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Flashcards · 9701 · AS Topic 5
Chemical energetics (AS Level) flashcards
13 cards on AS Topic 5, subtopics 5.1 to 5.2. Pick a subtopic or study them all.
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5.1 Enthalpy change
- Sign of ΔH for exothermic and endothermic reactions
- Exothermic: ΔH negative, heat released to the surroundings, products lower in energy. Endothermic: ΔH positive, heat absorbed, products higher in energy.
- Standard enthalpy change of formation, ΔHf⦵
- The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. It is zero for any element in its standard state.
- Standard enthalpy change of combustion, ΔHc⦵
- The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.
- Standard enthalpy change of neutralisation
- The enthalpy change when an acid and an alkali react to form one mole of water (under standard conditions).
- Bond enthalpy, and why calculations using it are approximate
- The energy required to break one mole of a specified bond in the gaseous state. Values are means over many compounds and apply only to gases, so results are approximate.
- Standard conditions used in 9701
- 298 K and 101 kPa, with all substances in their standard states.
- Calorimetry: the two equations, and what m is
- q = mcΔT, with m the mass of water or solution heated (not the fuel) and c = 4.18 J g⁻¹ K⁻¹. ΔH = −q ÷ n, with n the moles of the limiting reactant. Convert J to kJ.
- Why is an experimental enthalpy of combustion usually less exothermic than the data-book value?
- Heat lost to the surroundings, incomplete combustion, evaporation of the fuel, heat absorbed by the apparatus, non-standard conditions.
5.2 Hess's law
- Hess's law
- The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
- ΔHr from formation data, and from combustion data
- Formation: ΔHr = ΣΔHf(products) − ΣΔHf(reactants). Combustion: ΔHr = ΣΔHc(reactants) − ΣΔHc(products). Multiply each value by its coefficient.
- Find ΔHf⦵(CH₄) from ΔHc: C(graphite) −394, H₂ −286, CH₄ −890 kJ mol⁻¹
- ΔHf = [ΔHc(C) + 2ΔHc(H₂)] − ΔHc(CH₄) = [−394 + 2(−286)] − (−890) = −966 + 890 = −76 kJ mol⁻¹
- ΔH from bond enthalpies; H–H 436, Cl–Cl 242, H–Cl 431 kJ mol⁻¹. Find ΔH for H₂ + Cl₂ → 2HCl.
- ΔH = Σ(bonds broken) − Σ(bonds made) = (436 + 242) − (2 × 431) = 678 − 862 = −184 kJ mol⁻¹
- Why do Hess cycles let you find enthalpy changes that cannot be measured directly?
- Because ΔH is independent of route, an unmeasurable change (such as ΔHf of methane) can be found from measurable ones (such as enthalpies of combustion) that link the same start and end.