Calculate the enthalpy change for the process "CC"I_(4(g)) + C_((g)) + 4CI_((g)) and calculate bond enthalpy of C - CI in "CC"_(4(g)) Delta_("vap")H^( Theta ) ("CC"I_(4) ) = 30.5 "kJ mol"^(-1) Delta_(f) H^( Theta ) ("CCI"_(4) ) = -135.5 "kJ mol"^(-1) Delta_(a) H^( Theta ) (C) = 715.0 "kJ mol"^(-1) where Delta_(a) H^( Theta ) is enthalpy of atomisationDelta_(a) H^( Theta ) (CI_(2) ) = 242 "kJ mol"^(-1)

Calculate the enthalpy change for the process "CC"I_(4(g)) + C_((g)) +

1.	Calculate the enthalpy change for the process "CC"I_(4(g)) + C_((g)) + 4CI_((g)) and calculate bond enthalpy of C - CI in "CC"_(4(g)) Delta_("vap")H^( Theta ) ("CC"I_(4) ) = 30.5 "kJ mol"^(-1) Delta_(f) H^( Theta ) ("CCI"_(4) ) = -135.5 "kJ mol"^(-1) Delta_(a) H^( Theta ) (C) = 715.0 "kJ mol"^(-1) where Delta_(a) H^( Theta ) is enthalpy of atomisationDelta_(a) H^( Theta ) (CI_(2) ) = 242 "kJ mol"^(-1)
Answer» SOLUTION :As per the given information. (i) `"CCI"_(4(l)) to "CCI"_(4(G)) , Delta H = 30.5 "kJ/mol"` (ii) `C_((s)) + 2CI_(2(g)) to "CCI"_(4(l)), DeltaH = -135.5 "kJ /mol"` (III) `C_((s)) to C_((g)) , Delta H = 715 "kJ/mol"` (iv) `CI_(2(g)) to 2CI_((g)) , Delta H = 242 "kJ/mol"` Now, for `"CCI"_(4(g)) to C_((g)) + 4CI_((g))` `DeltaH= [715 + 2(242)] - [30.5 +(-135.5)]` `=1304 "kJ/mol"` `therefore` Bond enthalpy of `C- CI=(1304)/(4)=326 "kJ/mol"`