Chemistry: Atoms First

# 13.3Shifting Equilibria: Le Châtelier’s Principle

Chemistry: Atoms First13.3 Shifting Equilibria: Le Châtelier’s Principle

### Learning Objectives

By the end of this section, you will be able to:
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Châtelier’s principle

As we saw in the previous section, 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 13.8 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.

Figure 13.8 (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. $Fe 3+ ( a q ) + SCN − ( a q ) ⇌ Fe ( SCN ) 2+ ( a q ) . Fe 3+ ( a q ) + SCN − ( a q ) ⇌ Fe ( SCN ) 2+ ( a q ) .$ (c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN as the white solid AgSCN. $Ag + ( a q ) + SCN − ( a q ) ⇌ AgSCN ( s ) . Ag + ( a q ) + SCN − ( a q ) ⇌ AgSCN ( s ) .$ 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:

$H2(g)+I2(g)⇌2HI(g)Kc=50.0at400°CH2(g)+I2(g)⇌2HI(g)Kc=50.0at400°C$

The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with [H2] = [I2] = 0.221 M 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. This gives:

$Qc=[HI]2[H2][I2]=(1.692)2(0.374)(0.153)=50.0=KcQc=[HI]2[H2][I2]=(1.692)2(0.374)(0.153)=50.0=Kc$

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 Pressure on Equilibrium

Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for Kc) or partial pressure (for KP). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.

As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.

Consider what happens when we increase the pressure on a system in which NO, O2, and NO2 are at equilibrium:

$2NO(g)+O2(g)⇌2NO2(g)2NO(g)+O2(g)⇌2NO2(g)$

The formation of additional amounts of NO2 decreases the total number of molecules in the system because each time two molecules of NO2 form, a total of three molecules of NO and O2 are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of NO2 into NO and O2, which tends to restore the pressure.

Now consider this reaction:

$N2(g)+O2(g)⇌2NO(g)N2(g)+O2(g)⇌2NO(g)$

Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.

#### Effect of Change in Temperature on Equilibrium

Changing concentration or pressure perturbs 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.

$H2(g)+I2(g)⇌2HI(g)ΔH=−9.4kJ(exothermic)H2(g)+I2(g)⇌2HI(g)ΔH=−9.4kJ(exothermic)$

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

$H2(g)+I2(g)⇌2HI(g)+heatH2(g)+I2(g)⇌2HI(g)+heat$

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

$N2O4(g)⇌2NO2(g)ΔH=57.20kJN2O4(g)⇌2NO2(g)ΔH=57.20kJ$

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

$heat+N2O4(g)⇌2NO2(g)heat+N2O4(g)⇌2NO2(g)$

At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown NO2 molecules. If, however, 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 NO2 decreases, causing the brown color to fade.

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