A reaction has H = -50\ kJ mol^-1 and S = -100\ J K^-1\,mol^-1. Calculate G at 298\ K and state if it is feasible. [2 marks]

CCEA CCEA 2022-style practice: A reaction has H = +25.0\ kJ mol^-1 and S = +120\ J K^-1\,mol^-1. Determine whether it is feasible at 298\ K and find the temperature above which it becomes feasible.

Feasibility depends on the free energy change, G = H - T S, which must be negative (or zero) for a feasible reaction. Convert S to kJ: 120\ J K^-1\,mol^-1 = 0.120\ kJ K^-1\,mol^-1. At 298\ K: G = 25.0 - (298 × 0.120) = 25.0 - 35.8 = -10.8\ kJ mol^-1. Since G is negative, the reaction is feasible at 298\ K. Feasibility limit when G = 0: T = H S = 25.00.120 = 208\ K. Markers reward (1) the G = H - T S equation, (2) converting S units, (3) the value of G and the feasible conclusion, (4) the threshold temperature 208\ K.

Northern IrelandChemistrySyllabus dot point

What makes a reaction feasible, and how do we build Born-Haber cycles?

Lattice enthalpy and Born-Haber cycles, enthalpies of solution and hydration, entropy and the second law, and the use of free energy change to decide the feasibility of a reaction.

A CCEA A-Level Chemistry answer on thermodynamics, covering lattice enthalpy and Born-Haber cycles, enthalpies of solution and hydration, entropy and the second law, and using the free energy change to decide whether a reaction is feasible.

Generated by Claude Opus 4.813 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 and free energy
Examples in context
Try this

What this dot point is asking

CCEA wants you to define lattice enthalpy, construct Born-Haber cycles, link enthalpies of solution and hydration, define entropy and the second law, and use $\Delta G = \Delta H - T\Delta S$ to judge feasibility.

Lattice enthalpy and Born-Haber cycles

Enthalpies of solution and hydration

Smaller, more highly charged ions have more exothermic hydration enthalpies because of stronger ion-dipole attraction to water.

Entropy and free energy

Born-Haber for magnesium chloride step

For $\text{MgCl}_2$ , $\Delta H_f = -642$ , atomisation of $\text{Mg}$ $= +148$ , $IE_1 + IE_2$ of $\text{Mg} = +2186$ , $2\times$ atomisation of $\text{Cl} = +244$ , $2\times$ electron affinity of $\text{Cl} = -698\ \text{kJ mol}^{-1}$ . Calculate the lattice enthalpy.

Step 1: Write the cycle

$\Delta H_f = \Delta H_{at}(\text{Mg}) + (IE_1+IE_2) + 2\Delta H_{at}(\text{Cl}) + 2EA + \Delta H_{latt}$

Step 2: Sum the known steps

$+148 + 2186 + 244 - 698 = +1880\ \text{kJ mol}^{-1}$

Step 3: Solve for the lattice enthalpy

$-642 = 1880 + \Delta H_{latt}$ , so $\Delta H_{latt} = -642 - 1880 = -2522\ \text{kJ mol}^{-1}$ . The large negative value reflects the $2+$ charge on magnesium.

Examples in context

Whether a salt dissolves and whether the process feels hot or cold both come from this analysis: ammonium nitrate dissolving is endothermic ( $\Delta H_{sol} > 0$ ) because the lattice energy needed slightly exceeds the hydration energy released, which is why instant cold packs use it. The feasibility test explains why the thermal decomposition of calcium carbonate, endothermic with a positive $\Delta S$ (a gas is released), only becomes feasible at high temperature once $T\Delta S$ outweighs $\Delta H$ .

Try this

Q1. Define the lattice enthalpy of formation. [1 mark]

Cue. The enthalpy change when one mole of ionic solid forms from its gaseous ions.

Q2. Write the equation linking enthalpy of solution, lattice enthalpy of dissociation and hydration enthalpies. [1 mark]

Cue. $\Delta H_{sol} = \Delta H_{latt,diss} + \sum \Delta H_{hyd}$ .

Q3. A reaction has $\Delta H = -50\ \text{kJ mol}^{-1}$ and $\Delta S = -100\ \text{J K}^{-1}\,\text{mol}^{-1}$ . Calculate $\Delta G$ at $298\ \text{K}$ and state if it is feasible. [2 marks]

Cue. $\Delta G = -50 - (298 \times -0.100) = -50 + 29.8 = -20.2\ \text{kJ mol}^{-1}$ ; feasible.

Exam-style practice questions

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

CCEA 20205 marksUse the following data to calculate the lattice enthalpy of sodium chloride by a Born-Haber cycle: enthalpy of formation

-411

; atomisation of

\text{Na}

+107

; first ionisation energy of

\text{Na}

+496

; atomisation of

\text{Cl}

+122

; first electron affinity of

\text{Cl}

-349\ \text{kJ mol}^{-1}

Show worked answer →

Apply Hess law around the Born-Haber cycle. The enthalpy of formation equals the sum of the steps to gaseous ions plus the lattice enthalpy (formation of the lattice from gaseous ions):

$\Delta H_f = \Delta H_{at}(\text{Na}) + IE_1(\text{Na}) + \Delta H_{at}(\text{Cl}) + EA_1(\text{Cl}) + \Delta H_{latt}$

Substitute the numbers (all in $\text{kJ mol}^{-1}$ ):

$-411 = (+107) + (+496) + (+122) + (-349) + \Delta H_{latt}$

$-411 = +376 + \Delta H_{latt}$

$\Delta H_{latt} = -411 - 376 = -787\ \text{kJ mol}^{-1}$

Markers reward (1) the correct Hess-law construction, (2) using the right signs (ionisation positive, electron affinity negative), (3) the arithmetic, (4) the negative value, (5) units $\text{kJ mol}^{-1}$ .

CCEA 20224 marksA reaction has

\Delta H = +25.0\ \text{kJ mol}^{-1}

and

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

. Determine whether it is feasible at

298\ \text{K}

and find the temperature above which it becomes feasible.

Show worked answer →

Feasibility depends on the free energy change, $\Delta G = \Delta H - T\Delta S$ , which must be negative (or zero) for a feasible reaction.

Convert $\Delta S$ to $\text{kJ}$ : $120\ \text{J K}^{-1}\,\text{mol}^{-1} = 0.120\ \text{kJ K}^{-1}\,\text{mol}^{-1}$ .

At $298\ \text{K}$ : $\Delta G = 25.0 - (298 \times 0.120) = 25.0 - 35.8 = -10.8\ \text{kJ mol}^{-1}$ . Since $\Delta G$ is negative, the reaction is feasible at $298\ \text{K}$ .

Feasibility limit when $\Delta G = 0$ : $T = \dfrac{\Delta H}{\Delta S} = \dfrac{25.0}{0.120} = 208\ \text{K}$ .

Markers reward (1) the $\Delta G = \Delta H - T\Delta S$ equation, (2) converting $\Delta S$ units, (3) the value of $\Delta G$ and the feasible conclusion, (4) the threshold temperature $208\ \text{K}$ .

Related dot points

Sources & how we know this

CCEA GCE Chemistry specification — CCEA (2016)