Calculate the enthalpy change for the process
`"CC"l_(4)(g)toC(g)+4CI(g)`
and calculate bond enthalpy of `C-Cl in C Cl_(4)(g).`
`Delta_("vap")H^(theta)("C C"l_(4))=30.5 kJ mol^(-1)`.
`Delta_(f)H^(theta)(CCl_(4))=-135.5 kJ mol^(-1)`.
`Delta_(0)H^(theta)(C)=715.0 kJ mol^(-1)`, where `Delta_(a)H^(theta)` is enthalpy of atomisation.
`Delta_(a)H^(theta)(Cl_(2))=242 kJ mol^(-1)`.

Explanation

`CCl_(4(1)) to CCl_(4(g)) to C_(g) +4Cl_(g)`
`triangle_("vap")H^@CCl_4 ="30.5 KJ mol"^(-1)`
`triangle_(r)H^@ (CCl_(4))=-"135.5 KJ mol"^(-1)`
`triangleH^@(C)="715.0 KJ mol"^(-1)`
`triangle_(r)H^@ (Cl_2)="242 KJ mol"^(-1)`
`CCl_(4(l)) to CCl_(4(g)) to C_(g) +4Cl_(g) (therefore Cl_2 overset(242)to 2Cl, 2Cl_2 overset(2(2+2))to 4Cl)`
`"+30.5 -135.5 715 2(242)"`
`triangle_(r)H^@=715+484+135.5-30.5`
`triangle_r H^@="1334.5-30.5KJ mol"^(-1)=1304KJ`
C-Cl bond enthalpy `=(1304)/(4)="325KJ mol"^(-1)`