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Chemistry LibreTexts

7.1: The Concept of Dynamic Equilibrium

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Learning Objectives
  • To understand what is meant by chemical equilibrium.

In the last chapter, we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibrium, the point at which the composition of the system no longer changes with time.

imageedit_5_3015884689.jpg
Figure 7.1.1: Dinitrogen tetroxide is a powerful oxidizer that reacts spontaneously upon contact with various forms of hydrazine, which makes the pair a popular propellant combination for rockets. Nitrogen dioxide at −196 °C, 0 °C, 23 °C, 35 °C, and 50 °C. (NO2) converts to the colorless dinitrogen tetroxide (N2O4) at low temperatures, and reverts to NO2 at higher temperatures. (CC BY-SA 3.0; Eframgoldberg).

Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (N2O4) to nitrogen dioxide (NO2). You may recall that NO2 is responsible for the brown color we associate with smog. When a sealed tube containing solid N2O4 (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of NO2 appears (Figure 7.1.1). The reaction can be followed visually because the product (NO2) is colored, whereas the reactant (N2O4) is colorless:

N2O4(g)colorlesskfkr2NO2(g)redbrown

The double arrow indicates that both the forward reaction

N2O4(g)kf2NO2(g)

and reverse reaction

2NO2(g)krN2O4(g)

occurring simultaneously (i.e, the reaction is reversible). However, this does not necessarily mean the system is equilibrium as the following chapter demonstrates.

Figure 7.1.2 shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of NO2 were zero, then it increases as the concentration of N2O4 decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no N2O4 but an initial NO2 concentration twice the initial concentration of N2O4 (Figure 7.1.2a), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure 7.1.2b). Thus equilibrium can be approached from either direction in a chemical reaction.

imageedit_7_4442889682.jpg
Figure 7.1.2: The Composition of N2O4/NO2 Mixtures as a Function of Time at Room Temperature. (a) Initially, this idealized system contains 0.0500 M gaseous N2O4 and no gaseous NO2. The concentration of N2O4 decreases with time as the concentration of NO2 increases. (b) Initially, this system contains 0.1000 M NO2 and no N2O4. The concentration of NO2 decreases with time as the concentration of N2O4 increases. In both cases, the final concentrations of the substances are the same: [N2O4] = 0.0422 M and [NO2] = 0.0156 M at equilibrium. (CC BY-SA-NC; Anonymous by request)

Figure 7.1.3 shows the forward and reverse reaction rates for a sample that initially contains pure NO2. Because the initial concentration of N2O4 is zero, the forward reaction rate (dissociation of N2O4) is initially zero as well. In contrast, the reverse reaction rate (dimerization of NO2) is initially very high (2.0×106M/s), but it decreases rapidly as the concentration of NO2 decreases. As the concentration of N2O4 increases, the rate of dissociation of N2O4 increases—but more slowly than the dimerization of NO2—because the reaction is only first order in N2O4 (rate = kf[N2O4], where kf is the rate constant for the forward reaction in Equations ??? and ???). Eventually, the forward and reverse reaction rates become identical, kf=kr, and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.

imageedit_10_2117565046.jpg
Figure 7.1.3: The Forward and Reverse Reaction Rates as a Function of Time for the N2O4(g)⇌2NO2(g) System Shown in Part (b) in Figure 7.1.2. (CC BY-SA-NC; Anonymous by request)

The rate of dimerization of NO2 (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of N2O4 is zero, the rate of the dissociation reaction (forward reaction) at t=0 is also zero. As the dimerization reaction proceeds, the N2O4 concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of N2O4 and NO2 no longer change.

At equilibrium, the forward reaction rate is equal to the reverse reaction rate.

Example 7.1.1

The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation:

2AB

where the blue circles are A and the purple ovals are B. Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium?

In reaction system 1 theree are four purple ovals at t3. In reaction system 2 there are size purple ovals at t3. In reaction system systems there are six ovals at t2 and t3.

Given: three reaction systems

Asked for: relative time to reach chemical equilibrium

Strategy:

Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium.

Solution:

In systems 1 and 3, the concentration of A decreases from t0 through t2 but is the same at both t2 and t3. Thus systems 1 and 3 are at equilibrium by t3. In system 2, the concentrations of A and B are still changing between t2 and t3, so system 2 may not yet have reached equilibrium by t3. Thus system 2 took the longest to reach chemical equilibrium.

Exercise 7.1.1

In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium?

In reaction system 1 there are seven orange ovals at t3. In reaction system two there are four orange ovals at t3. In reaction system three there are three orange ovals at t3.
Answer

system 2

A Video Introduction to Dynamic Equilibrium: Introduction to Dynamic Equilibrium(opens in new window) [youtu.be]

Summary

At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached.


7.1: The Concept of Dynamic Equilibrium is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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