Using H_f values CO_2 = -394, H_2O = -286, C_2H_5OH = -278 (kJ per mol), find H_c of ethanol. (C_2H_5OH + 3O_2 → 2CO_2 + 3H_2O.) [3 marks]

A reaction has H = -90 kJ per mol and S = -200 J per K per mol. State whether it is feasible at 298 K. [3 marks]

SingaporeChemistrySyllabus dot point

How are enthalpy changes defined, measured and calculated, and what controls whether a reaction is energetically favourable?

Define standard enthalpy changes (formation, combustion, neutralisation, atomisation, lattice energy, hydration, solution), apply Hess's law and Born-Haber cycles, and use the relationship between enthalpy, entropy and Gibbs free energy to judge feasibility

A focused answer to the H2 Chemistry learning outcome on energetics. Standard enthalpy definitions, Hess's law cycles, Born-Haber cycles for lattice energy, the entropy change of a reaction, and using Gibbs free energy to decide feasibility.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking
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What this dot point is asking

SEAB wants you to define the standard enthalpy changes precisely, construct Hess's law and Born-Haber cycles to find values that cannot be measured directly (such as lattice energy and enthalpy of formation), and combine enthalpy with entropy through the Gibbs free energy equation to decide whether a reaction is feasible. This topic supplies the energetics arguments used throughout equilibrium and inorganic chemistry.

The answer

Standard enthalpy definitions

All "standard" values are quoted at $298$ K and $1$ bar, per mole, with elements in their standard states.

Enthalpy change of formation $\Delta H_f$ : one mole of a compound formed from its elements.
Enthalpy change of combustion $\Delta H_c$ : one mole of a substance burned completely in oxygen.
Enthalpy change of neutralisation $\Delta H_n$ : one mole of water formed when acid neutralises alkali.
Enthalpy change of atomisation $\Delta H_{at}$ : one mole of gaseous atoms formed from an element.
Lattice energy $\Delta H_{latt}$ : one mole of an ionic solid formed from its gaseous ions (always negative, exothermic).
Enthalpy change of hydration $\Delta H_{hyd}$ : one mole of gaseous ions dissolved in water (exothermic).
Enthalpy change of solution $\Delta H_{sol}$ : one mole of solid dissolved to infinite dilution.

Hess's law

The total enthalpy change for a reaction is independent of the route taken, because enthalpy is a state function. This lets you find an unknown $\Delta H$ by building a cycle through measurable steps. The two standard cycle forms:

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

\Delta H_{\text{reaction}} = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})

Born-Haber cycles

A Born-Haber cycle is a Hess cycle for an ionic compound that links its enthalpy of formation to atomisation, ionisation energy, electron affinity, and lattice energy. Rearranging the cycle gives the lattice energy, which cannot be measured directly:

\Delta H_f = \Delta H_{at}(\text{metal}) + \Delta H_{at}(\text{non-metal}) + \text{I.E.} + \text{E.A.} + \Delta H_{lattice}

The magnitude of lattice energy increases with greater ionic charge and smaller ionic radius (stronger electrostatic attraction).

Entropy

Entropy $S$ measures the disorder, or number of ways energy and particles can be arranged. It increases when: solids melt or dissolve, liquids vaporise, the number of gas molecules increases, or temperature rises.

\Delta S_{\text{reaction}} = \sum S(\text{products}) - \sum S(\text{reactants})

Gibbs free energy and feasibility

A reaction is feasible (spontaneous) when the Gibbs free energy change is negative:

\Delta G = \Delta H - T\Delta S

$\Delta H$	$\Delta S$	Feasibility
negative	positive	always feasible
positive	negative	never feasible
negative	negative	feasible at low $T$
positive	positive	feasible at high $T$

Note that $\Delta S$ values are usually in J per K per mol, so convert to kJ before combining with $\Delta H$ in kJ.

Worked example

For the thermal decomposition $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ , $\Delta H = +178$ kJ per mol and $\Delta S = +161$ J per K per mol. Determine the minimum temperature at which the reaction becomes feasible.

The reaction is feasible when $\Delta G \le 0$ :

\Delta G = \Delta H - T\Delta S \le 0 \implies T \ge \frac{\Delta H}{\Delta S}

Convert $\Delta S$ to kJ: $\Delta S = 0.161$ kJ per K per mol.

T \ge \frac{178}{0.161} = 1106\ \text{K}

So decomposition becomes feasible above about $1106$ K (roughly $833$ degrees Celsius). This is why limestone must be heated strongly in a lime kiln.

Try it: Gibbs free energy calculator - enter $\Delta H$ and $\Delta S$ to find $\Delta G$ at any temperature, or the feasibility threshold temperature.

Common traps

Mixing kJ and J: Enthalpy is in kJ but entropy is in J per K. Convert entropy to kJ before using $\Delta G = \Delta H - T\Delta S$ .
Forgetting lattice energy is exothermic: $\Delta H_{lattice}$ for formation from gaseous ions is negative; some textbooks define it as the reverse, so state your sign convention.
Reversing the Hess formula: Formation route subtracts reactants from products; combustion route subtracts products from reactants. Mixing them flips the sign.
Treating feasible as fast: $\Delta G < 0$ means thermodynamically feasible, not that the reaction is fast. Kinetics (activation energy) controls rate separately.
Ignoring the factor in cycles: Multiply each enthalpy by the number of moles in the equation (the factor of 2 for hydrogen is the classic slip).

Examples in context

Example 1. Why Group 2 carbonates decompose at different temperatures. Down Group 2, the cation gets larger, so its charge density falls and it polarises the carbonate ion less. The decomposition enthalpy becomes more endothermic down the group, so a higher temperature is needed to make $\Delta G$ negative. This is why $\text{MgCO}_3$ decomposes more readily than $\text{BaCO}_3$ , a result revisited in Inorganic Chemistry.

Example 2. Dissolving and the energetics balance. For a salt dissolving, $\Delta H_{sol} = \Delta H_{lattice\ (broken)} + \sum \Delta H_{hyd}$ . If hydration releases nearly as much energy as is needed to break the lattice, $\Delta H_{sol}$ is small, and the positive entropy of dissolving usually drives the process. SEAB uses this energy-cycle reasoning to explain why some salts dissolve endothermically yet spontaneously.

Try this

Q1. Define the standard enthalpy change of formation. [1 mark]

Cue. The enthalpy change when one mole of a compound is formed from its elements in their standard states at $298$ K and $1$ bar.

Q2. Using $\Delta H_f$ values $\text{CO}_2 = -394$ , $\text{H}_2\text{O} = -286$ , $\text{C}_2\text{H}_5\text{OH} = -278$ (kJ per mol), find $\Delta H_c$ of ethanol. ( $\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}$ .) [3 marks]

Cue. $\Delta H_c = [2(-394) + 3(-286)] - (-278) = -1646 + 278 = -1368$ kJ per mol.

Q3. A reaction has $\Delta H = -90$ kJ per mol and $\Delta S = -200$ J per K per mol. State whether it is feasible at $298$ K. [3 marks]

Cue. $\Delta G = -90 - 298(-0.200) = -90 + 59.6 = -30.4$ kJ per mol; negative, so feasible at $298$ K (but not at high $T$ ).

Exam-style practice questions

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

Specimen (9729)4 marksThe standard enthalpy changes of combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ per mol respectively. Use Hess's law to calculate the standard enthalpy change of formation of methane,

\text{CH}_4

Show worked answer →

Formation of methane: C(s) + 2H2(g) -> CH4(g).

Build a Hess cycle through the combustion products (CO2 and H2O).

$\Delta H_f = [\Delta H_c(\text{C}) + 2\Delta H_c(\text{H}_2)] - \Delta H_c(\text{CH}_4)$

$\Delta H_f = [(-394) + 2(-286)] - (-890)$

$\Delta H_f = [-394 - 572] + 890 = -966 + 890 = -76$ kJ per mol.

Markers reward a correct cycle or the combustion-route formula, correct use of the factor of 2 for hydrogen, and the final value with sign and units.

2023 (style)4 marksFor the reaction in which solid ammonium nitrate dissolves in water, the temperature of the solution falls. State the sign of the enthalpy change of solution, explain in terms of entropy why the process is still spontaneous at room temperature, and state how the spontaneity changes at lower temperature.

Show worked answer →

The temperature falls, so the process is endothermic: $\Delta H_{sol} > 0$ (positive).

Dissolving an ordered solid to give freely-moving aqueous ions increases disorder, so $\Delta S > 0$ (positive).

Feasibility uses $\Delta G = \Delta H - T\Delta S$ . With $\Delta H$ positive and $\Delta S$ positive, the $-T\Delta S$ term is negative and large enough at room temperature to make $\Delta G$ negative, so dissolving is spontaneous.

At lower temperature, $T\Delta S$ is smaller, so $-T\Delta S$ is less negative; $\Delta G$ becomes less negative and eventually positive, so the process becomes non-spontaneous.

Markers reward the positive $\Delta H$ , the positive $\Delta S$ with a disorder reason, the use of $\Delta G = \Delta H - T\Delta S$ , and the temperature dependence.