This page brings together the three types of energy cycle with fully worked examples.
Example 1: Using Formation Data
Calculate ΔH for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Given:
- ΔHf[CH₄] = −74.8 kJ mol⁻¹
- ΔHf[CO₂] = −393.5 kJ mol⁻¹
- ΔHf[H₂O] = −285.8 kJ mol⁻¹
- ΔHf[O₂] = 0 kJ mol⁻¹ (element)
\( \Delta H = \sum \Delta H_f(\text{prod}) - \sum \Delta H_f(\text{react}) \)
\( = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)] \)
\( = [-393.5 - 571.6] - [-74.8] \)
\( = -965.1 + 74.8 \)
\( = -890.3 \text{ kJ mol}^{-1} \)
Example 2: Born-Haber Cycle (NaCl)
Calculate the lattice enthalpy of NaCl
Given:
| ΔHf[NaCl] | = −411 kJ mol⁻¹ |
| ΔHat[Na] | = +107 kJ mol⁻¹ |
| ΔHat[Cl] | = +122 kJ mol⁻¹ |
| IE₁[Na] | = +496 kJ mol⁻¹ |
| EA₁[Cl] | = −349 kJ mol⁻¹ |
\( \Delta H_f = \Delta H_{at}(\text{Na}) + \Delta H_{at}(\text{Cl}) + IE_1 + EA_1 + \Delta H_{lat} \)
\( -411 = (+107) + (+122) + (+496) + (-349) + \Delta H_{lat} \)
\( -411 = +376 + \Delta H_{lat} \)
\( \Delta H_{lat} = -411 - 376 = -787 \text{ kJ mol}^{-1} \)
Example 3: Enthalpy of Solution
Calculate ΔHsol for KCl
Given:
- ΔHlat[KCl] = −711 kJ mol⁻¹
- ΔHhyd[K⁺] = −322 kJ mol⁻¹
- ΔHhyd[Cl⁻] = −363 kJ mol⁻¹
\( \Delta H_{sol} = -\Delta H_{lat} + \sum \Delta H_{hyd} \)
\( = -(-711) + [(-322) + (-363)] \)
\( = +711 - 685 \)
\( = +26 \text{ kJ mol}^{-1} \)
The positive value means dissolving KCl is endothermic — the solution cools down. This is used in instant cold packs.
Exam Strategy
Always draw the cycle before attempting the calculation. Label all known values with their signs, then use Hess's Law to find the unknown. The most common error is getting the sign wrong — pay close attention to the direction of each arrow.