6.5: Le Châtelier’s Principle

Last updated
Save as PDF

Page ID: 5711

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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

\[\mathrm{CH_3COOH}(aq)+\mathrm{H_2O}(l)\rightleftharpoons\mathrm{H_3O^+}(aq)+\mathrm{CH_3COO^-}(aq)\]

has an equilibrium constant of

\[K_\textrm a=\mathrm{\dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}}=1.75\times10^{-5}\tag{6.26}\]

Because equation 6.26 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 must have the same value of K_a.

If we add sodium acetate to a solution of acetic acid, the concentration of CH₃COO^– increases, suggesting an apparent increase in the value of K_a. Because K_a must remain constant, the concentration of all three species in equation 6.26 must change to restore K_a to its original value. In this case, a partial reaction of CH₃COO^– and H₃O⁺ decreases their concentrations, producing additional CH₃COOH and reestablishing the equilibrium.

The observation that a system at equilibrium responds to an external stress by reequilibrating in a manner that diminishes the stress, is formalized as Le Châtelier’s principle. One of the most common stresses to a system at equilibrium is to change the concentration of a reactant or product. We already have seen, in the case of adding sodium acetate to acetic acid, that 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 are directly applying the stress to the system. It is also possible to indirectly apply a concentration stress. Consider, for example, the solubility of AgCl.

\[\mathrm{AgCl}(s)\rightleftharpoons\mathrm{Ag^+}(aq)+\mathrm{Cl^-}(aq)\tag{6.27}\]

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

\[\mathrm{Ag^+}(aq)+\mathrm{2NH_3}(aq)\rightleftharpoons A\mathrm{g(NH_3)_2^+}(aq)\tag{6.28}\]

Adding ammonia decreases the concentration of Ag⁺ as the Ag(NH₃)₂⁺ complex forms. In turn, decreasing the concentration of Ag⁺ increases the solubility of AgCl as reaction 6.27 reestablishes its equilibrium position. Adding together reaction 6.27 and reaction 6.28 clarifies the effect of ammonia on the solubility of AgCl, by showing ammonia as a reactant.

\[\mathrm{AgCl}(s)+\mathrm{2NH_3}(aq)\rightleftharpoons A\mathrm{g(NH_3)_2^+}(aq)+\mathrm{Cl^-}(aq)\tag{6.29}\]

Note

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, K_sp, of course, remains unchanged.

Example 6.6

What happens to the solubility of AgCl if we add HNO₃ to the equilibrium solution defined by reaction 6.29?

Solution

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

\[\mathrm{HNO_3}(aq)+\mathrm{NH_3}(aq)\rightleftharpoons \mathrm{NH_4^+}(aq)+\textrm{NO}_3^-(aq)\]

Adding nitric acid lowers the concentration of ammonia. Decreasing ammonia’s concentration causes reaction 6.29 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.

The relationship between pressure and concentration can be deduced using the ideal gas law. Starting with PV = nRT, we solve for the molar concentration

\[\textrm{molar concentration}=\dfrac{n}{V}=\dfrac{P}{RT}\]

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 or concentrating the solution, then the concentrations of all reactants and products either decrease or increase. The effect of simultaneously changing the concentrations of all reactants and products is not as intuitively obvious as when changing 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 6.28). The equilibrium constant for this reaction is

\[\mathrm{\beta_2=\dfrac{[Ag(NH_3)_2^+]_{eq}}{[Ag^+]_{eq}[NH_3]_{eq}^2}}\tag{6.30}\]

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 6.30 is cut in half. The reaction quotient, Q, becomes

\[Q=\mathrm{\dfrac{0.5[Ag(NH_3)_2^+]_{eq}}{0.5[Ag^+]_{eq}(0.5)^2[NH_3]_{eq}^2}=\dfrac{0.5}{(0.5)^3}\times\dfrac{[Ag(NH_3)_2^+]_{eq}}{[Ag^+]_{eq}[NH_3]_{eq}^2}=4\beta_2}\]

Because Q is greater than β₂, equilibrium is reestablished by shifting the reaction to the left, decreasing the concentration of Ag(NH₃)₂⁺. Note that the new equilibrium position lies toward the side of the equilibrium reaction having 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 Ag(NH₃)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 producing the greatest number of particles, and decreasing volume always favors the direction producing 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.

Contributors

David Harvey (DePauw University)