The enthalpies of dissolution of BaCl_2(s) an | vedclass

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(A) According to Hess's Law,the enthalpy change for the reaction $BaCl_2(s) + 2H_2O(l) 	o BaCl_2 \cdot 2H_2O(s)$ can be calculated using the dissolut

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$-29.4 \ kJ \ mol^{-1}$

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(A) According to Hess's Law,the enthalpy change for the reaction $BaCl_2(s) + 2H_2O(l) 	o BaCl_2 \cdot 2H_2O(s)$ can be calculated using the dissolution enthalpies.Let $\Delta H_{hyd}$ be the enthalpy of hydration.We have the following processes:$1$) $BaCl_2(s) 	o BaCl_2(aq)$,$\Delta H_1 = -20.6 \ kJ \ mol^{-1}$$2$) $BaCl_2 \cdot 2H_2O(s) 	o BaCl_2(aq)$,$\Delta H_2 = 8.8 \ kJ \ mol^{-1}$The target reaction is: $BaCl_2(s) + 2H_2O(l) 	o BaCl_2 \cdot 2H_2O(s)$This can be written as: $(1) - (2)$$\Delta H_{hyd} = \Delta H_1 - \Delta H_2$$\Delta H_{hyd} = -20.6 - 8.8 = -29.4 \ kJ \ mol^{-1}$

$H-H$ bond energy	$431.37 \text{ kJ mol}^{-1}$
$C=C$ bond energy	$606.10 \text{ kJ mol}^{-1}$
$C-C$ bond energy	$336.49 \text{ kJ mol}^{-1}$
$C-H$ bond energy	$410.50 \text{ kJ mol}^{-1}$

The enthalpies of dissolution of $BaCl_2(s)$ and $BaCl_2 \cdot 2H_2O(s)$ are $-20.6 \ kJ \ mol^{-1}$ and $8.8 \ kJ \ mol^{-1}$ respectively. Calculate the enthalpy of hydration for the given reaction: $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$

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