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
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Solution
(i) Enthalpy of formation of C2H6=−83KJmol−1
2C(s)+3H2(g)⟶C2H6(g)....−83KJmol−1
(ii) Enthalpy of sublimation of C(s)⟶C(g)=719KJmol−1
Multiply by 2
∴2C(s)⟶2C(g)=1438KJmol−1
(iii) Enthalpy of bond dissociation of H2(g)=435KJmol−1
Multiply by 3
3H2(g)⟶6H(g)=1305KJmol−1
Adding equation (ii) & (iii) and subtract from equation (i)
2C(g)+6H(g)⟶C2H6(g)⟶−2826KJmol−1
This means that enthalpy of formation of 6 moles of C−H bonds and one mole of C−C bonds is 2826KJmol−1
Enthalpy of formation of 6 moles of C−H bonds is 6×414KJmol−1=2484KJmol−1