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

`-326 kJ mol^(-1)`

Explanation

The chemical equations implying to the given values of enthalpies are :
(i)`C CI_(4(l))rarrC CI_(4(g))Delta_("vap")H^(theta)=30.5 kJ "mol"^(-1)`
(ii) `C_((g))rarrC_((g))Delta_(a)H^(theta)=715.0 kJ "mol"^(-1)`
(iii) `CI_(2g)rarr2CI_((g))Delta_(a)H^(theta)=242 kJK "mol"^(-1)`
(iv) `C_((g))+4CI_((g))rarrC CI_(4(g))Delta_(f)H=-135.5 kJ "mol"^(-1)`
Enthalpy change for the given process `C CI_(4(g))rarrC_((g))+4CI_((g))'` can be calculated using the following algebraic calculations as :
Equation (ii) +2`xx` Equation (ii) -Equation (i) -Equation (iv)
`DeltaH=Delta_(a)H^(theta)(C)+2Delta_(a)H^(theta)(CI_(2))-Delta_("vap")H^(theta)-Delta_(f)H`
`=(715.0 kJ "mol"^(-1))+2(242 kJ "mol"^(-1))-(30.5 kJ "mol"^(-1)) -(-135.5 lkJ "mol"^(-1))`
`:. DeltaH=1304 kJ "mol"^(-1)`
Bond enthalpy of C-CI bond in `C CI_(4(g))`
`=(1304)/(4)kJ "mol"^(-1)`
=326 kJ `"mol"^(-1)`