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1.5 Chemical Equilibria

Learning Objectives

By the end of this section, you will be able to:

  • Describe the nature of equilibrium systems
  • Explain the dynamic nature of a chemical equilibrium
  • Describe the ways in which an equilibrium system can be stressed
  • Predict the response of a stressed equilibrium using Le Châtelier’s principle

The convention for writing chemical equations involves placing reactant formulas on the left side of a reaction arrow and product formulas on the right side. By this convention, and the definitions of “reactant” and “product,” a chemical equation represents the reaction in question as proceeding from left to right. Reversible reactions, however, may proceed in both forward (left to right) and reverse (right to left) directions. When the rates of the forward and reverse reactions are equal, the concentrations of the reactant and product species remain constant over time and the system is at equilibrium. The relative concentrations of reactants and products in equilibrium systems vary greatly; some systems contain mostly products at equilibrium, some contain mostly reactants, and some contain appreciable amounts of both.

Figure 1.5.1 illustrates fundamental equilibrium concepts using the reversible decomposition of colorless dinitrogen tetroxide to yield brown nitrogen dioxide, an elementary reaction described by the equation:

N2O4(g)2NO2(g)Note that a special double arrow is used to emphasize the reversible nature of the reaction.

A three-part diagram is shown and labeled, “a,” “b,” and “c.” In part a, at the top of the diagram, three beakers are shown, and each one contains a sealed tube. The tube in the left beaker is labeled “t equals 0.” It is full of a colorless gas which is connected to a zoom-in view of the particles in the tube by a downward-facing arrow. This particle view shows seven particles, each composed of two connected blue spheres. Each blue sphere is connected to two red spheres. This type of particle is labeled “N subscript 2 O subscript 4.” The tube in the middle beaker is labeled “pre-equilibrium.” It is full of a light brown gas which is connected to a zoom-in view of the particles in the tube by a downward-facing arrow. This particle view shows nine particles, five of which are composed of two connected blue spheres. Each blue sphere is connected to two red spheres. This type of particle is labeled “N subscript 2 O subscript 4.” The remaining four are composed of two red spheres connected to a blue sphere. This type of particle is labeled “N O subscript 2.” The tube in the right beaker is labeled “at equilibrium.” It is full of a brown gas which is connected to a zoom-in view of the particles in the tube by a downward-facing arrow. This particle view shows eleven particles, three of which are composed of two connected blue spheres. Each blue sphere is connected to two red spheres. The remaining eight are composed of two red spheres connected to a blue sphere. In part b, in the middle of the image, is one graph. This graph has a y-axis labeled, “Concentration,” and an x-axis labeled, “Time.” A red line labeled, “N O subscript 2,” begins in the bottom left corner of the graph at a point labeled, “0,” and rises near the highest point on the y-axis before it levels off and becomes horizontal. A blue line labeled, “N subscript 2 O subscript 4,” begins near the highest point on the y-axis and drops below the midpoint of the y-axis before leveling off. In part c, at the bottom of the image is another graph. This graph has a y-axis labeled, “Rate,” and an x-axis labeled, “Time.” A red line labeled, “k subscript f, [ N subscript 2 O subscript 4 ],” begins in the bottom left corner of the graph at a point labeled, “0,” and rises near the middle of the y-axis before it levels off and becomes horizontal. A blue line labeled, “k subscript f, [ N O subscript 2 ] superscript 2,” begins near the highest point on the y-axis and drops to the same point on the y-axis as the red line before leveling off. The point where both lines become horizontal is labeled, “Equilibrium achieved.”
Figure 1.5.1 (a) A sealed tube containing colorless N2O4 darkens as it decomposes to yield brown NO2. (b) Changes in concentration over time as the decomposition reaction achieves equilibrium. (c) At equilibrium, the forward and reverse reaction rates are equal.
For this elementary process, rate laws for the forward and reverse reactions may be derived directly from the reaction stoichiometry:
ratef=kf[N­2O4]
rater=kr[NO2]2As the reaction begins (t = 0), the concentration of the N2O4 reactant is finite and that of the NO2 product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. As time passes, N­2O4 is consumed and its concentration falls, while NO2 is produced and its concentration increases (Figure 1.5.1b). The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate (Figure 1.5.1c). This process continues until the forward and reverse reaction rates become equal, at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products (shaded areas of Figure 1.5.1b and Figure 1.5.1c). It’s important to emphasize that chemical equilibria are dynamic; a reaction at equilibrium has not “stopped,” but is proceeding in the forward and reverse directions at the same rate. This dynamic nature is essential to understanding equilibrium behavior as discussed in this and subsequent chapters of the text.

Note: The equilibrium of an uncatalyzed reaction is the same as that of a catalyzed reaction, only the reaction rate is different. The reaction will reach equilibrium much faster with a catalyst. Enzymes are catalysts in biosystems such as our body. You will learn more about enzymes in the later chapters of this textbook.

Two people are shown throwing juggling pins back and forth to one another.
Figure 1.5.2 A two-person juggling act illustrates the dynamic aspect of chemical equilibria. Each person is throwing and catching clubs at the same rate, and each holds a (approximately) constant number of clubs.

Equilibrium Constant

The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Q). For a reversible reaction described by

mA+nBxC+yD,the reaction quotient is derived directly from the stoichiometry of the balanced equation as

Qc=[C]x[D]y[A]m[B]nwhere the subscript c denotes the use of molar concentrations in the expression.

EXAMPLE 1.5.1

Writing Reaction Quotient Expressions

Write the concentration-based reaction quotient expression for each of the following reactions:

(a) 3O2(g)2O3(g)3O2(g)2O3(g) 

(b) N2(g)+3H2(g)2NH3(g)N2(g)+3H2(g)2NH3(g)

(c) 4NH3(g)+7O2(g)4NO2(g)+6H2O(g) 

Solution

(a) Qc=[O3]2[O2]3Qc=[O3]2[O2]3 (b) Qc=[NH3]2[N2][H2]3Qc=[NH3]2[N2][H2]3 (c) Qc=[NO2]4[H2O]6[NH3]4[O2]7Qc=[NO2]4[H2O]6[NH3]4[O2]7

Check Your Learning

Write the concentration-based reaction quotient expression for each of the following reactions:

(a) 2SO2(g)+O2(g)2SO3(g) 

(b) C4H8(g)2C2H4(g)

(c) 2C4H10(g)+13O2(g)8CO2(g)+10H2O(g) 

ANSWER:

(a) Qc=[SO3]2[SO2]2[O2];Qc=[SO3]2[SO2]2[O2]; 

(b) Qc=[C2H4]2[C4H8];Qc=[C2H4]2[C4H8]; 

(c) Qc=[CO2]8[H2O]10[C4H10]2[O2]13Qc=[CO2]8[H2O]10[C4H10]2[O2]13 

Le Châtelier’s Principle

A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Châtelier’s principleif an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.

Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.

Effect of a Change in Concentration

If an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction

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

The rate laws for the forward and reverse reactions are

forwardH2(g)+I2(g)2HI(g)ratef=kf[H2]m[I2]nreverse2HI(g)H2(g)+I2(g)rater=kr[HI]xforwardH2(g)+I2(g)2HI(g)ratef=kf[H2]m[I2]nreverse2HI(g)H2(g)+I2(g)rater=kr[HI]x

When this system is at equilibrium, the forward and reverse reaction rates are equal.

ratef=raterratef=rater If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction:

ratef>raterratef=rater

The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.

The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction

ratef<rater

and a temporary net reaction in the reverse direction to re-establish equilibrium.

As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,

Qc=[HI]2[H2][I2]=KcQc=[HI]2[H2][I2]=Kc

If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the new equilibrium mixture will have greater concentrations of H2 and I2 and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left.

Equilibrium and Soft Drinks

The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804) developed a method of infusing water with carbon dioxide to make carbonated water. Priestly’s approach involved production of carbon dioxidey reacting oil of vitriol (sulfuric acid) with chalk (calcium carbonate).

The carbon dioxide was then dissolved in water, reacting to produce hydrogen carbonate, a weak acid that subsequently ionized to yield bicarbonate and hydrogen ions:

dissolutionCO2(g)=CO2(aq)hydrolysisCO2(aq)+H2O(l)=H2CO3(aq)ionizationH2CO3(aq)=HCO3(aq)+H+(aq)dissolutionCO2(g)=CO2(aq)hydrolysisCO2(aq)+H2O(l)=H2CO3(aq)ionizationH2CO3(aq)=HCO3(aq)+H+(aq)These same equilibrium reactions are the basis of today’s soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace.

When a carbonated beverage container is opened, a hissing sound is heard as pressurized CO2 escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved CO2 and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the CO2(aq) concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as “flat.”

A bottle of soda sitting on the ground is shown with a large amount of fizz-filled liquid spewing out of the top.
Figure 1.5.3 Opening a soft-drink bottle lowers the CO2 pressure above the beverage, shifting the dissolution equilibrium and releasing dissolved CO2 from the beverage. (credit: modification of work by “D Coetzee”/Flickr)

Effect of a Change in Temperature

Consistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to re-establish equilibrium without any change in the value of the equilibrium constant, K. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.

To understand this phenomenon, consider the elementary reaction

AB

Since this is an elementary reaction, the rates laws for the forward and reverse may be derived directly from the balanced equation’s stoichiometry:

ratef=kf[A]rater=kr[B]When the system is at equilibrium,

ratef=raterSubstituting the rate laws into this equality and rearranging gives

kf[A]=kr[B][B][A]=kfkr=Kckf[A]=kr[B][B][A]=kfkr=KcThe equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). For more complex reactions involving multistep reaction mechanisms, a similar but more complex mathematical relation exists between the equilibrium constant and the rate constants of the steps in the mechanism. Regardless of how complex the reaction may be, the temperature-dependence of its equilibrium constant persists.

Predicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. For example, the decomposition of dinitrogen tetroxide is an endothermic (heat-consuming) process:

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

For purposes of applying Le Chatelier’s principle, heat (q) may be viewed as a reactant:

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

Raising the temperature of the system is akin to increasing the amount of a reactant, and so the equilibrium will shift to the right. Lowering the system temperature will likewise cause the equilibrium to shift left. For exothermic processes, heat is viewed as a product of the reaction and so the opposite temperature dependence is observed.

Section Summary

  • Reversible chemical reactions may proceed in both forward (left to right) and reverse (right to left) directions.
  • The system is at equilibrium when the rates of the forward and reverse reactions are equal.
    • At equilibrium, the concentrations of the reactant and product species remain constant over time (but not necessary the same).
  • Le Châtelier’s principle describes how the equilibrium is reestablished when the system is experienced a change of condition.
    • If an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.