University Chemistry — Year 2 · Bachelor Year 2

2Entropy and Free Energy of Reaction

A sports cold pack holds a bag of water and a few spoonfuls of ammonium nitrate. Squeeze it, the inner bag bursts, the salt dissolves, and within seconds the pack is cold enough to soothe a sprained ankle. The dissolution absorbs heat — it is endothermic — and yet it runs on its own, to completion. The enthalpy of the previous chapter cannot be the whole story: a reaction does not go only “downhill in heat”. The missing quantity is entropy, which the second law of thermodynamics, met in physics, makes the judge of every spontaneous change. This chapter tabulates entropies, computes the entropy of a reaction, and combines enthalpy and entropy into one function, the Gibbs energy, whose reaction derivative decides the direction of every reaction at fixed temperature and pressure.

You already know

From physics: the second law. Every system has a state function, its entropy $S$ (extensive); in a transformation of a closed system, $\dd S = \delta Q/T_{\text{ext}} + \delta S_{\text{c}}$ , where the created entropy $\delta S_{\text{c}}$ is zero for a reversible change and positive otherwise; for a closed system of fixed composition, $\dd U = T\dd S - p\dd V$ . Chapter 1: reaction quantities $\Delta_r X = (\partial X/\partial\xi)_{T,p}$ and standard reaction enthalpies. The Year 1 volume stated, as laws “that follow from the second law”, that a reaction evolves until $Q = K^\circ$ ; this chapter and the next two derive them.

A cold pack: when its inner bag is broken, ammonium nitrate dissolves in water and the pack cools. The dissolution is endothermic and spontaneous; its entropy explains why (). — A cold pack: when its inner bag is broken, ammonium nitrate dissolves in water and the pack cools. The dissolution is endothermic and spontaneous; its entropy explains why (Example 2.16).

2.1 Standard entropies and the third law

Unlike enthalpy, entropy has a natural zero.

Theorem 2.1 (Third law of thermodynamics)

The entropy of a pure, perfectly ordered crystal tends to zero as its temperature tends to $0\,\mathrm{K}$ .

Proof. Admitted at this level. ∎

Remark 2.2 (Where the zero comes from)

Walther Nernst stated in 1906 that reaction entropies between condensed phases vanish as $T \to 0$ ; Max Planck sharpened it into the statement above. Its origin is statistical: entropy counts the microscopic arrangements compatible with the state of a system, and a perfect crystal at $0\,\mathrm{K}$ has a single one. The counting is done in the Year 3 volume, with partition functions.

Definition 2.3 (Standard molar entropy)

The standard molar entropy $S^\circ_m(T)$ of a substance is the entropy of one mole of it in its standard state at temperature $T$ , the zero being that of the third law. Unlike $\Delta_f H^\circ$ , it is an absolute quantity, positive for every substance, elements included.

Proposition 2.4 (Absolute entropy from heat capacities)

Heating a substance at constant pressure $p^\circ$ from $0\,\mathrm{K}$ to $T$ ,

S^\circ_m(T) = \int_0^T\frac{C^\circ_{p,m}(T')}{T'}\,\dd T' + \sum_{\text{transitions}}\frac{\Delta_{\text{trs}}H^\circ}{T_{\text{trs}}} ,

the integral being taken in each phase over its range of temperature.

Proof. Heat the substance reversibly at constant pressure: $\delta S_{\text{c}} = 0$ and $\delta Q = \dd H = C_p\,\dd T$ , so $\dd S = C_p\,\dd T/T$ . At a transition (melting, boiling) the temperature stays at $T_{\text{trs}}$ while the heat $\Delta_{\text{trs}}H$ is received reversibly: the entropy jumps by $\Delta_{\text{trs}}H/T_{\text{trs}}$ . Add the pieces from the third-law zero. ∎

Standard molar entropy of zinc from the third-law zero. It grows with temperature as the metal stores heat, then jumps at the melting point by _ fusH/T_ fus and, much more, at the boiling point by _ vapH/T_b: a gas is far more disordered than a liquid. — Standard molar entropy of zinc from the third-law zero. It grows with temperature as the metal stores heat, then jumps at the melting point by $\Delta_{\text{fus}}H/T_{\text{fus}}$ and, much more, at the boiling point by $\Delta_{\text{vap}}H/T_b$ : a gas is far more disordered than a liquid.

The figure gives orders of magnitude worth remembering. At room temperature a metal or a hard crystal holds a few tens of $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , a liquid around a hundred, a gas one to three hundred; melting adds a little, boiling adds a lot. The standard molar entropies used in this volume are tabulated with the formation enthalpies.

2.2 Reaction entropy

Definition 2.5 (Reaction entropy)

The reaction entropy is $\Delta_r S = (\partial S/\partial\xi)_{T,p}$ , and the standard reaction entropy is $\Delta_r S^\circ(T) = \sum_i\nu_iS^\circ_{m,i}(T)$ , in $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ .

Because standard entropies are absolute, no formation reaction is needed: the elements count with their own entropies.

Example 2.6 (Three reaction entropies)

At $298.15\,\mathrm{K}$ :

$\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ : $\Delta_r S^\circ = 39.75 + 213.79 - 92.9 = +160.6\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , a gas made from a solid;
$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ : $\Delta_r S^\circ = 2(192.77) - 191.61 - 3(130.68) = -198.1\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , four molecules of gas becoming two;
$\ce{N2O4(g) -> 2NO2(g)}$ : $\Delta_r S^\circ = 2(240.03) - 304.38 = +175.7\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ .

Proposition 2.7 (The sign of a reaction entropy)

When gases take part, the sign of $\Delta_r S^\circ$ is in general that of $\Delta\nu_{\text{gas}} = \sum_{\text{gases}}\nu_i$ , and its size is about $100\,$ to $200\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ per mole of gas made or consumed. When $\Delta\nu_{\text{gas}} = 0$ , or no gas is involved, $\Delta_r S^\circ$ is small and its sign must be computed.

Justification. The molar entropy of a gas at $298\,\mathrm{K}$ is between 130 and $300\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , that of a solid or a liquid a few tens: making or consuming one mole of gas dominates every other contribution. This is a rule of thumb, not a theorem: the dissolution of a salt, with no gas, can still change the entropy by a hundred $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . ∎

Method 2.8 (Predicting the sign of $\Delta_r S^\circ$ )

Count $\Delta\nu_{\text{gas}}$ : if it is not zero, it gives the sign.
Otherwise compare the order of the states: dissolving a crystal, melting, breaking a molecule into two pieces raise the entropy; precipitating, freezing, joining two molecules into one lower it.
When the decision matters, compute $\sum_i\nu_iS^\circ_{m,i}$ .

The Year 1 volume met a consequence: an elimination makes more molecules than a substitution with the same reagents (the alkene, the leaving group and the protonated base, against the substitution product and the leaving group). Its reaction entropy is larger, and the entropy term, which grows with temperature as shown in the next section, is why heat favours elimination over substitution.

2.3 The Gibbs energy

Definition 2.9 (Gibbs energy)

The Gibbs energy (or free enthalpy) of a system is the state function $G = H - TS$ , extensive, in joules.

Definition 2.10 (Reaction Gibbs energy)

The reaction Gibbs energy is $\Delta_r G = (\partial G/\partial\xi)_{T,p}$ , and the standard reaction Gibbs energy is $\Delta_r G^\circ(T) = \sum_i\nu_iG^\circ_{m,i}(T)$ , both in $\mathrm{kJ}/\mathrm{mol}$ .

Proposition 2.11 (Enthalpy and entropy terms)

$\Delta_r G = \Delta_r H - T\Delta_r S$ , and $\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$ .

Proof. Differentiate $G = H - TS$ with respect to $\xi$ at fixed $T$ and $p$ : $T$ is a constant of the derivative. ∎

Proposition 2.12 (Differential of $G$ for a reacting system)

For a closed system in which one reaction takes place, at uniform $T$ and $p$ ,

\dd G = V\,\dd p - S\,\dd T + \Delta_r G\,\dd\xi .

Proof. $G(T, p, \xi)$ has the exact differential $\dd G = (\partial G/\partial T)_{p,\xi}\dd T + (\partial G/\partial p)_{T,\xi}\dd p + \Delta_r G\,\dd\xi$ . The first two partial derivatives are taken at fixed composition: they are those of a closed system of fixed composition, for which physics gives $\dd U = T\dd S - p\dd V$ , so that $\dd G = \dd U + p\dd V + V\dd p - T\dd S - S\dd T = V\dd p - S\dd T$ . Hence $(\partial G/\partial T)_{p,\xi} = -S$ and $(\partial G/\partial p)_{T,\xi} = V$ . ∎

2.4 The evolution criterion

Theorem 2.13 (Evolution criterion)

In a closed system kept at constant temperature and pressure, exchanging no work other than that of the pressure forces, a reaction can only advance in the direction that makes

\Delta_r G\,\dd\xi \leq 0 ,

with equality at equilibrium. If $\Delta_r G < 0$ the reaction runs forward, if $\Delta_r G > 0$ it runs backward, and the system is at equilibrium when $\Delta_r G = 0$ .

Proof. Take an infinitesimal change at uniform $T$ and $p$ , equal to those of the surroundings. The first law with pressure work only gives $\dd U = \delta Q - p\dd V$ ; the second gives $\delta Q = T\dd S - T\delta S_{\text{c}}$ . Then

\dd G = \dd U + p\dd V + V\dd p - T\dd S - S\dd T = -T\,\delta S_{\text{c}} + V\dd p - S\dd T .

Comparing with Proposition 2.12, at fixed $T$ and $p$ : $\Delta_r G\,\dd\xi = -T\,\delta S_{\text{c}}$ . Since $\delta S_{\text{c}} \geq 0$ , $\Delta_r G\,\dd\xi \leq 0$ , and $\delta S_{\text{c}} = 0$ (a reversible, equilibrium change) needs $\Delta_r G = 0$ . ∎

Remark 2.14 (What the criterion says, and what it does not)

The entropy created by a reaction at constant $T$ and $p$ is $-\Delta_r G\, \dd\xi/T$ : the Gibbs energy is the second law, restricted to the conditions of the laboratory and written with quantities of the system alone. It says in which direction a reaction may go, never how fast: diamond, for which the conversion to graphite has $\Delta_r G^\circ < 0$ , lasts for ever on a ring. And it assumes no other work: a cell that gives electrical work, or an electrolyser that receives it, obeys a modified criterion (Chapter 9).

Definition 2.15 (Exergonic and endergonic)

A reaction is exergonic in a given state when $\Delta_r G < 0$ and endergonic when $\Delta_r G > 0$ . With every species in its standard state these words apply to the sign of $\Delta_r G^\circ$ .

The criterion uses $\Delta_r G$ , the value in the actual state of the system; $\Delta_r G^\circ$ refers to every species in its standard state, which is rarely the state of a real mixture. The link between them, through the composition, is the subject of Chapter 3, and gives the Year-1 law $Q = K^\circ$ in Chapter 4. For reactions between pure solids and liquids, and gases at $1\,\mathrm{bar}$ , the two coincide.

Example 2.16 (The cold pack)

For $\ce{NH4NO3(s) -> NH4+(aq) + NO3-(aq)}$ at $298.15\,\mathrm{K}$ , standard states: $\Delta_r H^\circ = -339.87 + 365.56 = +25.69\,\mathrm{kJ}/\mathrm{mol}$ and $\Delta_r S^\circ = 259.8 - 151.08 = +108.7\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , so $\Delta_r G^\circ = 25.69 - 298.15 \times 0.1087 = -6.7\,\mathrm{kJ}/\mathrm{mol}$ : the dissolution is endothermic and exergonic. The ions, free in the water, are more disordered than in the crystal, and the entropy term $T\Delta_r S^\circ = 32.4\,\mathrm{kJ}/\mathrm{mol}$ outweighs the enthalpy cost.

History — Josiah Willard Gibbs

Josiah Willard Gibbs (1839–1903), professor of mathematical physics at Yale, published between 1876 and 1878 On the Equilibrium of Heterogeneous Substances, some three hundred pages in the transactions of a small academy, where the free energy, the chemical potential and the phase rule of the next chapters all appear. Few chemists read it until it was translated in the 1890s. (Photograph: public domain, Wikimedia Commons.)

2.5 Standard reaction Gibbs energies and temperature

Definition 2.17 (Ellingham approximation)

The Ellingham approximation treats $\Delta_r H^\circ$ and $\Delta_r S^\circ$ as independent of the temperature over an interval containing no change of state. Then $\Delta_r G^\circ(T) = \Delta_r H^\circ - T\Delta_r S^\circ$ is a straight line in $T$ , of slope $-\Delta_r S^\circ$ .

Proposition 2.18 (Temperature derivatives)

\frac{\dd\Delta_r G^\circ}{\dd T} = -\Delta_r S^\circ, \qquad \frac{\dd}{\dd T}\!\left(\frac{\Delta_r G^\circ}{T}\right) = -\frac{\Delta_r H^\circ}{T^2}\ \ \text{(Gibbs--Helmholtz)}, \qquad \frac{\dd\Delta_r S^\circ}{\dd T} = \frac{\Delta_r C^\circ_p}{T}.

In particular the Ellingham approximation holds when $\Delta_r C^\circ_p$ is small.

Proof. With every species in its standard state, $G^\circ(T, \xi)$ has $(\partial G^\circ/\partial T)_\xi = -S^\circ$ (Proposition 2.12). By Schwarz’s theorem,

\frac{\dd\Delta_r G^\circ}{\dd T} = \partial_\xi\partial_T G^\circ = -\partial_\xi S^\circ = -\Delta_r S^\circ .

Then $\frac{\dd}{\dd T}(\Delta_r G^\circ/T) = \frac{-T\Delta_r S^\circ - \Delta_r G^\circ}{T^2} = -\frac{\Delta_r H^\circ}{T^2}$ . Finally, $\dd S^\circ_m = C^\circ_{p,m}\,\dd T/T$ for each species (Proposition 2.4), and the same combination gives the last relation. If $\Delta_r C^\circ_p$ is small, $\Delta_r H^\circ$ (Kirchhoff) and $\Delta_r S^\circ$ hardly change. ∎

Definition 2.19 (Inversion temperature)

When $\Delta_r H^\circ$ and $\Delta_r S^\circ$ have the same sign, the inversion temperature of the reaction is the temperature where $\Delta_r G^\circ$ changes sign; in the Ellingham approximation, $T_i = \Delta_r H^\circ/\Delta_r S^\circ$ .

The enthalpy and entropy terms of CaCO3(s) -> CaO(s) + CO2(g) in the Ellingham approximation. Below the inversion temperature the enthalpy cost wins and limestone is stable under 1\, bar of carbon dioxide; above it the entropy term wins and limestone decomposes. — The enthalpy and entropy terms of $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ in the Ellingham approximation. Below the inversion temperature the enthalpy cost wins and limestone is stable under $1\,\mathrm{bar}$ of carbon dioxide; above it the entropy term wins and limestone decomposes.

The four cases of _r G = _r H - T _r S (Ellingham approximation). Combustions are in the upper left box; the decomposition of limestone and the cold pack in the upper right; the synthesis of ammonia in the lower left. — The four cases of $\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$ (Ellingham approximation). Combustions are in the upper left box; the decomposition of limestone and the cold pack in the upper right; the synthesis of ammonia in the lower left.

Method 2.20 (A standard reaction Gibbs energy at temperature $T$ )

From the tables at $298.15\,\mathrm{K}$ , compute $\Delta_r H^\circ$ and $\Delta_r S^\circ$ .
Check that no species changes state between $298\,\mathrm{K}$ and $T$ ; otherwise add the transition (enthalpy and entropy) or start from the new state.
In the Ellingham approximation, $\Delta_r G^\circ(T) = \Delta_r H^\circ - T\Delta_r S^\circ$ .
Where $\Delta_r C^\circ_p$ is not small or the interval is long, use tabulated $\Delta_f G^\circ(T)$ directly.

Example 2.21 (Two tests of the approximation)

For limestone, $\Delta_r C^\circ_p = 42.80 + 37.13 - 81.88 = -2.0\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ : the approximation is excellent, and $\Delta_r G^\circ = 178.3 - T \times 0.1606$ gives $T_i = 1110\,\mathrm{K}$ . For ammonia, $\Delta_r G^\circ(700) \approx -91.9 + 700 \times 0.1981 = +46.8\,\mathrm{kJ}/\mathrm{mol}$ , while the tables give $2\Delta_f G^\circ(\ce{NH3}, 700) = +54.4\,\mathrm{kJ}/\mathrm{mol}$ : here $\Delta_r C^\circ_p = -44\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ and the error is $8\,\mathrm{kJ}/\mathrm{mol}$ , a factor of 4 on the equilibrium constant at that temperature.

2.6 Exercises

Exercise 2.1 ★

Predict the sign of $\Delta_r S^\circ$ for: (a) $\ce{2H2(g) + O2(g) -> 2H2O(l)}$ ; (b) $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ ; (c) $\ce{N2O4(g) -> 2NO2(g)}$ ; (d) $\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$ ; (e) $\ce{H2O(s) -> H2O(l)}$ ; (f) $\ce{H2(g) + Cl2(g) -> 2HCl(g)}$ .

Solution

Solution of Exercise 2.1.

(a) Three moles of gas become liquid: $\Delta_r S^\circ < 0$ . (b) A gas is made: $> 0$ . (c) One gas molecule gives two: $> 0$ . (d) Ions in solution form a crystal: $< 0$ . (e) Melting: $> 0$ . (f) $\Delta\nu_{\text{gas}} = 0$ : small, and here positive ( $+20\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ from the tables, $\ce{HCl}$ being less symmetrical than $\ce{H2}$ and $\ce{Cl2}$ ).

Exercise 2.2 ★

Compute $\Delta_r S^\circ$ and $\Delta_r G^\circ$ at $298.15\,\mathrm{K}$ for the synthesis of ammonia, $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ , with $\Delta_r H^\circ = -91.9\,\mathrm{kJ}/\mathrm{mol}$ and the entropies of the chapter. Is it exergonic?

Solution

Solution of Exercise 2.2.

$\Delta_r S^\circ = 2(192.77) - 191.61 - 3(130.68) = -198.1\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ; $\Delta_r G^\circ = -91.9 - 298.15 \times (-0.1981) = -32.8\,\mathrm{kJ}/\mathrm{mol}$ : exergonic at $298\,\mathrm{K}$ , though the entropy term ( $+59.1\,\mathrm{kJ}/\mathrm{mol}$ ) works against it.

Exercise 2.3 ★

Compute $\Delta_r G^\circ(298.15\,\mathrm{K})$ of $\ce{H2(g) + 1/2O2(g) -> H2O(l)}$ from $\Delta_f H^\circ(\ce{H2O}, \text{l}) = -285.83\,\mathrm{kJ}/\mathrm{mol}$ and the standard entropies $\ce{H2(g)}$ 130.68, $\ce{O2(g)}$ 205.15, $\ce{H2O(l)}$ $69.95\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . Why is this value the standard Gibbs energy of formation of liquid water?

Solution

Solution of Exercise 2.3.

$\Delta_r S^\circ = 69.95 - 130.68 - \tfrac12(205.15) = -163.3\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ; $\Delta_r G^\circ = -285.83 + 298.15 \times 0.1633 = -237.14\,\mathrm{kJ}/\mathrm{mol}$ . It is the formation reaction of liquid water (one mole from the elements in their reference states), so its standard reaction Gibbs energy is by definition $\Delta_f G^\circ(\ce{H2O}, \text{l})$ ; the tables give the same value.

Exercise 2.4 ★

Use the figure of the entropy of zinc to read its standard molar entropy at $298\,\mathrm{K}$ and at $1000\,\mathrm{K}$ , and its jumps at the melting and boiling points. Deduce its enthalpies of fusion and of vaporisation.

Solution

Solution of Exercise 2.4.

About $42\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ at $298\,\mathrm{K}$ and $87\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ at $1000\,\mathrm{K}$ (liquid). Jumps: $10.6\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ at $692.7\,\mathrm{K}$ and $97.7\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ at $1180.2\,\mathrm{K}$ . $\Delta_{\text{fus}}H = 10.6 \times 692.7 = 7.3\,\mathrm{kJ}/\mathrm{mol}$ and $\Delta_{\text{vap}}H = 97.7 \times 1180.2 = 115\,\mathrm{kJ}/\mathrm{mol}$ .

Exercise 2.5 ★★

For $\ce{H2O(l) -> H2O(g)}$ at $298.15\,\mathrm{K}$ , compute $\Delta_r H^\circ$ , $\Delta_r S^\circ$ and $\Delta_r G^\circ$ from the tables of the chapter. Explain why $\Delta_r S^\circ \neq \Delta_r H^\circ/T$ at $298\,\mathrm{K}$ , what the positive $\Delta_r G^\circ$ means, and estimate in the Ellingham approximation the temperature at which liquid water and its vapour at $1\,\mathrm{bar}$ are in equilibrium. Compare with the boiling point.

Solution

Solution of Exercise 2.5.

$\Delta_r H^\circ = 44.00\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta_r S^\circ = 188.84 - 69.95 = 118.9\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , $\Delta_r G^\circ = 44.00 - 298.15 \times 0.1189 = +8.55\,\mathrm{kJ}/\mathrm{mol}$ . At $298\,\mathrm{K}$ the change is not reversible (liquid water and vapour at $1\,\mathrm{bar}$ are not in equilibrium), so $\Delta S \ne Q/T$ : $\Delta_r H^\circ/T = 147.6\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . The positive $\Delta_r G^\circ$ means that at $298\,\mathrm{K}$ water does not evaporate into an atmosphere of pure vapour at $1\,\mathrm{bar}$ : the vapour condenses. The two are in equilibrium where $\Delta_r G^\circ = 0$ : $T = 44\,004/118.9 = 370\,\mathrm{K}$ , within $3\,\mathrm{K}$ of the normal boiling point ( $373\,\mathrm{K}$ ): the Ellingham approximation is good over $75\,\mathrm{K}$ .

Exercise 2.6 ★★

For $\ce{N2O4(g) -> 2NO2(g)}$ , $\Delta_r H^\circ = 57.1\,\mathrm{kJ}/\mathrm{mol}$ and $\Delta_r S^\circ = 175.7\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . Compute the inversion temperature, and explain why a sealed tube of the brown gas $\ce{NO2}$ pales when it is cooled in ice.

Solution

Solution of Exercise 2.6.

$T_i = 57\,100/175.7 = 325\,\mathrm{K}$ . Below it $\Delta_r G^\circ > 0$ and colourless $\ce{N2O4}$ is favoured; cooling the tube shifts the gas towards $\ce{N2O4}$ and the brown colour of $\ce{NO2}$ fades (the quantitative link with the composition is in Chapter 4).

Exercise 2.7 ★★

A reaction has $\Delta_r G^\circ = -12.0\,\mathrm{kJ}/\mathrm{mol}$ at $300\,\mathrm{K}$ and $-4.0\,\mathrm{kJ}/\mathrm{mol}$ at $400\,\mathrm{K}$ . In the Ellingham approximation, find $\Delta_r H^\circ$ and $\Delta_r S^\circ$ , and check the Gibbs–Helmholtz relation.

Solution

Solution of Exercise 2.7.

$\Delta_r S^\circ = -\dd\Delta_r G^\circ/\dd T = -(-4.0 + 12.0)/100 = -0.080\,\mathrm{kJ}/(\mathrm{K}\,\mathrm{mol}) = -80\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ and $\Delta_r H^\circ = \Delta_r G^\circ + T\Delta_r S^\circ = -12.0 + 300 \times (-0.080) = -36.0\,\mathrm{kJ}/\mathrm{mol}$ . Check: $\Delta_r G^\circ/T$ goes from $-0.0400$ to $-0.0100\,\mathrm{kJ}/(\mathrm{K}\,\mathrm{mol})$ , slope $3.00 \times 10^{-4}$ per kelvin, equal to $-\Delta_r H^\circ/T^2 = 36.0/346^2 = 3.00 \times 10^{-4}$ at the geometric mean temperature $\sqrt{300 \times 400} = 346\,\mathrm{K}$ .

Exercise 2.8 ★★

Compute $\Delta_r G^\circ$ of the formation of methane, $\ce{C(s) + 2H2(g) -> CH4(g)}$ , at $298.15\,\mathrm{K}$ , from $\Delta_f H^\circ = -74.87\,\mathrm{kJ}/\mathrm{mol}$ and the entropies $\ce{C}$ (graphite) 5.74, $\ce{H2}$ 130.68, $\ce{CH4}$ $186.25\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . Above which temperature does methane become unstable with respect to its elements under the standard pressure?

Solution

Solution of Exercise 2.8.

$\Delta_r S^\circ = 186.25 - 5.74 - 2(130.68) = -80.8\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , $\Delta_r G^\circ = -74.87 + 298.15 \times 0.0808 = -50.8\,\mathrm{kJ}/\mathrm{mol}$ , the tabulated $\Delta_f G^\circ$ of methane. Both terms having the same sign, methane becomes unstable above $T_i = 74\,870/80.8 \approx 926\,\mathrm{K}$ (Ellingham approximation): hot methane cracks to carbon and hydrogen, a route to both.

Exercise 2.9 ★★

A haloalkane reacts with a base by substitution or by elimination. Write both reactions for 2-bromopropane and hydroxide. With $\Delta_r H^\circ(\text{elimination}) - \Delta_r H^\circ(\text{substitution}) = +40\,\mathrm{kJ}/\mathrm{mol}$ and $\Delta_r S^\circ(\text{elimination}) - \Delta_r S^\circ(\text{substitution}) = +90\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ (orders of magnitude), find above which temperature the elimination becomes the more exergonic of the two, and explain the role of the number of molecules.

Solution

Solution of Exercise 2.9.

Substitution: $\ce{CH3CHBrCH3 + OH- -> CH3CH(OH)CH3 + Br-}$ ; elimination: $\ce{CH3CHBrCH3 + OH- -> CH2=CHCH3 + H2O + Br-}$ . The difference of the two standard reaction Gibbs energies is $\Delta(\Delta_r G^\circ) = 40 - T \times 0.090$ , negative above $40\,000/90 \approx 440\,\mathrm{K}$ . Elimination makes one more molecule than substitution: its entropy is larger, and the entropy term grows in proportion to $T$ .

Exercise 2.10 ★★★

Crystalline carbon monoxide does not reach zero entropy at $0\,\mathrm{K}$ : each molecule can sit as CO or OC almost at random, the two orientations having nearly the same energy. Admitting Boltzmann’s formula $S = k_B\ln W$ , where $W$ is the number of arrangements (counted in the Year 3 volume), compute the residual molar entropy. Why does the third law not apply to this crystal?

Solution

Solution of Exercise 2.10.

Each of the $N_A$ molecules has 2 orientations: $W = 2^{N_A}$ , so $S = k_B\ln 2^{N_A} = R\ln 2 = 5.76\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . The crystal is not perfectly ordered: at low temperature the molecules are frozen in their random orientations and cannot reach the single ordered arrangement the third law assumes.

Exercise 2.11 ★★★

Show that if $\Delta_r C^\circ_p$ is a constant $c$ , $\Delta_r S^\circ(T) = \Delta_r S^\circ(T_0) + c\ln(T/T_0)$ and $\Delta_r G^\circ(T) = \Delta_r H^\circ(T_0) - T\Delta_r S^\circ(T_0) + c\,[T - T_0 - T\ln(T/T_0)]$ . Apply it to limestone ( $c = -2.0\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ) at $1100\,\mathrm{K}$ , and measure the error of the Ellingham approximation.

Solution

Solution of Exercise 2.11.

Integrate $\dd\Delta_r S^\circ/\dd T = c/T$ : $\Delta_r S^\circ(T) = \Delta_r S^\circ(T_0) + c\ln(T/T_0)$ ; Kirchhoff gives $\Delta_r H^\circ(T) = \Delta_r H^\circ(T_0) + c(T - T_0)$ ; subtract $T\Delta_r S^\circ(T)$ . For limestone at $1100\,\mathrm{K}$ : Ellingham $178.32 - 1100 \times 0.16064 = 1.62\,\mathrm{kJ}/\mathrm{mol}$ ; correction $c[T - T_0 - T\ln(T/T_0)] = -0.00195 \times (801.85 - 1436.3) = +1.24\,\mathrm{kJ}/\mathrm{mol}$ : $\Delta_r G^\circ = 2.86\,\mathrm{kJ}/\mathrm{mol}$ . The error, $1.2\,\mathrm{kJ}/\mathrm{mol}$ , is less than one per cent of $\Delta_r H^\circ$ , but near the inversion temperature it decides the sign.

Exercise 2.12 ★★★

Titanium is made from its oxide, rutile, through its chloride. (a) Compute $\Delta_r H^\circ$ and $\Delta_r S^\circ$ at $298\,\mathrm{K}$ of $\ce{TiO2(s) + 2Cl2(g) -> TiCl4(g) + O2(g)}$ , with $\ce{TiO2}$ ( $-944.0$ $\mathrm{kJ}/\mathrm{mol}$ , $50.62\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ) and $\ce{TiCl4(g)}$ ( $-763.2$ , 353.2), and its $\Delta_r G^\circ$ at $1000\,\mathrm{K}$ in the Ellingham approximation. (b) Carbon is added: $\ce{C(s) + O2(g) -> CO2(g)}$ , $\Delta_r H^\circ = -393.5\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta_r S^\circ = +2.9\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . Compute $\Delta_r G^\circ(1000\,\mathrm{K})$ of the sum of the two reactions and explain how the second drives the first.

Solution

Solution of Exercise 2.12.

(a) $\Delta_r H^\circ = -763.2 + 944.0 = +180.8\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta_r S^\circ = 353.2 + 205.15 - 50.62 - 2(223.08) = +61.6\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ; $\Delta_r G^\circ(1000) = 180.8 - 61.6 = +119.2\,\mathrm{kJ}/\mathrm{mol}$ : endergonic, no chlorination. (b) For $\ce{C + O2 -> CO2}$ , $\Delta_r G^\circ(1000) = -393.5 - 2.9 = -396.4\,\mathrm{kJ}/\mathrm{mol}$ . The sum, $\ce{TiO2(s) + 2Cl2(g) + C(s) -> TiCl4(g) + CO2(g)}$ , has $\Delta_r G^\circ = -277.2\,\mathrm{kJ}/\mathrm{mol}$ : carbon consumes the dioxygen made by the first reaction, and the coupled reaction is strongly exergonic.

2.7 Problem: The Lime Kiln

Problem 2.1

Weekend problem — the entropy and enthalpy of a decomposition, its Gibbs energy against temperature, the evolution criterion in a kiln, and the energy and carbon dioxide of a tonne of lime

Lime, $\ce{CaO}$ , is made by heating limestone, $\ce{CaCO3}$ , in a kiln. Data at $298.15\,\mathrm{K}$ : $\Delta_f H^\circ$ ( $\mathrm{kJ}/\mathrm{mol}$ ) $\ce{CaCO3(s)}$ $-1206.92$ , $\ce{CaO(s)}$ $-635.09$ , $\ce{CO2(g)}$ $-393.51$ ; $S^\circ_m$ ( $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ) 92.9, 39.75, 213.79; $C^\circ_{p,m}$ ( $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ) 81.88, 42.80, 37.13. Molar masses: $\ce{CaO}$ $56.08\,\mathrm{g}/\mathrm{mol}$ , $\ce{CO2}$ $44.01\,\mathrm{g}/\mathrm{mol}$ , $\ce{CH4}$ $16.04\,\mathrm{g}/\mathrm{mol}$ ; combustion of methane, water as vapour, $\Delta_r H^\circ = -802.3\,\mathrm{kJ}/\mathrm{mol}$ .

Part I — At room temperature.

Write the decomposition of limestone.
Compute $\Delta_r H^\circ(298\,\mathrm{K})$ . Is the reaction exothermic?
Compute $\Delta_r S^\circ(298\,\mathrm{K})$ and explain its sign.
Compute $\Delta_r G^\circ(298\,\mathrm{K})$ .
Is limestone stable at room temperature under $1\,\mathrm{bar}$ of carbon dioxide? Why do limestone cliffs last?
Compute $\Delta_r C^\circ_p$ , and say what it implies for the temperature dependence of $\Delta_r H^\circ$ and $\Delta_r S^\circ$ .

Part II — Heating.

Write $\Delta_r G^\circ(T)$ in the Ellingham approximation.
Compute it at $1000\,\mathrm{K}$ and at $1300\,\mathrm{K}$ .
Compute the inversion temperature $T_i$ .
Check the Gibbs–Helmholtz relation on this expression.
Estimate, with the formula of Exercise 2.11, the change of $\Delta_r G^\circ$ at $T_i$ caused by $\Delta_r C^\circ_p$ , and the resulting shift of $T_i$ .

Part III — The kiln. In this part the carbon dioxide above the solids is at $1\,\mathrm{bar}$ , so that $\Delta_r G = \Delta_r G^\circ$ .

What does the evolution criterion say at $1000\,\mathrm{K}$ ? And at $1300\,\mathrm{K}$ ?
What happens to a mixture of lime and carbon dioxide at $1\,\mathrm{bar}$ cooled below $T_i$ ?
How much entropy is created per mole of limestone decomposed at $1300\,\mathrm{K}$ ?
Kilns are swept by the combustion gases, poorer in carbon dioxide. Guess, before the next chapter, whether this helps the decomposition.

Part IV — A tonne of lime.

Compute the amount of lime in one tonne.
Compute the heat absorbed by the reaction per tonne of lime (take $\Delta_r H^\circ$ as constant).
Compute the mass of carbon dioxide released by the limestone per tonne of lime.
The heat comes from burning methane with a useful efficiency of 50 %. Compute the mass of methane burnt per tonne of lime.
Compute the carbon dioxide from this methane.
Compute the total carbon dioxide per tonne of lime, and the share due to the limestone itself.
Why can a cleaner fuel reduce only part of these emissions?
State the temperature above which limestone decomposes under $1\,\mathrm{bar}$ of carbon dioxide.

Solution

Solution of Problem 2.1.

1. $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ . 2. $\Delta_r H^\circ = -635.09 - 393.51 + 1206.92 = +178.3\,\mathrm{kJ}/\mathrm{mol}$ : endothermic. 3. $\Delta_r S^\circ = 39.75 + 213.79 - 92.9 = +160.6\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , positive because a gas is formed from a solid. 4. $\Delta_r G^\circ = 178.32 - 298.15 \times 0.16064 = +130.4\,\mathrm{kJ}/\mathrm{mol}$ . 5. Yes: $\Delta_r G < 0$ is needed for decomposition, and here it is strongly positive; the carbon dioxide of the air, far below $1\,\mathrm{bar}$ , lowers $\Delta_r G$ but by far less than $130\,\mathrm{kJ}/\mathrm{mol}$ (next chapter). 6. $\Delta_r C^\circ_p = 42.80 + 37.13 - 81.88 = -1.95\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , tiny: $\Delta_r H^\circ$ and $\Delta_r S^\circ$ hardly change with $T$ . 7. $\Delta_r G^\circ(T) = 178.32 - 0.16064\,T$ ( $\mathrm{kJ}/\mathrm{mol}$ ). 8. $+17.7\,\mathrm{kJ}/\mathrm{mol}$ at $1000\,\mathrm{K}$ ; $-30.5\,\mathrm{kJ}/\mathrm{mol}$ at $1300\,\mathrm{K}$ . 9. $T_i = 178\,320/160.64 = 1110\,\mathrm{K}$ . 10. $\Delta_r G^\circ/T = 178.32/T - 0.16064$ , whose derivative is $-178.32/T^2 = -\Delta_r H^\circ/T^2$ . 11. At $T_i$ , $c[T_i - T_0 - T_i\ln(T_i/T_0)] = -0.00195 \times (811.9 - 1459.4) = +1.26\,\mathrm{kJ}/\mathrm{mol}$ ; $\Delta_r G^\circ$ reaches zero $1260/160.6 \approx 8\,\mathrm{K}$ higher, at about $1118\,\mathrm{K}$ : less than one per cent. 12. At $1000\,\mathrm{K}$ , $\Delta_r G > 0$ : the decomposition cannot proceed (lime would absorb carbon dioxide). At $1300\,\mathrm{K}$ , $\Delta_r G < 0$ : limestone decomposes. 13. $\Delta_r G$ becomes positive: lime takes up the carbon dioxide and turns back into carbonate. 14. $\delta S_{\text{c}} = -\Delta_r G\,\dd\xi/T = 30\,506/1300 = 23.5\,\mathrm{J}/\mathrm{K}$ per mole. 15. Yes: Chapter 3 will show that $\Delta_r G = \Delta_r G^\circ + RT\ln(p_{\ce{CO2}}/p^\circ)$ , lowered when the pressure of carbon dioxide is below $1\,\mathrm{bar}$ , so that the decomposition starts below $T_i$ . 16. $n = 10^6/56.08 = 1.783 \times 10^{4}\,\mathrm{mol}$ . 17. $1.783 \times 10^4 \times 178.32 = 3.18 \times 10^{6}\,\mathrm{kJ}$ , that is $3.18\,\mathrm{GJ}$ per tonne. 18. $1.783 \times 10^4 \times 44.01 = 785\,\mathrm{kg}$ of $\ce{CO2}$ . 19. Heat to supply $3.18 \times 10^6/0.50 = 6.36 \times 10^{6}\,\mathrm{kJ}$ , so $6.36 \times 10^6/802.3 = 7.93 \times 10^{3}\,\mathrm{mol}$ of methane, $127\,\mathrm{kg}$ . 20. $7.93 \times 10^3 \times 44.01 = 349\,\mathrm{kg}$ of $\ce{CO2}$ . 21. $785 + 349 = 1134\,\mathrm{kg}$ per tonne of lime, 69 % of it from the limestone. 22. The limestone’s carbon dioxide is set by the stoichiometry, whatever heats the kiln; only the fuel’s share can be cut. 23. Limestone decomposes under $1\,\mathrm{bar}$ of carbon dioxide above its inversion temperature, $\boldsymbol{T_i \approx 1.11 \times 10^{3}\,\mathrm{K}}$ (about $840\,{}^{\circ}\mathrm{C}$ ).

Terms defined in this chapter

See all 852 terms in the glossary