specific bond in a specific molecule (e.g. the first C−H bond in methane). Mean bond energy is an average value across many compounds containing that bond type. The Data Booklet provides mean bond energies; use them for approximate calculations and acknowledge they introduce some error.
Why do Born-Haber cycles use gaseous ions rather than aqueous ions? Lattice energy is defined as the enthalpy change when one mole of ionic solid forms from its constituent gaseous ions at infinite separation. Using gaseous ions isolates the lattice formation step from solvation effects, allowing the pure electrostatic interaction to be calculated via Hess' law.
When is a reaction spontaneous despite being endothermic? When the entropy increase (positive ΔS) is large enough that TΔS>ΔH, making ΔG negative. This typically occurs at high temperatures. A classic example is the thermal decomposition of calcium carbonate: endothermic but spontaneous above a threshold temperature because the release of gaseous COX2 greatly increases entropy.
Struggling with Chemical Energetics? Our H2 Chemistry tuition programme covers this topic with structured practice, Paper 4 practical drills, and worked exam solutions.
