Le Chatelier's Principle
When you stand in front of a stalactite in the Cacahuamilpa caves, you see rock growing drop by drop. How can a reversible reaction "decide" to deposit rock right on the ceiling of the cave? The answer lies in a simple principle: when an equilibrium is disturbed, it reacts to counteract the disturbance. At the end of the page we will come back to the cave.
A disturbed equilibrium
In a dynamic equilibrium, the forward and reverse reactions occur at the same rate, and the concentrations stay constant. But that state is not indestructible: if you change the conditions, the system leaves equilibrium and evolves until it reaches a new one.
Think of an open-air market with two lines of people crossing from one side to the other. If more people suddenly arrive on one side, more people cross to the other side until the flow evens out again. The system "compensates" for the excess.
We will always use this convention: shifting to the right means the forward reaction is favored (more products form); to the left, the reverse reaction (more reactants form).
Changes in concentration
If you add a substance, the system consumes part of it: it shifts toward the opposite side from where you added it. If you remove a substance, the system produces more of it: it shifts toward the side where you removed it.
| Disturbance | Shift |
|---|---|
| [reactant] increases | Right (→) |
| [reactant] decreases | Left (←) |
| [product] increases | Left (←) |
| [product] decreases | Right (→) |
Why? Using collision theory: more reactant means more effective collisions for the forward reaction, which speeds up for a moment until the reverse reaction catches up with the new products. Removing a product does the opposite: the reverse reaction loses collisions and the forward reaction gains.
In the Cacahuamilpa caverns (Guerrero), stalactites grow thanks to a natural equilibrium: water dissolves the calcium carbonate in the rock in the presence of…
In the Cacahuamilpa caverns (Guerrero), stalactites grow thanks to a natural equilibrium: water dissolves the calcium carbonate in the rock in the presence of $\text{CO}_2$, $\text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{Ca}(\text{HCO}_3)_2$, and when the drop reaches the inside of the cavern some of the $\text{CO}_2$ escapes into the air. Explain, using Le Chatelier's principle, why the escape of $\text{CO}_2$ causes solid $\text{CaCO}_3$ to be deposited (the stalactite).
Changes in pressure and volume
Pressure only matters when gases are involved. If you compress the container (increase the pressure), the molecules are packed more tightly and the system tries to reduce that pressure: it shifts toward the side with fewer moles of gas. If you increase the volume (the pressure drops), it shifts toward the side with more moles of gas.
Why? The pressure of a gas depends on the number of particles colliding with the walls: fewer moles of gas, fewer collisions, less pressure.
| Disturbance | Shift |
|---|---|
| Pressure increases (V decreases) | Toward the side with fewer moles of gas |
| Pressure decreases (V increases) | Toward the side with more moles of gas |
| Equal moles of gas on both sides | No shift |
In the industrial synthesis of ammonia, $\text{N}2 + 3\ \text{H}2 \rightleftharpoons 2\ \text{NH}3$, predict using Le Chatelier's principle what effect increasing the…
In the industrial synthesis of ammonia, $\text{N}_2 + 3\ \text{H}_2 \rightleftharpoons 2\ \text{NH}_3$, predict using Le Chatelier's principle what effect increasing the pressure in the reactor has on the ammonia yield. Justify your answer by counting the moles of gas on each side of the equation.
Changes in temperature
Temperature is the only disturbance that changes the value of the equilibrium constant $K_c$. To predict the direction of the shift, treat heat as a reactant or a product:
-
Exothermic reaction ($\Delta H < 0$): it releases heat. Write it as $\text{reactants} \rightleftharpoons \text{products} + \text{heat}$.
-
Endothermic reaction ($\Delta H > 0$): it absorbs heat. $\text{reactants} + \text{heat} \rightleftharpoons \text{products}$.
So heating is "adding heat" and cooling is "removing heat," and the same logic applies as with concentration.
| Forward reaction | Heat | Cool |
|---|---|---|
| Exothermic | Left (←) | Right (→) |
| Endothermic | Right (→) | Left (←) |
Why? When heated, the system favors the reaction that absorbs heat and so "cools" itself; when cooled, it favors the reaction that releases heat.
Nitrogen dioxide, which is brown, is in equilibrium with dinitrogen tetroxide, which is colorless: $2\ \text{NO}2 \rightleftharpoons \text{N}2\text{O}4$, with…
Nitrogen dioxide, which is brown, is in equilibrium with dinitrogen tetroxide, which is colorless: $2\ \text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_4$, with $\Delta H < 0$ (exothermic). A sealed tube containing this mixture is placed first in hot water and then in crushed ice. Predict in which bath the gas turns darker and in which it turns paler, and justify your answer with Le Chatelier's principle, treating heat as one more participant in the reaction.
A catalyst does not shift the equilibrium
A catalyst lowers the activation energy of the forward and the reverse reactions equally. Both speed up in the same proportion, so equilibrium is reached faster, but the final concentrations are the same and $K_c$ does not change.
That is why an iron catalyst is used in an ammonia plant: it does not increase the yield, but it allows equilibrium to be reached in minutes instead of years.
Summary of disturbances
| Disturbance | The system shifts... | Does $K_c$ change? |
|---|---|---|
| Add reactant | → (right) | No |
| Add product | ← (left) | No |
| Raise the pressure | Toward fewer moles of gas | No |
| Heat (endothermic) | → | Yes, it increases |
| Heat (exothermic) | ← | Yes, it decreases |
| Catalyst | No shift | No |
Back to the beginning: in the Cacahuamilpa cave, water with $\text{CO}_2$ dissolves rock in $\text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{Ca}(\text{HCO}_3)_2$. When the drop reaches the inside and $\text{CO}_2$ escapes into the air, a reactant decreases: the equilibrium shifts to the left and solid $\text{CaCO}_3$ is deposited. Drop by drop, that small disturbance builds the stalactite.