Why Hess's Law Works

Hess's law is not an empirical rule — it is a necessary consequence of enthalpy being a state function. State functions depend only on the initial and final state of a system, not on the path taken. Therefore, if reaction A → C can go directly or via A → B → C, the total $\Delta H$ must be the same. This allows enthalpy to be calculated by any convenient combination of known reactions.

Method 1: Using Standard Formation Enthalpies

$\Delta H_{rxn}^\circ = \sum\Delta H_f^\circ(\text{products}) - \sum\Delta H_f^\circ(\text{reactants})$

The reference point: $\Delta H_f^\circ = 0$ for any element in its standard state (graphite for C, $O_{2}$(g) for oxygen, etc.). Formation enthalpies can be positive (e.g., $\Delta H_f^\circ(NO, g) = +91$ kJ/mol) or negative (e.g., $\Delta H_f^\circ(CO_2, g) = -393.5$ kJ/mol).

Example: For $CH_{4}$(g) + $2O_{2}$(g) → $CO_{2}$(g) + $2H_{2}O$(l): $\Delta H = [-393.5 + 2(-285.8)] - [-74.8 + 2(0)] = -965.1 + 74.8 = -890.3$ kJ/mol

Method 2: Using Bond Enthalpies

$\Delta H = \sum(BE_{\text{broken}}) - \sum(BE_{\text{formed}})$

Breaking bonds absorbs energy (endothermic, positive); forming bonds releases energy (exothermic, negative). Counting bonds carefully is critical — draw Lewis structures for each species.

Example: $C_{2}H_{6}$(g) → $C_{2}H_{4}$(g) + $H_{2}$(g): Broken = 1(C-C) + 6(C-H) = 2831 kJ; Formed = 1(C=C) + 4(C-H) + 1(H-H) = 2703 kJ; $\Delta H = +128$ kJ/mol.

Key Enthalpy Type Signs