1.2 Energy cycles

Breaking bonds absorbs energy; forming bonds releases energy. Count every bond with its stoichiometric multiplicity before applying the signed sum.

For H₂ + Cl₂ → 2HCl, break one H–H and one Cl–Cl bond, then form two H–Cl bonds. Average bond enthalpies give an estimate because the tabulated value averages that bond across different gaseous molecules.

Worked example — bond enthalpies: for $\ce{C2H4(g) + HBr(g) -> C2H5Br(g)}$, the local course book gives $\ce{C-H}=414$, $\ce{C=C}=614$, $\ce{H-Br}=366$, $\ce{C-C}=346$ and $\ce{C-Br}=285\,\mathrm{kJ\,mol^{-1}}$. $\Delta H=[4(414)+614+366]-[5(414)+346+285]=2636-2701=-65\,\mathrm{kJ\,mol^{-1}}$. The negative estimate means the bonds formed release more energy than the bonds broken absorb; it remains approximate because the values are gaseous averages.

Hess's law states that enthalpy change is independent of reaction pathway. Enthalpy values can therefore be combined through a balanced cycle.

Reverse an entire balanced equation by changing the sign of ΔH; scale every coefficient and ΔH by the same factor; then add equations and cancel identical species in identical physical states. Never change a chemical subscript, formula or state symbol merely to force cancellation. The surviving equation must exactly match the target before enthalpies are summed.

Treat chemical equations like algebra: reverse a step and reverse its ΔH sign; multiply all coefficients and ΔH by the same factor; then add and cancel species. The surviving overall equation must exactly match the target before the enthalpies are summed.

Worked example — Hess's law: target $\ce{C(s)+2H2(g)+1/2O2(g)->CH3OH(l)}$. Use $\ce{C+O2->CO2}$, $\Delta H=-394\,\mathrm{kJ\,mol^{-1}}$; double $\ce{H2+1/2O2->H2O(l)}$ to give $-572\,\mathrm{kJ\,mol^{-1}}$; reverse methanol combustion to give $+726\,\mathrm{kJ\,mol^{-1}}$. After cancelling $\ce{CO2}$ and $\ce{H2O}$, $\Delta H=-394-572+726=-240\,\mathrm{kJ\,mol^{-1}}$ for the target equation.