Mission 1: Carbon and hydrogen don't combine cleanly to make methane. So how does the data booklet know its ΔH⦵f?

From the core lesson

• Hess's law: ΔH is the same whatever the route from start to finish.
• Reverse an equation: change the sign of ΔH. Multiply an equation: multiply its ΔH.
• An arrow followed against its direction in a cycle contributes −ΔH.
• First ionization energy: X(g) → X⁺(g) + e⁻, always endothermic.

Standard conditions and standard states

• ⦵ means standard conditions: 100 kPa, a stated temperature (usually 298 K), solutions 1 mol dm⁻³.
• Standard state: the pure, most stable form under those conditions.
• Examples: C(s, graphite), O₂(g), Br₂(l), H₂O(l), Na(s).

Standard enthalpy of formation

• ΔH⦵f: enthalpy change when 1 mol of a compound forms from its elements in their standard states.
• So ΔH⦵f of an element in its standard state is zero.
• Use fractions to make exactly 1 mol: 2Al(s) + 1½O₂(g) → Al₂O₃(s).
• Start from elements, never from gaseous atoms such as C(g) or H(g).
• Values are in the data booklet.

Standard enthalpy of combustion

• ΔH⦵c: enthalpy change when 1 mol of a substance burns completely in oxygen, under standard conditions.
• Complete: carbon ends as CO₂(g), hydrogen as H₂O(l), liquid in its standard state.
• ΔH⦵c is always negative. Elements have ΔH⦵c values too: C(s) and H₂(g) burn.
• Values are in the data booklet.

Using formation data

• Reactants and products can both be made from the same elements.
• Hess's law then gives ΔH⦵ = ΣΔH⦵f(products) − ΣΔH⦵f(reactants) (data booklet).
• Multiply each value by its coefficient in the equation.
• Elements in their standard states count as zero.

Worked example: decomposing limestone

• CaCO₃(s) → CaO(s) + CO₂(g)
• ΔH⦵f / kJ mol⁻¹: CaCO₃(s) −1207.6 and CaO(s) −634.9 (literature values); CO₂(g) −394 (data booklet).
• ΔH⦵ = [−634.9 + (−394)] − [−1207.6]
• = −1028.9 + 1207.6 = +178.7 kJ mol⁻¹. Endothermic: keep this value for Mission 4.
• The core lesson quotes +178 kJ mol⁻¹: the same value, rounded.

Using combustion data

• Reactants and products both burn to the same CO₂ and H₂O.
• So the order flips: ΔH⦵ = ΣΔH⦵c(reactants) − ΣΔH⦵c(products) (data booklet).
• Worked: C₂H₄(g) + H₂(g) → C₂H₆(g). ΔH⦵c / kJ mol⁻¹ (data booklet): C₂H₄ −1411; H₂ −286; C₂H₆ −1561.
• ΔH⦵ = [−1411 + (−286)] − [−1561] = −1697 + 1561 = −136 kJ mol⁻¹.
• Elements that burn, such as C and H₂, are not zero here; O₂, CO₂ and H₂O(l) cannot burn, so they are.
• Data sources differ slightly: 1.8's ΔH⦵f route gives −1562 for ethane; the data booklet gives −1561.

Mission 2: No one can pull a salt crystal apart into gaseous ions and measure the energy. So how do we know the lattice enthalpy of NaCl?

Steps that make gaseous ions

• Enthalpy of atomization: element in its standard state → 1 mol of gaseous atoms. Always endothermic.
• For a metal it is sublimation: Na(s) → Na(g).
• For chlorine it is half the bond enthalpy: ½Cl₂(g) → Cl(g), ½ × 242 = +121 kJ mol⁻¹.
• For bromine it includes vaporization: ½Br₂(l) → Br(g).
• Ionization energies then remove electrons from gaseous atoms: endothermic.

Electron affinity and lattice enthalpy

• First electron affinity: 1 mol of electrons added to 1 mol of gaseous atoms: Cl(g) + e⁻ → Cl⁻(g).
• First electron affinities of Cl and O are exothermic.
• Second electron affinity, O⁻(g) + e⁻ → O²⁻(g), is endothermic: the ion repels the electron.
• Lattice enthalpy (IB): 1 mol of ionic solid → gaseous ions: NaCl(s) → Na⁺(g) + Cl⁻(g).
• It is positive: separating the ions absorbs energy.

Reading a Born–Haber cycle

• A Born–Haber cycle is Hess's law for an ionic solid.
• Two routes from the elements to the solid: directly (ΔH⦵f) or through gaseous ions.
• Arrows up are endothermic; arrows down are exothermic.
• Both routes give the same total, so one unknown step can be found.
• You will be given the cycle or the data; you won't have to build a full cycle.

Worked example: lattice enthalpy of NaCl

• Data / kJ mol⁻¹: ΔH⦵f(NaCl) −411 and atomization Na +107 (literature values); Cl +121 (½ × 242); IE₁(Na) +496 and EA₁(Cl) −349 (data booklet).
• Route through ions: 107 + 121 + 496 − 349 − L = ΔH⦵f = −411
• L = 107 + 121 + 496 − 349 + 411 = +786 kJ mol⁻¹.
• The data booklet lists 790: values from different sources differ slightly.
• Only the lattice step releases enough energy to make ΔH⦵f negative.

Divalent ions: calcium chloride

• Ca²⁺ needs IE₁ + IE₂. CaCl₂ has two Cl⁻, so double the chlorine steps.
• Data / kJ mol⁻¹: ΔH⦵f(CaCl₂) −796, atomization Ca +178 and IE₂ +1145 (literature values); IE₁ +590, Cl +121 and EA₁(Cl) −349 (data booklet).
• 178 + 590 + 1145 + 2(121) + 2(−349) − L = −796
• L = 1457 + 796 = +2253 kJ mol⁻¹: far larger than NaCl. (The data booklet lists 2271.)
• For an O²⁻ ion, add EA₁ (exothermic) and EA₂ (endothermic).

Mission 3: Why does perfume spread across a room but never gather back into the bottle?

Entropy is dispersal

• Entropy, S: a measure of how dispersed the matter and energy in a system are.
• More ways to arrange the particles and share out the energy: higher entropy.
• Under the same conditions: gas > liquid > solid.
• A gas spreading into more space is an entropy increase.

Predicting the sign of ΔS

• Look first at moles of gas: more gas formed → ΔS positive; less gas → negative.
• Melting, vaporization, sublimation: ΔS positive. Freezing, condensation: negative.
• Dissolving an ionic solid usually raises S. Dissolving a gas lowers it.
• Solids and liquids count for little when the amount of gas changes.

Calculating ΔS⦵

• Standard entropies, S⦵, are absolute values: positive for every pure substance, elements included.
• ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants), each times its coefficient. Units: J K⁻¹ mol⁻¹.
• Worked: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). S⦵ / J K⁻¹ mol⁻¹: CH₄ 186, CO₂ 214, H₂O(l) 70 (data booklet); O₂ 205.2 (literature value).
• ΔS⦵ = [214 + 2(70)] − [186 + 2(205.2)] = 354 − 596.4 = −242.4 J K⁻¹ mol⁻¹.
• Negative, as predicted: 3 mol of gas become 1 mol.

Mission 4: Why does limestone need a kiln at about 900 °C to turn into lime?

The surroundings count too

• A change alters the entropy of the system and of its surroundings.
• Heat released (ΔH < 0) disperses energy into the surroundings: their entropy rises.
• ΔS(surroundings) = −ΔH/T: the same heat matters more at low temperature.
• A change is spontaneous when the total entropy increases.

Gibbs energy

• ΔG⦵ = ΔH⦵ − TΔS⦵ (data booklet) combines both entropy changes in one value.
• When ΔG is negative (constant pressure), the change is spontaneous.
• Units: ΔH⦵ and ΔG⦵ in kJ mol⁻¹; ΔS⦵ in J K⁻¹ mol⁻¹, so divide ΔS⦵ by 1000.
• T is in kelvin.
• Spontaneous says nothing about rate.

Worked example: limestone at two temperatures

• CaCO₃(s) → CaO(s) + CO₂(g): ΔH⦵ = +178.7 kJ mol⁻¹ (Mission 1); ΔS⦵ = +0.1604 kJ K⁻¹ mol⁻¹ (Mission 3).
• 298 K: ΔG⦵ = 178.7 − 298(0.1604) = 178.7 − 47.8 = +130.9 kJ mol⁻¹. Not spontaneous.
• 1200 K: ΔG⦵ = 178.7 − 1200(0.1604) = 178.7 − 192.5 = −13.8 kJ mol⁻¹. Spontaneous.
• Assumes ΔH⦵ and ΔS⦵ do not change with temperature.

When ΔG changes sign

• At the switch-over, ΔG⦵ = 0, so T = ΔH⦵/ΔS⦵ (ΔS⦵ in kJ).
• Limestone: T = 178.7 ÷ 0.1604 = 1114 K (841 °C).

Rearranging, and ΔG = 0 at a change of state

• Rearrange for any unknown: ΔH⦵ = ΔG⦵ + TΔS⦵; ΔS⦵ = (ΔH⦵ − ΔG⦵)/T.
• At the melting point, solid and liquid are in equilibrium: ΔG = 0.
• So ΔS = ΔH/T for melting or boiling at that temperature.

Mission 5: If ΔG⦵ is negative, why doesn't a reaction go all the way to products?

ΔG⦵ and ΔG are different

• ΔG⦵: a fixed value for the reaction at a given temperature, with every species in its standard state.
• ΔG: for the actual mixture; it changes as the composition changes.
• ΔG = ΔG⦵ + RT ln Q (data booklet). Q has the form of K but uses the concentrations now.
• R = 8.31 J K⁻¹ mol⁻¹, so work in J mol⁻¹.
• ln is the natural logarithm, not log₁₀.

Downhill to equilibrium

• Few products: Q < K, ΔG is negative, net forward reaction.
• As products build up, Q rises and ΔG becomes less negative.
• At equilibrium Q = K and ΔG = 0: no net change either way.
• Q > K: ΔG is positive, so net reverse reaction.

Linking ΔG⦵ and K

• At equilibrium, 0 = ΔG⦵ + RT ln K, so ΔG⦵ = −RT ln K (data booklet).
• ΔG⦵ negative → K > 1; positive → K < 1; zero → K = 1.
• Worked: ethanoic acid + ethanol ⇌ ethyl ethanoate + water, K ≈ 4.0 at 298 K (literature value).
• ΔG⦵ = −8.31 × 298 × ln 4.0 = −3433 J mol⁻¹ = −3.43 kJ mol⁻¹.
• Rearranged: K = e^(−ΔG⦵/RT).

Worked example: which way will a mixture go?

• Esterification at 298 K, ΔG⦵ = −3433 J mol⁻¹. Mixture / mol dm⁻³: ester 2.0, water 2.0, acid 0.40, alcohol 1.0.
• Q = [ester][water] / ([acid][alcohol]) = (2.0 × 2.0) / (0.40 × 1.0) = 10.
• ΔG = −3433 + 8.31 × 298 × ln 10 = −3433 + 5702 = +2269 J mol⁻¹ (+2.27 kJ mol⁻¹).
• ΔG is positive (Q > K): net reverse reaction until ΔG = 0.

The big ideas

• ΔH⦵f: 1 mol of compound from its elements in standard states. ΔH⦵c: 1 mol burned completely.
• Formation data: products − reactants. Combustion data: reactants − products.
• Born–Haber: two routes from elements to ionic solid; IB lattice enthalpy is positive.
• Entropy is dispersal; moles of gas decide the sign; S⦵ of an element is not zero.
• ΔG⦵ = ΔH⦵ − TΔS⦵ (ΔS⦵ in kJ); switch-over at T = ΔH⦵/ΔS⦵.
• ΔG = ΔG⦵ + RT ln Q falls to zero at equilibrium, where ΔG⦵ = −RT ln K.