2.5.9 · D5Thermodynamics (Chemical)
Question bank — Bond enthalpies — estimating ΔH_rxn from bond energies
First, SEE the one sentence — built one step at a time
Before you attack the traps, watch the master equation get drawn. The three panels below build the energy hill left-to-right: (1) the reactants sitting at a level, (2) climbing up to free gas-phase atoms as we break every bond, (3) sliding back down as new bonds form to give products. Read the panels like a flip-book.
Now for the deeper why: the next figure shows two different routes from reactants to products drawn on the same enthalpy axis, and derives — not asserts — why they must give equal .
And because almost every "spot the error" trap below is a miscount of bonds, keep this Lewis-count picture in view.
Figure s03 — Bond inventory for methane combustion, . Left (coral heading): reactants — one central carbon with 4 mint hydrogens (4 C–H bonds) and two lavender O=O double bonds. Right (green heading): products — the O=C=O with two C=O double bonds, and two water molecules each with two O–H bonds (4 O–H total). The label under each species states the bond tally you must feed into the equation; miscounting these is the #1 error.
True or false — justify
Every item: decide true/false, then give the reason. The reason is the real answer.
Bond enthalpy can be negative for very stable bonds.
False — bond enthalpy is the energy to break a bond, and breaking always costs energy, so it is always positive. Stability shows up as a large positive value, not a negative one.
Forming a bond is an endothermic process.
False — forming a bond releases energy (exothermic); atoms fall into a lower-energy state when they bond, so energy leaves as heat.
A tabulated C–H bond enthalpy is exact for methane.
False — tables list an average over many molecules at 298 K; the true C–H value in methane differs slightly from the value in, say, ethanol, so any table-based answer is only approximate.
The master equation gives an exact if all species are gaseous.
False — even for all-gas reactions it stays approximate, because the bond enthalpies used are averages, not molecule-specific values.
Because enthalpy is a state function, the atomic (fully-broken) path gives the same as the real reaction.
True — path-independence (Hess's Law) means we may route reactants → free atoms → products, and the total equals that of the direct route (see Figure s02).
A reaction that only breaks bonds and forms none must be endothermic.
True — with no bonds formed there is no energy paid back, so ; pure atomization is always endothermic.
If the products have more total bonds than the reactants, the reaction is automatically exothermic.
False — number of bonds is not the whole story; a few strong bonds can outweigh many weak ones. What matters is total bond energy formed vs broken, not bond count.
Bond enthalpy and standard enthalpy of formation always give identical .
False — Standard Enthalpy of Formation uses experimental, molecule-specific data and is more accurate; bond enthalpies use averages, so the two usually differ by tens of kJ/mol.
Using bond enthalpies (which treat water as gas) for a reaction that actually produces liquid water gives a release smaller in magnitude than the true experimental value.
True — the bond method stops at gaseous water; condensing to liquid releases extra energy the method never counts, so the real (liquid-product) heat release is larger in magnitude than the bond estimate.
Tabulated bond enthalpies are temperature- and phase-independent constants of nature.
False — they are averages quoted at 298 K for gas-phase species; at other temperatures they shift slightly, and they simply do not apply to liquids or solids, which is part of the 10–40 kJ/mol scatter.
Spot the error
Each line states a claim with a hidden mistake; the answer names and fixes it. When in doubt, re-count using Figure s03.
"."
Sign reversed — it must be broken − formed. Formation releases energy so it carries the minus sign; writing it backwards flips the sign of every answer.
"For , count 2 O–H bonds formed."
Undercounted — two water molecules have 4 O–H bonds (Figure s03 shows 2 per water × 2 waters). Forgetting the stoichiometric coefficient is the classic slip; draw Lewis Structures to count.
"Breaking H₂ releases 436 kJ/mol because 436 is a big number."
Direction error — 436 kJ/mol (the average gas-phase H–H value at 298 K) is the energy required to break H–H (climbing the hill), not released. Breaking always costs energy regardless of how large the number is.
"CO₂ has one C=O bond, so count it once."
Miscount — has two C=O bonds (see the product side of Figure s03). Missing a bond in the product understates energy released and pushes too positive.
"We can plug in bond enthalpies for a reaction in aqueous solution directly."
Invalid — tabulated bond enthalpies apply only to gas-phase species at 298 K. Dissolving or condensing adds enthalpy the method never accounts for.
"The reaction is exothermic, so ."
Backwards — exothermic means more energy is released forming than spent breaking, so , giving .
"Since C=C (614) > C–C (348), the double bond is more than twice as strong as a single bond."
False comparison — . A double bond is stronger than one single bond but weaker than two, which is exactly why hydrogenation of ethene is exothermic.
"A dative (coordinate) bond doesn't need counting because no new atoms bonded."
Wrong — a dative bond (e.g. the extra N–B bond in ) is a real shared-pair bond; forming it releases energy and must be counted like any other bond, even though both electrons came from one atom.
Why questions
Why is a bond enthalpy always defined in the gas phase?
So the number captures only the bond-breaking energy, with no intermolecular forces or phase-change enthalpy mixed in; gas-phase atoms are the clean, isolated reference state (the top of the energy hill).
Why can we legally use the imaginary "break everything into atoms" path?
Because enthalpy is a state function (Hess's Law), the between reactants and products depends only on the two levels, not the route — Figure s02 shows both paths sharing the same start and end levels, so they must give the same total.
Why does resonance make the bond-energy estimate wrong for molecules like benzene?
Resonance Energy makes the real molecule more stable (lower energy) than any single Lewis structure predicts, so summing localized bond enthalpies overestimates the true energy needed to break it.
Why is "broken − formed" the right order and not the reverse?
Breaking is the energy in (positive, the climb) and forming is the energy out (negative, the descent); the reaction's net enthalpy is what you put in minus what you get back, i.e. broken − formed.
Why are combustion reactions of hydrocarbons reliably exothermic by this method?
They convert C–H and weak-ish bonds into very strong C=O and O–H bonds; far more energy is released forming those than is spent breaking the fuel and O₂, so .
Why does the bond method sometimes disagree with Standard Enthalpy of Formation data by 10–40 kJ/mol?
Because it uses averaged, 298 K gas-phase bond enthalpies and ignores resonance, ring strain, and phase effects — all real energy contributions the formation-based route captures.
Why must you draw Lewis structures before using the master equation?
To know exactly which bonds (and how many, including double/triple and dative bonds) exist in each species; miscounting bonds is the single biggest source of error, as Figure s03 makes visible.
Edge cases
For a reaction where reactant and product bond types are identical in count and kind, what is by this method?
Zero — broken and formed sums cancel exactly; any real nonzero heat would come only from effects (resonance, phase) the bond method cannot see.
An isomerization like a simple rearrangement often gives near-zero bond-energy . Why might reality differ?
If the same bonds appear on both sides the estimate is ~0, but strain, resonance, or subtle environment differences give a real nonzero the averages miss.
What does the bond-enthalpy method predict for atomizing an already-atomic gas, e.g. Ne(g) → Ne(g)?
— there are no bonds to break or form, so both sums are zero; a noble-gas atom is a degenerate, no-bond case.
Can this method handle a reaction involving an ionic solid like NaCl(s)?
Poorly — ionic solids have lattice energies, not gas-phase covalent bonds; bond-enthalpy tables don't apply, and you'd use lattice/formation data instead.
When a mechanism goes through radical intermediates (e.g. Cl• or CH₃•), how should you count bonds for the overall ΔH?
Count only the bonds broken in the reactants and formed in the final products; the radical steps break and re-make bonds internally, and since is a state function those intermediate bonds cancel out over the full path.
Is a single half-arrow homolysis (A–B → A• + B•) endothermic or exothermic?
Endothermic — homolysis just breaks one bond into two radicals with nothing formed, so ; radicals are high-energy species sitting up the hill.
If two reactions have the same , must they release heat at the same rate?
No — is about total energy (thermodynamics), not speed; rate is governed by kinetics (activation energy), which bond enthalpies alone do not determine.
For dissociation of a diatomic like H₂ → 2H, is positive, negative, or zero?
Positive — only bond-breaking occurs (no bond formed), so kJ/mol; pure dissociation is endothermic.
Connections
- Hinglish version
- Hess's Law — justifies the atomic-path trick behind every item.
- Standard Enthalpy of Formation — the accurate alternative these traps contrast with.
- Exothermic vs Endothermic Reactions — the sign logic tested throughout.
- Lewis Structures — the antidote to every miscounting trap.
- Resonance Energy — source of several "why is it only an estimate" answers.
- Enthalpy and the First Law of Thermodynamics