ZIMSEC A Level · Chemistry · Form 5, Form 6
Enthalpy changes
ΔH is negative for exothermic and positive for endothermic reactions. Standard conditions: 298 K, 100 kPa (or 1 atm), 1 mol dm⁻³.
Key terms: enthalpy of formation (ΔHf), combustion (ΔHc), neutralisation, atomisation, and bond enthalpy (average energy to break one mole of a bond in gaseous molecules).
Calorimetry
q = mcΔT; ΔH = −q ÷ n.
Worked example: 50.0 cm³ of 1.00 mol dm⁻³ HCl and 50.0 cm³ of 1.00 mol dm⁻³ NaOH are mixed; the temperature rises by 6.8 °C. Find ΔH of neutralisation (c = 4.18 J g⁻¹ K⁻¹).
q = 100 × 4.18 × 6.8 = 2 842 J; n(water) = 0.0500 mol
ΔH = −2 842 ÷ 0.0500 = −56 800 J mol⁻¹ ≈ −56.8 kJ mol⁻¹
Hess's law
The total enthalpy change is independent of the route taken. ΔHr = ΣΔHf(products) − ΣΔHf(reactants), or ΔHr = ΣΔHc(reactants) − ΣΔHc(products).
Bond enthalpies
ΔH = Σ(bonds broken) − Σ(bonds formed). Values are averages, so results are approximate.
Lattice energy and Born–Haber cycles (Form 6)
Lattice energy is the enthalpy change when one mole of an ionic solid forms from gaseous ions. It is more exothermic for smaller ions with higher charges.
Entropy and Gibbs energy (Form 6)
ΔG = ΔH − TΔS. A reaction is feasible when ΔG is negative.
Check yourself
- State Hess's law.
- Bonds broken total +2 648 kJ and bonds formed total −3 466 kJ. Find ΔH.
- When is a reaction feasible?
Show answers
- The enthalpy change for a reaction is independent of the route, provided the start and end conditions are the same.
- 2 648 − 3 466 = −818 kJ mol⁻¹
- When ΔG = ΔH − TΔS is negative.
Questions about this topic (0)
No questions about this topic yet.
Sign in or create a free account to join in.