The standard heat of formation, in ${\text{kcalmol}}^{-1}$, of $\mathrm{Ba}{}^{2+}$ is: [Given: standard heat of formation of $\mathrm{SO}{}_4{}^{2-}$ ion (aq) $=-216{\text{kcalmol}}^{-1}$, standard heat of crystallisation of $\mathrm{BaSO}{}_4(\mathrm{s})=-4.5{\text{kcalmol}}^{-1}$, standard heat of formation of $\mathrm{BaSO}{}_4(\mathrm{s})=-349{\text{kcalmol}}^{-1}$]

Solution & Explanation

\textbf{Answer:} (C) $-128.5$ \textbf{Solution:} For crystallisation $\mathrm{Ba}{}^{2+}(\mathrm{aq})+\mathrm{SO}{}_4{}^{2-}(\mathrm{aq})\to \mathrm{BaSO}{}_4(\mathrm{s})$, by Hess's law: $\Delta H_{crys}={\Delta }_{f}H(\mathrm{BaSO}{}_4,s)-[{\Delta }_{f}H(\mathrm{Ba}{}^{2+})+{\Delta }_{f}H(\mathrm{SO}{}_4{}^{2-})]$. Substituting: $-4.5=-349-[{\Delta }_{f}H(\mathrm{Ba}{}^{2+})+(-216)]$. Solving: ${\Delta }_{f}H(\mathrm{Ba}{}^{2+})=-349+216+4.5=-128.5{\text{kcalmol}}^{-1}$. Mapped: source lowercase answer 'c' → option C.