The enthalpy of formation of hypothetical CaCl(s) is found to be - 180 kJ `mol^(-1)` and that of `CaCI_(2)` (s) is -800 kJ `mol^(-1)`. Calculate `Delta_(f)H^(@)` for the disproportionation reaction:
`2CaCI(s) to CaCI_(2)(s) + Ca(s)`

Explanation

The give equations are :
(i) `Ca(s) + Cl_(2)(s) to CaCl_2(s) Delta_f H^(@) = -795kJ "mol"^(-1)`
(ii) `Ca(s) + (1)/(2)Cl_2(g) to CaCl(s) Delta_fH^(@)=-188 kJ " mol"^(-1)`
(iii) `or 2Ca(s) + Cl_2(g) to 2CaCl(s) DeltaH^(@) = -376 kJ " mol"^(-1)`
Subtracting (iii) from (i)
`Delta_r H^(@)` for disproportion reaction is
`2CaCl(s) to CaCl_(2) (s) + Ca(s)`
`Delta_rH^(@) =-795-(-376)`
`=-419k J`