# 6.5: Le Châtelier’s Principle

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At a temperature of 25oC, acetic acid’s dissociation reaction

$\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber$

has an equilibrium constant of

$K_{\mathrm{a}}=\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=1.75 \times 10^{-5} \label{6.1}$

Because equation \ref{6.1} has three variables—[CH3COOH], [CH3COO], and [H3O+]—it does not have a unique mathematical solution. Nevertheless, although two solutions of acetic acid may have different values for [CH3COOH], [CH3COO], and [H3O+], each solution has the same value of Ka.

If we add sodium acetate to a solution of acetic acid, the concentration of CH3COO increases, which suggests there is an increase in the value of Ka; however, because Ka must remain constant, the concentration of all three species in equation \ref{6.1} must change to restore Ka to its original value. In this case, a partial reaction of CH3COOand H3O+ decreases their concentrations, increases the concentration of CH3COOH, and reestablishes the equilibrium.

The observation that a system at equilibrium responds to an external action by reequilibrating itself in a manner that diminishes that action, is formalized as Le Châtelier’s principle. One common action is to change the concentration of a reactant or product for a system at equilibrium. As noted above for a solution of acetic acid, if we add a product to a reaction at equilibrium the system responds by converting some of the products into reactants. Adding a reactant has the opposite effect, resulting in the conversion of reactants to products.

When we add sodium acetate to a solution of acetic acid, we directly apply the action to the system. It is also possible to apply a change concentration indirectly. Consider, for example, the solubility of AgCl.

$\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \label{6.2}$

The effect on the solubility of AgCl of adding AgNO3 is obvious, but what is the effect if we add a ligand that forms a stable, soluble complex with Ag+? Ammonia, for example, reacts with Ag+ as shown here

$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \label{6.3}$

Adding ammonia decreases the concentration of Ag+ as the $$\text{Ag(NH}_3)_2^+$$ complex forms. In turn, a decrease in the concentration of Ag+ increases the solubility of AgCl as reaction \ref{6.2} reestablishes its equilibrium position. Adding together reaction \ref{6.2} and reaction \ref{6.3} clarifies the effect of ammonia on the solubility of AgCl, by showing ammonia as a reactant.

$\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)+\mathrm{Cl}^{-}(a q) \label{6.4}$

So what is the effect on the solubility of AgCl of adding AgNO3? Adding AgNO3 increases the concentration of Ag+ in solution. To reestablish equilibrium, some of the Ag+ and Cl react to form additional AgCl; thus, the solubility of AgCl decreases. The solubility product, Ksp, of course, remains unchanged.

Example $$\PageIndex{1}$$

What happens to the solubility of AgCl if we add HNO3 to the equilibrium solution defined by reaction \ref{6.4}?

Solution

Nitric acid is a strong acid, which reacts with ammonia as shown here

$\mathrm{HNO}_{3}(a q)+\mathrm{NH}_{3}(a q)\rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \nonumber$

Adding nitric acid lowers the concentration of ammonia. Decreasing ammonia’s concentration causes reaction \ref{6.4} to move from products to reactants, decreasing the solubility of AgCl.

Increasing or decreasing the partial pressure of a gas is the same as increasing or decreasing its concentration. Because the concentration of a gas depends on its partial pressure, and not on the total pressure of the system, adding or removing an inert gas has no effect on a reaction’s equilibrium position.

We can use the ideal gas law to deduce the relationship between pressure and concentration. Starting with PV = nRT, we solve for the molar concentration

$M=\frac{n}{V}=\frac{P}{R T} \nonumber$

Of course, this assumes that the gas is behaving ideally, which usually is a reasonable assumption under normal laboratory conditions.

Most reactions involve reactants and products dispersed in a solvent. If we change the amount of solvent by diluting the solution, then the concentrations of all reactants and products must increase; conversely, if we allow the solvent to evaporate partially, then the concentration of the solutes must increase. The effect of simultaneously changing the concentrations of all reactants and products is not intuitively as obvious as when we change the concentration of a single reactant or product. As an example, let’s consider how diluting a solution affects the equilibrium position for the formation of the aqueous silver‐amine complex (reaction \ref{6.3}). The equilibrium constant for this reaction is

$\beta_{2}=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]_{\mathrm{eq}}}{\left[\mathrm{Ag}^{+}\right]_{\mathrm{eq}}\left[\mathrm{NH}_{3}\right]_{\mathrm{eq}}^{2}} \label{6.5}$

where we include the subscript “eq” for clarification. If we dilute a portion of this solution with an equal volume of water, each of the concentration terms in equation \ref{6.5} is cut in half. The reaction quotient, Qr, becomes

$Q_r=\frac{0.5\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]_{\mathrm{eq}}}{0.5\left[\mathrm{Ag}^{+}\right]_{\mathrm{eq}}(0.5)^{2}\left[\mathrm{NH}_{3}\right]_{\mathrm{eq}}^{2}}=\frac{0.5}{(0.5)^{3}} \times \frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]_{\mathrm{eq}}}{\left[\mathrm{Ag}^{+}\right]_{\mathrm{eq}}\left[\mathrm{NH}_{3}\right]_{\mathrm{eq}}^{2}}=4 \beta_{2} \label{6.6}$

Because Qr is greater than $$\beta_2$$, equilibrium is reestablished by shifting the reaction to the left, decreasing the concentration of $$\text{Ag(NH}_3)_2^+$$. Note that the new equilibrium position lies toward the side of the equilibrium reaction that has the greatest number of solute particles (one Ag+ ion and two molecules of NH3 versus a single metal‐ligand complex). If we concentrate the solution of $$\text{Ag(NH}_3)_2^+$$ by evaporating some of the solvent, equilibrium is reestablished in the opposite direction. This is a general conclusion that we can apply to any reaction. Increasing volume always favors the direction that produces the greatest number of particles, and decreasing volume always favors the direction that produces the fewest particles. If the number of particles is the same on both sides of the reaction, then the equilibrium position is unaffected by a change in volume.

This page titled 6.5: Le Châtelier’s Principle is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.