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Calculate the enthalpy change for the pr...

Calculate the enthalpy change for the process
`C Cl_(4)(g) rarr C(g)+4Cl(g)`
and calculate bond enthalpy of `C-Cl` in `C Cl_(4)(g)`.
`Delta_(vap)H^(Θ)(C Cl_(4))=30.5 kJ mol^(-1)`
`Delta_(f)H^(Θ)(C Cl_(4))=-135.5 kJ mol^(-1)`
`Delta_(a)H^(Θ)(C )=715.0 kJ mol^(-1)`, where `Delta_(a)H^(Θ)` is enthalpy of atomisation
`Delta_(a)H^(Θ)(Cl_(2))=242 kJ mol^(-1)`

Text Solution

Verified by Experts

The enthalpy change for the reaction:
`C CI_(4)(g) to C(g) + 4 CI(g)`
`DeltaH^(@) = Delta_(alpha)H^(@)(C) + 4 Delta_(alpha)H^(@)(CI) - Delta_(f)H^(@) (C CI_(4))(g)`
`Delta_(alpha)H^(@)(C) = 715.0 kJ mol^(-1)`
`Delta_(alpha)H^(@)(CI_(2)) = 242 kJ mol^(-1)`
Let us first calculate `Delta_(f)H^(@)[C CI_(4)(g)]`
`C(s) + 2CI_(2)(g) to C CI_(4)(l) DeltaH^(@) = -135.5 kJ mol^(-1)`
`C CI_(4)(l) to C CI_(4)(g) DeltaH^(@) = 30.5 kJ mol^(-1)`
Adding `C(s) + 2Cl_(2)(g) to C CI_(4)(g)`
`therefore DeltaH^(@) = 715.0 + 2 xx 242 - (-105.0) = 1304.0 kJ mol^(-1)`
This enthalpy change corresponds to breaking of four C - CI bonds
`therefore` Bond enthalpy of C - CI bond = `(1304.0)/(4) = 326 kJ mol^(-1)`.
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