Hess cycles and bond enthalpies

What you’ll learn

• Combine equations and enthalpy changes using Hess’s law.
• Use formation or combustion data consistently.
• Estimate a gas-phase reaction enthalpy from bond enthalpies.

Understand the idea

Enthalpy is a state function, so the change between the same initial and final states is independent of route. A Hess cycle joins reactions whose equations add to the required equation. Reverse a reaction and reverse its enthalpy sign; multiply its coefficients and multiply its enthalpy by the same factor.

Formation data give reaction enthalpy as the total for products minus the total for reactants, including coefficients. A combustion cycle instead takes both sides to a common set of fully oxidised products; draw the arrows before deciding the signs. States matter because vaporising or condensing a substance changes enthalpy.

Breaking a covalent bond requires energy, while forming one releases energy. Subtract the total energy of bonds formed from the total energy needed to break bonds. Mean bond enthalpies are gas-phase averages over different compounds, so results are estimates and require care if the reaction contains liquids or solids.

Key formulae

$$\Delta H_{\mathrm{target}}=\sum\Delta H_{\mathrm{steps}}$$

$$\Delta_rH^\circ=\sum\nu\Delta_fH^\circ(\mathrm{products})-\sum\nu\Delta_fH^\circ(\mathrm{reactants})$$

$$\Delta H\approx\sum E(\mathrm{bonds\ broken})-\sum E(\mathrm{bonds\ formed})$$

Worked example

Given $\mathrm{C(s)+O_2(g)\rightarrow CO_2(g)}$, $\Delta H=-394\ \mathrm{kJ\,mol^{-1}}$, and $\mathrm{CO(g)+\tfrac12O_2(g)\rightarrow CO_2(g)}$, $\Delta H=-283\ \mathrm{kJ\,mol^{-1}}$, find the enthalpy for $\mathrm{C(s)+\tfrac12O_2(g)\rightarrow CO(g)}$.

1. Keep the first equation as written.
2. Reverse the second equation, so its enthalpy becomes $+283\ \mathrm{kJ\,mol^{-1}}$.
3. Add and cancel carbon dioxide and half a mole of oxygen to obtain the required equation.
4. Add the enthalpies: $-394+283=-111\ \mathrm{kJ\,mol^{-1}}$.

Answer: $-111\ \mathrm{kJ\,mol^{-1}}$.