Le Châtelier’s Principle

Derived from Shifting Equilibria: Le Châtelier’s Principle by OpenStax

Page by: OpenStax Chemistry: Atom First

Summary

Reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient (Q) is equal to the equilibrium constant (K). We next address what happens when a system at equilibrium is disturbed so that Q is no longer equal to K. If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of Q will no longer equal the value of K. To re-establish equilibrium, the system will either shift toward the products (if Q < K) or the reactants (if Q > K) until Q returns to the same value as K.

This process is described by Le Châtelier’s principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in Q; the reaction will shift to re-establish Q = K.

Predicting the Direction of a Reversible Reaction

Le Châtelier’s principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of Q and K for the system to predict the changes.

Effect of Change in Concentration on Equilibrium

A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium.

The stress on the system in Figure 1 is the reduction of the equilibrium concentration of SCN (lowering the concentration of one of the reactants would cause Q to be larger than K). As a consequence, Le Châtelier’s principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN part way back to its original concentration and increasing the concentration of Fe3+ above its initial equilibrium concentration.

Three capped test tubes held vertically in clamps are shown in pictures labeled, “a,” “b,” and “c.” The test tube in picture a is half filled with a clear, orange liquid. The test tube in picture b is half filled with a dark, burgundy liquid. The test tube in picture c is half filled with a slightly cloudy, orange liquid.
(a) The test tube contains 0.1 M Fe3+. (b) Thiocyanate ion has been added to solution in (a), forming the red Fe(SCN)2+ ion, <span id="MathJax-Element-67-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 13px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="Fe3+(aq)+SCN(aq)Fe(SCN)2+(aq).Fe3+(aq)+SCN(aq)Fe(SCN)2+(aq).“>Fe3+(aqSCN(aq⇌ Fe(SCN)2+(aq)(c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN as the white solid AgSCN, <span id="MathJax-Element-68-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 13px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="Ag+(aq)+SCN(aq)AgSCN(s).Ag+(aq)+SCN(aq)AgSCN(s).“>Ag+(aqSCN(aq⇌ AgSCN(s). The decrease in the SCN concentration shifts the first equilibrium in the solution to the left, decreasing the concentration (and lightening color) of the Fe(SCN)2+ (credit: modification of work by Mark Ott).

The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:

<span id="MathJax-Element-69-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="H2(g)+I2(g)2HI(g)Kc=50.0at400°CH2(g)+I2(g)2HI(g)Kc=50.0at400°C“>H2(gI2(g⇌ 2HI(gKc=50.0 at 400°C

The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with [H2] = [I2] = 0.221 and [HI] = 1.563 M is at equilibrium; for this mixture, Qc = Kc = 50.0. If H2 is introduced into the system so quickly that its concentration doubles before it begins to react (new [H2] = 0.442 M), the reaction will shift so that a new equilibrium is reached, at which [H2] = 0.374 M, [I2] = 0.153 M and [HI] = 1.692 M. <span id="MathJax-Element-70-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="Qc=[HI]2[H2][I2]=(1.692)2(0.374)(0.153)=50.0=KcQc=[HI]2[H2][I2]=(1.692)2(0.374)(0.153)=50.0=Kc“>

We have stressed this system by introducing additional H2. The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess H2, reducing the amount of uncombined I2 and forming additional HI.

Effect of Change in Temperature on Equilibrium

Changing concentration or pressure disrupts an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect. A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Châtelier’s principle.

When hydrogen reacts with gaseous iodine, heat is evolved.

<span id="MathJax-Element-73-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="H2(g)+I2(g)2HI(g)ΔH=−9.4kJ(exothermic)H2(g)+I2(g)2HI(g)ΔH=−9.4kJ(exothermic)“>H2(gI2(g⇌ 2HI(g)     Δ−9.4kJ (exothermic)

Because this reaction is exothermic, we can write it with heat as a product.

<span id="MathJax-Element-74-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="H2(g)+I2(g)2HI(g)+heatH2(g)+I2(g)2HI(g)+heat“><span id="MathJax-Element-73-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="H2(g)+I2(g)2HI(g)ΔH=−9.4kJ(exothermic)H2(g)+I2(g)2HI(g)ΔH=−9.4kJ(exothermic)“>
H
2
(gI2(g⇌ 2HI(g
heat

Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress and there is an increase in the concentration of H2 and I2 and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat and favors the formation of hydrogen iodide.

When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium constant. At the new equilibrium the concentration of HI has increased and the concentrations of H2 and I2 decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.

Temperature affects the equilibrium between NO2 and N2O4 in this reaction

<span id="MathJax-Element-75-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="N2O4(g)2NO2(g)ΔH=57.20kJN2O4(g)2NO2(g)ΔH=57.20kJ“>
N
2
O4(g⇌ 2NO2(g)       Δ= + 57.20 kJ (endothermic)

The positive ΔH value tells us that the reaction is endothermic and could be written

<span id="MathJax-Element-76-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="heat+N2O4(g)2NO2(g)heat+N2O4(g)2NO2(g)“>
heat 
+<span id="MathJax-Element-75-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: center;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="N2O4(g)2NO2(g)ΔH=57.20kJN2O4(g)2NO2(g)ΔH=57.20kJ“>N2O4(g⇌ 2NO2(g)

At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown NO2 molecules. However, if we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless N2O4 increases and the concentration of brown NOdecreases, causing the brown color to fade.

Key Concepts and Summary

Systems at equilibrium can be disturbed by changes to temperature and concentration. The system’s response to these disturbances is described by Le Châtelier’s principle: The system will respond in a way that counteracts the disturbance.

Effects of Disturbances of Equilibrium and K
Disturbance Observed Change as Equilibrium is Restored Direction of Shift Effect on K
reactant added added reactant is partially consumed toward products none
product added added product is partially consumed toward reactants none
temperature increase heat is absorbed toward products for endothermic, toward reactants for exothermic changes
temperature decrease heat is given off toward reactants for endothermic, toward products for exothermic changes

Chemistry End of Chapter Exercises

  1.  What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
  2. Suggest four ways in which the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the following equation:
    N2(g2H2(g⇌ N2H4(g)       ΔH=95kJ
  3. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:
    P4(g6H2(g⇌ 4PH3(gΔH=110.5kJ
  4. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
    (a) <span id="MathJax-Element-84-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="2NH3(g)N2(g)+3H2(g)ΔH=92kJ2NH3(g)N2(g)+3H2(g)ΔH=92kJ“>2NH3(g⇌ N2(g3H2(g)            Δ= + 92 kJ
    (b) <span id="MathJax-Element-85-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="N2(g)+O2(g)2NO(g)ΔH=181kJN2(g)+O2(g)2NO(g)ΔH=181kJ“>N2(gO2(g⇌ 2NO(g)               Δ= + 181 kJ
    (c) <span id="MathJax-Element-86-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="2O3(g)3O2(g)ΔH=−285kJ2O3(g)3O2(g)ΔH=−285kJ“>2O3(g⇌ 3O2(g)                           Δ− 285 kJ
    (d) <span id="MathJax-Element-87-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="CaO(s)+CO2(g)CaCO3(s)ΔH=−176kJCaO(s)+CO2(g)CaCO3(s)ΔH=−176kJ“>CaO(sCO2(g⇌ CaCO3(s)     Δ− 176 kJ
  5. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
    (a) <span id="MathJax-Element-88-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="2H2O(g)2H2(g)+O2(g)ΔH=484kJ2H2O(g)2H2(g)+O2(g)ΔH=484kJ“>2H2O(g⇌ 2H2(gO2(g)       Δ= + 484 kJ
    (b) <span id="MathJax-Element-89-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="N2(g)+3H2(g)2NH3(g)ΔH=−92.2kJN2(g)+3H2(g)2NH3(g)ΔH=−92.2kJ“>N2(g3H2(g⇌ 2NH3(g)       Δ− 92.2 kJ
    (c) <span id="MathJax-Element-90-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="2Br(g)Br2(g)ΔH=−224kJ2Br(g)Br2(g)ΔH=−224kJ“>2Br(g⇌ Br2(g)                         Δ− 224kJ
    (d) <span id="MathJax-Element-91-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml="H2(g)+I2(s)2HI(g)ΔH=53kJH2(g)+I2(s)2HI(g)ΔH=53kJ“>H2(gI2(s⇌ 2HI(g)              Δ= + 53kJ
  6. Ammonia is a weak base that reacts with water according to this equation:
    NH3(aqH2O(l⇌ NH4+(aqOH(aq)
    Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water?
    (a) Addition of NaOH
    (b) Addition of HCl
    (c) Addition of NH4Cl
  7. Acetic acid is a weak acid that reacts with water according to this equation:
    CH3CO2H(aqH2O(aq⇌ H3O+(aqCH3CO2(aq)
    Will any of the following increase the percent of acetic acid that reacts and produces CH3CO2 ion?
    (a) Addition of HCl
    (b) Addition of NaOH
    (c) Addition of NaCH3CO2

Glossary

Le Châtelier’s principle
when a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance
position of equilibrium
concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe conditions before a disturbance)
stress
change to a reaction’s conditions that may cause a shift in the equilibrium

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