TheIBAxolotl
Bond enthalpy
Bond breaking is endothermic; bond making is exothermic. The bond enthalpy is the energy needed to break one mole of a given bond in gaseous molecules. Values in data tables are mostly averages across many compounds.
\[\Delta H \approx \sum \text{(bonds broken)} - \sum \text{(bonds formed)}\]
If the bonds formed are stronger than those broken, the reaction is exothermic.
Worked example
Estimate \(\Delta H\) for H\(_2\)(g) + Cl\(_2\)(g) → 2HCl(g), using typical values: H–H 436, Cl–Cl 242, H–Cl 431 kJ mol\(^{-1}\).
- Broken: \(436 + 242 = 678\) kJ
- Formed: \(2 \times 431 = 862\) kJ
- \(\Delta H = 678 - 862 = -184\) kJ mol\(^{-1}\)
(In the exam, use the values given in your data booklet.)
Why bond enthalpy answers are approximate
- Average values do not match the exact bond environment in a specific molecule.
- They apply only to gases; if water is formed as a liquid, the energy of condensation is missing.
So values from bond enthalpies differ from those found using enthalpies of formation.
Hess's law
The total enthalpy change for a reaction is the same whatever route is taken, provided the starting and finishing states are the same. This follows from conservation of energy.
Rules when combining equations:
- Reverse an equation → change the sign of \(\Delta H\).
- Multiply an equation → multiply \(\Delta H\) by the same factor.
- Add the equations so unwanted species cancel; add the \(\Delta H\) values the same way.
Example: C(s) + O\(_2\)(g) → CO\(_2\)(g), \(\Delta H = -394\) kJ; CO(g) + ½O\(_2\)(g) → CO\(_2\)(g), \(\Delta H = -283\) kJ. Reversing the second and adding gives C(s) + ½O\(_2\)(g) → CO(g), \(\Delta H = -394 + 283 = -111\) kJ mol\(^{-1}\).
Formation and combustion data (HL)
- Standard enthalpy of formation: one mole of compound formed from its elements in standard states. For elements in their standard states it is zero.
- Standard enthalpy of combustion: one mole of substance burned completely in oxygen.
\[\Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})\]
\[\Delta H = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})\]
Worked example (HL)
Find \(\Delta H_c\) of methane: CH\(_4\)(g) + 2O\(_2\)(g) → CO\(_2\)(g) + 2H\(_2\)O(l). \(\Delta H_f\) values: CH\(_4\) −74.8, CO\(_2\) −393.5, H\(_2\)O(l) −285.8 kJ mol\(^{-1}\); O\(_2\) is 0.
\[\Delta H = [-393.5 + 2(-285.8)] - [-74.8] = -965.1 + 74.8 = -890.3\ \text{kJ mol}^{-1}\]
Born–Haber cycles (HL)
A Born–Haber cycle is a Hess's law cycle that finds the lattice enthalpy of an ionic compound, which cannot be measured directly. Steps for NaCl (kJ mol\(^{-1}\)):
| Step | Value |
|---|---|
| Atomise Na(s) → Na(g) | +107 |
| Ionise Na(g) → Na\(^+\)(g) | +496 |
| Atomise ½Cl\(_2\)(g) → Cl(g) | +121 |
| Electron affinity Cl(g) → Cl\(^-\)(g) | −349 |
| Formation of NaCl(s) | −411 |
Lattice enthalpy (NaCl(s) → Na\(^+\)(g) + Cl\(^-\)(g)):
\[\Delta H_\text{latt} = -(-411) + 107 + 496 + 121 + (-349) = +786\ \text{kJ mol}^{-1}\]
Common mistakes
- Doing "formed − broken" instead of "broken − formed".
- Forgetting to multiply bond enthalpies by the number of bonds (CH\(_4\) has four C–H).
- Not reversing the sign when reversing an equation.
- (HL) Giving an element in its standard state a non-zero \(\Delta H_f\).
- (HL) Using the full Cl–Cl bond enthalpy in a Born–Haber cycle instead of half.
Exam tips
- Draw displayed formulas before counting bonds.
- Explain differences between bond-enthalpy and data-table values using "average values" and "gaseous state".
- In Hess's law questions, write the target equation first and tick off species as they cancel.
- (HL) In Born–Haber cycles, label each arrow and check that all upward arrows are endothermic.