Thermochemistry and Energetics
This domain tests enthalpy change: sign conventions, calorimetry calculations, Hess's law cycles, standard enthalpy of formation and combustion data, mean bond enthalpies, and the entropy and Gibbs free energy link to whether a reaction is feasible.
Enthalpy basics
- ΔH is negative for exothermic reactions and positive for endothermic ones; standard conditions are 298 K and 100 kPa.
- ΔHf° for an element in its standard state is zero by definition.
- ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants). For combustion data the sign convention reverses: ΔH = ΣΔHc(reactants) − ΣΔHc(products).
Calorimetry
- q = mcΔT; for dilute aqueous solutions assume density 1.00 g cm−3 and c = 4.18 J g−1 K−1, so the mass is the total mass of solution.
- ΔH per mole = −q/n, converted to kJ mol−1.
Hess's law and bond enthalpies
- Any route between the same initial and final states gives the same ΔH; construct cycles, including Born–Haber cycles for ionic compounds.
- ΔH = Σ bonds broken − Σ bonds formed. Mean bond enthalpies are averages over many compounds, so answers differ slightly from formation-data values.
Feasibility
- ΔG = ΔH − TΔS; a reaction is feasible when ΔG is negative. Entropy rises with more moles of gas, higher temperature and greater disorder; a reaction with positive ΔH can become feasible at high T if ΔS is positive.
Worked example
50.0 cm3 of 1.00 M HCl is mixed with 50.0 cm3 of 1.00 M NaOH and the temperature rises by 6.8 K. q = 100 g × 4.18 × 6.8 = 2842 J. n = 0.0500 mol of water formed. ΔH = −2.842/0.0500 = −56.8 kJ mol−1; the negative sign follows from the temperature rise.
Traps that cost marks
- Omitting the minus sign, or using only one solution's mass in a neutralisation calculation.
- Expecting the textbook value exactly; heat losses make the measured value less exothermic.
- Getting the bond enthalpy sign the wrong way round.
- Assigning a non-zero ΔHf° to an element.
- Forgetting that ΔG, not ΔH alone, decides feasibility.