In thermodynamics, the enthalpy change of a chemical reaction remains constant regardless of the pathway taken to complete the reaction. This principle was articulated by Germain Hess, a Swiss-born Russian chemist and physician, in 1840 and is now known as Hess's law of constant heat summation [1]. The law states that for a given set of reactants and products at identical initial and final conditions, the total enthalpy change is independent of the number or nature of intermediate steps.
This invariance arises because enthalpy is a state function—its value depends solely on the current state of the system, not on how it reached that state. The first law of thermodynamics underpins this behavior by asserting conservation of energy: energy can neither be created nor destroyed but only transformed or transferred. Consequently, the enthalpy change in a system due to a reaction at constant pressure is equal to the heat absorbed (or the negative of the heat released), which can be determined by calorimetry for many reactions.
The practical utility of Hess's law lies in enabling calculation of enthalpy changes for reactions where direct measurement is challenging. By decomposing complex reactions into sequences with known enthalpy changes, one can algebraically sum these values to obtain the overall enthalpy change.
Mathematically, if individual reactions with enthalpy changes are combined such that their net equation corresponds to the target reaction, then
\[
\Delta H_{\text{net}} = \sum_i {\Delta H_i}
\]
where each term represents an intermediate step’s enthalpy change. The sign convention follows that exothermic reactions have negative enthalpy changes (\( \Delta H < 0\)) and endothermic ones positive values. This allows prediction about spontaneity in conjunction with entropy considerations, since some reactions with positive enthalpy can still be spontaneous due to favorable entropy increases.
Standard enthalpies of formation provide a foundational reference for applying Hess's law quantitatively. Defined as the enthalpy change when one mole of compound forms from its elements in their standard states, these values are tabulated and commonly used. Elements in their standard states have an enthalpy of formation of zero.
The standard enthalpy change for any reaction can be calculated using:
\[
\Delta H_{\text{reaction}}^{\circ} =
\sum_i a_i
\Delta_f H_{\text{products},i}^{\circ} -
\sum_j b_j
\Delta_f H_{\text{reactants},j}^{\circ}
\]
Here, coefficients \(a_i,b_j\) represent stoichiometric factors for products and reactants respectively, while the superscript ∘ denotes standard state conditions.
This formulation conceptualizes any reaction as two fictive processes: decomposition of reactants into elemental forms,
\[
\Delta H_{RE}^{\circ} = -
\sum_j b_j
\Delta_f H_{\text{reactants},j}^{\circ}
\]
and formation of products from those elements,
\[
\Delta H_{EP}^{\circ} =
\sum_i a_i
\Delta_f H_{\text{products},i}^{\circ}
.
\]
Summing these yields total reaction enthalpy.
Consider combustion pathways involving graphite carbon oxidizing to carbon dioxide and carbon monoxide:
Direct combustion:
\[
C_{\text{graphite}} + O_{2} \rightarrow CO_{2}(g),
\Delta H = -393.5\,\text{kJ/mol}
.\]
And via intermediate formation:
\[
C_{\text{graphite}} + \frac{1}{2} O_{2} \rightarrow CO(g),
\Delta H = -110.5\,\text{kJ/mol}
,
\]
followed by oxidation:
\[
CO(g) + \frac{1}{2} O_{2} \rightarrow CO_{2}(g),
\Delta H = -283.0\,\text{kJ/mol}
.
\]
Summing these latter two steps reproduces the direct combustion enthalpy exactly,
\[
-110.5 + (-283.0) = -393.5\,\text{kJ/mol},
\]
validating Hess's law experimentally and confirming that total heat released is path-independent in this case [1].
A more involved example concerns synthesizing boron oxide from elemental boron and oxygen using several intermediate reactions:
Given data include:
\[
B_{2}O_{3}(s) + 3H_{2}O(g) \rightarrow 3O_{2}(g) + B_{2}H_{6}(g),
\Delta H = 2035\,\text{kJ/mol},
\]
\[
H_{2}O(l) \rightarrow H_{2}O(g),
\Delta H = 44\,\text{kJ/mol},
\]
\[
H_{2}(g) + \frac{1}{2} O_{2}(g) \rightarrow H_{2}O(l),
\Delta H = -286\,\text{kJ/mol},
\]
\[
2B(s) + 3H_{2}(g) \rightarrow B_{2}H_{6}(g),
\Delta H = 36\,\text{kJ/mol}.
\]
Reversing and scaling these equations appropriately leads to the target reaction \(2B(s) + \frac{3}{2} O_{2}(g) \rightarrow B_{2}O_{3}(s)\) with a total \(\Delta H = -1273\,\text{kJ/mol}\).
[1] https://en.wikipedia.org/wiki/Hess%27s_law
[2] https://www.revisiondojo.com/blog/hess-s-law-explained-simply
[3] https://www.chemistrystudent.com/cie-a-level/5-chemical-energetics...
[4] https://myedspace.co.uk/myresources/a-level/chemistry/aqa/revision...
[5] https://www.firgelliauto.com/blogs/calculators/hesss-law-calculato...
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