For an ionic solid , where X is monovalent, the enthalpy of formation of the solid from and is 1.5 t — Thermodynamics and Thermochemistry Chemistry Question
Question
For an ionic solid $\text{MX}_2$, where X is monovalent, the enthalpy of formation of the solid from $\text{M(s)}$ and $\text{X}_2\text{(g)}$ is 1.5 times the electron gain enthalpy of $\text{X(g)}$. The first and second ionization enthalpies of the metal (M) are 1.2 and 2.8 times of the enthalpy of sublimation of $\text{M(s)}$. The bond dissociation enthalpy of $\text{X}_2\text{(g)}$ is 0.8 times the first ionization enthalpy of metal and it is also equal to one-fifth of the magnitude of lattice enthalpy of $\text{MX}_2$. If the electron gain enthalpy of $\text{X(g)}$ is $-96\text{ kcal/mol}$, then what is the enthalpy of sublimation (in kcal/mol) of the metal (M)?
💡 Solution & Explanation
Using the Born-Haber cycle for $\text{MX}_2$: $\Delta H_f = \Delta H_{\text{sub}}(\text{M}) + \text{IE}_1 + \text{IE}_2 + \Delta H_{\text{diss}}(\text{X}_2) + 2\Delta H_{\text{eg}}(\text{X}) + \Delta H_{\text{lattice}}$. Let $S = \Delta H_{\text{sub}}(\text{M})$. Given $\Delta H_{\text{eg}} = -96\text{ kcal/mol}$, $\Delta H_f = 1.5(-96) = -144\text{ kcal/mol}$. $\text{IE}_1 = 1.2S$, $\text{IE}_2 = 2.8S$. $\Delta H_{\text{diss}}(\text{X}_2) = 0.8(\text{IE}_1) = 0.8(1.2S) = 0.96S$. $\Delta H_{\text{lattice}}$ magnitude is $5 \times 0.96S = 4.8S \implies \Delta H_{\text{lattice}} = -4.8S$. Substituting values: $-144 = S + 1.2S + 2.8S + 0.96S + 2(-96) - 4.8S \implies -144 = 1.16S - 192 \implies 1.16S = 48 \implies S = 48 / 1.16 = 41.379\text{ kcal/mol}$.