0 1 NaN Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Learning Objective To understand what is meant by 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 (N 2 O 4 ) to nitrogen dioxide (NO 2 ). You may recall from Chapter 14 that NO 2 is responsible for the brown color we associate with smog. When a sealed tube containing solid N 2 O 4 (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of NO 2 appears ( Figure 15.1.1 ). The reaction can be followed visually because the product (NO 2 ) is colored, whereas the reactant (N 2 O 4 ) is colorless: \[N_2O_4{(g)} colorless \rightleftharpoons 2NO_{2(g)} \; red-brown \tag{15.1.1}\] The double arrow indicates that both the forward and reverse reactions are occurring simultaneously; it is read “is in equilibrium with.” Figure 15.1.1 The N 2 O 4 (g) ⇌ 2 NO 2 (g) System at Different Temperatures (left) At dry ice temperature (−78.4°C), the system contains essentially pure solid N 2 O 4 , which is colorless. (center) As the system is warmed above the melting point of N 2 O 4 (−9.3°C), the N 2 O 4 melts and then evaporates, and some of the vapor dissociates to red-brown NO 2 . (right) Eventually the sample reaches room temperature, and a mixture of gaseous N 2 O 4 and NO 2 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. Figure 15.1.2 shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of NO 2 were zero, then it increases as the concentration of N 2 O 4 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 N 2 O 4 but an initial NO 2 concentration twice the initial concentration of N 2 O 4 in part (a) in Figure 15.1.2 , in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition, as shown in part (b) in Figure 15.1.2 . Thus equilibrium can be approached from either direction in a chemical reaction. Figure 15.1.2 The Composition of N 2 O 4 /NO 2 Mixtures as a Function of Time at Room Temperature (a) Initially, this idealized system contains 0.0500 M gaseous N 2 O 4 and no gaseous NO 2 . The concentration of N 2 O 4 decreases with time as the concentration of NO 2 increases. (b) Initially, this system contains 0.1000 M NO 2 and no N 2 O 4 . The concentration of NO 2 decreases with time as the concentration of N 2 O 4 increases. In both cases, the final concentrations of the substances are the same: [N 2 O 4 ] = 0.0422 M and [NO 2 ] = 0.0156 M at equilibrium. Figure 15.1.3 shows the forward and reverse reaction rates for a sample that initially contains pure NO 2 . Because the initial concentration of N 2 O 4 is zero, the forward reaction rate (dissociation of N 2 O 4 ) is initially zero as well. In contrast, the reverse reaction rate (dimerization of NO 2 ) is initially very high (2.0 × 10 6 M/s), but it decreases rapidly as the concentration of NO 2 decreases. (Recall from Chapter 14 that the reaction rate of the dimerization reaction is expected to decrease rapidly because the reaction is second order in NO 2 : rate = k r [NO 2 ] 2 , where k r is the rate constant for the reverse reaction shown in Equation 15.1.1 .) As the concentration of N 2 O 4 increases, the rate of dissociation of N 2 O 4 increases—but more slowly than the dimerization of NO 2 —because the reaction is only first order in N 2 O 4 (rate = k f [N 2 O 4 ], where k f is the rate constant for the forward reaction in Equation 15.1.1 ). 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 15.1.3 The Forward and Reverse Reaction Rates as a Function of Time for the N 2 O 4 (g) ⇌ 2 NO 2 (g) System Shown in Part (b) in Figure 15.1.2 The rate of dimerization of NO 2 (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of N 2 O 4 is zero, the rate of the dissociation reaction (forward reaction) at t = 0 is also zero. As the dimerization reaction proceeds, the N 2 O 4 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 N 2 O 4 and NO 2 no longer change. Note the Pattern At equilibrium, the forward reaction rate is equal to the reverse reaction rate. Example 15.1.1 The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation <math display="inline" xml:id="av_1.0-15_m004"><semantics><mrow><mn>2</mn><mtext>A</mtext><mo>⇌</mo><mtext>B,</mtext></mrow></semantics></math> 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 t 0 through t 2 but is the same at both t 2 and t 3 . Thus systems 1 and 3 are at equilibrium by t 3 . In system 2, the concentrations of A and B are still changing between t 2 and t 3 , so system 2 may not yet have reached equilibrium by t 3 . 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 Summary 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. Key Takeaway At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Conceptual Problems What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the concentrations or amounts of the reactants and the products? Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example. Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of NaCl in water. What is occurring on a microscopic level? What is happening on a macroscopic level? Which of these systems exists in a state of chemical equilibrium? oxygen and hemoglobin in the human circulatory system iodine crystals in an open beaker the combustion of wood the amount of 14 C in a decomposing organism Answer Both forward and reverse reactions occur but at the same rate. Na + and Cl − ions continuously leave the surface of an NaCl crystal to enter solution, while at the same time Na + and Cl − ions in solution precipitate on the surface of the crystal. Contributors Anonymous Modified by Joshua B. Halpern