The enthalpy of dissolution of `BaCl_(2)(s) and BaCl_(2).2H_(2)O` are -20.6 and 8.8 KJ `mol^(-1)` respectively . Calculate enthalpy of hydration forgiven reaction:
`BaCl_(2)(s)+2H_(2)OtoBaCl_(2).2H_(2)O(s)`

Options

A. 29.4 kJ
B. `-29.4 kJ`
C. `-11.8 kJ`
D. `38.2 kJ`

Answer

B

Explanation

(i) `BaCl_2(s)+aq rarr BaCl_2(aq). Delta H =- 20.6 kJ `
(ii) `BaCl_2. 2H_2 O(s)+aq rarr BaCl_2(aq),Delta H= +8.8 kJ `
Eqn. (i) - Eqn. (ii) gives
`BaCl_2(s) +2H_2O(s) + aq rarr BaCl_2 (s)`
`Delta H =- 20.6 -8.8 =-29.4 kJ `