Part 4: Bond enthalpies and fuel comparisons

Count the bonds changed by a reaction, estimate its enthalpy, and explain why an averaged gas-phase model can differ from a state-specific measured value.

Bond enthalpy and mean bond enthalpy

Bond enthalpy is the enthalpy needed to break one mole of a specified covalent bond in gaseous species by homolytic cleavage. The fragments retain one electron each from the bond. Bond breaking is endothermic, so bond enthalpies are positive. For a diatomic molecule, H₂(g) → 2H(g) directly represents breaking H–H bonds; in a larger molecule a single cleavage may first produce radicals rather than separate every atom.

The energy of a bond depends on its molecular environment. A mean bond enthalpy averages values for a bond type over different environments. Even successive C–H removals from methane need different energies because each step starts from a different species. A table's average C–H value is useful for estimates, not an exact universal energy attached to every C–H bond.

Break reactant bonds, then form product bonds

Imagine separating the gaseous reactants into atoms and then assembling the gaseous products. Energy in = sum of bond enthalpies for bonds broken; energy out = sum for bonds formed. The net estimate is broken minus formed. This atom route is a Hess construction, not a claim that the actual reaction first destroys every molecule into free atoms.

Draw structural formulae or count bonds carefully, including all equation coefficients. O₂ has one O=O bond per molecule, CO₂ has two C=O bonds, and one H₂O has two O–H bonds. A double bond has its own bond enthalpy; it is not simply twice the single-bond value. Unchanged bonds may be cancelled when using the same average value on both sides.

Original worked hydrogenation calculation

For CH₂=CH₂(g) + H₂(g) → CH₃CH₃(g), use supplied mean values E(C=C) = 612, E(H–H) = 436, E(C–C) = 348 and E(C–H) = 412 kJ mol⁻¹. The four original C–H bonds can cancel, leaving one C=C and one H–H broken, with one C–C and two C–H formed.

Breaking total = 612 + 436 = 1048 kJ mol⁻¹. Formation total = 348 + 2(412) = 1172 kJ mol⁻¹. ΔH ≈ 1048 − 1172 = −124 kJ mol⁻¹. The result is exothermic because forming the product bonds releases more than breaking the reactant bonds requires. Counting only the broken double bond would miss both H–H breaking and new C–H formation.

For a combustion example, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), supplied values C–H 413, O=O 498, C=O in CO₂ 805 and O–H 463 give broken = 4(413) + 2(498) = 2648; formed = 2(805) + 4(463) = 3462. ΔH ≈ −814 kJ mol⁻¹. These are explicitly supplied teaching values; use the data in the actual question and do not substitute a generic carbonyl C=O value when a CO₂-specific value is given.

Rearrange to find an unknown mean bond enthalpy

For ½N₂(g) + 1½H₂(g) → NH₃(g), suppose ΔH = −46.0 kJ mol⁻¹, E(N≡N) = 944 and E(H–H) = 436 kJ mol⁻¹. If the mean N–H enthalpy in this estimate is x, write −46.0 = ½(944) + 1½(436) − 3x. The bond count is three N–H bonds per NH₃, despite the fractional reactant coefficients.

Breaking total = 472 + 654 = 1126. Rearrange: 3x = 1126 + 46.0 = 1172, so x = 390.667 ≈ 391 kJ mol⁻¹. A negative x would contradict endothermic bond breaking and should trigger a sign check.

If methane atomisation CH₄(g) → C(g) + 4H(g) requires 1660 kJ mol⁻¹, dividing by four gives a mean of 415 kJ per mole of C–H bonds for that complete atomisation. The balanced one-quarter equation is ¼CH₄(g) → ¼C(g) + H(g); do not leave one whole carbon atom on the right.

Evaluate the model separately from the experiment

Mean bond enthalpies describe gaseous species. If a target forms liquid water or burns a liquid fuel, vaporisation/condensation enthalpies are additional steps. Mean values also differ from molecule-specific values. These are theoretical model limitations; 'heat lost from the cup' is instead an experimental limitation and cannot explain why two purely calculated values disagree.

Successive members of an alcohol homologous series commonly have increasingly negative molar combustion enthalpies because an extra CH₂ group changes a similar set of bonds each time. More bonds must be broken as well as more product bonds formed; explain the net balance. The change is approximately regular, not an exact constant law for all states and isomers.

Compare fuels on a consistent basis. A fuel releasing 1500 kJ mol⁻¹ with M = 60.0 g mol⁻¹ releases 25.0 kJ g⁻¹; one releasing 900 kJ mol⁻¹ with M = 30.0 g mol⁻¹ releases 30.0 kJ g⁻¹. The first wins per mole but not per gram. Practical fuel choice also involves storage, combustion products, source, cost and safety. A high energy value alone does not settle the comparison.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources