Let's consider the following equations:

Formation of $\mathrm{SO}_4^{2-}$:

$ \mathrm{S} + 2 \mathrm{O}_2 + 2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_4^{2-} \quad \Delta \mathrm{H}_{\mathrm{f}} = -216 \mathrm{kcal/mol} $

Crystallization of $\mathrm{BaSO}_4(\mathrm{s})$:

$ \mathrm{Ba}^{2+}(\mathrm{g}) + \mathrm{SO}_4^{2-}(\mathrm{g}) \longrightarrow \mathrm{BaSO}_4(\mathrm{s}) \quad \Delta \mathrm{H}_{\text{crystallisation}} = -4.5 \mathrm{kcal/mol} $

Formation of $\mathrm{BaSO}_4(\mathrm{s})$:

$ \mathrm{Ba} + \mathrm{S} + 2 \mathrm{O}_2 \longrightarrow \mathrm{BaSO}_4(\mathrm{s}) \quad \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{BaSO}_4) = -349 \mathrm{kcal/mol} $

We need to find the standard heat of formation for $ \mathrm{Ba}^{2+} $.

To determine this, apply the following relationship derived from the equations:

$ \mathrm{Ba}(\mathrm{s}) \longrightarrow \mathrm{Ba}^{2+}(\mathrm{g}) + 2 \mathrm{e}^{-} $

Using the data provided:

Start from equation (3), then subtract equations (1) and (2):

$ -349 - (-4.5) - (-216) $

Simplifying the expression:

$ -349 + 4.5 + 216 $

$ = -349 + 220.5 $

$ = -128.5 \, \mathrm{kcal/mol} $

This calculation shows that the standard heat of formation for $ \mathrm{Ba}^{2+} $ is $-128.5 \, \mathrm{kcal/mol}$.