Energetics and Thermochemistry
Energetics and Thermochemistry
The exam tests standard enthalpy definitions, Hess's law cycles, bond enthalpies, calorimetry and, at the top end, entropy and Gibbs free energy as the test of feasibility.
Definitions and signs
Standard conditions are 298 K and 100 kPa. Learn them precisely: enthalpy of formation (one mole of compound from elements in standard states), combustion (complete combustion of one mole), neutralisation (one mole of water formed from acid and base), atomisation (one mole of gaseous atoms formed), lattice enthalpy (one mole of ionic solid formed from gaseous ions), hydration and solution. Exothermic means negative delta H; bond breaking is endothermic, bond making exothermic.
Hess's law and cycles
Enthalpy change is independent of route. Build a cycle with the elements in standard states at the top or bottom and apply Hess's law to the arrows. Bond enthalpies give only average values and apply strictly to gaseous species, so they are less accurate than formation data.
Worked example
Find the standard enthalpy of formation of methane from combustion data: C(s) + O2(g) -> CO2(g), delta H = -393 kJ mol-1; H2(g) + 1/2 O2(g) -> H2O(l), delta H = -286 kJ mol-1; CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), delta H = -890 kJ mol-1. Formation means C(s) + 2H2(g) -> CH4(g). Follow the route: burn the elements to CO2 and 2H2O, then reverse the methane combustion. delta Hf = -393 + 2(-286) - (-890) = -965 + 890 = -75 kJ mol-1.
Calorimetry and entropy
For a solution experiment, q = mc(delta T), moles = concentration x volume, and delta H = -q/moles, with m in grams and c = 4.18 J K-1 g-1. Feasibility uses delta G = delta H - T delta S; a reaction is feasible when delta G is negative, so an endothermic reaction can be feasible if delta S is positive and T is high. Mind the units: entropy is in J K-1 mol-1, so convert to kJ.
Traps
Reversing a cycle without changing the sign; confusing combustion and formation; ignoring heat loss to the surroundings in calorimetry; forgetting the negative sign in delta H = -q/n; and assuming a negative delta H guarantees a spontaneous reaction.
Key ideas
- Standard conditions: 298 K, 100 kPa, stated physical states.
- Hess's law: route does not matter, sign reversal does.
- Bond enthalpies are average and apply to gaseous species only.
- q = mc(delta T); delta H = -q/n for calorimetry.
- Feasibility requires delta G = delta H - T delta S < 0.