Enthalpy of solution ( Delta H ) for BaCl_(2). 2H_(2) O and BaCl_(2) are 8.8 and -20.6 kJ mol^(-1) respectively. Calculate the heatof hydration of BaCl_(2) to BaCl_(2). 2H_(2)O.

Enthalpy of solution ( Delta H ) for BaCl_(2). 2H_(2) O and BaCl_(2) a

1.	Enthalpy of solution ( Delta H ) for BaCl_(2). 2H_(2) O and BaCl_(2) are 8.8 and -20.6 kJ mol^(-1) respectively. Calculate the heatof hydration of BaCl_(2) to BaCl_(2). 2H_(2)O.
Answer» Solution :We are given (i) `BaCl_(2).2H_(2)O(s) +aq rarr BaCl_(2)(aq), Delta_(sol) H^(@) = 8.8 kJ mol^(-1)` (II) `BaCl_(2)(s)+aq rarr BaCl_(2)(aq),Delta_(sol) H^(@) = - 20.6kJ mol^(-1)` We AIM at `BaCl_(2) (s) + 2H_(2)O rarr BaCl_(2).2H_(2)O(s) ,Delta_(hyd) H^(@) = ?`....(iii) Equation (ii) MAYBE written in two steps as `BaCl_(2) (s) + 2H_(2)O rarr BaCl_(2).2H_(2)O(s), DeltaH= Delta_(r)H_(1)^(@)(say) `....(iv) `BaCl_(2).2H_(2)O(s) +aq rarr BaCl_(2)(aq), DeltaH= Delta _(r) H_(2)^(@) `(say).....(v) Then accordingto Hess's law, `Delta_(r) H_(1)^(@)+Delta_(r)H_(2)^(@)= 20.6kJ` But`Delta_(r) H_(2)^(@) = 8.8 kJ mol^(-1)``[ :'`Equation (i) `=`Equation (v) ] `:. Delta_(r) H_(10^(@)) = - 20.6- 8.8 = - 29.4 kJ mol^(-1)` But Equation (iii) `=` Equation (iv) Hence, the heat of hydration of `BaCl_(2)` `Delta_(hyd) H^(@) =- 29.4 kJ mol^(-1)` Alternativly, the DISSOLUTION in one step or in two stepsmay be represented as shown in the Fig. Applying Hess's law,`Delta H = Delta H_(1) + Delta H _(2)` `- 20.6 = Delta H_(1) = 8.8` or `Delta H _(1) = -20.6 - 8.8 kJ = - 29.4 kJ`