State the condition on G for a reaction to be feasible. [1 mark]

AQA AQA 2018-style practice: A reaction has H = +30\ kJ mol^-1 and S = +120\ J K^-1mol^-1. Determine the minimum temperature at which the reaction becomes feasible.

A reaction is feasible when G ≤ 0. Using G = H - T S, the system becomes feasible when G = 0. Set G = 0: 0 = H - T S, so T = H S. Convert S to kJ K^-1mol^-1: 120\ J = 0.120\ kJ. Then T = 300.120 = 250\ K. Above 250\ K the T S term outweighs H, making G negative and the reaction feasible. Markers reward the unit conversion and the answer 250\ K.

EnglandChemistrySyllabus dot point

What really decides whether a reaction is feasible?

Born-Haber cycles and lattice enthalpies, enthalpies of solution and hydration, entropy change, and Gibbs free energy DeltaG = DeltaH - T DeltaS as the test of feasibility.

A focused answer to AQA A-Level Chemistry 3.1.8, covering Born-Haber cycles and lattice enthalpies, enthalpies of solution and hydration, entropy change, and Gibbs free energy as the measure of reaction feasibility.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Lattice enthalpy and Born-Haber cycles
Enthalpies of solution and hydration
Entropy
Gibbs free energy and feasibility
Try this

What this dot point is asking

AQA wants you to define lattice enthalpy, construct and use Born-Haber cycles, link enthalpies of solution to hydration and lattice enthalpies, define and calculate entropy change, and use Gibbs free energy to judge feasibility.

Lattice enthalpy and Born-Haber cycles

Lattice enthalpy cannot be measured directly, so a Born-Haber cycle uses Hess's law to find it from measurable steps: enthalpy of formation, enthalpies of atomisation, ionisation energies, and electron affinities. A more negative lattice enthalpy means stronger ionic bonding, favoured by higher ionic charge and smaller ionic radius (both increase the charge density, so the ions attract more strongly). Comparing the experimental Born-Haber value with a theoretical value calculated by assuming a perfectly ionic model gives evidence of covalent character: if the experimental value is significantly more exothermic than the theoretical one, the bonding is partly covalent (the cation polarises the anion), which is common for compounds of small, highly charged cations with large, polarisable anions.

Enthalpies of solution and hydration

When an ionic solid dissolves:

Entropy

The biggest contribution to $\Delta S$ is usually the change in the number of moles of gas, because gases are far more disordered than liquids or solids. A reaction that produces more gas molecules than it consumes has a large positive $\Delta S$ ; one that consumes gas (such as the Haber process) has a negative $\Delta S$ . This matters for feasibility, because the entropy term $T\Delta S$ grows with temperature: an endothermic reaction with a positive $\Delta S$ becomes feasible only above a certain temperature, while an exothermic reaction with a negative $\Delta S$ becomes infeasible above a certain temperature.

Gibbs free energy and feasibility

A reaction's feasibility depends on both enthalpy and entropy:

Worked example: feasibility temperature

A reaction has $\Delta H = +50\ \text{kJ mol}^{-1}$ and $\Delta S = +100\ \text{J K}^{-1}\text{mol}^{-1}$ . Find the temperature above which it is feasible.

Step 1: Set the feasibility condition

A reaction is feasible when $\Delta G \leq 0$ . The boundary between feasible and not feasible is the temperature where $\Delta G = 0$ exactly. Setting $\Delta G = 0$ in the Gibbs equation and rearranging for $T$ gives us the minimum temperature required.

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

Step 2: Convert $\Delta S$ to consistent units

$\Delta H$ is in $\text{kJ mol}^{-1}$ but $\Delta S$ is given in $\text{J K}^{-1}\text{mol}^{-1}$ . Both must be in the same energy unit before dividing. Divide $\Delta S$ by 1000 to convert from joules to kilojoules.

\Delta S = 100\ \text{J K}^{-1}\text{mol}^{-1} = 0.100\ \text{kJ K}^{-1}\text{mol}^{-1}

Step 3: Calculate the minimum feasibility temperature

Substituting the converted values gives the temperature at which $\Delta G$ crosses zero. Above this temperature the $T\Delta S$ term outweighs $\Delta H$ , making $\Delta G$ negative and the reaction feasible.

T = \frac{50}{0.100} = 500\ \text{K}

Final answer: The reaction becomes feasible above 500 K.

Try this

Q1. State the condition on $\Delta G$ for a reaction to be feasible. [1 mark]

Cue. $\Delta G \leq 0$ (negative or zero).

Q2. Explain why dissolving a solid increases entropy. [1 mark]

Cue. The ordered lattice breaks up and ions become free in solution, increasing disorder.

Exam-style practice questions

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

AQA 20184 marksA reaction has

\Delta H = +30\ \text{kJ mol}^{-1}

and

\Delta S = +120\ \text{J K}^{-1}\text{mol}^{-1}

. Determine the minimum temperature at which the reaction becomes feasible.

Show worked answer →

A reaction is feasible when $\Delta G \leq 0$ . Using $\Delta G = \Delta H - T\Delta S$ , the system becomes feasible when $\Delta G = 0$ .

Set $\Delta G = 0$ : $0 = \Delta H - T\Delta S$ , so $T = \dfrac{\Delta H}{\Delta S}$ .

Convert $\Delta S$ to $\text{kJ K}^{-1}\text{mol}^{-1}$ : $120\ \text{J} = 0.120\ \text{kJ}$ . Then $T = \dfrac{30}{0.120} = 250\ \text{K}$ .

Above $250\ \text{K}$ the $T\Delta S$ term outweighs $\Delta H$ , making $\Delta G$ negative and the reaction feasible. Markers reward the unit conversion and the answer $250\ \text{K}$ .

AQA 20214 marksUse a Born-Haber cycle to calculate the lattice enthalpy of formation of sodium chloride, given: enthalpy of formation

-411

, atomisation of sodium

+107

, atomisation of chlorine

+122

, first ionisation energy of sodium

+496

, electron affinity of chlorine

-349

(all in

\text{kJ mol}^{-1}

Show worked answer →

By Hess's law, the enthalpy of formation equals the sum of the steps from elements to gaseous ions plus the lattice enthalpy of formation:

$\Delta_f H = \Delta_{at}H(\text{Na}) + \Delta_{at}H(\text{Cl}) + IE_1(\text{Na}) + EA(\text{Cl}) + \Delta_{lattice}H$ .

Rearrange: $\Delta_{lattice}H = \Delta_f H - [\Delta_{at}H(\text{Na}) + \Delta_{at}H(\text{Cl}) + IE_1(\text{Na}) + EA(\text{Cl})]$ .

$\Delta_{lattice}H = -411 - [107 + 122 + 496 + (-349)] = -411 - 376 = -787\ \text{kJ mol}^{-1}$ .

Markers reward the correct cycle, the rearrangement, careful handling of the signs (especially the negative electron affinity), and the answer of about $-787\ \text{kJ mol}^{-1}$ .

Related dot points

Sources & how we know this

AQA A-level Chemistry (7405) specification — AQA (2015)