University Chemistry — Year 2 · Bachelor Year 2

4Equilibrium Constants and Shifts

An ammonia plant runs its converter at about $450\,{}^{\circ}\mathrm{C}$ and $200\,\mathrm{bar}$ . At room temperature the equilibrium of $\ce{N2 + 3H2 <=> 2NH3}$ lies so far towards ammonia that a mixture of the elements could in principle be almost completely converted; yet the engineers heat it, which pushes the equilibrium back, and then compress it to two hundred atmospheres, which costs a great deal of energy. The reasons are a compromise between how far a reaction can go and how fast it goes. This chapter turns the Gibbs energy of the previous chapters into the equilibrium constant of the Year 1 volume, shows how that constant changes with temperature, counts how many conditions an experimenter may choose freely, and predicts in which direction an equilibrium moves when the temperature, the pressure or the composition is changed.

You already know

Chapter 2: the evolution criterion $\Delta_r G\,\dd\xi \leq 0$ and the Gibbs–Helmholtz relation; Chapter 3: $\Delta_r G = \Delta_r G^\circ + RT\ln Q$ . The Year 1 volume: the quotient $Q = \prod_i a_i^{\nu_i}$ , the standard equilibrium constant $K^\circ$ as the value of $Q$ at equilibrium, the law of mass action and the rule “ $Q < K^\circ$ : forward”, all stated there as laws; progress tables; final states of one or two reactions. It also promised that the choice of temperature and pressure for ammonia and for sulfur trioxide would be explained here.

An ammonia plant at night. In its converter, nitrogen and hydrogen pass over an iron catalyst at a few hundred degrees and a few hundred bar; the choice of those conditions is the subject of the weekend problem.

4.1 The equilibrium constant from the Gibbs energy

Theorem 4.1 (Standard equilibrium constant)

At equilibrium $\Delta_r G = 0$ , and the reaction quotient takes the value

K^\circ(T) = \exp\!\left(-\frac{\Delta_r G^\circ(T)}{RT}\right), \qquad \Delta_r G^\circ(T) = -RT\ln K^\circ(T) .

It depends on the temperature alone.

Proof. The evolution criterion makes $\Delta_r G = 0$ the condition of equilibrium. With $\Delta_r G = \Delta_r G^\circ + RT\ln Q$ (Proposition 3.13), equilibrium means $\ln Q_{\text{eq}} = -\Delta_r G^\circ/(RT)$ , a function of $T$ only since $\Delta_r G^\circ$ is. ∎

Corollary 4.2 (The laws of the Year 1 volume)

$\Delta_r G = RT\ln(Q/K^\circ)$ . Hence a system with $Q < K^\circ$ evolves forward, one with $Q > K^\circ$ backward, and one with $Q = K^\circ$ is at equilibrium (the law of mass action).

Proof. Substitute $\Delta_r G^\circ = -RT\ln K^\circ$ in $\Delta_r G = \Delta_r G^\circ + RT\ln Q$ ; the sign of $\Delta_r G$ is that of $\ln(Q/K^\circ)$ , and the evolution criterion requires $\Delta_r G\,\dd\xi \leq 0$ . ∎

Example 4.3 (Three constants at $298\,\mathrm{K}$ )

Ammonia synthesis, $\Delta_r G^\circ = -32.8\,\mathrm{kJ}/\mathrm{mol}$ : $K^\circ = \exp(32\,800/(8.314 \times 298.15)) = 5.6 \times 10^{5}$ (the tables, with more digits, give $5.4 \times 10^{5}$ ). The dissociation of $\ce{N2O4}$ , $\Delta_r G^\circ = +4.7\,\mathrm{kJ}/\mathrm{mol}$ : $K^\circ = 0.15$ . The decomposition of limestone, $\Delta_r G^\circ = +130.4\,\mathrm{kJ}/\mathrm{mol}$ : $K^\circ = 1.5 \times 10^{-23}$ , and the equilibrium pressure of carbon dioxide above limestone at room temperature is $1.5 \times 10^{-23}$ bar, nothing at all. A change of $5.7\,\mathrm{kJ}/\mathrm{mol}$ in $\Delta_r G^\circ$ at $298\,\mathrm{K}$ multiplies or divides $K^\circ$ by ten.

The Gibbs energy of the whole system shows the same thing as a landscape: it is lowest at the equilibrium extent, where its slope, $\Delta_r G$ , is zero.

Gibbs energy of a closed system of perfect gases at 298\, K and 1\, bar, starting from 1\, mol of N2O4, against the extent of N2O4 <=> 2NO2. Although _r G > 0 (dashed line, pure species), mixing the two gases lowers G, and the minimum lies at = 0.189, where Q = K = 0.15. To its left the slope _r G = G/ is negative, to its right positive. — Gibbs energy of a closed system of perfect gases at $298\,\mathrm{K}$ and $1\,\mathrm{bar}$ , starting from $1\,\mathrm{mol}$ of $\ce{N2O4}$ , against the extent of $\ce{N2O4 <=> 2NO2}$ . Although $\Delta_r G^\circ > 0$ (dashed line, pure species), mixing the two gases lowers $G$ , and the minimum lies at $\xi = 0.189$ , where $Q = K^\circ = 0.15$ . To its left the slope $\Delta_r G = \partial G/\partial\xi$ is negative, to its right positive.

4.2 Temperature: the van ’t Hoff equation

Theorem 4.4 (van ’t Hoff equation)

\frac{\dd\ln K^\circ}{\dd T} = \frac{\Delta_r H^\circ(T)}{RT^2}

(the van ’t Hoff equation): the equilibrium constant of an endothermic reaction increases with the temperature, that of an exothermic reaction decreases.

Proof. Write $\ln K^\circ = -\Delta_r G^\circ/(RT)$ and use the Gibbs–Helmholtz relation of Proposition 2.18,

\frac{\dd}{\dd T}\left(\frac{\Delta_r G^\circ}{T}\right) = -\frac{\Delta_r H^\circ}{T^2} ;

divide by $-R$ . ∎

Proposition 4.5 (Integrated form)

If $\Delta_r H^\circ$ is constant between $T_1$ and $T_2$ (Ellingham approximation),

\ln\frac{K^\circ(T_2)}{K^\circ(T_1)} = -\frac{\Delta_r H^\circ}{R}\left( \frac{1}{T_2} - \frac{1}{T_1}\right),

and $\ln K^\circ$ is a straight line of slope $-\Delta_r H^\circ/R$ against $1/T$ .

Proof. Integrate $\dd\ln K^\circ = (\Delta_r H^\circ/R)\,\dd T/T^2$ , noting that $\dd T/T^2 = -\dd(1/T)$ . ∎

The constant of N2 + 3H2 <=> 2NH3 from the tables (dots, 298 to 1000\, K) and the van ’t Hoff line drawn with _r H at 298\, K (dashed). The reaction is exothermic: K falls from 5 × 105 to 3 × 10-7. At high temperature the dots fall below the line: _r H becomes more negative as T rises (). — The constant of $\ce{N2 + 3H2 <=> 2NH3}$ from the tables (dots, 298 to $1000\,\mathrm{K}$ ) and the van ’t Hoff line drawn with $\Delta_r H^\circ$ at $298\,\mathrm{K}$ (dashed). The reaction is exothermic: $K^\circ$ falls from $5 \times 10^{5}$ to $3 \times 10^{-7}$ . At high temperature the dots fall below the line: $\Delta_r H^\circ$ becomes more negative as $T$ rises (Example 1.21).

Method 4.6 (An equilibrium constant at another temperature)

From tables at $298\,\mathrm{K}$ , compute $\Delta_r H^\circ$ , $\Delta_r S^\circ$ , and $K^\circ(298)$ .
Either use the integrated van ’t Hoff equation, or compute $\Delta_r G^\circ(T) = \Delta_r H^\circ - T\Delta_r S^\circ$ and then $K^\circ(T)$ : the two are the same approximation.
For a wide interval or a large $\Delta_r C_p^\circ$ , prefer tabulated $\Delta_f G^\circ(T)$ ; an error of $5.7\,\mathrm{kJ}/\mathrm{mol}$ at $298\,\mathrm{K}$ (or $13\,\mathrm{kJ}/\mathrm{mol}$ at $700\,\mathrm{K}$ ) is a factor of ten on $K^\circ$ .

History — Jacobus Henricus van ’t Hoff

Jacobus Henricus van ’t Hoff (1852–1911) gave in 1884, in his Studies in Chemical Dynamics, the relation between the temperature and the equilibrium constant that bears his name, and the next year the law of osmotic pressure of Chapter 3; ten years earlier he had proposed the tetrahedral carbon atom. He received the first Nobel Prize in Chemistry, in 1901. (Photograph by N. Perscheid, 1904: public domain, Wikimedia Commons.)

4.3 Variance

How many intensive quantities can an experimenter fix freely before the equilibrium state is completely determined?

Definition 4.7 (Variance)

The independent intensive parameters of a system at equilibrium are a set of intensive quantities (temperature, pressure, mole fractions in each phase) that can be chosen independently and that, once chosen, fix all the others. Their number is the variance $v$ of the system.

Theorem 4.8 (Phase rule)

For a system of $n$ species distributed among $\varphi$ phases, linked by $r$ independent equilibrium reactions and $s$ additional relations between intensive quantities fixed by the preparation (a stoichiometric feed, electroneutrality),

v = c + 2 - \varphi, \qquad c = n - r - s .

Proof. Intensive variables: $T$ , $p$ , and in each phase the mole fractions of the $n$ species, which sum to 1, so $n - 1$ free ones per phase: in all $2 + \varphi(n - 1)$ . Relations: each species is at equilibrium between the phases, $\varphi - 1$ equalities of chemical potentials per species, $n(\varphi - 1)$ in all; each reaction gives $Q = K^\circ(T)$ , $r$ relations; and $s$ additional relations. Subtract: $v = 2 + \varphi(n - 1) - n(\varphi - 1) - r - s = n - r - s + 2 - \varphi$ . (A species absent from a phase removes one variable and one relation together.) ∎

Method 4.9 (Computing a variance)

List the species, with their phase, and count $n$ and $\varphi$ (each pure solid is a phase; all gases form one phase).
Count the independent reactions $r$ among them.
Count the extra relations $s$ : amounts in stoichiometric proportion (only if the relation is between quantities of the same phase), electroneutrality of an ionic solution.
$v = n - r - s + 2 - \varphi$ .

Example 4.10 (Three variances)

Limestone in a closed vessel, $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ : $n = 3$ , $r = 1$ , $\varphi = 3$ (two solids, one gas): $v = 1$ . Choose $T$ , and the pressure of carbon dioxide is fixed: $p = K^\circ(T)\,p^\circ$ . A gas mixture of $\ce{N2}$ , $\ce{H2}$ and $\ce{NH3}$ at equilibrium: $n = 3$ , $r = 1$ , $\varphi = 1$ , $v = 3$ ( $T$ , $p$ and one composition variable); made from ammonia alone, or from a stoichiometric feed, $\ce{N2}$ and $\ce{H2}$ are in the ratio 1 : 3, $s = 1$ and $v = 2$ : at given $T$ and $p$ the composition is fixed, as the weekend problem uses.

4.4 Shifting an equilibrium

Definition 4.11 (Shift of an equilibrium)

When a system at equilibrium is perturbed (a change of temperature, of pressure, or of an amount), it is no longer at equilibrium in general and evolves to a new one. The shift of the equilibrium is forward if the reaction advances ( $\dd\xi > 0$ ) and backward otherwise.

Method 4.12 (Predicting a shift)

Just after the perturbation, compute (or estimate the sign of the change of) $Q$ and $K^\circ$ . If now $Q < K^\circ$ , the shift is forward; if $Q > K^\circ$ , backward; if the equality still holds, there is no shift.

A perturbation moves Q (or K, if the temperature changes); the system then shifts until Q = K again, forward if Q has become too small, backward if too large. — A perturbation moves $Q$ (or $K^\circ$ , if the temperature changes); the system then shifts until $Q = K^\circ$ again, forward if $Q$ has become too small, backward if too large.

Proposition 4.13 (Temperature)

At constant pressure, raising the temperature shifts an equilibrium in the endothermic direction (forward if $\Delta_r H^\circ > 0$ ), lowering it in the exothermic direction.

Proof. At fixed composition and pressure, $Q$ does not change with $T$ (for ideal systems); by van ’t Hoff, $K^\circ$ increases with $T$ if $\Delta_r H^\circ > 0$ . Just after a rise of $T$ , then, $Q < K^\circ$ : forward. ∎

Proposition 4.14 (Pressure)

At constant temperature and composition, raising the total pressure of a gas-phase equilibrium shifts it in the direction that lowers the number of gas molecules ( $\Delta\nu_{\text{gas}} < 0$ forward). If $\Delta\nu_{\text{gas}} = 0$ , the pressure has no effect.

Proof. With $p_i = x_ip$ , $Q = \prod_i x_i^{\nu_i}\,(p/p^\circ)^{\Delta\nu_{\text{gas}}}$ : at fixed composition, raising $p$ multiplies $Q$ by $(p'/p)^{\Delta\nu_{\text{gas}}}$ , which is $< 1$ if $\Delta\nu_{\text{gas}} < 0$ . Then $Q < K^\circ$ : forward. Condensed phases contribute nothing. ∎

Proposition 4.15 (Adding a constituent or an inert gas)

Adding a small amount of a gaseous constituent $j$ to a gas-phase equilibrium changes $\ln Q$ by

\left(\frac{\partial\ln Q}{\partial n_j}\right) = \frac{\nu_j}{n_j} - \frac{\Delta\nu_{\text{gas}}}{n_{\text{tot}}}\ \ \text{at constant } T, p ; \qquad \frac{\nu_j}{n_j}\ \ \text{at constant } T, V .

Adding an inert gas has no effect at constant volume, and at constant pressure acts as a lowering of the pressure.

Proof. At constant $T$ , $p$ : $Q = \prod_i n_i^{\nu_i}\,(p/(p^\circ n_{\text{tot}}))^{ \Delta\nu_{\text{gas}}}$ , whose logarithmic derivative with respect to $n_j$ is $\nu_j/n_j - \Delta\nu_{\text{gas}}/n_{\text{tot}}$ (zero first term for an inert gas, $\nu_j = 0$ ). At constant $T$ , $V$ : $p_i = n_iRT/V$ and $Q = \prod_i n_i^{ \nu_i}\,(RT/(p^\circ V))^{\Delta\nu_{\text{gas}}}$ , whose derivative is $\nu_j/n_j$ alone. ∎

Remark 4.16 (A reactant added that drives the reaction backward)

In ammonia synthesis, $\nu(\ce{N2}) = -1$ and $\Delta\nu_{\text{gas}} = -2$ : adding nitrogen at constant $T$ and $p$ changes $\ln Q$ by $-1/n_{\ce{N2}} + 2/n_{\text{tot}}$ , which is positive when $x(\ce{N2}) > 1/2$ . In a mixture already rich in nitrogen, adding more nitrogen dilutes the hydrogen so much that ammonia decomposes. The rule “adding a reactant pushes forward” is safe at constant volume, not always at constant pressure.

Proposition 4.17 (Le Chatelier’s principle)

The propositions above are summed up by Le Chatelier’s principle: a system at equilibrium, perturbed in temperature or pressure, shifts in the direction that opposes the perturbation (it absorbs heat when heated, reduces its gas volume when compressed).

Proof. It restates Proposition 4.13 (the endothermic direction absorbs heat) and Proposition 4.14 (fewer gas molecules occupy less volume at the same $T$ and $p$ ). For additions of matter at constant pressure the principle is not reliable (Remark 4.16): compute $Q$ . ∎

Remark 4.18 (When the equilibrium ends)

A shift can end not in a new equilibrium but in the disappearance of a phase. Heated in a closed vessel, limestone decomposes until $p(\ce{CO2}) = K^\circ(T)\,p^\circ$ ; if there is not enough of it to reach that pressure, it disappears entirely and $Q$ stays below $K^\circ$ : the system is out of equilibrium for good, as the Year 1 volume already observed.

History — Henry Le Chatelier

Henry Le Chatelier (1850–1936), a mining engineer and chemist who worked on cements, metallurgy and pyrometry, stated in 1884 the “law of displacement of equilibrium” named after him. He had also tried, around 1901, to make ammonia from its elements under pressure; an explosion in his apparatus ended the attempt. (Photograph: unknown author, CC BY-SA 4.0, Wikimedia Commons.)

4.5 Optimising a synthesis

Ammonia

The synthesis is exothermic and reduces four molecules of gas to two: by the propositions above, low temperature and high pressure favour ammonia. At $1\,\mathrm{bar}$ and room temperature the equilibrium is almost complete, but the reaction does not run at all: the triple bond of nitrogen is broken only on a catalyst, and the iron catalyst works only at several hundred degrees. Plants therefore choose:

a temperature high enough for the rate, as low as the catalyst allows (around $700\,\mathrm{K}$ );
a high pressure to recover the yield lost to the temperature, limited by the cost of compression and of thick-walled vessels;
a loop: the gas leaving the converter is cooled, the ammonia condenses and is removed, and the unconverted gas is recycled with fresh feed; a small purge removes the argon and methane that would otherwise accumulate.

Definition 4.19 (Single-pass conversion, recycle, purge)

In a process with recycling, the single-pass conversion is the fraction of a reactant converted in one passage through the reactor; the unconverted part, separated from the product, is returned to the reactor inlet as the recycle; a purge is a small stream withdrawn continuously from the recycle to keep inert species from accumulating.

The ammonia loop. Each pass converts only part of the gas; ammonia is condensed out, and the rest is recompressed and sent back with the fresh feed. The purge keeps argon and methane, which do not react, from building up.

History — Haber and Bosch

Fritz Haber (1868–1934) measured the equilibrium of ammonia under pressure and, in 1909, made it flow continuously from a laboratory converter over an osmium catalyst. Carl Bosch (1874–1940) turned it into an industry, with steel vessels that resisted hot hydrogen under pressure and a cheap iron catalyst; the first plant started in 1913. Haber received the Nobel Prize in 1918, Bosch in 1931. (Nobel Foundation portraits: public domain, Wikimedia Commons.)

Sulfur trioxide

In the contact process, $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ ( $\Delta_r H^\circ = -197.8\,\mathrm{kJ}/\mathrm{mol}$ for the equation as written) runs over a vanadium oxide catalyst, active only when hot, in several adiabatic beds. In each bed the heat released raises the temperature, which lowers $K^\circ$ ; the conversion stops when the gas reaches the equilibrium curve. Cooling between beds brings it back to a temperature where the equilibrium allows more conversion.

$Equilibrium conversion of SO2 to SO3 against temperature (black) for a model feed of 10\,\% SO2, 11\,\% O2 at 1\, bar, and the path of the gas through three adiabatic beds (red, rising in temperature as the reaction heats the gas), cooled between beds (blue).$

Equilibrium conversion of

\ce{SO2}

\ce{SO3}

against temperature (black) for a model feed of

10\,\%

\ce{SO2}

11\,\%

\ce{O2}

1\,\mathrm{bar}

, and the path of the gas through three adiabatic beds (red, rising in temperature as the reaction heats the gas), cooled between beds (blue).

4.6 Exercises

Exercise 4.1 ★

Compute $K^\circ$ at $298\,\mathrm{K}$ for a reaction with $\Delta_r G^\circ = -32.8\,\mathrm{kJ}/\mathrm{mol}$ , then for $\Delta_r G^\circ = +32.8\,\mathrm{kJ}/\mathrm{mol}$ . What change of $\Delta_r G^\circ$ multiplies $K^\circ$ by 100?

Solution

Solution of Exercise 4.1.

$K^\circ = \exp(32\,800/(8.314 \times 298.15)) = 5.6 \times 10^{5}$ , and for $+32.8\,\mathrm{kJ}/\mathrm{mol}$ its inverse, $1.8 \times 10^{-6}$ . A factor of 100 needs $\Delta(\Delta_r G^\circ) = -RT\ln 100 = -11.4\,\mathrm{kJ}/\mathrm{mol}$ .

Exercise 4.2 ★

For ammonia synthesis $K^\circ(298) = 5.4 \times 10^{5}$ and $\Delta_r H^\circ = -91.9\,\mathrm{kJ}/\mathrm{mol}$ . Estimate $K^\circ$ at $400\,\mathrm{K}$ with the integrated van ’t Hoff equation; the tables give 35.6.

Solution

Solution of Exercise 4.2.

$\ln K^\circ(400) = \ln(5.4 \times 10^5) + (91\,880/8.314)(1/400 - 1/298.15) = 13.20 - 9.44 = 3.76$ , $K^\circ \approx 43$ . The tables give 35.6: over $100\,\mathrm{K}$ the constant- $\Delta_r H^\circ$ approximation is already off by 20 % (the reaction becomes more exothermic as $T$ rises).

Exercise 4.3 ★

Compute the variance of: (a) liquid water in equilibrium with its vapour; (b) $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ in a closed vessel; (c) $\ce{N2}$ , $\ce{H2}$ , $\ce{NH3}$ at equilibrium, arbitrary feed; (d) $\ce{C(s) + H2O(g) <=> CO(g) + H2(g)}$ .

Solution

Solution of Exercise 4.3.

(a) $n = 1$ , $r = 0$ , $\varphi = 2$ : $v = 1$ (the vapour pressure is a function of $T$ ). (b) $v = 3 - 1 + 2 - 3 = 1$ . (c) $v = 3 - 1 + 2 - 1 = 3$ . (d) Four species, one reaction, two phases (carbon and the gas): $v = 4 - 1 + 2 - 2 = 3$ ; if the gas is made only from carbon and steam, $n(\ce{CO}) = n(\ce{H2})$ in the gas, $s = 1$ and $v = 2$ .

Exercise 4.4 ★

For $\ce{N2O4(g) <=> 2NO2(g)}$ , $\Delta_r H^\circ > 0$ . In which direction is the equilibrium shifted by heating at constant pressure, by compressing at constant temperature, by adding argon at constant volume?

Solution

Solution of Exercise 4.4.

Heating: endothermic direction, towards $\ce{NO2}$ (forward). Compressing: towards fewer gas molecules, $\ce{N2O4}$ (backward). Argon at constant volume: no change, the partial pressures of the reacting gases are unchanged.

Exercise 4.5 ★★

For $\ce{N2(g) + O2(g) <=> 2NO(g)}$ , $\Delta_r H^\circ = 180.6\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta_r S^\circ = 24.8\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . Compute $K^\circ$ at $298\,\mathrm{K}$ and at $2000\,\mathrm{K}$ (Ellingham approximation), and the mole fraction of $\ce{NO}$ in air ( $x(\ce{N2}) = 0.78$ , $x(\ce{O2}) = 0.21$ , little consumed) at equilibrium at $2000\,\mathrm{K}$ . Why do engines make this pollutant, and why does it survive in the cooled exhaust?

Solution

Solution of Exercise 4.5.

At $298\,\mathrm{K}$ : $\Delta_r G^\circ = 180.6 - 298.15 \times 0.0248 = 173.2\,\mathrm{kJ}/\mathrm{mol}$ , $K^\circ = 4.5 \times 10^{-31}$ . At $2000\,\mathrm{K}$ : $\Delta_r G^\circ = 180.6 - 2000 \times 0.0248 = 131.0\,\mathrm{kJ}/\mathrm{mol}$ , $K^\circ = 3.8 \times 10^{-4}$ . With $\Delta\nu_{\text{gas}} = 0$ , $x(\ce{NO})^2 = K^\circ x(\ce{N2}) x(\ce{O2})$ : $x(\ce{NO}) = \sqrt{3.8 \times 10^{-4} \times 0.78 \times 0.21} = 7.9 \times 10^{-3}$ , almost one per cent. Engines burn at such temperatures; as the gases cool, $K^\circ$ falls enormously, but the decomposition of $\ce{NO}$ is very slow: the hot equilibrium is frozen in the exhaust (until a catalyst removes it).

Exercise 4.6 ★★

An esterification $\mathrm{A + B \rightleftharpoons E + W}$ in the liquid phase has $K^\circ = 4.0$ (activities taken as mole fractions). Starting from $1.0\,\mathrm{mol}$ of acid A, compute the amount of ester with $1.0\,\mathrm{mol}$ , then $3.0\,\mathrm{mol}$ of alcohol B. Explain, with $Q$ and $K^\circ$ , why removing the water as it forms drives the reaction to completion.

Solution

Solution of Exercise 4.6.

With $\xi$ mol of ester and a constant total amount, $Q = \xi^2/[(1 - \xi)(n_B - \xi)]$ . For $n_B = 1.0$ : $\xi^2/(1 - \xi)^2 = 4$ , $\xi = 0.67\,\mathrm{mol}$ . For $n_B = 3.0$ : $3\xi^2 - 16\xi + 12 = 0$ , $\xi = 0.90\,\mathrm{mol}$ . Removing water keeps $Q$ below $K^\circ$ whatever $\xi$ : the reaction keeps going forward until the acid is consumed.

Exercise 4.7 ★★

At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$ , $K^\circ = 0.148$ for $\ce{N2O4 <=> 2NO2}$ . Compute the degree of dissociation of $1\,\mathrm{mol}$ of $\ce{N2O4}$ . Then $1\,\mathrm{mol}$ of argon is added at constant total pressure: compute the new degree of dissociation, and explain. What happens if the argon is added at constant volume?

Solution

Solution of Exercise 4.7.

Without argon: amounts $1 - \alpha$ , $2\alpha$ , total $1 + \alpha$ ; $Q = 4\alpha^2/(1 - \alpha^2) = 0.148$ , $\alpha = \sqrt{0.148/4.148} = 0.189$ . With $1\,\mathrm{mol}$ argon at $1\,\mathrm{bar}$ : total $2 + \alpha$ , $Q = 4\alpha^2/[(1 - \alpha)(2 + \alpha)] = 0.148$ , so $4.148\alpha^2 + 0.148\alpha - 0.296 = 0$ , $\alpha = 0.250$ : diluting at constant total pressure lowers the partial pressures, which favours the side with more gas molecules. At constant volume, the partial pressures of the reacting gases do not change: $\alpha$ stays 0.189.

Exercise 4.8 ★★

Ammonium chloride sublimes by dissociating, $\ce{NH4Cl(s) <=> NH3(g) + HCl(g)}$ . Compute the variance when the gas is made only by the solid, and when some ammonia has been added. What does each case mean for the experimenter?

Solution

Solution of Exercise 4.8.

Only the solid: $n = 3$ , $r = 1$ , $s = 1$ ( $p(\ce{NH3}) = p(\ce{HCl})$ ), $\varphi = 2$ : $v = 3 - 1 - 1 + 2 - 2 = 1$ ; choose $T$ and the pressure is fixed (a “dissociation pressure”). With added ammonia, $s = 0$ and $v = 2$ : $T$ and one pressure can be chosen; the product $p(\ce{NH3})\,p(\ce{HCl})$ is fixed by $T$ .

Exercise 4.9 ★★

The tables give, for ammonia synthesis, $K^\circ = 0.0993$ at $500\,\mathrm{K}$ and $K^\circ = 1.72 \times 10^{-3}$ at $600\,\mathrm{K}$ . Deduce $\Delta_r H^\circ$ in that interval and compare with its value at $298\,\mathrm{K}$ .

Solution

Solution of Exercise 4.9.

$\Delta_r H^\circ = -R\ln(K_2/K_1)/(1/T_2 - 1/T_1) = -8.314 \times \ln(0.0173)/(1/600 - 1/500) = -101\,\mathrm{kJ}/\mathrm{mol}$ , more negative than at $298\,\mathrm{K}$ ( $-91.9\,\mathrm{kJ}/\mathrm{mol}$ ), as Kirchhoff’s law predicted ( $\Delta_r C_p^\circ < 0$ ).

Exercise 4.10 ★★★

Prove the counter-example of Remark 4.16: write $Q$ of ammonia synthesis with the amounts and the total pressure, compute $\partial\ln Q/\partial n(\ce{N2})$ at constant $T$ and $p$ , and find the condition on $x(\ce{N2})$ for which adding nitrogen shifts the equilibrium backward. Can the same happen at constant volume?

Solution

Solution of Exercise 4.10.

$Q = \dfrac{n_{\ce{NH3}}^2\,n_{\text{tot}}^2}{n_{\ce{N2}}n_{\ce{H2}}^3} \left(\dfrac{p^\circ}{p}\right)^2$ . At constant $T$ , $p$ : $\partial\ln Q/\partial n_{\ce{N2}} = -1/n_{\ce{N2}} + 2/n_{\text{tot}}$ , positive when $n_{\ce{N2}} > n_{\text{tot}}/2$ , that is $x(\ce{N2}) > 1/2$ . Then $Q$ exceeds $K^\circ$ and the equilibrium shifts backward. At constant volume, $\partial\ln Q/\partial n_{\ce{N2}} = -1/n_{\ce{N2}} < 0$ : always forward.

Exercise 4.11 ★★★

Limestone is heated to $1100\,\mathrm{K}$ in an empty closed vessel of $10.0\,\mathrm{L}$ , where $\Delta_r G^\circ = 1.62\,\mathrm{kJ}/\mathrm{mol}$ for its decomposition (Chapter 2). Compute $K^\circ$ and the equilibrium pressure of carbon dioxide. Find the final state for $0.050\,\mathrm{mol}$ and for $0.200\,\mathrm{mol}$ of limestone.

Solution

Solution of Exercise 4.11.

$K^\circ = \exp(-1620/(8.314 \times 1100)) = 0.84$ : $p(\ce{CO2}) = 0.84\,\mathrm{bar}$ at equilibrium, that is $n = pV/RT = 0.84 \times 10^5 \times 0.0100/(8.314 \times 1100) = 0.092\,\mathrm{mol}$ of gas. With $0.050\,\mathrm{mol}$ of limestone, all of it decomposes ( $p = 0.46\,\mathrm{bar} < K^\circ p^\circ$ ): no equilibrium, a break of equilibrium. With $0.200\,\mathrm{mol}$ , equilibrium: $0.092\,\mathrm{mol}$ of $\ce{CO2}$ and $\ce{CaO}$ , $0.108\,\mathrm{mol}$ of $\ce{CaCO3}$ left.

Exercise 4.12 ★★★

Read on the contact-process figure the conversion and the temperature at the outlet of each bed. Explain why a single adiabatic bed, however long, could not exceed about 68 % conversion with this feed, why the gas is cooled between beds rather than the catalyst kept cold throughout, and what limits the conversion of the last bed.

Solution

Solution of Exercise 4.12.

Bed 1 leaves at about $890\,\mathrm{K}$ and 68 % conversion, bed 2 at about $785\,\mathrm{K}$ and 92 %, bed 3 at about $725\,\mathrm{K}$ and 98 %. In one adiabatic bed the gas heats up as it reacts until it meets the equilibrium curve, which at that temperature allows only about 68 %. A cold catalyst throughout is not an option because the catalyst works only when hot; cooling between beds keeps the rate and moves the equilibrium limit up. The last bed is limited by the equilibrium at the lowest temperature at which the catalyst still works (plants then absorb the trioxide and pass the gas once more over a catalyst).

4.7 Problem: The Ammonia Loop

Problem 4.1

Weekend problem — the equilibrium constant of ammonia synthesis and its temperature dependence, the composition at the converter’s conditions, the shifts caused by pressure, temperature and inert gases, and the loop

Ammonia is made by $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ from a stoichiometric feed. Data: at $298.15\,\mathrm{K}$ , $\Delta_f H^\circ(\ce{NH3}) = -45.94\,\mathrm{kJ}/\mathrm{mol}$ , $S^\circ_m$ ( $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ) $\ce{NH3}$ 192.77, $\ce{N2}$ 191.61, $\ce{H2}$ 130.68; the tables give $\Delta_f G^\circ(\ce{NH3}) = +27.19\,\mathrm{kJ}/\mathrm{mol}$ at $700\,\mathrm{K}$ and $+38.66\,\mathrm{kJ}/\mathrm{mol}$ at $800\,\mathrm{K}$ . Gases are perfect; $R = 8.314\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ .

Part I — The constant.

Compute $\Delta_r H^\circ$ , $\Delta_r S^\circ$ and $\Delta_r G^\circ$ at $298\,\mathrm{K}$ , and $K^\circ(298)$ .
Compute $K^\circ$ at $700\,\mathrm{K}$ and at $800\,\mathrm{K}$ from the tables.
Deduce from these two values the mean $\Delta_r H^\circ$ between $700\,\mathrm{K}$ and $800\,\mathrm{K}$ , and compare with question 1.
Estimate $K^\circ(723\,\mathrm{K})$ from the two tabulated values with the van ’t Hoff equation.
Compute $K^\circ(723\,\mathrm{K})$ in the Ellingham approximation from the data at $298\,\mathrm{K}$ , and comment on the difference.

Part II — The equilibrium in the converter.

From a feed of $1\,\mathrm{mol}$ $\ce{N2}$ and $3\,\mathrm{mol}$ $\ce{H2}$ , write the amounts at extent $\xi$ and the total amount.
Show that, with $x$ the mole fraction of ammonia, the mole fractions of $\ce{N2}$ and $\ce{H2}$ are $(1-x)/4$ and $3(1-x)/4$ .
Show that $\dfrac{x}{(1-x)^2} = \dfrac{\sqrt{27}}{16}\,\dfrac{p}{p^\circ}\, \sqrt{K^\circ}$ .
Compute $x$ at $723\,\mathrm{K}$ and $1\,\mathrm{bar}$ .
Compute $x$ at $723\,\mathrm{K}$ and $200\,\mathrm{bar}$ .
Compute the variance of the system and explain why $T$ and $p$ fix the composition.

Part III — Shifts.

Without computing, in which direction does the equilibrium move when the pressure rises? When the temperature rises?
Argon enters with the air used to make the nitrogen. At constant total pressure, how does its presence shift the equilibrium?
A mixture at equilibrium has $x(\ce{N2}) = 0.30$ . Does adding a little nitrogen at constant $T$ and $p$ shift it forward or backward?
Would the answer change at constant volume?
Before the gas returns to the converter, its ammonia is condensed out. What does this do to $Q$ at the converter inlet?

Part IV — The loop.

The gas leaves the converter with $18\,\%$ of ammonia, below the equilibrium value. Why does a real converter not reach equilibrium?
Compute the single-pass conversion of hydrogen corresponding to $x(\ce{NH3}) = 0.18$ from a stoichiometric feed.
The unconverted gas is recycled. In steady operation, $1.00\,\mathrm{mol}$ of fresh $\ce{N2}$ enters per unit time: how much ammonia leaves, and how much nitrogen passes through the converter per unit time?
Why must a little gas be purged from the recycle?
Why is the converter not run at $500\,\mathrm{K}$ , where the equilibrium is far more favourable?
State the equilibrium mole fraction of ammonia at $723\,\mathrm{K}$ and $200\,\mathrm{bar}$ for a stoichiometric feed, in the perfect-gas model.

Solution

Solution of Problem 4.1.

1. $\Delta_r H^\circ = -91.88\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta_r S^\circ = 2(192.77) - 191.61 - 3(130.68) = -198.1\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ , $\Delta_r G^\circ = -32.8\,\mathrm{kJ}/\mathrm{mol}$ , $K^\circ = 5.6 \times 10^{5}$ . 2. $K^\circ(700) = \exp(-54\,380/(8.314 \times 700)) = 8.8 \times 10^{-5}$ ; $K^\circ(800) = \exp(-77\,320/(8.314 \times 800)) = 8.9 \times 10^{-6}$ . 3. $\Delta_r H^\circ = -R\ln(K_{800}/K_{700})/(1/800 - 1/700)$ , that is $-106\,\mathrm{kJ}/\mathrm{mol}$ , more negative than at $298\,\mathrm{K}$ . 4. $\ln K^\circ(723) = \ln(8.75 \times 10^{-5}) + 12\,777 \times (1/723.15 - 1/700)$ , with $12\,777 = 106\,230/8.314$ : $K^\circ = 4.9 \times 10^{-5}$ . 5. $\Delta_r G^\circ(723) = -91.88 + 723.15 \times 0.1981 = 51.4\,\mathrm{kJ}/\mathrm{mol}$ , $K^\circ = 1.9 \times 10^{-4}$ : four times too large, because $\Delta_r H^\circ$ and $\Delta_r S^\circ$ change over $425\,\mathrm{K}$ ( $\Delta_r C_p^\circ \approx -44\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ). 6. $\ce{N2}$ $1 - \xi$ , $\ce{H2}$ $3 - 3\xi$ , $\ce{NH3}$ $2\xi$ , total $4 - 2\xi$ . 7. $x = 2\xi/(4 - 2\xi)$ , so $1 - x = (4 - 4\xi)/(4 - 2\xi)$ ; the nitrogen fraction $(1 - \xi)/(4 - 2\xi) = (1 - x)/4$ and the hydrogen fraction is three times larger. 8.

K^\circ = \frac{x^2}{\frac{1-x}{4}\left(\frac{3(1-x)}{4}\right)^3} \left(\frac{p^\circ}{p}\right)^2 = \frac{256}{27}\,\frac{x^2}{(1 - x)^4} \left(\frac{p^\circ}{p}\right)^2 ;

take the square root. 9. $a = (\sqrt{27}/16) \times \sqrt{4.9 \times 10^{-5}} \times 1 = 2.27 \times 10^{-3}$ ; $x/(1 - x)^2 = a$ gives $x = 0.0023$ . 10. $a = 0.455$ ; $ax^2 - (2a + 1)x + a = 0$ : $x = 0.25$ . 11. $n = 3$ , $r = 1$ , $s = 1$ (ratio 1 : 3 kept by the reaction), $\varphi = 1$ : $v = 2$ . Once $T$ and $p$ are chosen, nothing is left free. 12. Pressure: forward ( $\Delta\nu_{\text{gas}} = -2$ ). Temperature: backward (exothermic). 13. An inert gas at constant total pressure lowers the partial pressures of the reacting gases, as a pressure drop would: backward. 14. $\partial\ln Q/\partial n_{\ce{N2}} = (-1/0.30 + 2)/n_{\text{tot}} < 0$ : $Q$ falls below $K^\circ$ , forward. 15. No: at constant volume $\partial\ln Q/\partial n_{\ce{N2}} = -1/n_{\ce{N2}} < 0$ whatever the composition: forward. 16. With almost no ammonia, $Q$ is far below $K^\circ$ at the inlet: the gas enters far from equilibrium and reacts forward. 17. The gas spends too short a time on the catalyst for the equilibrium to be reached; a longer contact would cost a larger, more expensive converter for a small gain. 18. $x = 2\xi/(4 - 2\xi) = 0.18$ gives $\xi = 0.305\,\mathrm{mol}$ per mole of $\ce{N2}$ fed: single-pass conversion of hydrogen (and of nitrogen) $0.305$ , 31 %. 19. In steady operation all the fresh feed is eventually converted: $2.00\,\mathrm{mol}$ of ammonia leave per unit time; the converter must pass $1.00/ 0.305 = 3.3\,\mathrm{mol}$ of $\ce{N2}$ per unit time, the rest being recycled. 20. The argon (and methane) that enter with the feed never react and are not removed with the liquid ammonia: without a purge they would accumulate in the loop and lower the partial pressures of the reactants. 21. At $500\,\mathrm{K}$ the iron catalyst is too slow: the equilibrium would be favourable but would never be approached in an industrial time. 22. At $723\,\mathrm{K}$ and $200\,\mathrm{bar}$ , stoichiometric feed, perfect gases: $\boldsymbol{x(\ce{NH3}) \approx 0.25}$ .

Terms defined in this chapter

See all 852 terms in the glossary