CHEMISTRY THE CENTRAL SCIENCE

15 CHEMICAL EQUILIBRIUM

15.6 APPLICATIONS OF EQUILIBRIUM CONSTANTS

We have seen that the magnitude of K indicates the extent to which a reaction proceeds. If K is very large, the equilibrium mixture contains mostly substances on the product side of the equation for the reaction. (That is, the reaction proceeds far to the right.) If K is very small (that is, much less than 1), the equilibrium mixture contains mainly substances on the reactant side of the equation. The equilibrium constant also allows us to (1) predict the direction in which a reaction mixture achieves equilibrium and (2) calculate equilibrium concentrations of reactants and products.

Predicting the Direction of Reaction

For the formation of NH3 from N2 and H2 (Equation 15.6), Kc = 0.105 at 472 °C. Suppose we place 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of NH3 in a 1.00-L container at 472 °C. How will the mixture react to reach equilibrium? Will N2 and H2 react to form more NH3, or will NH3decompose to N2 and H2?

To answer this question, we substitute the starting concentrations of N2, H2, and NH3 into the equilibrium-constant expression and compare its value to the equilibrium constant:

To reach equilibrium, the quotient [NH3]2/[N2][H2]3 must decrease from the starting value of 0.500 to the equilibrium value of 0.105. Because the system is closed, this change can happen only if [NH3] decreases and [N2] and [H2] increase. Thus, the reaction proceeds toward equilibrium by forming N2 and H2 from NH3; that is, the reaction as written in Equation 15.6 proceeds from right to left.

This approach can be formalized by defining a quantity called the reaction quotient. The reaction quotientQ, is a number obtained by substituting reactant and product concentrations or partial pressures at any point during a reaction into an equilibrium-constant expression. Therefore, for the general reaction

the reaction quotient in terms of molar concentrations is

(A related quantity Qp can be written for any reaction that involves gases by using partial pressures instead of concentrations.)

Although we use what looks like the equilibrium-constant expression to calculate the reaction quotient, the concentrations we use may or may not be the equilibrium concentrations. For example, when we substituted the starting concentrations into the equilibrium-constant expression of Equation 15.22, we obtained Qc = 0.500 whereas Kc = 0.105. The equilibrium constant has only one value at each temperature. The reaction quotient, however, varies as the reaction proceeds.

Of what use is Q? One practical thing we can do with Q is tell whether our reaction really is at equilibrium, which is an especially valuable option when a reaction is very slow. We can take samples of our reaction mixture as the reaction proceeds, separate the components, and measure their concentrations. Then we insert these numbers into Equation 15.23 for our reaction. To determine whether or not we are at equilibrium, or in which direction the reaction proceeds to achieve equilibrium, we compare the values of Qc and Kc or Qp and Kp. Three possible situations arise:

• Q = K: The reaction quotient equals the equilibrium constant only if the system is at equilibrium.

• Q > K: The concentration of products is too large and that of reactants too small. Substances on the right side of the chemical equation react to form substances on the left; the reaction proceeds from right to left to approach equilibrium.

• Q < K: The concentration of products is too small and that of reactants too large. The reaction achieves equilibrium by forming more products; it proceeds from left to right.

These relationships are summarized in FIGURE 15.8.

FIGURE 15.8 Predicting the direction of a reaction by comparing Q and K at a given temperature.

SAMPLE EXERCISE 15.10 Predicting the Direction of Approach to Equilibrium

At 448 °C the equilibrium constant Kc for the reaction

is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 × 10–2 mol of HI, 1.0 × 10–2 mol of H2, and 3.0 × 10–2 mol of I2 in a 2.00-L container.

SOLUTION

Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium.

Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed.

Solve

The initial concentrations are

The reaction quotient is therefore

Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.

PRACTICE EXERCISE

At 1000 K the value of Kp for the reaction 2 SO3(g 2 SO2(g) + O2(g) is 0.338. Calculate the value for Qp, and predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are PSO3 = 0.16 atm; PSO2 = 0.41 atm; PO2 = 2.5 atm.

Answer: Qp = 16; Qp > Kp, and so the reaction will proceed from right to left, forming more SO3.

Calculating Equilibrium Concentrations

Chemists frequently need to calculate the amounts of reactants and products present at equilibrium in a reaction for which they know the equilibrium constant. The approach in solving problems of this type is similar to the one we used for evaluating equilibrium constants: We tabulate initial concentrations or partial pressures, changes in those concentrations or pressures, and final equilibrium concentrations or partial pressures. Usually we end up using the equilibrium-constant expression to derive an equation that must be solved for an unknown quantity, as demonstrated in Sample Exercise 15.11.

SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations

For the Haber process, N2(g) + 3 H2(g 2 NH3(g), Kp = 1.45 × 10 at 500 °C. In an equilibrium mixture of the three gases at 500 °C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?

SOLUTION

Analyze We are given an equilibrium constant, Kp, and the equilibrium partial pressures of two of the three substances in the equation (N2 and H2), and we are asked to calculate the equilibrium partial pressure for the third substance (NH3).

Plan We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that we know. Then we can solve for the only unknown in the equation.

Solve We tabulate the equilibrium pressures:

Because we do not know the equilibrium pressure of NH3, we represent it with x. At equilibrium the pressures must satisfy the equilibrium-constant expression:

We now rearrange the equation to solve for x:

Check We can always check our answer by using it to recalculate the value of the equilibrium constant:

PRACTICE EXERCISE

At 500 K the reaction PCl5(g PCl3(g) + Cl2(g) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture?

Answer: 1.22 atm

In many situations we know the value of the equilibrium constant and the initial amounts of all species. We must then solve for the equilibrium amounts. Solving this type of problem usually entails treating the change in concentration as a variable. The stoichiometry of the reaction gives us the relationship between the changes in the amounts of all the reactants and products, as illustrated in Sample Exercise 15.12. The calculations frequently involve the quadratic formula, as you will see in this exercise.

SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrations from Initial Concentrations

A 1.000-L flask is filled with 1.000 mol of H2(g) and 2.000 mol of I2(g) at 448 °C. The value of the equilibrium constant Kc for the reaction

at 448 °C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter?

SOLUTION

Analyze We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container and are asked to calculate the equilibrium concentrations of all species.

Plan In this case we are not given any of the equilibrium concentrations. We must develop some relationships that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial concentrations.

Solve First, we note the initial concentrations of H2 and I2:

Second, we construct a table in which we tabulate the initial concentrations:

Third, we use the stoichiometry of the reaction to determine the changes in concentration that occur as the reaction proceeds to equilibrium. The H2 and I2 concentrations will decrease as equilibrium is established and that of HI will increase. Let's represent the change in concentration of H2by x. The balanced chemical equation tells us the relationship between the changes in the concentrations of the three gases. For each x mol of H2 that reacts, x mol of I2 are consumed and 2x mol of HI are produced:

Fourth, we use initial concentrations and changes in concentrations, as dictated by stoichiometry, to express the equilibrium concentrations. With all our entries, our table now looks like this:

Fifth, we substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x:

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this solution. We then use x = 0.935 to find the equilibrium concentrations:

Check We can check our solution by putting these numbers into the equilibrium-constant expression to assure that we correctly calculate the equilibrium constant:

Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions to the equation will give you a value that leads to negative concentrations and thus is not chemically meaningful. Reject this solution to the quadratic equation.

PRACTICE EXERCISE

For the equilibrium PCl5(g PCl3(g) + Cl2(g), the equilibrium constant Kp is 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature?

Answer: PPCl5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm