2.5.11 · D5Thermodynamics (Chemical)
Question bank — Enthalpy of combustion, neutralization, hydration, solution
Symbol legend & conditions (read this first)
Before any trap, here is exactly what each symbol means. Every one is a (an enthalpy change) measured under standard conditions, defined once below.
Look at the picture before the traps — it shows the "break then wrap" energy story that most of these questions turn on.
True or false — justify
State true/false AND the reason. A bare "true" scores zero — the reason is the whole point.
Enthalpy of combustion can sometimes be positive.
False. Complete combustion always forms stronger bonds (in , ) than it breaks, so heat is always released — is always negative.
Enthalpy of neutralization is always .
False. That value holds only for a strong acid with a strong base (both fully dissociated). Weak acids/bases spend energy ionizing first, so comes out smaller.
Enthalpy of solution is always exothermic because water "hugs" the ions.
False. Dissolving is lattice-breaking () plus hydration (). If breaking the lattice costs more than hydration releases, is positive (e.g. ). See the energy diagram above.
Enthalpy of hydration is always negative.
True. Water dipoles are attracted to a bare gaseous ion, forming ion–dipole bonds; forming bonds always releases energy.
Lattice enthalpy of breaking (solid → gaseous ions) is endothermic.
True. You must pull apart ions held by strong electrostatic attraction, which costs energy, so this direction is positive.
All four special enthalpies are measured at constant pressure.
True. They are all values, and at constant pressure (the derivation is in the legend above) — that's why we call them enthalpy changes.
Burning 2 moles of methane gives an enthalpy change equal to .
False. is defined per 1 mole of fuel; 2 moles releases .
A more negative means weaker ion–water attraction.
False. It's the opposite — more negative means stronger attraction (more energy released), driven by higher charge density .
For a strong acid + strong base, the spectator ions (, ) contribute to .
False. Spectators start and end as free hydrated ions, unchanged, so they contribute nothing; only has a net enthalpy change.
If two salts have identical lattice enthalpies, the one with the more exothermic hydration dissolves with more heat released.
True. With lattice terms equal, , so a more negative hydration makes more negative (more exothermic).
Spot the error
Each line has a common mistake in it. Say what's wrong and fix it.
" is the combustion reaction of carbon."
Wrong product. Complete combustion requires excess giving ; is incomplete combustion, so this is not the standard reaction.
", so kJ."
Wrong state of water. Standard combustion needs products in their standard states → at 298 K, 1 bar, not . Using gaseous water under-counts by the latent heat of vaporization.
" because they oppose each other."
Wrong sign handling. Signs come from a fixed thermodynamic convention: heat absorbed by the system is positive, heat released is negative. Breaking the lattice absorbs heat, so is ; hydration releases heat, so is . Adding them () already builds in the opposition — subtracting would double-count it.
"Acetic acid + NaOH releases exactly kJ per mole of water."
Wrong — acetic acid is weak. Some energy is absorbed to ionize it first, so the released heat is less than kJ.
"Total hydration enthalpy of NaCl = only."
Incomplete. Hydration is summed over all ions, so it's ; both cations and anions get hydrated.
" cold packs prove dissolving violates the first law by absorbing heat."
Wrong conclusion. Endothermic dissolving is perfectly allowed — it's just (lattice cost exceeds hydration release); it's driven by the large entropy increase, no law is broken.
"Hydration enthalpy of is less negative than that of because aluminium is heavier."
Wrong reasoning and wrong result. It's charge density, not mass, that matters; has higher charge and smaller radius, so its hydration is much more negative.
"."
Reversed. For a formation-from-combustion cycle it's (see Hess's Law and Thermochemical Cycles); flipping the sign flips the answer.
Why questions
These probe the mechanism, not the number.
Why is strong-acid–strong-base neutralization a constant value regardless of which strong acid or base?
Because they are fully dissociated in dilute water, the only chemical event is — identical every time, so identical .
Why does a weak acid give out less heat on neutralization?
It is only partly ionized, so extra energy (endothermic) is first spent to ionize the remaining acid before neutralization; this partly cancels the heat released.
Why must combustion products be in their standard states?
So the value is reproducible and tabulatable. Standard states (1 bar, 298 K, most stable form) fix the reference so anyone measuring the same reaction gets the same , letting us combine values via Hess's law.
Why is hydration enthalpy always exothermic while solution enthalpy can go either way?
Hydration is only bond-forming (ion–dipole), which always releases energy. Solution includes lattice-breaking (endothermic) too, so the net sign depends on which term wins — exactly the balance drawn in the energy diagram above.
Why does smaller ionic radius make hydration more exothermic?
A smaller radius concentrates the same charge in less space, so charge density rises, water dipoles are pulled in harder, and more energy is released.
Why can we add up and to get even though dissolving doesn't go through gaseous ions?
Enthalpy is a state function — it depends only on start and end states, not the path — so the imagined "break lattice then hydrate" route gives the same as the real direct route (see Lattice Enthalpy and Born–Haber Cycle).
Why do endothermic salts like still dissolve spontaneously?
Dissolving increases disorder hugely (entropy ); at constant , can be negative even with (see Spontaneity and Gibbs Free Energy).
Why is the natural quantity for all these reactions?
They happen in open beakers at constant atmospheric pressure. From the first law with , at constant we get (see First Law of Thermodynamics).
Edge cases
The scenarios where the "usual" rule needs care.
Combustion of : what is the product and its state?
. Even though there's no carbon, complete combustion still needs the product in its standard state — liquid water at 298 K, 1 bar.
Neutralizing a weak acid with a weak base — is larger or smaller than ?
Smaller (often much smaller), because ionization energy is spent on both the weak acid and the weak base before water forms.
A salt where exactly — what is ?
Zero. The heat to break the lattice is exactly repaid by hydration, so dissolving releases/absorbs no net heat (dissolution is then entropy-driven).
Comparing of vs — which is more exothermic and why?
. It has a much smaller radius, hence higher charge density, so water is attracted more strongly and more energy is released.
Comparing of vs (similar size) — which wins?
, because its charge is vs ; higher charge dominates, giving a far more negative hydration enthalpy.
A "combustion" that produces and soot — is its the standard ?
No. That's incomplete combustion; the standard demands excess oxygen so all carbon ends as .
Dissolving a gas (e.g. gas) in water — does the lattice-enthalpy step apply?
No lattice to break, since there's no ionic solid; the relation is built for ionic solids, so it doesn't apply unchanged here.
What does "infinite dilution / large excess solvent" mean, and why does the definition demand it?
It means so much water that adding still more releases negligible extra heat — the ions are effectively isolated and never "feel" each other. The definition fixes this so has one reproducible value rather than drifting with concentration.
If you double the volume of water but keep 1 mole of solute, does change?
Not meaningfully once you're near infinite dilution — you're already on the flat part of the enthalpy-vs-dilution curve, so further dilution releases negligible extra heat.
Why do tabulated values quote a temperature (298 K), and what happens at a different ?
Because itself depends on temperature (Kirchhoff's law): , the difference in heat capacities between products and reactants. If , the same reaction has a different at, say, 350 K — so the temperature must be stated.
Recall One-line self-test
Cover every answer above and re-derive the reasons cold. If any answer you gave was "true/false" with no because, you haven't earned it yet — go back to the parent note.
Connections — how each ties into these traps
- Hess's Law and Thermochemical Cycles — the state-function logic behind adding and the sign trap.
- Standard Enthalpy of Formation — the "combustion → formation" reversal error in Spot the error.
- Lattice Enthalpy and Born–Haber Cycle — where the sign of lattice-breaking in the solution traps comes from.
- Bond Enthalpies — why combustion is always exothermic (stronger bonds formed) in the True/False section.
- First Law of Thermodynamics — the derivation used to justify "all measured at constant pressure."
- Spontaneity and Gibbs Free Energy — why dissolves despite (the entropy edge case).