The enthlpy of formation of hypothetical `CaCl_((s))` theoretically found to be `-188 kJ mol^(-1` and the `Delta_(f)H^(Θ)` for `CaCl_(2(s))` is `-795 kJ mol^(-1)`. Calculate the `Delta_(f)H^(Θ)` for the disproportionation reaction.
`2 CaCl_((s))rarrCaCl_(2(s))+Ca_((s))`

Explanation

Given, `Delta_(f)H^(Θ) "for" CaCl=-188 kJ mol^(-1)`
`Delta_(f)H^(Θ) "for" CaCl=-795 kJ mol^(-1)`
`Ca_((s))+(1)/(2)Cl_(2(s)) rarr CaCI_((s)) Delta_(f)H_(1)^(Θ)=-188 kJ mol^(-1)`…(i)
`Ca_((s))+Cl_(2(g))rarrCaCl_(2(s))Delta_(f)H_(2)^(Θ)=-795 kJ mol^(-1)`…(ii)
To calculate `Delta_(f)H_(3)^(Θ)` for disproportionation reaction:
`2CaCl_((s))rarrCa_((s))+CaCl_(2(s)), Delta_(f)H_(3)^(Θ)=?`...(iii)
`Delta_(f)H_(3)^(Θ)=Delta_(f)H_(2)^(Θ)-2Delta_(f)H_(1)^(Θ)`
`=-795-(2xx-188)=-419 kJ mol^(-1)`
Hence, `Delta_(f)H^(Θ)` for disproportionation of `CaCl_((s))` is `-419 kJ mol^(-1)`.