Question

Calculate the enthalpy of hydrogenation of ethylene, given that the enthalpy of combustion of enthylene, hydrogen and ethane are – 1410.0, – 286.2 and – 1560.6 kJ mol^–1 respectively at 298 K.

Answer 1

We are given (i) C₂H₄ (g) + 3O₂ (g) → 2CO₂ (g) + 2H₂O

(l) ΔH = – 1410 kJ mol^–1

(ii) H₂ (g)+ 1/2 O₂(g)→ H₂O (l), ΔH = – 286.2 kJ mol^–1

(iii) C₂H₆ (g) + 31/2 O₂(g)→ 2CO₂ (g)+ 3H₂O (l), ΔH 

= – 1560.6 kJ mol^–1

We aim at: C₂H₄ + H₂ (g) → C₂H₆ (g), ΔH = ?

Equation (i) + Equation (ii) – Equation (iii) gives

C₂H₄ (g) + H₂ (g) → C₂H₆ (g),

ΔH = – 1410.0 + (–286.2) – (1560.6) = – 135.6 kJ mol^–1