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Chemistry · Ch 5 — Thermodynamics

Hess's Law of Constant Heat Summation

5.4(e)

Hess's Law of Constant Heat Summation

Hess's Law of Constant Heat Summation

Enthalpy is a state function. Therefore, the change in enthalpy is independent of the path between the initial state (reactants) and the final state (products). In other words, the enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps.

Hess's Law

If a reaction takes place in several steps, then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Let us understand the importance of this law with an example. Consider the enthalpy change for:

C(graphite, s)+12O2(g)→CO(g);ΔrH∘=?\text{C(graphite, }s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g); \quad \Delta_r H^\circ = ?

Although CO(g) is the major product, some CO2_2 gas is always produced in this reaction. Therefore, we cannot measure the enthalpy change for this reaction directly. However, we can find other reactions involving related species and calculate the desired enthalpy change.

Consider these reactions:

C(graphite, s)+O2(g)→CO2(g);ΔrH∘=−393.5 kJ mol−1(i)\text{C(graphite, }s) + \text{O}_2(g) \rightarrow \text{CO}_2(g); \quad \Delta_r H^\circ = -393.5\ \text{kJ mol}^{-1} \quad \text{(i)}

CO(g)+12O2(g)→CO2(g);ΔrH∘=−283.0 kJ mol−1(ii)\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g); \quad \Delta_r H^\circ = -283.0\ \text{kJ mol}^{-1} \quad \text{(ii)}

To get one mole of CO(g) on the right, we reverse equation (ii). When we reverse, heat is absorbed instead of being released, so we change the sign of ΔrH∘\Delta_r H^\circ:

CO2(g)→CO(g)+12O2(g);ΔrH∘=+283.0 kJ mol−1(iii)\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2}\text{O}_2(g); \quad \Delta_r H^\circ = +283.0\ \text{kJ mol}^{-1} \quad \text{(iii)}

Adding equation (i) and (iii):

C(graphite, s)+O2(g)→CO2(g);ΔrH∘=−393.5 kJ mol−1\text{C(graphite, }s) + \text{O}_2(g) \rightarrow \text{CO}_2(g); \quad \Delta_r H^\circ = -393.5\ \text{kJ mol}^{-1}

CO2(g)→CO(g)+12O2(g);ΔrH∘=+283.0 kJ mol−1\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2}\text{O}_2(g); \quad \Delta_r H^\circ = +283.0\ \text{kJ mol}^{-1}

C(graphite, s)+12O2(g)→CO(g);ΔrH∘=−110.5 kJ mol−1‾\overline{\text{C(graphite, }s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g); \quad \Delta_r H^\circ = -110.5\ \text{kJ mol}^{-1}}

In general, if the enthalpy of an overall reaction A →\rightarrow B along one route is ΔrH\Delta_r H, and ΔrH1,ΔrH2,ΔrH3,…\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \ldots represent enthalpies of reactions leading to the same product B along another route, then:

ΔrH=ΔrH1+ΔrH2+ΔrH3+…\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \ldots …

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