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 productionis used in agricultural fertilizers.Le Chatelier was on the cusp of realizing a process for thesynthesis of ammonia butwas beaten to the punch by Fritz Haberand Carl Bosch, two German scientists,in 1910.The Haber-Bosch process(or Haber process) converts atmospheric nitrogen into ammonia using high temperaturesand pressures. This process is still used today to produce ammonia on a wide scale.A staggering 176 million tonsof ammonia is produced every year and roughly 50% of the world’s food production relieson these types of fertilizers.The Haber-Bosch process is one of the most importantsynthesis reactions in modern times.However, the process is extremely energy intensiverequiring 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 tremendousgrowth in agricultural output and gave rise to an explosion in the world’spopulation.
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 Educationdemonstrates this12.
LE CHATELIER AND THE SYNTHESIS OF AMMONIA
To the Editor
Dear Sir:
In the excellent article by Professor Silverman onHenry Le Chatelier in the December issue of thisjournal, there is one statement to which I must takeexception. On page 556 we read, “It is not strangethat he [Le Chatelier] should have accomplished thesynthesis of ammonia from the elements in 1901, anticipating Fritz Haber,who is usually the only one mentioned in connection with the process.” Now LeChatelier himself in his last book “De la Methode dansles Sciences Experimentales,” published in 1936, devotes three pages(pp.73-6) to this synthesis in whichhe says that he tried to accomplish the direct union ofhydrogen and nitrogen under a pressure of 200 atm. at atemperature of 600 ° in the presence of metallic iron. Aterrific explosion occurred which nearly killed an assistant.Some time later Le Chatelier found that the explosion was due to the presenceof air in the apparatusused. And thus it was left for Haber to succeed wherea 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 disarmingfrankness, tells us, “I let the discovery of the ammoniasynthesis slip through my hands. It was the greatestblunder of my scientific career. I should have realizedthis synthesis five years before Haber…..”
In calling attention to this statement I may say thatthose of your readers who are familiar with French willfind a great many interesting observations and reflections in this scientifictestament 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 reachequilibrium.
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 quotientis now larger and Q > K. The reaction will therefore proceed left (consume products andproduce 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 aproduct 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 temperaturewas 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 volumeof the vessel that the gas was contained in), which way will the reaction proceedto reach equilibrium?
What exhibits more pressure in a unit volume, a gas containing 11 moles of particlesor a gas containing 10 moles of particles? Clearly it is the gas containing the larger numberof particles!
Applying that concept here leads us to conclude that the reaction will proceed rightsince there are only 10 moles of gaseous products for the reaction vs.the 11 molesof gaseous reactants!
Decreasing the pressure (increasing the volume) of the reaction will cause thereaction to shift to the left!
4.6.4 Adding a Catalyst
A catalyst will not affect the equilibrium of a reaction.
References
(12)
Silverman, A. Le Chatelier and the Synthesis of Ammonia. J. Chem. Ed. 1938, 15 (6), 289. https://doi.org/10.1021/ed015p289.3.
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