Calculate the enthalpy change for the process C Cl_(4)(g) rarrC(g) +4Cl(g) and calculate bond enthalpyofC-Cl in C Cl_(4)(g) Given :V_(vap) H^(@) (C C l_(4)) =30.5 kJ mol^(-1), Delta_(f)H^(@) (C C l_94))= -135.5 kJ mol^(-1) Delta_(a)H^(@) ( C ) = 715.0 kJ mol^(-1)whereDelta_(a)H^(@)is enthalpyof atomisation Delta_(a) H^(@)( Cl_(2)) = 242 kJ mol^(-1)

Calculate the enthalpy change for the process C Cl_(4)(g) rarrC(g) +4C

1.	Calculate the enthalpy change for the process C Cl_(4)(g) rarrC(g) +4Cl(g) and calculate bond enthalpyofC-Cl in C Cl_(4)(g) Given :V_(vap) H^(@) (C C l_(4)) =30.5 kJ mol^(-1), Delta_(f)H^(@) (C C l_94))= -135.5 kJ mol^(-1) Delta_(a)H^(@) ( C ) = 715.0 kJ mol^(-1)whereDelta_(a)H^(@)is enthalpyof atomisation Delta_(a) H^(@)( Cl_(2)) = 242 kJ mol^(-1)
Answer» Solution :The given data imply as under `:(i)C Cl_(4) (l) rarr C C l_(4)(g) , DeltaH= 30.5 kJ mol^(-1)` (II)`C(s) + 2Cl_(2)(g) rarrC C l_(4)(l) , DeltaH =-135.5 kJ mol^(-1)` (iii) `C(s)rarr C(g) , DeltaH= 715 .0 kJ mol^(-1)` (IV) `Cl_(2)(g) rarr2Cl(g), DeltaH = 242 kJ mol^(-1)` AIM`:C C l_(4)(g)rarr C(g) + 4Cl(g) , DeltaH = ?` Eqn. (iii)`+2 xx` Eqn. (iv) - Eqn. (i). Eqn. (ii) gives the requiredequation with `DeltaH=715.0+ 2 ( 242)-30.5 - ( - 135.5) kJ mol^(-1) = 1304 kJ mol^(-1)` Bond enthalpy ofC-Cl in `C C l_(4)` ( average value)`= ( 1304)/( 4) = 326 kJ mol^(-1)`