Topic 12

Topic 12 on a page

Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.

Values and conversions

Quantity	Value
entropy units	J K⁻¹ mol⁻¹
enthalpy units	kJ mol⁻¹, converted to J mol⁻¹ before dividing by T
temperature	K = °C + 273
gas constant, R	8.31 J mol⁻¹ K⁻¹
entropy of a perfect crystal at 0 K	0

The equations

To find	Use	Page
entropy change of the system	$\Delta S_\text{system} = \Sigma S(\text{products}) - \Sigma S(\text{reactants})$	Entropy change of the system
entropy change of the surroundings	$\Delta S_\text{surroundings} = -\Delta H / T$	Surroundings and total entropy
total entropy change	$\Delta S_\text{total} = \Delta S_\text{system} + \Delta S_\text{surroundings}$	Surroundings and total entropy
feasibility temperature	$T = \Delta H / \Delta S_\text{system}$ ( $\Delta H$ in J)	The feasibility temperature
lattice energy	$\Delta_\text{f}H - (\Sigma\Delta_\text{at}H + \Sigma\text{IE} + \Sigma\text{EA})$	Born-Haber calculations
enthalpy of solution	$\Delta_\text{sol}H = -(\text{lattice energy}) + \Sigma\Delta_\text{hyd}H$	Enthalpy of solution and hydration
solubility test	$\Delta S_\text{total} = \Delta S_\text{system} - \Delta_\text{sol}H / T$	Will it dissolve?

A route map of Topic 12 in two rows: block 12A on entropy and feasibility, block 12B on lattice energy and solubility, joined by the same total-entropy rule. — Block 12A decides whether a change happens. Block 12B uses the same rule for dissolving.

Definitions to learn word for word

Entropy: a measure of how dispersed the particles and the energy of a system are. More ways of arranging them means higher entropy.
Standard enthalpy change of atomisation: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.
First electron affinity: the enthalpy change when each atom in one mole of gaseous atoms gains one electron to form one mole of gaseous 1− ions.
Lattice energy: the enthalpy change when one mole of an ionic solid is formed from its gaseous ions.
Enthalpy change of solution: the enthalpy change when one mole of an ionic solid dissolves in enough water to form an infinitely dilute solution.
Enthalpy change of hydration of an ion: the enthalpy change when one mole of gaseous ions dissolves in enough water to form an infinitely dilute solution.

Methods

Sign of $\Delta S_\text{system}$ : count moles of gas first, then compare states (gas > liquid > solid) and numbers of particles.
Feasibility: $\Delta S_\text{system}$ , then $\Delta S_\text{surroundings}$ in J K⁻¹ mol⁻¹, then the sum. Positive: feasible.
Feasible above or below a temperature: set $\Delta S_\text{total} = 0$ , so $T = \Delta H / \Delta S_\text{system}$ .
Born-Haber cycle: elements to gaseous atoms, atoms to gaseous ions, gaseous ions to solid. The formation arrow is the long route.
Solution cycle: break the lattice (positive), then hydrate each ion (negative). Multiply by the number of each ion.
Charge and radius: higher charge and smaller radius give more exothermic lattice energy and hydration enthalpy.

The ten traps

Leaving $\Delta H$ in kJ when dividing by T in K, so the answer is 1000 times too small.
Using $\Delta S_\text{surroundings} = +\Delta H / T$ , with the wrong sign.
Forgetting that the entropy of the reaction depends on the coefficients (2 mol of NH₃, not 1).
Saying "feasible" when you mean "fast". Feasible means thermodynamically possible, with no information about rate.
Writing $\Delta S_\text{total}$ in kJ and $\Delta S_\text{system}$ in J in the same sum.
Atomisation of chlorine written as $\ce{Cl2 -> 2Cl}$ instead of $\ce{1/2 Cl2 -> Cl}$ .
Missing the "× 2" or the second ionisation energy in a Born-Haber cycle for MgCl₂, or giving oxygen's second electron affinity the wrong sign.
Using the lattice energy with its own sign as the first step of a solution cycle. Breaking the lattice is the reverse.
Explaining a trend in hydration or lattice energy with "more electrons" instead of charge and radius.
Explaining a Group 2 solubility trend with only one of the two terms (lattice energy or hydration).

Mixed test

Try these with no notes, then check the answers.

Predict the sign of $\Delta S_\text{system}$ for (a) $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ and (b) $\ce{2H2(g) + O2(g) -> 2H2O(l)}$ . Explain each.
$\ce{N2O4(g) -> 2NO2(g)}$ has $\Delta H$ = +57 kJ mol⁻¹ and $\Delta S_\text{system}$ = +176 J K⁻¹ mol⁻¹. Is it feasible at 298 K? Above what temperature is it feasible?
For $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ , $S$ values / J K⁻¹ mol⁻¹: CaCO₃ 92.9, CaO 39.7, CO₂ 213.6. Calculate $\Delta S_\text{system}$ . Then find the temperature above which the decomposition is feasible, with $\Delta H$ = +178 kJ mol⁻¹.
Diamond changes to graphite with a positive $\Delta S_\text{total}$ , yet diamonds last for ever. Explain.
Calculate the first electron affinity of chlorine in KCl. Data / kJ mol⁻¹: $\Delta_\text{f}H$ −437, lattice energy −718, $\Delta_\text{at}H$ (K) +89, $\Delta_\text{at}H$ (Cl) +122, first IE(K) +419.
Calculate $\Delta_\text{sol}H$ of NaI. Data / kJ mol⁻¹: lattice energy −704, $\Delta_\text{hyd}H$ (Na⁺) −406, $\Delta_\text{hyd}H$ (I⁻) −293.
AgCl has an experimental lattice energy of −915 kJ mol⁻¹ and a theoretical value of −833 kJ mol⁻¹. Calculate the percentage difference and explain what it shows.
Explain why Ba(OH)₂ is more soluble in water than Mg(OH)₂.

Answers cover these until you have tried the questions

(a) Positive: a gas forms from a solid, so the particles are much more dispersed. (b) Negative: three moles of gas become two moles of liquid, so there is a large loss of dispersal.
$\Delta S_\text{surroundings}$ = −57 000 ÷ 298 = −191 J K⁻¹ mol⁻¹. $\Delta S_\text{total}$ = 176 − 191 = −15 J K⁻¹ mol⁻¹: negative, so not feasible at 298 K. Feasible above $T$ = 57 000 ÷ 176 = 324 K.
$\Delta S_\text{system}$ = (39.7 + 213.6) − 92.9 = +160.4 J K⁻¹ mol⁻¹. $T$ = 178 000 ÷ 160.4 = 1110 K (837 °C).
A positive $\Delta S_\text{total}$ means the change is thermodynamically feasible, but the activation energy is very high (strong covalent bonds must break), so the rate at room temperature is negligible. Diamond is kinetically stable but thermodynamically unstable.
Sum without EA = 89 + 122 + 419 = 630. $\Delta_\text{f}H$ = 630 + EA + (−718), so −437 = −88 + EA, giving EA = −437 + 88 = −349 kJ mol⁻¹.
$\Delta_\text{sol}H$ = +704 + (−406) + (−293) = 704 − 699 = +5 kJ mol⁻¹.
Difference = 915 − 833 = 82 kJ mol⁻¹; 82 ÷ 915 × 100 = 9%. The experimental value is more exothermic than the theoretical value, so there is covalent character: Ag⁺ polarises the Cl⁻ ion, which adds extra attraction.
Down Group 2 both the lattice energy and the cation's hydration enthalpy become less exothermic. The hydroxide ion is small, so the lattice energy falls faster than the hydration enthalpy. $\Delta_\text{sol}H$ becomes more exothermic, and solubility increases.