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Q.56) The standard heat of formation, in $\mathrm{kcal} / \mathrm{mol}$ of $\mathrm{Ba}^{2+}$ is :
[Given : standard heat of formation of $\mathrm{SO}_4^{2-}$ ion $(\mathrm{aq})=-216 \mathrm{kcal} / \mathrm{mol}$, standard heat of crystallisation of
$\mathrm{BaSO}_4(\mathrm{~s})=-4.5 \mathrm{kcal} / \mathrm{mol}$, standard heat of formation of $\left.\mathrm{BaSO}_4(\mathrm{~s})_5=-349 \mathrm{kcal} / \mathrm{mol}\right]$

A) +220.5

B) -128.5

C) -133.0

D) +133

Answers (1)

best_answer

Given that 

$\Delta \mathrm{Hf}^{\circ}\left(\mathrm{BaSO}_4, \mathrm{~s}\right)=-349 \mathrm{kcal} / \mathrm{mol}$
$\Delta \mathrm{Hf}^{\circ}\left(\mathrm{SO}_4{ }^{2-}, \mathrm{aq}\right)=-216 \mathrm{kcal} / \mathrm{mol}$
$\Delta \operatorname{Hcryst}^{\circ}\left(\mathrm{BaSO}_4\right)=-4.5 \mathrm{kcal} / \mathrm{mol}$

Apply Hess’s Law, consider the following reaction:

$\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_4^{2-}(a q) \longrightarrow \mathrm{BaSO}_4(s)$

Enthalpy change for this reaction is the heat of crystallization:

$
\Delta H_{\mathrm{rxn}}=\Delta H_{\mathrm{cryst}}=-4.5 \mathrm{kcal} / \mathrm{mol}
$

Using Hess's law:

$
\Delta H_f\left(\mathrm{BaSO}_4\right)=\Delta H_f\left(\mathrm{Ba}^{2+}\right)+\Delta H_f\left(\mathrm{SO}_4^{2-}\right)+\Delta H_{\text {cryst }}
$
Rearranged to solve for $\Delta H_f\left(\mathrm{Ba}^{2+}\right)$ :

$
\Delta H_f\left(\mathrm{Ba}^{2+}\right)=\Delta H_f\left(\mathrm{BaSO}_4\right)-\Delta H_f\left(\mathrm{SO}_4^{2-}\right)-\Delta H_{\mathrm{cryst}}
$

$\begin{gathered}\Delta H_f\left(\mathrm{Ba}^{2+}\right)=(-349)-(-216)-(-4.5) \\ \Delta H_f\left(\mathrm{Ba}^{2+}\right)=-349+216+4.5 \\ \Delta H_f\left(\mathrm{Ba}^{2+}\right)=-128.5 \mathrm{kcal} / \mathrm{mol}\end{gathered}$

Hence the correct option is (B)

Posted by

Saumya Singh

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