1.2 Energy cycles

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.

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.

A formation equation must produce exactly one mole of the compound from elements in their standard states, so fractional coefficients may be necessary. A combustion equation must burn exactly one mole completely in O₂; keep the stated standard state of water because changing H₂O(l) to H₂O(g) changes ΔH°.

Worked scaling example: $\Delta H_c^\circ(\ce{C2H6})=-1560\,\mathrm{kJ\,mol^{-1}}$ refers to complete combustion of one mole of ethane. For the corresponding balanced combustion of two moles, multiply the complete equation and its enthalpy by two: $\Delta H^\circ=2(-1560)=-3120\,\mathrm{kJ}$. Do not report $\mathrm{kJ\,mol^{-1}}$ for the two-mole equation unless the result is normalized back to one mole.

For formation data, imagine every reactant and product connected to the same elements: product sums minus reactant sums gives the target. For combustion data the paths run toward common combustion products, reversing the subtraction. Write coefficients beside every tabulated value before calculating.

Worked example — formation data for pentane combustion: $\ce{C5H12(l)+8O2(g)->5CO2(g)+6H2O(l)}$. Using $\Delta H_f^\circ[\ce{C5H12(l)}]=-173$, $\Delta H_f^\circ[\ce{CO2(g)}]=-394$, $\Delta H_f^\circ[\ce{H2O(l)}]=-286$ and $\Delta H_f^\circ[\ce{O2(g)}]=0\,\mathrm{kJ\,mol^{-1}}$, $\Delta H^\circ=5(-394)+6(-286)-[-173+8(0)]=-3513\,\mathrm{kJ\,mol^{-1}}$. The negative sign identifies exothermic combustion.

Build the cycle from physical steps and electron accounting: atomize each element, ionize the metal the required number of times, add electrons to the non-metal, then form the lattice. For a 2− ion include two electron-affinity terms, and confirm whether the supplied lattice enthalpy is defined for formation or dissociation before assigning its sign.

Worked example — KBr lattice enthalpy: use the DP lattice-dissociation convention $\ce{KBr(s)->K+(g)+Br-(g)}$. Following the alternative path in the local cycle, $\Delta H_\mathrm{lattice}^\circ=-(-392)+89+419+112-325=+687\,\mathrm{kJ\,mol^{-1}}$. The formation enthalpy is reversed, atomization and ionization are positive, and the first electron affinity is negative. The positive result is consistent with separating a solid lattice into gaseous ions.