Tính entanpy tạo thành (${\Delta }_f{H}_{298}^0$) của acetaldehyde (CH${}_3$CHO):
Cho các dữ kiện như sau:

$${\text{H}}_{2(g)}+\frac{1}{2}{\text{O}}_{2(g)}⟶{\text{H}}_2{\text{O}}_{(l)}{\Delta }_f{H}_{298}^0=-286\text{ kJ/mol}(1)$$

$${\text{C}}_{(s)}+{\text{O}}_{2(g)}⟶{\text{CO}}_{2(g)}{\Delta }_f{H}_{298}^0=-394\text{ kJ/mol}(2)$$

$$2{\text{CH}}_3{\text{CHO}}_{(g)}+5{\text{O}}_{2(g)}⟶4{\text{H}}_2{\text{O}}_{(l)}+4{\text{CO}}_{2(g)}{\Delta }_r{H}_{298}^0=-2388\text{ kJ/mol}(3)$$