Question

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)`

Answer 1

The give data imply as under:
i. `C Cl_(4)(l)rarr C Cl_(4)(g), DeltaH=30.5 kJ mol^(-1)`
ii. `C(s)+2Cl_(2)(g)rarr C Cl_(4)(l), DeltaH=-135.5 kJ mol^(-1)`
iii. `C(s) rarr C(g), DeltaH=715.0 kJ mol^(-1)`
iv. `Cl_(2)(g) rarr 2Cl(g), DeltaH=242 kJ mol^(-1)`
Aim: `C Cl_(4)(g) rarr C(g)+4Cl(g), DeltaH=?`
Equation (iii)`+2xx` Equation (iv)-Equation (i)-Equation (ii) gives the required equation with
`DeltaH=715.0+2(242)-30.5-(-135.5) kJ mol^(-1)`
`=1304 kJ mol^(-1)`
Bond enthalopy of `C-Cl` in `C Cl_(4)` (average value)
`=1304/4=326 kJ mol^(-1)`