The standard heat of formation, in $\text{kcal\,mol}^{-1}$, of $\ce{Ba^2+}$ is: [Given: standard heat of formation of $\ce{SO4^2-}$ ion (aq) $= -216\ \text{kcal\,mol}^{-1}$, standard heat of crystallisation of $\ce{BaSO4(s)} = -4.5\ \text{kcal\,mol}^{-1}$, standard heat of formation of $\ce{BaSO4(s)} = -349\ \text{kcal\,mol}^{-1}$]

Solution & Explanation

\textbf{Answer:} (C) $-128.5$ \textbf{Solution:} For crystallisation $\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}$, by Hess's law: $\Delta H_{crys} = \Delta_f H(\ce{BaSO4},s) - \big[\Delta_f H(\ce{Ba^2+}) + \Delta_f H(\ce{SO4^2-})\big]$. Substituting: $-4.5 = -349 - \big[\Delta_f H(\ce{Ba^2+}) + (-216)\big]$. Solving: $\Delta_f H(\ce{Ba^2+}) = -349 + 216 + 4.5 = -128.5\ \text{kcal\,mol}^{-1}$. Mapped: source lowercase answer 'c' → option C.