Flashcards · 9701 · A Level Topic 23

Chemical energetics (A Level) flashcards

21 cards on A Level Topic 23, subtopics 23.1 to 23.4. Pick a subtopic or study them all.

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Lattice energy, ΔHlatt: The enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. Always exothermic (negative).
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 one mole of gaseous atoms each gain an electron to form one mole of gaseous 1− ions.
Why is the second electron affinity endothermic?: The electron is being added to an already negative ion, so energy is needed to overcome the repulsion.
Two factors that make a lattice energy more exothermic: Higher ionic charges (this dominates) and smaller ionic radii: lattice energy is proportional to (charge⁺ × charge⁻) ÷ (r⁺ + r⁻). MgO is far more exothermic than NaCl.
Born–Haber cycle: the Hess equation for ΔHf of an ionic solid: ΔHf = Σ(atomisation) + Σ(ionisation energies) + Σ(electron affinities) + ΔHlatt. Watch multipliers: Cl₂ → 2Cl needs 2 × atomisation; Mg²⁺ needs the first and second IE.

Enthalpy change of hydration: The enthalpy change when one mole of gaseous ions dissolves in an excess of water. Always exothermic.
Enthalpy change of solution: The enthalpy change when one mole of solute dissolves in a solvent to form an infinitely dilute solution. It can be positive or negative.
Equation linking ΔHsol, lattice energy and hydration enthalpies: ΔHsol = −ΔHlatt + ΔHhyd(cation) + ΔHhyd(anion)
ΔHlatt(NaCl) = −787, ΔHhyd(Na⁺) = −406, ΔHhyd(Cl⁻) = −364 kJ mol⁻¹. Calculate ΔHsol.: ΔHsol = −(−787) + (−406) + (−364) = 787 − 770 = +17 kJ mol⁻¹
How do ionic charge and radius affect hydration enthalpy?: Higher charge and smaller radius make it more exothermic: stronger attraction between the ion and the polar water molecules.

Entropy, S: A measure of the dispersal of energy and particles: the number of possible arrangements (disorder) of a system. Units J K⁻¹ mol⁻¹.
Order of entropy for the three states: Gas ≫ liquid > solid.
How do you predict the sign of ΔS for a reaction?: More moles of gas produced: positive. Fewer moles of gas: negative. Melting, boiling and usually dissolving a solid: positive.
How do you calculate ΔS from standard entropies?: ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants), multiplying each by its coefficient. The answer is in J K⁻¹ mol⁻¹.
Effect of raising temperature on the entropy of a substance: Entropy increases: the particles have more energy and more ways of arranging it.

Gibbs equation and the condition for feasibility: ΔG = ΔH − TΔS (T in K; convert ΔS to kJ K⁻¹ mol⁻¹). A reaction is feasible when ΔG is negative (or zero).
Feasibility for each combination of signs of ΔH and ΔS: ΔH −, ΔS +: always feasible. ΔH +, ΔS −: never. ΔH −, ΔS −: at low temperature. ΔH +, ΔS +: at high temperature.
Temperature at which a reaction becomes feasible: Set ΔG = 0: T = ΔH ÷ ΔS (with ΔS in kJ K⁻¹ mol⁻¹ if ΔH is in kJ mol⁻¹).
A reaction has a large negative ΔG but nothing is observed. Why?: Feasible does not mean fast: the activation energy may be so high that the rate is negligible.
ΔH = +178 kJ mol⁻¹, ΔS = +0.161 kJ K⁻¹ mol⁻¹. Minimum temperature for feasibility?: T = ΔH ÷ ΔS = 178 ÷ 0.161 ≈ 1106 K (about 1110 K).