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The enthalpy of dissolution of ${BacI}_{2} (s)$ and ${BaCl}_{2} \cdot 2 H_{2} O (s)$ are $- 20.6$ and $8.8 KJ {mol}^{- 1}$ respectively. Calculate enthalpy of hydration for given reaction:
${BaCl}_{2} (s) + 2 H_{2} O ⟶ {BaCl}_{2} \cdot 2 H_{2} O (s)$

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$- 29.4 kJ$

$- 35.4 kJ$

$- 24.4 kJ$

$- 15 .2 kJ$

answer is A.

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Detailed Solution

Enthalpy of hydration = enthalpy of dissolution of ${BacI}_{2} (s)$ enthalpy of dissolution of ${BaCl}_{2} \cdot 2 H_{2} O (s)$

Enthalpy of hydration

$- 20.6 - 8.8 = - 29.4 kJ$

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