wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Enthalpy of formation of C2H6=83kJ/mol
Enthalpy of sublimation of graphite = 719kJ/mol
Enthalpy of bond dissociation of H2=435kJ/mol
C - H bond enthalpy = 414kJ/mol
Calculate the bond enthalpy of C - C

Open in App
Solution

(i) Enthalpy of formation of C2H6=83KJmol1
2C(s)+3H2(g)C2H6(g)....83KJmol1
(ii) Enthalpy of sublimation of C(s)C(g)=719KJmol1
Multiply by 2
2C(s)2C(g)=1438KJmol1
(iii) Enthalpy of bond dissociation of H2(g)=435KJmol1
Multiply by 3
3H2(g)6H(g)=1305KJmol1
Adding equation (ii) & (iii) and subtract from equation (i)
2C(g)+6H(g)C2H6(g)2826KJmol1
  • This means that enthalpy of formation of 6 moles of CH bonds and one mole of CC bonds is 2826KJmol1
  • Enthalpy of formation of 6 moles of CH bonds is 6×414KJmol1=2484KJmol1
Enthalpy of formation of CC bond is
28262484=342KJmol1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Bond Parameters
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon