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Chemistry · Ch 7 — Thermodynamics

Hess's Law of Constant Heat Summation

7.8

Hess's Law of Constant Heat Summation

Because ΔU\Delta U and ΔH\Delta H are both functions of a system's STATE, the heat evolved or absorbed by a given reaction can only depend on its initial and final states -- never on the particular path or sequence of intermediate steps used to get from one to the other. This generalisation is known as Hess's law, stated as: the enthalpy change of a reaction -- whether measured at constant volume or constant pressure -- is the same whether the reaction takes place in a single step or via multiple steps, provided the initial and final states are identical.

Symbolically, if a reaction from A to B can be reached directly (enthalpy change ΔHr\Delta H_r) OR via an indirect route through intermediate states X and Y (enthalpy changes ΔH1\Delta H_1, ΔH2\Delta H_2, ΔH3\Delta H_3 for the three legs of the indirect path), then:

ΔHr=ΔH1+ΔH2+ΔH3\Delta H_r = \Delta H_1+\Delta H_2+\Delta H_3

Application of Hess's law. Its main practical value is calculating the enthalpy of reactions that are DIFFICULT to measure directly -- for example, it is very difficult to cleanly measure the heat of combustion of graphite reacting to give pure CO alone, since some of the CO invariably keeps oxidising further to CO2_2 before you can isolate the measurement. However, the enthalpy of graphite oxidising all the way to CO2_2, and of CO oxidising to CO2_2, can each be measured cleanly and separately: −393.5-393.5 kJ and −283.5-283.5 kJ respectively. …

Figure hess-cycle-generalHess's law cycle (general)

What this figure shows. A square cycle: A at top-left connects to B at top-right via a rightward arrow labelled ΔHr\Delta H_r (the direct, single-step path); A also connects down to X via ΔH1\Delta H_1, X connects right to Y via ΔH2\Delta H_2, and Y connects up to B via ΔH3\Delta H_3 (the indirect, three-step path) -- illustrating $\Delta H_r=\Delta H_1+\Delta H_ …

Figure hess-cycle-graphite-coHess's law applied: graphite to CO

What this figure shows. A triangular cycle: C(graphite) at top-left goes directly to CO2_2(g) at top-right via ΔH1\Delta H_1 (using O2_2(g), −393.5-393.5 kJ); alternatively C(graphite) goes down to CO(g) at bottom via ΔH2=X\Delta H_2=X (using 12\tfrac{1}{2}O2_2(g)), and CO(g) goes up-right to CO2_2(g) via ΔH3\Delta H_3 (using another 12\tfrac{1}{2}O2_2(g), −283.5-283.5 kJ) -- solved as $X=-110 …