Le Chatelier's principle

One way compares two experiments with different protocols, 'no-feedback, driver only' and 'post-feedback', for the imposition of the change ${\displaystyle \Delta L.}$ The 'no-feedback, driver only' experimental protocol prevents 'feedback' by enforcing that ${\displaystyle \Delta X=0}$, and it observes the 'no-feedback' response ${\displaystyle \Delta M.}$ The 'post-feedback' experimental protocol allows the 'moderating' variable response ${\displaystyle \delta X}$, with the response of interest ${\displaystyle \delta M}$. Provided that the observed response is indeed that ${\displaystyle \delta X\neq 0,}$ then the principle states that

${\displaystyle |\delta M|<|\Delta M|}$.

In other words, change in the 'moderating' state variable ${\displaystyle X}$ moderates the effect of the imposed change in ${\displaystyle L}$ on the responding conjugate variable ${\displaystyle M}$.[3][4]

The other way also compares two experiments with different protocols, 'post-feedback' and 'no-driver, feedback only', for the imposition of the change ${\displaystyle \Delta L.}$ The response of prime interest, ${\displaystyle \delta M}$, is manifest in the 'post-feedback' protocol, when the response ${\displaystyle \Delta X}$ of the 'moderating' or 'feedback' variable is measured. This time, the 'no-driver, feedback only' experimental protocol enforces that ${\displaystyle \Delta L=0,}$ with the driving variable ${\displaystyle L}$ held fixed; it also imposes a change ${\displaystyle \Delta X}$ in the 'moderating' or 'feedback' response variable, and measures the change ${\displaystyle \delta ^{\prime }M.}$ The principle then states that if the 'moderating' response is indeed that ${\displaystyle \Delta X\neq 0,}$ then the signs of ${\displaystyle \delta M}$ and ${\displaystyle \delta ^{\prime }M}$ are opposite.

Again, in other words, change in the 'feedback' state variable ${\displaystyle X}$ opposes the effect of the imposed change in ${\displaystyle L}$ on the responding conjugate variable ${\displaystyle M.}$[5]

Changing the concentration of a chemical will shift the equilibrium to the side that would counter that change in concentration. The chemical system will attempt to partly oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent, and yield of products will be altered corresponding to the impact on the system.

This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.

CO + 2 H₂ ⇌ CH₃OH

Suppose we were to increase the concentration of CO in the system. Using Le Chatelier's principle, we can predict that the concentration of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to "fill the gap" and favor the side where the species was reduced. This observation is supported by the collision theory. As the concentration of CO is increased, the frequency of successful collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if the desired product is not thermodynamically favored, the end-product can be obtained if it is continuously removed from the solution.

The effect of a change in concentration is often exploited synthetically for condensation reactions (i.e., reactions that extrude water) that are equilibrium processes (e.g., formation of an ester from carboxylic acid and alcohol or an imine from an amine and aldehyde). This can be achieved by physically sequestering water, by adding desiccants like anhydrous magnesium sulfate or molecular sieves, or by continuous removal of water by distillation, often facilitated by a Dean-Stark apparatus.

The reversible reaction N₂O₄(g) ⇌ 2NO₂(g) is endothermic, so the equilibrium position can be shifted by changing the temperature.
When heat is added and the temperature increases, the reaction shifts to the right and the flask turns reddish brown due to an increase in NO₂. This demonstrates Le Chatelier's principle: the equilibrium shifts in the direction that consumes energy.
When heat is removed and the temperature decreases, the reaction shifts to the left and the flask turns colorless due to an increase in N₂O₄: again, according to Le Chatelier's principle.

The effect of changing the temperature in the equilibrium can be made clear by 1) incorporating heat as either a reactant or a product, and 2) assuming that an increase in temperature increases the heat content of a system. When the reaction is exothermic (ΔH is negative and energy is released), heat is included as a product, and when the reaction is endothermic (ΔH is positive and energy is consumed), heat is included as a reactant. Hence, whether increasing or decreasing the temperature would favor the forward or the reverse reaction can be determined by applying the same principle as with concentration changes.

Take, for example, the reversible reaction of nitrogen gas with hydrogen gas to form ammonia:

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)    ΔH = −92 kJ mol⁻¹

Because this reaction is exothermic, it produces heat:

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) + heat

If the temperature were increased, the heat content of the system would increase, so the system would consume some of that heat by shifting the equilibrium to the left, thereby producing less ammonia. More ammonia would be produced if the reaction were run at a lower temperature, but a lower temperature also lowers the rate of the process, so, in practice (the Haber process) the temperature is set at a compromise value that allows ammonia to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable.

In exothermic reactions, an increase in temperature decreases the equilibrium constant, K, whereas in endothermic reactions, an increase in temperature increases K.

Le Chatelier's principle applied to changes in concentration or pressure can be understood by giving K a constant value. The effect of temperature on equilibria, however, involves a change in the equilibrium constant. The dependence of K on temperature is determined by the sign of ΔH. The theoretical basis of this dependence is given by the Van 't Hoff equation.

The equilibrium concentrations of the products and reactants do not directly depend on the total pressure of the system. They may depend on the partial pressure of the products and reactants, but if the number of moles of gaseous reactants is equal to the number of moles of gaseous products, pressure has no effect on equilibrium.

Changing total pressure by adding an inert gas at constant volume does not affect the equilibrium concentrations (see Effect of adding an inert gas below).

Changing total pressure by changing the volume of the system changes the partial pressures of the products and reactants and can affect the equilibrium concentrations (see §Effect of change in volume below).

Changing the volume of the system changes the partial pressures of the products and reactants and can affect the equilibrium concentrations. With a pressure increase due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable[7] and with a pressure decrease due to an increase in volume, the side with more moles is more favorable. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical equation.

Considering the reaction of nitrogen gas with hydrogen gas to form ammonia:

N₂ + 3 H₂4 moles ⇌ 2 NH₃2 moles    ΔH = −92kJ mol⁻¹

Note the number of moles of gas on the left-hand side and the number of moles of gas on the right-hand side. When the volume of the system is changed, the partial pressures of the gases change. If we were to decrease pressure by increasing volume, the equilibrium of the above reaction will shift to the left, because the reactant side has a greater number of moles than does the product side. The system tries to counteract the decrease in partial pressure of gas molecules by shifting to the side that exerts greater pressure. Similarly, if we were to increase pressure by decreasing volume, the equilibrium shifts to the right, counteracting the pressure increase by shifting to the side with fewer moles of gas that exert less pressure. If the volume is increased because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.

An inert gas (or noble gas), such as helium, is one that does not react with other elements or compounds. Adding an inert gas into a gas-phase equilibrium at constant volume does not result in a shift.[7] This is because the addition of a non-reactive gas does not change the equilibrium equation, as the inert gas appears on both sides of the chemical reaction equation. For example, if A and B react to form C and D, but X does not participate in the reaction: ${\displaystyle {\ce {{\mathit {a}}A{}+{\mathit {b}}B{}+{\mathit {x}}X<=>{\mathit {c}}C{}+{\mathit {d}}D{}+{\mathit {x}}X}}}$. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium. If, however, the volume is allowed to increase in the process, the partial pressures of all gases would be decreased resulting in a shift towards the side with the greater number of moles of gas. The shift will never occur on the side with fewer moles of gas. It is also known as Le Chatelier's postulate.

A catalyst increases the rate of a reaction without being consumed in the reaction. The use of a catalyst does not affect the position and composition of the equilibrium of a reaction, because both the forward and backward reactions are sped up by the same factor.

For example, consider the Haber process for the synthesis of ammonia (NH₃):

N₂ + 3 H₂ ⇌ 2 NH₃

In the above reaction, iron (Fe) and molybdenum (Mo) will function as catalysts if present. They will accelerate any reactions, but they do not affect the state of the equilibrium.