Exercise 6 | Thermochemistry

Given that:

ΔH0f (H (g)) = 218.0 kJ.mol-1

ΔH0f (C (g)) = 716.7 kJ.mol-1

ΔH0f (CH4 (g)) = - 74.60 kJ.mol-1


Calculate the average molar bond enthalpy of a carbon-hydrogen bond in a CH4 molecule.

CH4 (g) → C (g) + 4 H (g)              


ΔH0rxn = ΔH0f (C (g)) + 4 ΔH0f (H (g)) - ΔH0f (CH4 (g))

and

ΔH0rxn = Σ Hbonds broken – Σ Hbonds formed

⇒ ΔH0rxn = 4 Hbonds(C-H)


So, ΔH0rxn = ΔH0f (C (g)) + 4 ΔH0f (H (g)) - ΔH0f (CH4 (g)) = 4 Hbonds(C-H)

⇒ Hbonds(C-H) = 14 [ ΔH0f (C (g)) + 4 ΔH0f (H (g)) - ΔH0f (CH4 (g)) ]

⇒ Hbonds(C-H) = 14 (716.7 + 4 x 218.0 + 74.6)

⇒ Hbonds(C-H) = 415.8 kJ.mol-1