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Chemistry · Ch 8 — Thermodynamics

Hess's Law of Constant Heat Summation

8.7

Hess's Law of Constant Heat Summation

Hess's law of constant heat summation states that the total enthalpy change accompanying a chemical reaction is the same regardless of whether the reaction occurs in a single step or through any number of intermediate steps, provided the initial reactants and final products are the same in each case.

This is not an independent empirical law so much as a direct and inevitable consequence of enthalpy being a state function: since ΔH\Delta H depends only on the initial and final states, and not on the path taken between them, it does not matter whether that path is the single overall reaction or a sequence of hypothetical intermediate reactions — the total ΔH\Delta H must come out identical either way, exactly as ΔU\Delta U came out identical along the two different paths in the worked example at the start of this chapter.

Hess's law is put to practical use by taking a set of reactions with known enthalpy changes (usually enthalpies of formation or combustion, both directly measurable) and combining them algebraically — adding, subtracting, reversing, or multiplying through by a constant — so that they add up to the target reaction whose enthalpy change is not directly measurable or not yet known. When a reaction is reversed, the sign of its ΔH\Delta H is reversed; when a reaction is multiplied by a constant factor, its ΔH\Delta H is multiplied by the same factor.

As a worked illustration: to find the enthalpy of formation of carbon monoxide, C(s)+12O2(g)→CO(g)\text{C}(s) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) — a reaction that cannot be carried out cleanly in the laboratory, because burning carbon in a limited oxygen supply always produces some CO2\text{CO}_2 alongside the CO\text{CO} — combine two reactions that can be measured directly:

C(s)+O2(g)→CO2(g),ΔH1=−393.5 kJ mol−1\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), \quad \Delta H_1 = -393.5\ \text{kJ mol}^{-1}

CO(g)+12O2(g)→CO2(g),ΔH2=−283.0 kJ mol−1\text{CO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g), \quad \Delta H_2 = -283.0\ \text{kJ mol}^{-1} …