Enthalpies of solution of `BaCI_(2). 2H_(2)O` and `BaCI_(2)` are 8.8 and `-20.6 kJ mol^(-1)` respectively.Calculate the heat of hydration of `BaCI_(2). 2H_(2)O`

Explanation

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) may be 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 according to 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`