From the following data : Enthalpy of formation of `CH_(3)CN`=87.86 kJ/mol Enthalpy of formation of `C_(2)H_(6)=-` 83.68 kJ/mol Enthalpy of sublimatio
From the following data :
Enthalpy of formation of `CH_(3)CN`=87.86 kJ/mol
Enthalpy of formation of `C_(2)H_(6)=-` 83.68 kJ/mol
Enthalpy of sublimation of graphite =719.65 kJ/mol
enthalpy of dissociation of nitrogen =945.58 kJ/mol
Enthalpy of dissociation of `H_(2) =` 435.14 kJ/mol
`C-H` bond enthalpy =414.22 kJ/mol
Calculate the bond enthalpy of (i) `C-C` , (ii) `C equiv N`
1 Answers
Correct Answer - (i) `343.58" kJ/mol"^(-1)`; (ii) `891.2" kJ/mol"^(-1)`
(i) `2C(s)+3H_(2) overset(DeltaH)(rarr) H-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-H" "`[x=bond enethalpy of `C-C`]
`-83.68=1439.3+1305.42-[(414.22xx6)+x]`
`x=343.58`
(i) `2C(s)+3/2 H_(2)+1/2 N_(2) rarr CH_(3) -C equiv N" "` [x= bond enthalpy of `C equiv N`]
`87.86=1439.3+652.71+472.79-[343.08+x+1242.66]`
`x=891.2`