Thermochemistry and Energetics
This domain tests enthalpy bookkeeping: defining standard enthalpy changes correctly, measuring heat in the laboratory, and using Hess's law and bond enthalpies to find values that cannot be measured directly.
Definitions and standard conditions
Standard conditions are 298 K and 100 kPa. ΔHf° is the enthalpy change when one mole of a compound forms from its elements in their standard states; ΔHc° is complete combustion of one mole of a substance; ΔHneut° is per mole of water formed. Elements in their standard states have ΔHf° = 0, but a compound such as C2H5OH does not.
Calorimetry
q = mcΔT, where m is the mass of the solution being heated (approximately the volume in cm³, since density ≈ 1 g cm−3) and c = 4.18 J g−1 K−1 for water. ΔH = −q/n, the negative sign showing that heat released raises the temperature. Use a bomb calorimeter for combustion and account for heat capacity of the apparatus when given.
Hess's law and cycles
Enthalpy is a state function, so the total change is independent of route. For a reaction, ΔHr = ΣΔHf(products) − ΣΔHf(reactants). Use combustion data similarly: ΔHr = ΣΔHc(reactants) − ΣΔHc(products). Born–Haber cycles combine atomisation, ionisation, electron affinity and lattice enthalpy.
Bond enthalpies
Bond breaking is endothermic (+), bond making exothermic (−). ΔH = Σ bonds broken − Σ bonds formed. Mean bond enthalpies are averages over many compounds, so answers differ from values derived from ΔHf data.
Worked example
50.0 cm³ of 1.00 M HCl is mixed with 50.0 cm³ of 1.00 M NaOH; ΔT = 6.8 K. q = 100 × 4.18 × 6.8 = 2842 J. n(H2O) = 0.0500 mol, so ΔH = −2842/0.0500 = −56.8 kJ mol−1.
Traps that cost marks
- Omitting the negative sign for exothermic reactions.
- Using the mass of solute rather than the total solution mass in q = mcΔT.
- Assuming heat loss is zero; calculated values are usually numerically smaller than accepted ones for this reason.
- Reversing a Hess cycle and forgetting to change the sign.
- Treating mean bond enthalpies as exact.