Calculate the bond energy of C–H bond, given that the heat of formation of CH₄, heat of sublimation of carbon and heat of dissociation of H₂ are –74.8, + 719.6 and 435.4 kJ mol^–1 respectively. 

Here, we are given

C(s) + 2H₂(g) → CH₄(g), Δ_rH° = −74.8 kJ ... (i)

C(s) → C(g), Δ_rH° = +719.6 kJ ... (ii)

H₂(g) → 2H(g), Δ_rH° = +435.4 kJ ... (iii)

We aim at:  CH₄(g) → C(g) + 4H(g) → Δ_rH = Δ_fH ... (iv)

Equation (ii) + 2 × equation (iii) − Equation (i) gives

C(s) + 2H₂(g) → C(g) + 4H(g)

−C(s) − 2H₂(g)   − CH₄(g)

0 = C(g) + 4H(g) − CH₄(g), Δ_rH° = 719.6 + 2(435.4) − (−74.8)

or                       CH₄(g) → C(g) + 4H(g)  ΔH = +1665.2 kJ

This gives the enthalpy of dissociation of four moles of C−H bonds (called enthalpy of atomization).

Hence, bond energy for C−H bond (average value) i.e.,

Δ_C−HH° =1665.2/4 = 416.3 kJ mol⁻¹