Moving an equilibrium

A chemical plant is a machine for moving an equilibrium, and the three levers available to it push in directions that do not agree.

The previous lesson fixed the position of equilibrium with $Δ_{r} G^{ominus} = - R T \ln K$ , and ended with a hint: diluting dinitrogen tetroxide nearly doubled its dissociation while $K$ did not move at all. That is the pattern to generalise. Some changes move the composition and leave the constant alone; only one moves the constant.

Le Chatelier's principle, and why it needs care

Henri Le Chatelier stated the rule in 1884: a system at equilibrium, when disturbed, shifts so as to partially oppose the disturbance. Add a reactant and it is consumed; heat an exothermic reaction and it runs backward; compress a mixture and it moves toward fewer molecules.

The principle is genuinely useful and genuinely unreliable, and it is worth knowing which parts of it are which. It gives a direction and never a magnitude, so it cannot tell you whether a shift is worth engineering. It is stated in terms of "the disturbance", which is not well defined when several things change at once. And it has honest counterexamples: add nitrogen to an ammonia equilibrium that is already very rich in nitrogen and the mole fraction of ammonia goes down, not up, because the dilution of the hydrogen outweighs the mass action of the added nitrogen.

Everything the principle gets right follows from two facts already established, so the safe procedure is to use those instead. First, $K$ depends on temperature and on nothing else. Second, the system moves so as to return $Q$ to $K$ . Every case below is one of those two.

Adding a reagent: $Q$ moves and $K$ does not

Pour more of a reactant into a mixture at equilibrium and its activity rises, so $Q$ falls below $K$ , so $Δ_{r} G$ becomes negative and the reaction runs forward until $Q$ has climbed back. The new composition is not the old one plus the addition; it is a fresh equilibrium calculation with the disturbed mixture as the starting point.

Example. The mixture $+ \to 2 HI$ at $698$ K, with $K = 54.3$ , sits at $[] = [] = 0.213$ and $[HI] = 1.573$ mol dm⁻³. Now add $1.000$ mol dm⁻³ of hydrogen. Find the new equilibrium.

Immediately after the addition, $Q = {1.573}^{2} / (1.213 \times 0.213) = 9.58$ , well below $54.3$ , so the reaction goes forward. Let $x$ be the amount of iodine consumed:

\frac{(1.573 + 2 x)^{2}}{(1.213 - x) (0.213 - x)} = 54.3

Solving gives $x = 0.152$ mol dm⁻³, so the new concentrations are $[] = 1.061$ , $[] = 0.061$ and $[HI] = 1.877$ mol dm⁻³. The added hydrogen converted most of the remaining iodine, taking it from $0.213$ to $0.061$ : this is the standard industrial trick of driving a reaction to completion in the expensive reagent by flooding it with the cheap one.

Now you. Without solving anything, say what happens to that same equilibrium if hydrogen iodide is continuously removed as it forms, and why a plant might do that.

Answer

Removing the product lowers $Q$ below $K$ continuously, so the reaction never reaches equilibrium and keeps running forward. In principle it can be driven to any conversion, limited only by the rate. This is why ammonia is condensed out of the recycle loop and why esterifications are run with the water distilled off.

Squeezing a gas mixture

Compression is the interesting case, because the amounts of substance do not change at all. What changes is the activity of each gas, which is its partial pressure, and the activities appear in $Q$ with different powers on the two sides.

Write the constant in terms of mole fractions and total pressure. Since $p_{i} = x_{i} P$ , each activity carries a factor of $P$ , and collecting them gives

K = K_{x} {(\frac{P}{p^{ominus}})}^{Δ n}

where $K_{x}$ is the same quotient in mole fractions and $Δ n$ is the change in moles of gas. So if $Δ n$ is positive, raising $P$ must lower $K_{x}$ , and the mixture shifts toward the reactants; if $Δ n$ is zero, pressure does nothing whatever. That is Le Chatelier's pressure rule, derived, with the exception it always forgets attached.

Example. For $\to 2 N$ at $298$ K, $K = 0.150$ . Find the degree of dissociation $α$ at total pressures of $1.00$ bar and $10.0$ bar.

Starting from one mole of the dimer, $α$ moles dissociate to give $2 α$ of the monomer and $1 - α$ of the dimer, totalling $1 + α$ . The mole fractions are $2 α / (1 + α)$ and $(1 - α) / (1 + α)$ , so

K = \frac{4 α^{2}}{1 - α^{2}} \cdot \frac{P}{p^{ominus}}

using the difference of two squares to tidy the denominator. At $P = 1.00$ bar, $4 α^{2} / (1 - α^{2}) = 0.150$ , giving $α^{2} = 0.150 / 4.150$ and $α = 0.190$ . At $P = 10.0$ bar the left side must equal $0.0150$ , giving $α = 0.061$ . Compressing tenfold cuts the dissociation from $19$ per cent to $6$ .

Now you. Find $α$ for the same equilibrium at $P = 0.100$ bar.

Answer

Now $4 α^{2} / (1 - α^{2}) = 0.150 / 0.100 = 1.50$ , so $α^{2} = 1.50 / 5.50 = 0.273$ and $α = 0.522$ . At a tenth of a bar the gas is more than half dissociated, and the tube looks distinctly brown.

One trap is worth stating plainly. Pumping an inert gas such as argon into a vessel of fixed volume raises the total pressure and shifts nothing at all, because the partial pressures of the reacting species are unchanged and $Q$ is built from those. Adding argon while holding the total pressure fixed, so the vessel expands, is a genuine dilution and shifts the equilibrium exactly as expansion does. The total pressure is not what the system responds to.

Temperature is the only lever that moves $K$

Combine the two expressions for the standard free energy change, $Δ_{r} G^{ominus} = Δ_{r} H^{ominus} - T Δ_{r} S^{ominus}$ and $Δ_{r} G^{ominus} = - R T \ln K$ , and divide through by $- R T$ :

\ln K = - \frac{Δ_{r} H^{ominus}}{R T} + \frac{Δ_{r} S^{ominus}}{R}

Everything about the temperature dependence is in that one line. Treating $Δ_{r} H^{ominus}$ and $Δ_{r} S^{ominus}$ as constant over the range of interest and writing it at two temperatures, the entropy term cancels and leaves the van 't Hoff equation:

\ln \frac{K_{2}}{K_{1}} = - \frac{Δ_{r} H^{ominus}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})

Read the sign. For an endothermic reaction $Δ H$ is positive, so raising $T$ makes $1 / T$ smaller and the bracket negative, and $K$ increases. For an exothermic one $K$ falls. Le Chatelier's temperature rule, with the additional information that the size of the shift is set by $Δ H$ and by nothing else: a reaction with a small enthalpy change is nearly indifferent to temperature however favourable it is.

Example. $\to 2 N$ has $K = 0.150$ at $298.15$ K and $Δ_{r} H^{ominus} = 2 (33.2) - 9.16 = + 57.2$ kJ mol⁻¹. Find $K$ at $350$ K.

\ln \frac{K_{2}}{K_{1}} = - \frac{57240}{8.314} (\frac{1}{350} - \frac{1}{298.15}) = - 6886 \times (- 4.969 \times 10^{- 4}) = 3.42

So $K_{2} = 0.150 \times e^{3.42} = 4.59$ . A rise of $52$ K multiplies the constant by thirty, which is why the sealed tube of nitrogen dioxide used in lecture demonstrations goes from pale to deep brown when dropped in hot water.

Now you. Ammonia synthesis has $K = 5.0 \times 10^{5}$ at $298.15$ K and $Δ_{r} H^{ominus} = - 91.8$ kJ mol⁻¹. Estimate $K$ at $700$ K.

Answer

$\ln (K_{2} / K_{1}) = (91800 / 8.314) (1 / 700 - 1 / 298.15) = 11042 \times (- 1.925 \times 10^{- 3}) = - 21.26$ , so $K_{2} = 5.0 \times 10^{5} \times e^{- 21.26} = 2.9 \times 10^{- 4}$ . Nine orders of magnitude lost, which is the central problem of the ammonia industry.

A thermometer that measures an enthalpy

The relation runs backward as well, and this is how a great many reaction enthalpies are actually obtained. Measure $K$ at several temperatures, plot $\ln K$ against $1 / T$ , and the result is a straight line of slope $- Δ_{r} H^{ominus} / R$ and intercept $Δ_{r} S^{ominus} / R$ . Both quantities come out of composition measurements, with no calorimeter anywhere in the experiment.

Two points suffice in principle. Using the pair above, $\ln (4.59 / 0.150) = 3.42$ over an interval of $1 / 350 - 1 / 298.15 = - 4.969 \times 10^{- 4}$ K⁻¹ gives a slope of $- 6886$ K, so $Δ_{r} H^{ominus} = 6886 \times 8.314 = 57.2$ kJ mol⁻¹, recovering what we put in.

In practice a real plot uses many temperatures and its curvature is informative. A van 't Hoff plot is straight only while $Δ_{r} H^{ominus}$ is constant, and Kirchhoff's law from the second lesson says it is not, so any measurement over a few hundred kelvin bends slightly. The curvature is a measurement of $Δ_{r} C_{p}$ , which is otherwise awkward to get. This is the routine way biochemists obtain binding enthalpies for a protein and a ligand, where the reaction is too dilute to warm anything measurably.

Ammonia: three answers that disagree

Now put the three levers on one reaction. Nitrogen and hydrogen make ammonia with $Δ_{r} H^{ominus} = - 91.8$ kJ mol⁻¹ and $Δ n = - 2$ . So the equilibrium wants a low temperature, since the reaction is exothermic, and a high pressure, since it consumes gas. Taking $\ln K = 11042 / T - 23.91$ from the standard values and solving for the equilibrium mole fraction of ammonia from a stoichiometric feed gives:

Temperature	at 10 bar	at 200 bar
$473$ K (200 °C)	$53$ %	$87$ %
$573$ K (300 °C)	$20$ %	$68$ %
$673$ K (400 °C)	$7$ %	$46$ %
$773$ K (500 °C)	$2$ %	$28$ %

These are computed from thermodynamic data alone and agree with the measured equilibrium yields to within two or three percentage points across the whole table, which is a reasonable check on everything in the last three lessons.

Read the first column and the answer looks obvious: run it cold. At $473$ K and $10$ bar you keep half the feed as ammonia with a cheap vessel. Fritz Haber's problem, from 1905 onward, was that at $473$ K the reaction does not happen. Nitrogen's triple bond is $945$ kJ mol⁻¹ and nothing at $200$ °C breaks it at a usable rate. The equilibrium is inviting and inaccessible.

The industrial solution is a compromise on all three axes at once. Run hot enough for the iron catalyst to work, $400$ to $450$ °C, accept that this costs most of the equilibrium yield, and buy some of it back with pressure, $150$ to $250$ bar, which is expensive in steel and in compression work. Then, because even the equilibrium yield is not reached in one pass through the converter, condense the ammonia out of the product stream and recycle the unreacted gas, which is the trick from the second section of this lesson used to push the overall conversion above $95$ per cent. Carl Bosch's engineering of that loop between 1909 and 1913, and Alwin Mittasch's search across some twenty thousand catalyst samples, are what turned Haber's bench result into a process that now fixes more nitrogen than the entire biosphere.

What a yield chart cannot tell you

Every number in that table is a statement about where the mixture ends up, and not one of them is a statement about when. The $473$ K column is the proof: thermodynamics says a mixture of nitrogen and hydrogen at $200$ °C and $10$ bar should be half ammonia, and the experiment says that after a week it is not measurably anything.

Nothing in the last three lessons can be repaired to fix this, because free energy is a function of the end states and knows nothing about the route between them. The rate depends on the route: on which bonds have to break first, on how much energy a collision needs, on whether a surface is available to help. It is a separate theory with separate measurements, and it starts in the next lesson with the plainest possible question, how the composition of a reacting mixture changes with time.

Le Chatelier's principle, and why it needs care

Adding a reagent: Q moves and K does not

Squeezing a gas mixture

Temperature is the only lever that moves K

A thermometer that measures an enthalpy

Ammonia: three answers that disagree

What a yield chart cannot tell you

Adding a reagent: $Q$ moves and $K$ does not

Temperature is the only lever that moves $K$