Chemical Thermodynamics and Thermochemistry

Thermodynamics answers one question and refuses to answer another. It tells you whether a change can happen and how much energy it will release or absorb; it says nothing about how fast. Diamond turning into graphite is spontaneous by every thermodynamic test and yet takes geological time, which is why kinetics exists as a separate chapter. Most marks lost here come from three habits rather than from difficulty: forgetting that q and w depend on the path while ΔU and ΔH do not, dropping the sign convention halfway through a calculation, and treating "spontaneous" as though it meant "fast". Fix those three and the chapter becomes one of the most reliable scorers in the paper.

The First Law, State Functions and Work

The first law is conservation of energy written for a chemical system: ΔU = q + w, where q is heat absorbed by the system and w is work done on the system. With that convention, an expanding gas does work on its surroundings and w is negative. For expansion against a constant external pressure, w = −P_ext ΔV; for a reversible isothermal expansion of an ideal gas, w = −nRT ln(V₂/V₁).

The four standard processes
ProcessConditionKey results
IsothermalT constantΔU = 0 and ΔH = 0 for an ideal gas, so q = −w
Adiabaticq = 0ΔU = w, so the gas cools on expanding
IsochoricV constantw = 0, so ΔU = q_v — this is what a bomb calorimeter measures
IsobaricP constantΔH = q_p — this is what an open beaker measures
🎯 Why ΔU and ΔH are state functions but q and w are not
Think of a hill. Your change in altitude between two points depends only on where you started and finished — that is a state function. The distance you walked and the effort you spent depend entirely on the route, and two hikers reaching the same summit can have wildly different figures. Internal energy is altitude; heat and work are distance walked and effort spent. This is why an expansion from V₁ to V₂ has the same ΔU whether done reversibly or against vacuum, yet completely different q and w — reversible expansion delivers the maximum possible work, and free expansion against vacuum delivers none at all (P_ext = 0, so w = 0).

The bridge between the two calorimetric quantities is ΔH = ΔU + Δn_g RT, where Δn_g is the change in the number of moles of gas only. Solids and liquids are ignored because their volume change is negligible. So for a reaction with no change in gas moles, ΔH = ΔU exactly; for combustion of graphite, C(s) + O₂(g) → CO₂(g), Δn_g = 0 and again ΔH = ΔU. Similarly C_p − C_v = R for one mole of an ideal gas, which is the same statement in heat-capacity language.

Thermochemistry and Hess's Law

Because H is a state function, the enthalpy change of a reaction is the same however many steps you take to get there. That is Hess's law, and it is the single most productive tool in this chapter: it lets you compute an enthalpy you cannot measure from a set you can. The working form is ΔH_reaction = Σ ΔH_f(products) − Σ ΔH_f(reactants), where ΔH_f of any element in its standard state is defined as zero.

📖 Hess's law is what makes the enthalpy of formation of CO measurable
You cannot burn carbon to give only CO — some CO₂ always forms, so the reaction C + ½O₂ → CO can never be run cleanly in a calorimeter. But two reactions can: the complete combustion of carbon to CO₂ (−393.5 kJ/mol) and the combustion of CO to CO₂ (−283.0 kJ/mol). Subtracting gives ΔH_f(CO) = −393.5 − (−283.0) = −110.5 kJ/mol, a number no calorimeter ever read directly. This is the pattern behind almost every Hess's-law question: the target reaction is the one that cannot be performed, and you assemble it from ones that can.
Enthalpy terms and their fixed signs
TermMeaningSign
Enthalpy of combustionComplete burning of one mole in excess O₂Always negative
Bond dissociation enthalpyBreaking one mole of bonds in the gas phaseAlways positive — breaking costs energy
Lattice enthalpy (formation sense)Gaseous ions coming together into a crystalNegative; positive if defined as dissociation
Enthalpy of neutralisationStrong acid with strong baseConstant at −57.1 kJ/mol — it is just H⁺ + OH⁻ → H₂O
Enthalpy of formationOne mole from elements in standard statesEither sign; zero for an element itself
⚠️ Weak acids give a smaller neutralisation enthalpy, not the standard −57.1
The value −57.1 kJ/mol is constant for strong acid plus strong base because in every such case the only chemistry happening is H⁺ + OH⁻ → H₂O; the spectator ions do nothing. With a weak acid such as acetic acid, part of the released energy is consumed in ionising the acid first, so the measured enthalpy is numerically less than 57.1 — typically about −55.4 kJ/mol for acetic acid. Questions ask you to explain a value of −55 kJ/mol and the expected answer is "the acid is weak and its ionisation is endothermic", not an experimental error.

For lattice enthalpies, which cannot be measured directly either, the Born–Haber cycle is Hess's law applied to ionic solid formation: sublimation of the metal, its ionisation, dissociation of the non-metal, its electron gain, and finally lattice formation must together equal the enthalpy of formation of the salt. Given four of the five, the fifth follows. The step that most often carries the sign error is electron gain enthalpy, which is negative for the first electron of a halogen but positive for the second electron of oxygen.

Entropy, Gibbs Energy and Spontaneity

Entropy measures the number of microscopic arrangements available to a system, S = k ln W, and for a reversible change ΔS = q_rev/T. The second law states that the entropy of the universe — system plus surroundings — increases in any spontaneous process. Note carefully that the system's own entropy may fall, as it does when water freezes; what cannot fall is the total.

🎯 Why Gibbs energy exists at all
Applying the second law honestly means tracking the entropy of the surroundings as well as the system, which is impractical in a laboratory. Gibbs energy, G = H − TS, is an algebraic repackaging that folds the surroundings into system-only quantities: at constant T and P, ΔG = ΔH − TΔS, and ΔG < 0 is exactly equivalent to ΔS_universe > 0. So a chemist can decide spontaneity from measurements on the beaker alone. This is the whole reason G is defined, and it explains why the criterion holds only at constant temperature and pressure — the derivation assumed both.
ΔH and ΔS together decide spontaneity
ΔHΔSSpontaneous when
NegativePositiveAt all temperatures — both terms help
PositiveNegativeNever — both terms oppose
NegativeNegativeAt low T, where the TΔS penalty is small
PositivePositiveAt high T, where TΔS overtakes ΔH
⚠️ Spontaneous does not mean fast, and exothermic does not mean spontaneous
Two independent confusions, both regularly examined. Rate: the conversion of diamond to graphite has ΔG < 0 and is thermodynamically spontaneous, yet the activation barrier is so high that a diamond outlasts civilisations. Thermodynamics licenses the change; kinetics sets the timetable. Sign of ΔH: the dissolution of ammonium nitrate absorbs heat and still happens spontaneously, driven entirely by the large positive ΔS of a lattice dispersing into solution. Conversely a strongly exothermic reaction with a large negative ΔS can be non-spontaneous at high temperature. Never decide spontaneity from ΔH alone.

At equilibrium ΔG = 0, which gives the crossover temperature T = ΔH/ΔS for the two conditional rows of the table above — the temperature at which a reaction switches from non-spontaneous to spontaneous. Two further links tie this chapter to the next two. ΔG° = −RT ln K connects Gibbs energy to the equilibrium constant, so a negative ΔG° means K > 1. ΔG° = −nFE°_cell connects it to electrochemistry, so a spontaneous cell reaction is one with a positive cell potential. Both relations are examined as bridges rather than in isolation.

ℹ️ The third law and why absolute entropies are tabulated
The third law states that the entropy of a perfect crystal at 0 K is zero, because there is exactly one arrangement available and ln 1 = 0. That fixes an absolute zero for the entropy scale, which is why data books list S° values directly rather than only differences — unlike enthalpy, where only ΔH_f is tabulated and elements are assigned zero by convention. A practical consequence for the exam: S° of an element is not zero, so a reaction like C(s) + O₂(g) → CO₂(g) has a computable ΔS° even though its ΔH_f terms include zeros.

Key takeaways

  • ΔU = q + w with w = −P_ext ΔV; q and w depend on the path, ΔU and ΔH do not.
  • Isothermal ideal gas: ΔU = ΔH = 0 so q = −w. Adiabatic: q = 0 so ΔU = w and expansion cools the gas.
  • ΔH = ΔU + Δn_g RT — count only gas moles; a bomb calorimeter gives ΔU, an open beaker gives ΔH.
  • Hess's law lets you build an unmeasurable reaction from measurable ones, as for ΔH_f of CO.
  • Combustion enthalpy is always negative and bond dissociation enthalpy always positive.
  • Strong acid with strong base always gives −57.1 kJ/mol; a weak acid gives numerically less.
  • ΔG = ΔH − TΔS < 0 means spontaneous, and is equivalent to ΔS_universe > 0 at constant T and P.
  • Spontaneous is not fast, and exothermic is not automatically spontaneous — check both ΔH and ΔS.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. For an ideal gas expanding into a vacuum at constant temperature, the work done by the gas is:

    1. nRT ln(V₂/V₁)
    2. P_ext(V₂ − V₁)
    3. Zero
    4. Equal to the heat absorbed
    Show answer

    Answer: C — Zero

    Work against an external pressure is w = −P_ext ΔV, and expansion into a vacuum has P_ext = 0, so no work is done however large the volume change. The first option is the reversible isothermal result, which gives the maximum possible work and is offered because candidates apply it to any isothermal expansion without checking what the gas was pushing against.

  2. For which reaction is ΔH equal to ΔU?

    1. N₂(g) + 3H₂(g) → 2NH₃(g)
    2. C(s) + O₂(g) → CO₂(g)
    3. CaCO₃(s) → CaO(s) + CO₂(g)
    4. 2H₂(g) + O₂(g) → 2H₂O(l)
    Show answer

    Answer: B — C(s) + O₂(g) → CO₂(g)

    ΔH = ΔU + Δn_g RT, so the two are equal only when the number of moles of gas is unchanged. In C(s) + O₂(g) → CO₂(g) one mole of gas becomes one mole of gas, so Δn_g = 0; the solid carbon is ignored because its volume is negligible. The other three have Δn_g of −2, +1 and −3 respectively.

  3. A reaction has ΔH = +30 kJ/mol and ΔS = +100 J/K·mol. It becomes spontaneous above approximately:

    1. 30 K
    2. 300 K
    3. 3000 K
    4. It is spontaneous at all temperatures
    Show answer

    Answer: B — 300 K

    Setting ΔG = 0 gives T = ΔH/ΔS = 30000 J / 100 J·K⁻¹ = 300 K, above which TΔS exceeds ΔH and ΔG turns negative. The answer 30 K comes from dividing 30 by 100 without converting kJ to J — the single most frequent unit slip in this chapter, since ΔH is tabulated in kilojoules and ΔS in joules.

  4. Given ΔH_c(C) = −393.5 and ΔH_c(CO) = −283.0 kJ/mol, the enthalpy of formation of CO is:

    1. −676.5 kJ/mol
    2. −110.5 kJ/mol
    3. +110.5 kJ/mol
    4. −283.0 kJ/mol
    Show answer

    Answer: B — −110.5 kJ/mol

    By Hess's law, C + ½O₂ → CO is the combustion of carbon minus the combustion of CO: −393.5 − (−283.0) = −110.5 kJ/mol. The value −676.5 comes from adding the two instead of subtracting, which would describe carbon burning twice over. This route exists because CO can never be made cleanly from carbon and oxygen alone, so its formation enthalpy has to be assembled indirectly.

  5. The enthalpy of neutralisation of acetic acid by NaOH is numerically less than 57.1 kJ/mol because:

    1. Acetic acid is a weak acid and energy is used up in ionising it
    2. NaOH is a weak base
    3. Water is not formed in the reaction
    4. The reaction is endothermic overall
    Show answer

    Answer: A — Acetic acid is a weak acid and energy is used up in ionising it

    For a strong acid and strong base the only net change is H⁺ + OH⁻ → H₂O, giving a constant −57.1 kJ/mol. Acetic acid is only partly ionised, so part of the energy released by water formation is absorbed in ionising the remaining acid, leaving a smaller net value of about −55.4 kJ/mol. The reaction is still exothermic overall, so the last option describes something that does not happen.

  6. Which of the following is always positive?

    1. Enthalpy of combustion
    2. Bond dissociation enthalpy
    3. Enthalpy of formation
    4. Enthalpy of neutralisation
    Show answer

    Answer: B — Bond dissociation enthalpy

    Breaking a bond always costs energy, so bond dissociation enthalpy is invariably positive. Combustion and neutralisation are both always negative, and enthalpy of formation can take either sign — negative for CO₂ but positive for NO, which is why an unstable oxide can still exist. Note that bond formation enthalpy is the negative of dissociation, so the sign question hinges on which direction is named.

  7. Dissolution of NH₄NO₃ in water lowers the temperature of the solution, yet it happens spontaneously. The driving force is:

    1. A negative ΔH
    2. A large positive ΔS
    3. A negative ΔS
    4. A positive ΔG
    Show answer

    Answer: B — A large positive ΔS

    The cooling shows ΔH is positive, so enthalpy opposes the change; the process is spontaneous only because an ordered lattice dispersing into solution has a large positive ΔS, and at room temperature TΔS outweighs ΔH to make ΔG negative. A positive ΔG would mean non-spontaneous by definition, so that option contradicts the observation stated in the question.

  8. The conversion of diamond to graphite has ΔG° = −2.9 kJ/mol at 298 K. This means diamond:

    1. Rapidly turns into graphite at room temperature
    2. Is thermodynamically unstable but kinetically inert
    3. Is thermodynamically more stable than graphite
    4. Cannot be converted to graphite at any temperature
    Show answer

    Answer: B — Is thermodynamically unstable but kinetically inert

    A negative ΔG° means the conversion is spontaneous, so graphite is the thermodynamically stable form and diamond is not. But the rearrangement requires breaking the entire covalent network, an enormous activation barrier, so the rate is effectively zero at room temperature. This is the standard illustration that thermodynamics licenses a change without setting any timetable for it — the first option confuses the two.