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Thermodynamics
Hess's Law and Thermochemical Calculations
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- Enthalpy of Formation from Solution
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2025 1 question
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The standard heat of formation, in $\text{kcal/mol}$ of $\text{Ba}^{2+}$ is:
[Given: standard heat of formation of $\text{SO}_4^{2-}$ ion $(aq) = -216\text{ kcal/mol}$, standard heat of crystallization of $\text{BaSO}_4(s) = -4.5\text{ kcal/mol}$, standard heat of formation of $\text{BaSO}_4(s) = -349\text{ kcal/mol}$]Using Hess's law: $\Delta H_f^\circ[\text{BaSO}_4(s)] = \Delta H_f^\circ[\text{Ba}^{2+}(aq)] + \Delta H_f^\circ[\text{SO}_4^{2-}(aq)] + \Delta H_{\text{crys}}^\circ$. Thus $-349 = \Delta H_f^\circ[\text{Ba}^{2+}] - 216 - 4.5 \implies \Delta H_f^\circ[\text{Ba}^{2+}] = -128.5\text{ kcal/mol}$.
For crystallization of barium sulfate from its aqueous ions: $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$ with $\Delta H_{\text{crys}}^\circ = -4.5\text{ kcal/mol}$. Applying Hess's Law of constant heat summation: $\Delta H_{\text{crys}}^\circ = \Delta H_f^\circ[\text{BaSO}_4(s)] - \left\{\Delta H_f^\circ[\text{Ba}^{2+}(aq)] + \Delta H_f^\circ[\text{SO}_4^{2-}(aq)]\right\}$. Substituting given values: $-4.5 = -349 - \Delta H_f^\circ[\text{Ba}^{2+}(aq)] - (-216) \implies -4.5 = -133 - \Delta H_f^\circ[\text{Ba}^{2+}(aq)] \implies \Delta H_f^\circ[\text{Ba}^{2+}(aq)] = -133 + 4.5 = -128.5\text{ kcal/mol}$.