This question is about enthalpy changes and thermodynamic feasibility.
Theoretical values for enthalpies of lattice dissociation can be calculated using a perfect ionic model. State the meaning of the term perfect ionic model.
Table 1 shows some enthalpy changes:
| Enthalpy change | ΔH/kJ mol−1\Delta H / \text{kJ mol}^{-1}ΔH/kJ mol−1 |
|---|---|
| Enthalpy of atomisation of oxygen | +249+249+249 |
| Enthalpy of atomisation of potassium | +89+89+89 |
| Enthalpy of formation of potassium oxide | −363-363−363 |
| First ionisation energy of potassium | +419+419+419 |
| First electron affinity of oxygen | −141-141−141 |
| Second electron affinity of oxygen | +798+798+798 |
Use the data in Table 1 to calculate the enthalpy of lattice dissociation of potassium oxide, K2O\text{K}_2\text{O}K2O.
Explain why the enthalpy of lattice dissociation for magnesium oxide (MgO\text{MgO}MgO) is significantly greater than the enthalpy of lattice dissociation for calcium oxide (CaO\text{CaO}CaO).
Calculate the temperature, in ∘C^\circ\text{C}∘C, above which the following reaction becomes feasible:
KCl(s)→K(s)+12Cl2(g) \text{KCl(s)} \rightarrow \text{K(s)} + \frac{1}{2}\text{Cl}_2\text{(g)} KCl(s)→K(s)+21Cl2(g)where
ΔH=+437 kJ mol−1 \Delta H = +437\text{ kJ mol}^{-1} ΔH=+437 kJ mol−1 ΔS=+92.5 J K−1 mol−1 \Delta S = +92.5\text{ J K}^{-1}\text{ mol}^{-1} ΔS=+92.5 J K−1 mol−1
