Flashcards · 9620 · AS Topic 4

Energetics (AS Level) flashcards

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Enthalpy change, ΔH: The heat energy change of a reaction measured at constant pressure. Negative for exothermic (surroundings get hotter), positive for endothermic.
Standard enthalpy of combustion, ΔcH°: The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, with all reactants and products in their standard states.
Standard enthalpy of formation, ΔfH°: The enthalpy change when one mole of a compound is formed from its elements, all in their standard states, under standard conditions.
Standard conditions: 100 kPa and a stated temperature (usually 298 K), with each substance in its standard state.
Equation for the standard enthalpy of formation of ethanol: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)
ΔfH° of an element in its standard state: Zero: forming it from itself involves no change.

q = mcΔT: meaning and units: q: heat change (J). m: mass of the substance whose temperature changes, usually water or solution (g). c: specific heat capacity (J g⁻¹ K⁻¹). ΔT: temperature change (K).
Method to find a molar enthalpy change from calorimetry: q = mcΔT in J, convert to kJ, divide by moles of the limiting reagent or fuel burned. Add the sign last: temperature rise → ΔH negative.
0.92 g ethanol (Mr 46.0) heats 150 g water by 22.0 K (c = 4.18 J g⁻¹ K⁻¹). ΔcH?: q = 150 × 4.18 × 22.0 = 13 794 J = 13.8 kJ. n = 0.0200 mol. ΔcH = −13.8 ÷ 0.0200 = −690 kJ mol⁻¹.
Why is an experimental enthalpy of combustion less exothermic than the data-book value?: Heat lost to the surroundings and apparatus, incomplete combustion, fuel evaporating from the burner, and non-standard conditions.
Required practical 2: why extrapolate a temperature–time graph?: Extrapolating the cooling line back to the time of mixing gives ΔT allowing for heat lost while the reaction was happening.

Hess’s law: The enthalpy change for a reaction is independent of the route taken, provided the starting and finishing conditions are the same.
ΔH from enthalpies of formation: ΔH = ΣΔfH°(products) − ΣΔfH°(reactants)
ΔH from enthalpies of combustion: ΔH = ΣΔcH°(reactants) − ΣΔcH°(products)
Find ΔH for PbO(s) + CO(g) → Pb(s) + CO₂(g). ΔfH°: PbO −217, CO −111, CO₂ −394 kJ mol⁻¹: ΔH = −394 − (−217 + −111) = −394 + 328 = −66 kJ mol⁻¹ (Pb is an element: zero).
Rules for following a Hess cycle: Going with an arrow, add its ΔH. Going against an arrow, reverse its sign. Multiply each value by the number of moles in the equation.

Mean bond enthalpy: The enthalpy change for breaking one mole of a given type of bond in the gaseous state, averaged over many different molecules containing it.
ΔH from bond enthalpies: ΔH = Σ(bonds broken) − Σ(bonds formed). Breaking bonds is endothermic; making bonds is exothermic.
ΔH for N₂ + 3H₂ → 2NH₃ given N≡N 945, H–H 436, N–H 391 kJ mol⁻¹: Broken: 945 + 3 × 436 = 2253. Formed: 6 × 391 = 2346. ΔH = 2253 − 2346 = −93 kJ mol⁻¹
Why does a ΔH from mean bond enthalpies differ from a Hess’s law value?: Mean bond enthalpies are averages over many molecules, not the actual values in these compounds. The method also assumes all species are gases.