Tính ∆r các phản ứng đốt cháy hoàn toàn

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

∆_{r$H_{298}^o$} = 2 × ∆_{f$H_{298}^o$}(CO₂) + 2 × ∆_{f$H_{298}^o$}(H₂O) – 1 × ∆_{f$H_{298}^o$}(C₂H₄) – 3 × ∆_{f$H_{298}^o$}(O₂)

∆_{r$H_{298}^o$} = 2 × (-393,5) + 2 × (-241,8) – 1 × 52,4 – 3 × 0 = -1323 kJ

C₂H₆(g) + $\frac{7}{2}$O₂(g) → 2CO₂(g) + 3H₂O(g)

∆_{r$H_{298}^o$} = 2 × ∆_{f$H_{298}^o$}(CO₂) + 3 × ∆_{f$H_{298}^o$}(H₂O) – 1 × ∆_{f$H_{298}^o$}(C₂H₆) – $\frac{7}{2}$ × ∆_{f$H_{298}^o$}(O₂)

∆_{r$H_{298}^o$} = 2 × (-393,5) + 3 × (-241,8) – 1 × (-84) - $\frac{7}{2}$ × 0 = -1428,4 kJ

CO (g) + $\frac{1}{2}$O₂(g) → CO₂(g)

∆_{r$H_{298}^o$} = 1 × ∆_{f$H_{298}^o$}(CO₂) – 1 × ∆_{f$H_{298}^o$}(CO) – $\frac{1}{2}$ × ∆_{f$H_{298}^o$}(O₂)

∆_{r$H_{298}^o$} = 1 × (-393,5) - 1 × (-110,5) - $\frac{1}{2}$ × 0 = -283 kJ