- - J mol J K mol f 30400 T f = . K . SOLUTION : Given C H + 5O → 3CO + 4H O ∆ H C = − . kJ mol − −−−−−( ) C + O → CO ∆ H f = − .
kJ mol − −−−−−( ) H + O → H O ∆ H f = − . kJ mol − −−−−− ( ) 3C + 4H → C H ∆ H C = ? ( ) × ⇒ 3C + 3O → 3CO ∆ H f = − . kJ −−−−−( ) ( )× ⇒ H + 2O → 4H O ∆ H f = − .
kJ −−−−−( ) ( ) + ( ) − ( ) ⇒ 3C + 3O + 4H + 2O + 3CO + 4H O → 3CO + 4H O + C H + 5O ∆ H f = − . − . − (− . ) kJ 3C + H → C H ∆ H f = − .
kJ Standard heat of formation of propane is ∆ H f (C H ) = − . kJ . S. No Liquid Boiling points ( C) ΔH ( kJ mol − ) .
Ethanol . + . . Toluene .
+ . SOLUTION : For ethanol : Given : T b = . C = ( . + ) = .
K ΔH V (ethanol) = + . kJ mol − ∆ = ∆ S H b ∆ = + S kJ mol K S J mol K 42400 ΔS V = + . J K − mol − For Toluene : Given : T b = . C = ( .
+ ) = . K ΔH V (toluene) = + . kJ mol − ∆ = ∆ S H b ∆ = + S kJ mol K