Topic 12: Entropy and Energetics

The complete criterion combines the reacting system with its surroundings. A change is thermodynamically feasible in the forward direction when $\Delta S_{total}$ is positive at the stated temperature.

With many molecules and quanta, dispersed arrangements vastly outnumber concentrated ones. Natural change therefore tends towards macrostates compatible with more microscopic arrangements.

On an entropy–temperature graph, entropy rises within each phase and jumps upward at melting and boiling. For water at standard pressure these transitions occur near 273 K and 373 K.

A perfect crystal at 0 K has one perfectly ordered arrangement and is assigned zero entropy. Imperfections or temperatures above 0 K introduce additional arrangements.

When a partition is removed, gas molecules spread into the available volume. The dispersed state corresponds to vastly more arrangements than all molecules confined to one side, so entropy increases and spontaneous remixing is overwhelmingly favoured.

For sodium chloride, ions fixed in a lattice become mobile and dispersed in water; rearrangement/disruption of water structure also contributes, so $\Delta S_{system}$ is positive. Other salts must be judged from evidence because strong hydration can order water.

Add the signed values after converting them to the same units. For example, $\Delta S_{total}=-78.7$ and $\Delta S_{surroundings}=+157.4$ J K$^{-1}$ mol$^{-1}$ give $\Delta S_{system}=-236.1$ J K$^{-1}$ mol$^{-1}$.

Multiply each standard molar entropy by its balanced-equation coefficient, total the products, total the reactants, then subtract reactants from products. Preserve state symbols because the same substance has different entropy in different states.

If product entropies total 738.2 and reactant entropies total 401.9 J K$^{-1}$ mol$^{-1}$, then $\Delta S_{system}^{\circ}=+336.3$ J K$^{-1}$ mol$^{-1}$. The positive sign means the products have greater entropy.

Use temperature in kelvin and make units consistent: convert $\Delta H$ from kJ mol$^{-1}$ to J mol$^{-1}$ when combining with entropy in J K$^{-1}$ mol$^{-1}$. Then add the signed surroundings and system values.

For $\Delta H=-67.2$ kJ mol$^{-1}$ at 298 K, $\Delta S_{surroundings}=+225.5$ J K$^{-1}$ mol$^{-1}$. The minus sign in the equation reverses the sign of the exothermic enthalpy.

At the threshold use kelvin and identical energy units. Decide which side of the threshold is feasible from the signs, not from the algebra alone. The equivalent $\Delta G=\Delta H-T\Delta S_{system}$ approach is permitted but not required.

If $\Delta S_{system}=-120$ and $\Delta S_{surroundings}=+180$ J K$^{-1}$ mol$^{-1}$, then $\Delta S_{total}=+60$ J K$^{-1}$ mol$^{-1}$ and the forward reaction is feasible despite the system becoming more ordered.

A reaction can be thermodynamically feasible ($\Delta S_{total}>0$) but extremely slow because a large activation barrier makes the reactant kinetically stable. Heating or a catalyst can increase rate without changing the thermodynamic balance.

For MgCl$_2$, include atomisation of one Mg and two Cl atoms, first and second ionisation energies of Mg, twice the first electron affinity of Cl, then lattice energy. Every line needs correct formulae, charges, electrons, coefficients and state symbols.

Apply Hess’s law around the closed cycle and rearrange only after all arrows have signs. A step used in reverse contributes the negative of its stated value.

The theoretical value treats ions as spherical charges interacting electrostatically. Its magnitude becomes larger for higher ionic charges and smaller ionic radii because attraction is stronger at shorter distance.

NaF shows close experimental/theoretical values, while MgCl$_2$ shows a larger discrepancy. The higher-charge Mg$^{2+}$ distorts the larger Cl$^-$ electron cloud more strongly, adding covalent character beyond the simple ionic model.

A fluoride ion is small and difficult to polarise, so calcium fluoride is close to the ionic model. Iodide is larger and readily polarised by Ca$^{2+}$, so calcium iodide has more covalent character and a larger experimental–theoretical lattice-energy difference.

Hydration is normally exothermic because ion–dipole attractions form between ions and water. Enthalpy of solution may be positive or negative because separating the lattice costs energy while hydrating the ions releases energy.

For MX$_2$, include one cation hydration term and twice the anion hydration term. If a missing anion value is required, subtract the known lattice/solution/cation terms, then divide by two.

Label every energy level with formulae, charges and state symbols. Arrow direction controls sign: the course lattice energy arrow points from gaseous ions down to solid.

Lattice energy depends on both ions’ charges and their separation; hydration enthalpy is quoted for one ion and follows its charge density. Down a group, anions become larger, so both their hydration enthalpy and the corresponding lattice energy generally become less exothermic.

For NH$_4$Cl versus NH$_4$Br, Br$^-$ is larger: its hydration is less exothermic, but the NH$_4$Br lattice energy is also less exothermic. Because both effects weaken, their difference—and hence $\Delta_{sol}H$—may remain similar.

KCl has an endothermic $\Delta_{sol}H$ of about +17 kJ mol$^{-1}$ yet is soluble because its positive system entropy change is large enough to make $\Delta S_{total}$ positive at 298 K.

Use supplied entropy and enthalpy data to explain the Unit 2 trends: Group 2 hydroxides become more soluble down the group, whereas Group 2 sulfates become less soluble. Compare the total entropy balance for each salt rather than relying on lattice energy alone.