15.1 The Concept of Chemical Equilibrium

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 from the previous chapter 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 15.1 "The "). The reaction can be followed visually because the product (NO₂) is colored, whereas the reactant (N₂O₄) is colorless:

The double arrow indicates that both the forward and reverse reactions are occurring simultaneously; it is read “is in equilibrium with.”

Figure 15.1 The N₂O₄(g) ⇄ 2 NO₂(g) System at Different Temperatures

(left) At liquid nitrogen temperature (−196°C), the system contains essentially pure solid N₂O₄, which is colorless. As the system is warmed above the melting point of N₂O₄ (−9.3°C), the N₂O₄ melts and then evaporates, and some of the vapor dissociates to red-brown NO₂. (right) Eventually the sample reaches room temperature, and a mixture of gaseous N₂O₄ and NO₂ is present. The composition of the mixture and hence the color do not change further with time: the system has reached equilibrium at the new temperature. Image Credit: By Eframgoldberg (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons

Figure 15.2 "The Composition of N" 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 in part (a) in Figure 15.2 "The Composition of N", in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition, as shown in part (b) in Figure 15.2 "The Composition of N". Thus equilibrium can be approached from either direction in a chemical reaction.

Figure 15.2 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.

Figure 15.3 "The Forward and Reverse Reaction Rates as a Function of Time for the " 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 × 10⁶ M/s), but it decreases rapidly as the concentration of NO₂ decreases. (Recall from the previous chapter that the reaction rate of the dimerization reaction is expected to decrease rapidly because the reaction is second order in NO₂: rate = k_r[NO₂]², where k_r is the rate constant for the reverse reaction shown in Equation 15.1.) 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 = k_f[N₂O₄], where k_f is the rate constant for the forward reaction in Equation 15.1). Eventually, the forward and reverse reaction rates become identical, k_F = k_r, and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.

Figure 15.3 The Forward and Reverse Reaction Rates as a Function of Time for the ${\text{N}}_2{\text{O}}_4\text{(g)}\rightleftharpoons 2{\text{NO}}_2\text{(g)}$ System Shown in Part (b) in Figure 15.2 "The Composition of N"

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.

Example 1

The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation $2\text{A}\rightleftharpoons \text{B,}$ 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?

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

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?

Answer: system 2