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 sulfur | +279+279+279 |
| Enthalpy of atomisation of sodium | +107+107+107 |
| Enthalpy of formation of sodium sulfide | −365-365−365 |
| First ionisation energy of sodium | +496+496+496 |
| First electron affinity of sulfur | −200-200−200 |
| Second electron affinity of sulfur | +640+640+640 |
Use the data in Table 1 to calculate the enthalpy of lattice dissociation of sodium sulfide, Na2S\text{Na}_2\text{S}Na2S.
Explain why the enthalpy of lattice dissociation for magnesium oxide (MgO\text{MgO}MgO) is significantly greater than the enthalpy of lattice dissociation for strontium oxide (SrO\text{SrO}SrO).
Calculate the temperature, in ∘C^\circ\text{C}∘C, above which the following reaction becomes feasible:
LiCl(s)→Li(s)+12Cl2(g) \text{LiCl(s)} \rightarrow \text{Li(s)} + \frac{1}{2}\text{Cl}_2\text{(g)} LiCl(s)→Li(s)+21Cl2(g)where
ΔH=+408 kJ mol−1 \Delta H = +408\text{ kJ mol}^{-1} ΔH=+408 kJ mol−1 ΔS=+88.2 J K−1 mol−1 \Delta S = +88.2\text{ J K}^{-1}\text{ mol}^{-1} ΔS=+88.2 J K−1 mol−1
