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Flashcards · 9620 · A Level Topic 8

Thermodynamics (A Level) flashcards

14 cards on A Level Topic 8, subtopics 3.1.8.1 to 3.1.8.2. Pick a subtopic or study them all.

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3.1.8.1 Born–Haber cycles

Enthalpy of atomisation
The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. For example: ½Br₂(l) → Br(g). Always positive.
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. For example: Br(g) + e⁻ → Br⁻(g). Exothermic.
Lattice enthalpy of dissociation vs lattice enthalpy of formation
Dissociation: one mole of ionic solid → its gaseous ions (positive), e.g. SrBr₂(s) → Sr²⁺(g) + 2Br⁻(g). Formation: the reverse (negative). Same size, opposite sign: always say which you mean.
Enthalpy of hydration
The enthalpy change when one mole of gaseous ions becomes one mole of aqueous ions. For example: Sr²⁺(g) → Sr²⁺(aq). Exothermic.
Is the second electron affinity of oxygen exothermic or endothermic?
Endothermic: an electron is being added to a negative ion (O⁻), which repels it. The first electron affinity is exothermic.
Enthalpy of solution from lattice and hydration enthalpies
ΔsolH = lattice enthalpy of dissociation + Σ hydration enthalpies of all the ions (multiply by the number of each ion).
Born–Haber lattice enthalpy more exothermic than the perfect ionic model value. What does it show?
The compound has some covalent character: the cation polarises the anion, distorting its electron cloud. Most polarisation with a small, highly charged cation and a large anion.
ΔatH of hydrogen if the H–H bond enthalpy is +436 kJ mol⁻¹
+218 kJ mol⁻¹: atomisation is per mole of atoms formed, so it is half the bond enthalpy.

3.1.8.2 Gibbs free-energy change, ∆G, and entropy change, ∆S

Gibbs free-energy change equation
ΔG = ΔH − TΔS T in K. Divide ΔS (J K⁻¹ mol⁻¹) by 1000 to match ΔH in kJ mol⁻¹.
Condition for a reaction to be feasible
ΔG ≤ 0 (zero or negative).
Temperature at which a reaction becomes feasible
Set ΔG = 0: T = ΔH ÷ ΔS. Then use the signs to say whether it is feasible above or below that temperature.
When is ΔS positive?
When disorder increases: solid → liquid → gas, an increase in the number of moles of gas, or a solid or liquid forming a gas.
ΔH positive and ΔS positive: when is the reaction feasible?
Only above T = ΔH ÷ ΔS, because −TΔS becomes more negative as T rises.
Why does a negative ΔG not mean a reaction will be observed?
ΔG shows it is thermodynamically feasible, but says nothing about rate. A high activation energy can make it too slow to observe.

Revise Thermodynamics further