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

• Correct answer
  A

  -128.5
• B

  -133.0
• C

  +133.0
• D

  +220.5

Solution

$\begin{aligned} & \Delta \mathrm{H}_{\mathrm{crys}}\left(\mathrm{BaSO}_4\right)=\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{BaSO}_4\right)-\left[\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{Ba}^{+2}\right)+\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{SO}_4^{-2}\right)\right] \\ & -4.5=[-349]-\left[\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{Ba}^{+2}\right)+(-216)\right] \\ & \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{Ba}^{+2}\right)=-128.5 \mathrm{kCal} \mathrm{~mol}^{-1} \end{aligned}$

Hence, the correct answer is option: 'A'.