4.6 Le Chatelier’s Principle
Begin with a reaction at equilibrium. Now do something to it.
- Add/remove some reactant
- Add/remove some product
- Change the temperature
- Change the volume/pressure (for gaseous reactions)
It is no longer at equilibrium. What happens? Which way will the reaction proceed?
This principle is named after French chemist Henry Louis Le Chatelier.
Le Chatelier and Ammonia Synthesis
Ammonia is a highly produced inorganic chemical where 80% of its total production is used in agricultural fertilizers. Le Chatelier was on the cusp of realizing a process for the synthesis of ammonia but was beaten to the punch by Fritz Haber and Carl Bosch, two German scientists, in 1910. The Haber-Bosch process (or Haber process) converts atmospheric nitrogen into ammonia using high temperatures and pressures. This process is still used today to produce ammonia on a wide scale. A staggering 176 million tons of ammonia is produced every year and roughly 50% of the world’s food production relies on these types of fertilizers. The Haber-Bosch process is one of the most important synthesis reactions in modern times. However, the process is extremely energy intensive requiring pressures of 150-300 bar and 350-500 °C. 1.2% of the world’s total energy production is used annually for this process.
\[3\mathrm{H_2} + \mathrm{N_2} \longrightarrow 2\mathrm{NH_3}\]
The production of ammonia and its use in fertilizers allowed for a tremendous growth in agricultural output and gave rise to an explosion in the world’s population.
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Figure 4.1: The world’s population from 1900-2015. Source
Le Chatelier lamented not having fully realized the synthesis of ammonia. Published correspondence to the editor of the Journal of Chemical Education demonstrates this12.
LE CHATELIER AND THE SYNTHESIS OF AMMONIA
To the Editor
Dear Sir:
In the excellent article by Professor Silverman on Henry Le Chatelier in the December issue of this journal, there is one statement to which I must take exception. On page 556 we read, “It is not strange that he [Le Chatelier] should have accomplished the synthesis of ammonia from the elements in 1901, anticipating Fritz Haber, who is usually the only one mentioned in connection with the process.” Now Le Chatelier himself in his last book “De la Methode dans les Sciences Experimentales,” published in 1936, devotes three pages (pp. 73-6) to this synthesis in which he says that he tried to accomplish the direct union of hydrogen and nitrogen under a pressure of 200 atm. at a temperature of 600 ° in the presence of metallic iron. A terrific explosion occurred which nearly killed an assistant. Some time later Le Chatelier found that the explosion was due to the presence of air in the apparatus used. And thus it was left for Haber to succeed where a number of noted French chemists, including Thdnard, Sainte Claire Deville and even Berthelot had failed. At the end of his career Le Chatelier, with a disarming frankness, tells us, “I let the discovery of the ammonia synthesis slip through my hands. It was the greatest blunder of my scientific career. I should have realized this synthesis five years before Haber…..”
In calling attention to this statement I may say that those of your readers who are familiar with French will find a great many interesting observations and reflections in this scientific testament of a great Frenchman.
– H. S. van Klooster
4.6.1 Changing Concentrations
Consider the following reaction at equilibrium
\[\mathrm{Zn}(s) +2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(aq)\]
and its equilibrium expression
\[K = \dfrac{[\mathrm{ZnCl_2}][\mathrm{H_2}]}{[\mathrm{HCl}]^2}\]
Let us add, for example, some HCl to the mixture. The reaction is no longer at equilibrium, however, the reaction will proceed to once again reach equilibrium. Which way will the reaction go?
We must consider the reaction quotient, Q.
\[Q = \dfrac{[\mathrm{ZnCl_2}][\mathrm{H_2}]}{[\mathrm{HCl}]^2}\]
Upon the addition of HCl, the denominator in the expression increases, therefore, Q < K. The reaction will proceed right (consume reactants and produce products) to reach equilibrium.
Let us now add some product to this reaction that is at equilibrium. Perhaps we add a bit of ZnCl2. We notice that the numerator in the reaction quotient is now larger and Q > K. The reaction will therefore proceed left (consume products and produce reactants) to reach equilibrium once again.
Practice
Predict the direction of the given reaction (currently at equilibrium) for each of the following scenarios:
\[\mathrm{Zn}(s) +2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(aq)\]
- Add some HCl
- Add some Zn
- Remove some HCl
- Remove some Zn
- Add some ZnCl2
- Add some H2
- Remove some ZnCl2
- Remove some H2
Solution
\[\mathrm{Zn}(s) +2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(aq)\]
- Add some HCl - right
- Add some Zn - no shift
- Remove some HCl - left
- Remove some Zn - no shift
- Add some ZnCl2 - left
- Add some H2 - left
- Remove some ZnCl2 - right
- Remove some H2 - right
4.6.2 Changing Temperature
Exothermic Reaction
Consider the following reaction at equilibrium.
\[\mathrm{SO_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \rightleftharpoons \mathrm{SO_3}(g) \qquad \Delta H = -98.9~\mathrm{kJ~mol^{-1}}\]
The reaction is exothermic (ΔH < 0) meaning heat is generated (i.e. is a product of the reaction). If we treat heat (Δ) as a product of reaction, the reaction could be written as
\[\mathrm{SO_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \rightleftharpoons \mathrm{SO_3}(g) + \Delta\] Now we apply the same principles as we did with changing concentrations!
If the temperature of the reaction mixture was raised (i.e. heat is added), the reaction will shift left to consume the heat and reach equilibrium.
If the temperature of the reaction mixture was lowered (i.e. heat is removed), the reaction will shift right to produce heat and reach equilibrium.
Endothermic Reaction
Consider the following endothermic reaction at equilibrium.
\[\mathrm{H_2O}(g) \rightleftharpoons \mathrm{H_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \qquad \Delta H = 248.1~\mathrm{kJ~mol^{-1}}\]
We can write the reaction as follows (indicating heat, Δ, as a reactant)
\[\Delta + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{H_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g)\]
If the temperature was raised, the reaction will proceed to the right. If the temperature was lowered, the reaction will proceed to the left.
4.6.3 Changing Volume/Pressure
Recall from the ideal gas law that pressure and volume are inversely proportional.
Consider the following gaseous reaction at equilibrium
\[4\mathrm{NH}_{3}(g) + 7\mathrm{O}_{2}(g) \rightleftharpoons 4\mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\]
If the pressure of the reaction mixture was increased (perhaps by reducing the volume of the vessel that the gas was contained in), which way will the reaction proceed to reach equilibrium?
What exhibits more pressure in a unit volume, a gas containing 11 moles of particles or a gas containing 10 moles of particles? Clearly it is the gas containing the larger number of particles!
Applying that concept here leads us to conclude that the reaction will proceed right since there are only 10 moles of gaseous products for the reaction vs. the 11 moles of gaseous reactants!
Decreasing the pressure (increasing the volume) of the reaction will cause the reaction to shift to the left!