Question

Calculate the enthalpy of hydrogenation of ethylene from the following data.

Bond energies of C − H, C − C, C = C and H − H are 414, 347, 618 and 435 kJ mol⁻¹.

Answer 1

Given, E_C−H = 414 kJ mol⁻¹

E_C−C = 347 kJ mol⁻¹

E_C=C = 618 kJ mol⁻¹

E_H−H = 435 kJ mol⁻¹

ΔH_r = Σ (Bond energy)_r − Σ (Bond energy)_p

ΔH_r = (E_C=C + 4E_C−H + _EH−H) − (E_C−C + 6E_C−H)

ΔH_r = (618 + (4 × 414) + 435) − (347 + (6 × 414))

ΔH_r = 2709 − 2831

ΔH_r = −122 kJ mol⁻¹