Chemical Thermodynamics

A reaction has $\Delta G^{\circ }=+50$ kJ mol${}^{-1}$ at $298$ K. Does it happen at all?

Yes. It runs forward — just not very far.

$\Delta G^{\circ }$ is the free energy change when reactants in their standard states become products in their standard states, all at unit activity. It fixes the equilibrium constant and nothing else:

$\Delta G^{\circ }=-RT\ln K\Rightarrow K=e^{-50000/\left(8.314\times 298\right)}=1.7\times 10^{-9}$

What actually decides the direction at any moment is $\Delta G$, not $\Delta G^{\circ }$:

$\Delta G=\Delta G^{\circ }+RT\ln Q$

Start with pure reactants and $Q=0$, so $\ln Q\to -\infty$ and $\Delta G$ is hugely negative. The reaction proceeds forward, $Q$ rises, and it stops when $Q=K$.

A positive $\Delta G^{\circ }$ means an unfavourable equilibrium, never an impossible reaction. This distinction is the single most productive source of marks in the chapter, and the most common place to lose them.

1. Work depends on the path

For expansion against an external pressure,

$w=-\int P_{ext}dV$

and the value depends entirely on how the expansion is carried out. Two cases matter.

Reversible isothermal, where $P_{ext}$ tracks the gas pressure at every instant:

$w_{rev}=-nRT\ln \frac{V_2}{V_1}=-2.303nRT\log \frac{V_2}{V_1}$

Irreversible against a constant $P_{ext}$:

$w_{irr}=-P_{ext}\left(V_2-V_1\right)$

The reversible path always extracts the most work, because the gas pushes against the highest pressure it possibly can at every stage. For a free expansion into a vacuum, $P_{ext}=0$ and $w=0$ exactly.

Illustration 1

Two moles of an ideal gas at $300$ K expand from $5$ L to $20$ L, first reversibly and isothermally, then against a constant external pressure of $1$ atm. Compare the work done.

Reversible: $w=-2\left(8.314\right)\left(300\right)\ln 4=-4988\times 1.386=-6914$ J

Irreversible: $w=-\left(1.013\times 10^5\right)\left(15\times 10^{-3}\right)=-1520$ J

The reversible path delivers four and a half times as much work. Both start and end at the same state, so $\Delta U$ and $\Delta H$ are identical — only $q$ and $w$ differ, which is exactly what makes them path functions.

2. From $\Delta U$ to $\Delta H$

Enthalpy is defined so that the heat measured at constant pressure is a state function:

$\Delta H=\Delta U+\Delta \left(PV\right)=\Delta U+\Delta n_{g}RT$

where $\Delta n_g$ counts gaseous moles only. A bomb calorimeter measures $\Delta U$ at constant volume; a coffee-cup calorimeter measures $\Delta H$ at constant pressure. Converting between them needs only $\Delta n_g$.

Illustration 2

The combustion of benzene at $298$ K gives $\Delta U=-3263.9$ kJ mol${}^{-1}$ in a bomb calorimeter. Find $\Delta H$.

${\text{C}}_6{\text{H}}_6\left(l\right)+\frac{15}{2}{\text{O}}_2\left(g\right)\to 6{\text{CO}}_2\left(g\right)+3{\text{H}}_2\text{O}\left(l\right)$

$\Delta n_g=6-7.5=-1.5$

$\Delta H=-3263.9+\left(-1.5\right)\left(8.314\times 10^{-3}\right)\left(298\right)=-3263.9-3.7=-3267.6$ kJ mol${}^{-1}$

Only gases count. Liquid water on the product side contributes nothing to $\Delta n_g$, and forgetting to exclude it is the standard error here.

3. Thermochemistry: Hess, bonds and Born-Haber

Because enthalpy is a state function, any route from reactants to products gives the same $\Delta H$. Three constructions exploit that.

Hess's law lets reactions be added, reversed (changing the sign) and scaled (scaling $\Delta H$) to reach a target equation.

Bond enthalpies give an estimate for gas-phase reactions:

$\Delta H\approx \sum B{E}_{broken}-\sum B{E}_{formed}$

The difference between this estimate and the measured value is the resonance energy, which is why benzene is more stable than three isolated double bonds would suggest.

Born-Haber cycles obtain lattice enthalpies, which cannot be measured directly.

Illustration 3

Use the Born-Haber data for sodium chloride to find its lattice enthalpy: sublimation $+108$, ionisation $+496$, half-dissociation of chlorine $+121$, electron gain $-349$, formation $-411$ kJ mol${}^{-1}$.

$\Delta H_f=\Delta H_{sub}+IE+\frac{1}{2}D+\Delta H_{eg}+U$

$-411=108+496+121-349+U$

$U=-411-376=-787$ kJ mol${}^{-1}$

The lattice enthalpy is by far the largest term. Formation is exothermic only because the lattice pays back everything the three endothermic steps cost.

Illustration 4

The enthalpy of hydrogenation of cyclohexene is $-120$ kJ mol${}^{-1}$. That of benzene is $-208$ kJ mol${}^{-1}$. Find the resonance energy of benzene.

If benzene were simply cyclohexatriene, three isolated double bonds would give $3\times \left(-120\right)=-360$ kJ mol${}^{-1}$.

The measured value is only $-208$.

Resonance energy $=360-208=152$ kJ mol${}^{-1}$

Benzene releases less energy because it started lower. Delocalisation has already stabilised it by $152$ kJ mol${}^{-1}$ relative to the hypothetical localised structure.

4. Standard enthalpies and what calorimetry measures

Every tabulated value is defined against a convention: the standard enthalpy of formation of an element in its reference state is zero. From that,

$\Delta H_{rxn}^{\circ }=\sum \Delta H_f^{\circ }\left(\text{products}\right)-\sum \Delta H_f^{\circ }\left(\text{reactants}\right)$

Several named enthalpies recur, and each has a characteristic behaviour.

Enthalpy of neutralisation for a strong acid with a strong base is always close to $-57.1$ kJ mol${}^{-1}$, because the only reaction actually occurring is

${\text{H}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{H}}_2\text{O}\left(l\right)$

the spectator ions playing no part. A weak acid gives a less negative value, because part of the released energy is consumed in ionising it. Hydrofluoric acid is the exception, giving a more negative value, since the fluoride ion's exceptionally large hydration enthalpy more than pays for the ionisation.

Enthalpy of solution of an ionic solid is the sum of two large opposing terms:

$\Delta H_{sol}=\Delta H_{hydration}-U$

Because these are individually enormous and nearly equal, the difference is small and can fall on either side of zero — which is why some salts warm their solutions and others cool them.

Illustration 5

The enthalpy of neutralisation of acetic acid by sodium hydroxide is $-55.4$ kJ mol${}^{-1}$. Find the enthalpy of ionisation of acetic acid.

For a strong acid the value would be $-57.1$ kJ mol${}^{-1}$.

The shortfall is the energy absorbed in ionising the weak acid:

$\Delta H_{ionisation}=-55.4-\left(-57.1\right)=+1.7$ kJ mol${}^{-1}$

Positive, as it must be, since separating a covalently bound proton from acetate costs energy. The weaker the acid, the larger this correction and the less negative the measured neutralisation enthalpy.

Illustration 6

Find the enthalpy of solution of sodium chloride, given a lattice enthalpy of $-787$ kJ mol${}^{-1}$ and hydration enthalpies of $-406$ for ${\text{Na}}^{+}$ and $-377$ for ${\text{Cl}}^{-}$.

$\Delta H_{sol}=\left(-406-377\right)-\left(-787\right)=-783+787$

$\Delta H_{sol}=+4$ kJ mol${}^{-1}$

Slightly endothermic, which is why a solution of common salt cools very slightly on dissolving. Two terms near $800$ kJ mol${}^{-1}$ have cancelled to leave $4$, so a small error in either input would change the sign of the answer entirely.

5. Kirchhoff: enthalpy at a different temperature

Reaction enthalpies are tabulated at $298$ K, but reactions are run at other temperatures. Differentiating $\Delta H$ with respect to temperature gives

$\frac{d\left(\Delta H\right)}{dT}=\Delta C_p\Rightarrow \Delta H_{T_2}=\Delta H_{T_1}+\Delta C_p\left(T_2-T_1\right)$

where $\Delta C_p$ is the heat capacity of the products minus that of the reactants. The same relation holds for $\Delta U$ with $\Delta C_v$.

Illustration 7

A reaction has $\Delta H=-92.4$ kJ mol${}^{-1}$ at $298$ K and $\Delta C_p=-45$ J K${}^{-1}$ mol${}^{-1}$. Find $\Delta H$ at $500$ K.

$\Delta H_{500}=-92.4+\left(-45\times 10^{-3}\right)\left(500-298\right)$

$=-92.4-9.09=-101.5$ kJ mol${}^{-1}$

A negative $\Delta C_p$ makes an exothermic reaction more exothermic on heating. The products absorb less heat than the reactants per degree, so raising the temperature costs the products less.

6. Entropy, and the third law

Entropy is a state function measured by the reversible heat divided by temperature:

$\Delta S=\int \frac{d{q}_{rev}}{T}$

Four standard results cover almost every question:

The third law states that a perfect crystal has zero entropy at absolute zero, which is what makes absolute entropies — rather than merely changes — meaningful. Trouton's rule notes that $\Delta S_{vap}$ is close to $88$ J K${}^{-1}$ mol${}^{-1}$ for most liquids; the exceptions are hydrogen-bonded liquids such as water, whose ordered liquid state gives a larger value.

Illustration 8

Find the entropy change when $1$ mol of an ideal gas expands isothermally and reversibly to three times its volume, and when $18$ g of ice melts at $273$ K with $\Delta H_{fus}=6.0$ kJ mol${}^{-1}$.

Expansion: $\Delta S=R\ln 3=8.314\times 1.0986=9.13$ J K${}^{-1}$

Melting: $\Delta S=\frac{6000}{273}=22.0$ J K${}^{-1}$ mol${}^{-1}$

Both are positive, as disorder increases in each case. Note that the melting figure is far larger, because breaking a crystal lattice creates much more disorder than merely tripling the volume of a gas.

Illustration 9

Predict the sign of $\Delta S$ for each: ${\text{CaCO}}_3\left(s\right)\to \text{CaO}\left(s\right)+{\text{CO}}_2\left(g\right)$; ${\text{N}}_2\left(g\right)+3{\text{H}}_2\left(g\right)\to 2{\text{NH}}_3\left(g\right)$; ${\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_2\text{O}\left(s\right)$.

First: a gas is produced from a solid, $\Delta n_g=+1$, so $\Delta S$ is strongly positive.

Second: four moles of gas become two, $\Delta n_g=-2$, so $\Delta S$ is negative.

Third: liquid becomes an ordered solid, so $\Delta S$ is negative.

The change in gas moles dominates every time it is non-zero. Only when $\Delta n_g=0$ do the finer contributions from molecular complexity decide the sign.

7. Gibbs energy: what $\Delta G$ and $\Delta G^{\circ }$ each mean

$\Delta G=\Delta H-T\Delta S,\Delta G^{\circ }=-RT\ln K,\Delta G=\Delta G^{\circ }+RT\ln Q$

The three are constantly confused, so keep their jobs separate:

A reaction with a positive $\Delta G^{\circ }$ still proceeds forward from pure reactants; it simply stops early. A reaction with $\Delta G^{\circ }$ hugely negative goes essentially to completion.

Illustration 10

For a reaction $\Delta G^{\circ }=-13.8$ kJ mol${}^{-1}$ at $298$ K. Find $K$, and find $\Delta G$ when $Q=10^4$.

$\ln K=\frac{13800}{8.314\times 298}=5.57\Rightarrow K=262$

$\Delta G=-13800+8.314\times 298\times \ln \left(10^4\right)=-13800+22843=+9043$ J mol${}^{-1}$

Positive, so at that composition the reaction runs backwards. Since $Q=10^4$ far exceeds $K=262$, there is too much product and the system must consume some — exactly what Le Chatelier's principle would say qualitatively.

8. Spontaneity, temperature and coupling

Whether $\Delta G$ can be negative depends on the signs of $\Delta H$ and $\Delta S$, and on temperature.

In the two mixed cases the sign of $\Delta G$ flips at

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

which is exactly the temperature at which the process reaches equilibrium.

Coupling lets an unfavourable reaction be driven by a favourable one sharing a common species. Copper cannot be extracted by heating its oxide alone, but coupling that decomposition to the oxidation of carbon makes the pair spontaneous, which is the basis of smelting.

Illustration 11

For the decomposition of calcium carbonate, $\Delta H=+178$ kJ mol${}^{-1}$ and $\Delta S=+161$ J K${}^{-1}$ mol${}^{-1}$. Find the temperature above which it becomes spontaneous.

$T=\frac{\Delta H}{\Delta S}=\frac{178000}{161}=1106$ K

Above about $833^{\circ }$C the entropy term wins and limestone decomposes.

This is why lime kilns run near $900^{\circ }$C. The temperature is not arbitrary; it is set by the ratio of two thermodynamic quantities.

Illustration 12

A reaction has $\Delta H=-40$ kJ mol${}^{-1}$ and $\Delta S=-120$ J K${}^{-1}$ mol${}^{-1}$. Is it spontaneous at $298$ K, and at what temperature does it cease to be?

$\Delta G=-40000-298\left(-120\right)=-40000+35760=-4240$ J mol${}^{-1}$, so yes at $298$ K.

$T=\frac{40000}{120}=333$ K

Above $333$ K it reverses. Both terms are negative, so the enthalpy term wins only while the temperature is low enough to keep $T\Delta S$ small.

Illustration 13

Explain how carbon reduces copper(I) oxide even though the decomposition of that oxide alone is non-spontaneous.

Decomposition alone: $2{\text{Cu}}_2\text{O}\to 4\text{Cu}+{\text{O}}_2$ has $\Delta G^{\circ }$ strongly positive.

Oxidation of carbon: $2\text{C}+{\text{O}}_2\to 2\text{CO}$ has $\Delta G^{\circ }$ strongly negative, and becomes more so as temperature rises because $\Delta n_g=+1$.

Adding the two gives $2{\text{Cu}}_2\text{O}+2\text{C}\to 4\text{Cu}+2\text{CO}$ with a net negative $\Delta G^{\circ }$.

Free energies add exactly as enthalpies do, being state functions, which is what makes coupling legitimate rather than a trick.

Illustration 14

The equilibrium constant of a reaction doubles when the temperature rises from $300$ K to $310$ K. Estimate $\Delta H^{\circ }$.

Using $\ln \frac{K_2}{K_1}=-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$:

$\ln 2=-\frac{\Delta H^{\circ }}{8.314}\left(\frac{1}{310}-\frac{1}{300}\right)=\frac{\Delta H^{\circ }}{8.314}\times 1.075\times 10^{-4}$

$\Delta H^{\circ }=\frac{0.693\times 8.314}{1.075\times 10^{-4}}=5.36\times 10^4$ J mol${}^{-1}$, about $54$ kJ mol${}^{-1}$

Positive, so the reaction is endothermic, which is consistent with $K$ rising on heating — exactly what Le Chatelier predicts.

Summary

• $\Delta G^{\circ }$ fixes only $K$; $\Delta G=\Delta G^{\circ }+RT\ln Q$ decides direction now. A positive $\Delta G^{\circ }$ means an unfavourable equilibrium, not an impossible reaction.
• $w_{rev}=-nRT\ln \frac{V_2}{V_1}$ always exceeds $w_{irr}=-P_{ext}\Delta V$ in magnitude; free expansion gives $w=0$.
• $\Delta H=\Delta U+\Delta n_{g}RT$, counting gaseous moles only.
• Hess's law: reverse a reaction and change the sign; scale it and scale $\Delta H$.
• Bond enthalpy estimate minus measured value gives the resonance energy — $152$ kJ mol${}^{-1}$ for benzene.
• Born-Haber: $\Delta H_f=\Delta H_{sub}+IE+\frac{1}{2}D+\Delta H_{eg}+U$, and $U$ is the largest term by far.
• Strong acid with strong base always gives $-57.1$ kJ mol${}^{-1}$; a weak acid gives less, the shortfall being its ionisation enthalpy.
• $\Delta H_{sol}=\Delta H_{hydration}-U$: two enormous terms nearly cancelling, so the sign can go either way.
• Kirchhoff: $\Delta H_{T_2}=\Delta H_{T_1}+\Delta C_p\left(T_2-T_1\right)$.
• $\Delta S$: $\frac{\Delta H_{trans}}{T}$ for a phase change, $nR\ln \frac{V_2}{V_1}$ for expansion, $-nR\sum x_i\ln x_i$ for mixing.
• The third law gives absolute entropies; Trouton's rule puts $\Delta S_{vap}$ near $88$ J K${}^{-1}$ mol${}^{-1}$ except for hydrogen-bonded liquids.
• The sign of $\Delta S$ follows $\Delta n_g$ whenever it is non-zero.
• Four sign cases: both favourable means always spontaneous, both unfavourable never, and the two mixed cases flip at $T=\frac{\Delta H}{\Delta S}$.
• Calcium carbonate decomposes above $1106$ K, which is why lime kilns operate near $900^{\circ }$C.
• Coupling works because free energies add: an unfavourable step is driven by a favourable one sharing a species, as in smelting.