Chemical Thermodynamics

Ammonium nitrate dissolves in water and the beaker goes noticeably cold. So the process absorbs heat. Why does it happen at all?

Most students arrive believing exothermic means spontaneous. The belief is wrong, comfortable, and survives a whole year of study because most spontaneous reactions happen to be exothermic.

Ice melts at room temperature while absorbing heat. Ammonium nitrate cools its own solution. Both are spontaneous, both endothermic. Something other than enthalpy is driving them.

The organising principle: thermodynamics answers one question — will this go — and the answer is the sign of $\Delta G=\Delta H-T\Delta S$. Everything else in the chapter is machinery for getting those two terms.

A second idea runs alongside: state functions do not care how you got there. That single property is what makes Hess's law work, and why enthalpies of formation can be tabulated once and reused for reactions nobody has ever performed.

1. Systems, Surroundings and State Functions

A state function depends only on the current state, not the route. Internal energy, enthalpy, entropy, Gibbs energy, pressure, volume and temperature all qualify.

Trap. Heat and work are not state functions. They are path functions, meaningful for a process and not for a state. It is meaningless to ask how much heat a system contains.

Extensive properties depend on amount — mass, volume, $U$, $H$, $S$, heat capacity. Intensive ones do not — temperature, pressure, density, molar heat capacity, concentration. Quick test: divide the system in two. Whatever halves is extensive.

Processes. Isothermal holds $T$, isobaric holds $p$, isochoric holds $V$, adiabatic exchanges no heat. A reversible process passes through a continuous succession of equilibrium states and is an idealisation; every real process is irreversible.

Illustration 1

A gas goes from state A (2 atm, 1 L) to state B (1 atm, 2 L) by two routes.

• Route 1: expand at a constant 2 atm to 2 L, then drop the pressure to 1 atm at constant volume.
• Route 2: drop the pressure to 1 atm at constant volume first, then expand to 2 L.

Compare $w$, $q$ and $\Delta U$ for the two.

Work is done only during the expansion steps, since a constant-volume step moves nothing:

$w_1=-p\Delta V=-(2)(1)=-2\mathrm{L}\mathrm{atm}=-203\mathrm{J}$

$w_2=-(1)(1)=-1\mathrm{L}\mathrm{atm}=-101\mathrm{J}$

Twice as much work on the first route, from the same start to the same finish. Work is a path function.

But $U$ depends only on the state the gas is in, and both routes finish in state B, so $\Delta U$ is identical. The first law then forces the heat to differ by exactly as much as the work did:

$q=\Delta U-w\Rightarrow q_1-q_2=203-101=102\mathrm{J}$

That is the whole content of calling $U$ a state function. Neither $q$ nor $w$ is one, and either can be made almost anything by choosing a devious enough route — but their sum is pinned by the endpoints alone. It is why tabulating $\Delta U$ and $\Delta H$ is possible at all, and why tabulating $q$ would be meaningless.

2. The First Law

$\Delta U=q+w$

with $q$ the heat absorbed by the system and $w$ the work done on it.

Older books use the opposite convention for $w$, so check before borrowing any formula.

$w=-p_{ext}\Delta V,w_{rev}=-2.303nRT\log \frac{V_2}{V_1}$

Free expansion into a vacuum has $p_{ext}=0$ and does no work at all, however much the volume changes.

Illustration 2

One mole of ideal gas expands from 1 L to 10 L at 300 K, (a) reversibly and isothermally, (b) against a constant external pressure of 1 bar. Compare the work done by the gas.

$\text{(a)}w=-2.303(1)(8.314)(300)\log 10=-5744\text{J}$

$\text{(b)}w=-(10^5)(9\times 10^{-3})=-900\text{J}$

The same change of state, and the reversible route delivers 6.4 times as much work. Reversible work is the maximum obtainable — because at every instant the gas is pushing against the largest pressure it can still overcome, whereas the irreversible route wastes the difference.

3. Enthalpy

Most chemistry happens in open vessels at constant pressure, where the system expands and some energy leaks away as work. Internal energy is therefore inconvenient, and enthalpy is defined to absorb the problem:

$H=U+pV\Rightarrow \Delta H=q_p,\Delta U=q_v$

$\Delta H=\Delta U+\Delta n_{g}RT$

where $\Delta n_g$ counts gas moles only, products minus reactants. Solids and liquids are ignored because their volumes are negligible. When $\Delta n_g=0$ the two are equal, which is why the distinction is often invisible.

Illustration 3

For $\mathrm{C}(\mathrm{s})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\to \mathrm{CO}(\mathrm{g})$ at 298 K, $\Delta H=-110.5$ kJ. Find $\Delta U$.

$\Delta n_g=1-0.5=+0.5$

$\Delta U=\Delta H-\Delta n_{g}RT=-110.5-(0.5)(8.314\times 10^{-3})(298)\Rightarrow \Delta U=-111.7\text{kJ}$

Only the carbon is solid, so it contributes nothing to $\Delta n_g$ — counting it is the standard error. Note also that $RT$ in kilojoules is about 2.48 kJ mol⁻¹ at 298 K, so these corrections are always small; they matter for precision, not for sign.

Heat capacity

$C_p-C_v=R\text{(ideal gas)}$

The constant-pressure value is larger because heat supplied at constant pressure must also pay for the expansion work, whereas at constant volume all of it raises the temperature.

Illustration 4

5 mol of an ideal monatomic gas is heated from 300 K to 400 K at constant pressure. Find $q$, $\Delta U$ and $w$.

$C_{p,m}=\frac{5}{2}R=20.79\Rightarrow q_p=(5)(20.79)(100)=10.39\text{kJ}$

$\Delta U=n{C}_{v,m}\Delta T=(5)(12.47)(100)=6.24\text{kJ}$

$w=\Delta U-q=6.24-10.39=-4.16\text{kJ}$

Check independently: $w=-nR\Delta T=-(5)(8.314)(100)=-4.16$ kJ. Of the 10.39 kJ supplied, only 6.24 kJ raised the temperature — the other 4.16 kJ was spent pushing back the atmosphere. That gap is $C_p-C_v=R$.

Measuring it

A bomb calorimeter holds volume constant and measures $\Delta U$ directly. A coffee-cup calorimeter is open to the atmosphere, holds pressure constant, and measures $\Delta H$. Combustion data in tables comes from bomb calorimetry and is converted with $\Delta n_{g}RT$ — which is exactly why that relation appears in exams so often.

Illustration 5

Burning 0.500 g of benzoic acid ($M=122$ g mol⁻¹) in a bomb calorimeter of heat capacity 10.2 kJ K⁻¹ raises the temperature by 1.30 K. Find the molar enthalpy of combustion.

The calorimeter absorbs what the reaction releases:

$q=C\Delta T=(10.2)(1.29)=13.16\text{kJ released}$

$n=\frac{0.500}{122}=4.098\times 10^{-3}\text{mol}\Rightarrow \Delta U=-\frac{13.16}{4.098\times 10^{-3}}=-3211{\text{kJ mol}}^{-1}$

A bomb holds volume constant, so this is $\Delta U$, not $\Delta H$. To convert, count the gas moles in

${\mathrm{C}}_7{\mathrm{H}}_6{\mathrm{O}}_2(\mathrm{s})+\frac{15}{2}{\mathrm{O}}_2(\mathrm{g})\to 7C{\mathrm{O}}_2(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

$\Delta n_g=7-7.5=-0.5\Rightarrow \Delta H=-3211+(-0.5)(8.314\times 10^{-3})(298)=-3212{\text{kJ mol}}^{-1}$

The correction is barely 1 kJ in 3200, which is typical — but the step matters, because a bomb calorimeter never measures $\Delta H$ and tables always quote it. Note too that the benzoic acid and the water are condensed phases and contribute nothing to $\Delta n_g$.

4. Standard Enthalpy Changes

A standard state is the pure substance at 1 bar and the stated temperature, conventionally 298 K.

The enthalpy of formation of any element in its standard state is zero by definition — a convention, not a measurement, and it is what makes the whole table self-consistent. Combustion enthalpies are always negative; formation enthalpies may be either sign.

Solution enthalpy as a competition

${\Delta }_{sol}H={\Delta }_{lattice}H+{\Delta }_{hyd}H$

Breaking the lattice costs energy; hydrating the freed ions releases it.

Illustration 6

Sodium chloride has a lattice enthalpy of $+788$ and a hydration enthalpy of $-784$ kJ mol⁻¹. Find ${\Delta }_{sol}H$ and say what it predicts.

${\Delta }_{sol}H=788-784=+4{\text{kJ mol}}^{-1}$

Barely endothermic — dissolving salt in water produces almost no temperature change, which matches experience. But notice what produced that 4: two numbers near 800 that very nearly cancelled.

Both terms grow with ionic charge and shrink with ionic radius, so they move together, and predicting which wins from the ions alone is unreliable. Ammonium nitrate's lattice term wins slightly and the solution cools; anhydrous calcium chloride's hydration term wins and it warms. This is a case where the numbers must be looked up, not reasoned out.

5. Hess's Law

The enthalpy change is the same whether a reaction happens in one step or several — an immediate consequence of enthalpy being a state function, and the most useful single tool in the chapter.

${\Delta }_{r}H=\sum {\Delta }_{f}H(\text{products})-\sum {\Delta }_{f}H(\text{reactants})$

remembering to multiply each term by its stoichiometric coefficient. It lets you compute changes for reactions that cannot be run cleanly: methane cannot be made directly from carbon and hydrogen, but all three combustion enthalpies are easily measured.

Illustration 7

Find ${\Delta }_{r}H$ for $\mathrm{C}{\mathrm{H}}_4(\mathrm{g})+2{\mathrm{O}}_2(\mathrm{g})\to \mathrm{C}{\mathrm{O}}_2(\mathrm{g})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$, given ${\Delta }_{f}H$: CH${}_4$ $-74.8$, CO${}_2$ $-393.5$, H${}_2$O(l) $-285.8$ kJ mol⁻¹.

${\Delta }_{r}H=[(-393.5)+2(-285.8)]-[(-74.8)+2(0)]$

$=-965.1+74.8\Rightarrow {\Delta }_{r}H=-890.3{\text{kJ mol}}^{-1}$

Oxygen contributes zero because it is an element in its standard state. Note the water is specified as liquid — quoting the gaseous value instead would change the answer by 88 kJ, which is why the state symbol is never decoration.

Bond enthalpies

${\Delta }_{r}H=\sum (\text{bonds broken})-\sum (\text{bonds formed})$

Breaking costs, forming releases, so the subtraction runs that way round.

Trap. These are averages. The four C–H bonds in methane do not each cost 413 kJ mol⁻¹ to break; that is the mean over four successive dissociations and over many molecules. Bond enthalpy estimates are always approximate, unlike Hess's law calculations from formation data.

Illustration 8

Estimate $\Delta H$ for $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2\to \mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$, given C–H 413, Cl–Cl 242, C–Cl 328, H–Cl 431 kJ mol⁻¹.

Only one C–H bond breaks, not four — the other three survive intact:

$\text{broken}=413+242=655,\text{formed}=328+431=759$

$\Delta H=655-759\Rightarrow \Delta H=-104{\text{kJ mol}}^{-1}$

Counting every bond in every molecule instead of only those that change is the usual error here. Unchanged bonds appear on both sides and cancel, so leaving them out entirely is both faster and safer.

6. Spontaneity and Entropy

A spontaneous process occurs without continuous external help. It says nothing about speed: diamond turning into graphite is spontaneous and takes longer than the age of the Earth.

Entropy $S$ measures the number of ways the energy and particles of a system can be arranged.

$\Delta S=\frac{q_{rev}}{T}[{\text{J K}}^{-1}{\text{mol}}^{-1}]$

Trap. Entropy is in joules per kelvin per mole while enthalpy is in kilojoules per mole. Mixing them in $\Delta H-T\Delta S$ is the single commonest arithmetic error in this chapter.

Entropy rises on melting, vaporising, dissolving a solid, mixing, heating, and whenever a reaction produces more moles of gas than it consumes. Gases have far more entropy than liquids, and liquids somewhat more than solids — so counting gas moles on each side usually settles the sign at a glance.

Illustration 9

Predict the sign of $\Delta S$ for each: (a) $2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\to 2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$, (b) $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\to \mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$, (c) ${\mathrm{N}}_2(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})\to 2N{\mathrm{H}}_3(\mathrm{g})$.

Count only the gases. In (a) three moles of gas become a liquid, which is the largest drop available, and in (b) a gas appears from nothing but solids. Case (c) is the one worth noticing: ammonia synthesis has a negative $\Delta S$, so entropy actively opposes it — and yet it runs, because $\Delta H$ is large and negative.

The real criterion

$\Delta S_{total}=\Delta S_{system}+\Delta S_{surroundings}>0$

Correct but awkward, because it requires knowing what happens outside the system. Water freezing at $-10$ °C has a negative system entropy change and is still spontaneous, because the heat released raises the entropy of the surroundings by more.

7. Gibbs Energy

Gibbs energy repackages the second law entirely in terms of the system:

$\Delta G=\Delta H-T\Delta S$

Spontaneous when negative, at equilibrium when zero, non-spontaneous when positive. This is the same statement as $\Delta S_{total}>0$, rearranged so that only system properties appear — which is why chemists use $G$ rather than $S$.

The two mixed cases are where questions are set. When enthalpy and entropy pull opposite ways, temperature decides, because entropy enters multiplied by $T$. Setting $\Delta G=0$ gives the crossover:

$T=\frac{\Delta H}{\Delta S}$

Illustration 10

For $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\to \mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$, $\Delta H=+178$ kJ mol⁻¹ and $\Delta S=+161$ J K⁻¹ mol⁻¹. Find the decomposition temperature.

Both positive, so this is the "high temperature only" case:

$T=\frac{178000}{161}\Rightarrow T=1106\text{K}=833{}^{\circ }\text{C}$

Note the conversion of kJ to J — omitting it gives 1.1 K, an answer that should be rejected on sight. Industrial lime kilns run near 900 °C, comfortably above this crossover, which is exactly why they run that hot and not hotter.

What the G actually stands for

At constant $T$ and $p$, $-\Delta G$ is the maximum non-expansion work the process can deliver. That is why a reaction with a large negative $\Delta G$ can be harnessed to drive something useful, and why the relation between Gibbs energy and cell potential exists at all. A reaction at equilibrium has $\Delta G=0$ and can do no work — the thermodynamic statement of a dead battery.

Gibbs energy and the equilibrium constant

$\Delta G^{\circ }=-RT\ln K=-2.303RT\log K$

Trap. $\Delta G^{\circ }$ is fixed for a reaction at a given temperature and tells you where equilibrium lies. $\Delta G$ without the degree changes continuously as the reaction proceeds and reaches zero at equilibrium. Confusing them is the commonest conceptual error in the chapter.

Illustration 11

Show that at 298 K, every 5.7 kJ mol⁻¹ of $\Delta G^{\circ }$ corresponds to one factor of ten in $K$.

$\Delta G^{\circ }=-2.303RT\log K,2.303RT=2.303(8.314)(298)=5706\text{J}$

$\Rightarrow \Delta G^{\circ }=-5.71\log K{\text{(kJ mol}}^{-1})$

So $\Delta G^{\circ }=-5.7$ gives $K=10$; $-11.4$ gives $K=100$; $-17.1$ gives $K=1000$. A reaction only modestly downhill in energy is already overwhelmingly product-favoured, and one with $\Delta G^{\circ }=0$ has $K=1$ exactly. Worth carrying into Equilibrium and Electrochemistry as a sanity check on any answer.

Illustration 12

For ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2N{\mathrm{H}}_3$, $\Delta H^{\circ }=-92.4$ kJ and $\Delta S^{\circ }=-198.3$ J K⁻¹. Find the temperature above which the reaction stops being spontaneous, and explain why industry runs it far above that temperature anyway.

At 298 K, converting entropy to kilojoules:

$\Delta G^{\circ }=-92.4-(298)(-0.1983)=-92.4+59.1=-33.3\text{kJ}$

Spontaneous. At 700 K:

$\Delta G^{\circ }=-92.4+(700)(0.1983)=-92.4+138.8=+46.4\text{kJ}$

Not spontaneous. The crossover:

$T=\frac{92400}{198.3}=466\text{K}=193{}^{\circ }\text{C}$

Yet the industrial Haber process runs at about 700 K. Thermodynamics says that is the wrong side of the crossover — and industry does it anyway, because at 466 K the rate is hopeless and the plant would take years to reach that favourable equilibrium.

The compromise is to accept a poor equilibrium position and claw back yield by other means: 200 atm of pressure, which Le Chatelier favours because 4 moles of gas become 2, an iron catalyst to reach equilibrium quickly, and continuous removal of ammonia to keep pulling the reaction forward. This is the clearest case in the syllabus of thermodynamics and kinetics giving different advice, and of engineering having to satisfy both.

Beyond the JEE Main Syllabus

The JEE Main syllabus for this unit names the first and second laws and the Gibbs energy criterion. The third law, fixing the entropy of a perfect crystal at absolute zero as zero, is not named and does not appear in Main questions, though textbooks introduce it when defining absolute entropies.

Heat engines, efficiency, the Carnot cycle and refrigerators are Physics rather than Chemistry, and they were removed from the JEE Main Physics syllabus in the 2023 revision as well. Meeting them in a chemistry textbook means context, not examinable content.

Both remain examinable in JEE Advanced, where the Carnot cycle appears in Physics and where absolute entropy values rest on the third law. Worth reading if you are sitting both papers.

Summary

• Thermodynamics answers whether a reaction will go, and the answer is the sign of $\Delta G$ — never $\Delta H$ alone.
• Melting ice and dissolving ammonium nitrate are spontaneous and endothermic. That settles it.
• State functions ignore the route; heat and work do not, and asking how much heat a system "contains" is meaningless.
• $\Delta U=q+w$, with $q$ in and $w$ on the system positive. $w=-p_{ext}\Delta V$; free expansion does no work.
• Reversible work is the maximum — 5.74 kJ against 0.90 kJ for the same expansion.
• $\Delta H=q_p$, $\Delta U=q_v$, and $\Delta H=\Delta U+\Delta n_{g}RT$ counting gas moles only.
• $C_p-C_v=R$, because constant-pressure heating must also pay the expansion work.
• Bomb calorimeter gives $\Delta U$; coffee cup gives $\Delta H$.
• ${\Delta }_{f}H$ of an element in its standard state is zero by definition.
• ${\Delta }_{sol}H$ is a small difference between two large opposing terms, so its sign cannot be reasoned from the ions.
• Hess's law: $\sum {\Delta }_{f}H(\text{products})-\sum {\Delta }_{f}H(\text{reactants})$, coefficients included, state symbols respected.
• Bond enthalpies are averages and always approximate; count only the bonds that change.
• Spontaneous says nothing about speed — diamond to graphite is spontaneous and glacial.
• Entropy is in J K⁻¹ mol⁻¹ while enthalpy is in kJ mol⁻¹. Convert before subtracting.
• Count gas moles to get the sign of $\Delta S$ at a glance.
• Four cases: $-/+$ always, $+/-$ never, $-/-$ low T only, $+/+$ high T only, crossing at $T=\Delta H/\Delta S$.
• $-\Delta G$ is the maximum non-expansion work; equilibrium means $\Delta G=0$ and no work available.
• $\Delta G^{\circ }=-2.303RT\log K$, and at 298 K every 5.7 kJ mol⁻¹ is one factor of ten in $K$.
• $\Delta G^{\circ }$ is fixed and locates equilibrium; $\Delta G$ is the running value and hits zero there.