## 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 NH_{3} from N_{2} and H_{2} (Equation 15.6), *K** _{c}* = 0.105 at 472 °C. Suppose we place 2.00 mol of H

_{2}, 1.00 mol of N

_{2}, and 2.00 mol of NH

_{3}in a 1.00-L container at 472 °C. How will the mixture react to reach equilibrium? Will N

_{2}and H

_{2}react to form more NH

_{3}, or will NH

_{3}decompose to N

_{2}and H

_{2}?

To answer this question, we substitute the starting concentrations of N_{2}, H_{2}, and NH_{3} into the equilibrium-constant expression and compare its value to the equilibrium constant:

To reach equilibrium, the quotient [NH_{3}]^{2}/[N_{2}][H_{2}]^{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 [NH_{3}] decreases and [N_{2}] and [H_{2}] increase. Thus, the reaction proceeds toward equilibrium by forming N_{2} and H_{2} from NH_{3}; 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 quotient**, *Q, 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 *Q** _{p}* 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 *Q** _{c}* = 0.500 whereas

*K*

*= 0.105. The equilibrium constant has only one value at each temperature. The reaction quotient, however, varies as the reaction proceeds.*

_{c}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 *Q** _{c}* and

*K*

*or*

_{c}*Q*

*and*

_{p}*K*

*. Three possible situations arise:*

_{p}• *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 *K** _{c}* 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 H_{2}, and 3.0 **×** 10^{–2} mol of I_{2} 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, *Q** _{c}*. 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 *Q*_{c}**<** *K** _{c}*, the concentration of HI must increase and the concentrations of H

_{2}and I

_{2}must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.

**PRACTICE EXERCISE**

At 1000 K the value of **K*** _{p}* for the reaction 2 SO

_{3}(

*g*) 2 SO

_{2}(

*g*) + O

_{2}(

*g*) is 0.338. Calculate the value for

*Q*

*, and predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are*

_{p}*P*

_{SO}

*= 0.16 atm;*

_{3}*P*

_{SO}

*= 0.41 atm;*

_{2}*P*

_{O}*= 2.5 atm.*

_{2}*Answer:**Q** _{p}* = 16;

*Q*

_{p}**>**

*K*

*, and so the reaction will proceed from right to left, forming more SO*

_{p}_{3}.

**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, N_{2}(*g*) + 3 H_{2}(*g*) 2 NH_{3}(*g*), *K** _{p}* = 1.45

**×**10 at 500 °C. In an equilibrium mixture of the three gases at 500 °C, the partial pressure of H

_{2}is 0.928 atm and that of N

_{2}is 0.432 atm. What is the partial pressure of NH

_{3}in this equilibrium mixture?

**SOLUTION**

**Analyze** We are given an equilibrium constant, *K** _{p}*, and the equilibrium partial pressures of two of the three substances in the equation (N

_{2}and H

_{2}), and we are asked to calculate the equilibrium partial pressure for the third substance (NH

_{3}).

**Plan** We can set *K** _{p}* 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 NH_{3}, 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 PCl_{5}(*g*) PCl_{3}(*g*) + Cl_{2}(*g*) has K*p* = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl_{5} is 0.860 atm and that of PCl_{3} is 0.350 atm. What is the partial pressure of Cl_{2} 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 H_{2}(*g*) and 2.000 mol of I_{2}(*g*) at 448 °C. The value of the equilibrium constant *K** _{c}* for the reaction

at 448 °C is 50.5. What are the equilibrium concentrations of H_{2}, I_{2}, 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 H_{2} and I_{2}:

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 H_{2} and I_{2} concentrations will decrease as equilibrium is established and that of HI will increase. Let's represent the change in concentration of H_{2}by *x*. The balanced chemical equation tells us the relationship between the changes in the concentrations of the three gases. For each *x* mol of H_{2} that reacts, *x* mol of I_{2} are consumed and 2*x* 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 H_{2} and I_{2}. 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 PCl_{5}(*g*) PCl_{3}(*g*) + *C*l_{2}(*g*), the equilibrium constant *K** _{p}* is 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl

_{5}(

*g*) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl

_{5}, PCl

_{3}, and Cl

_{2}at this temperature?

*Answer:**P*_{PCl}_{5} = 0.967 atm, *P*_{PCl}_{3} = *P*_{Cl}* _{2}* = 0.693 atm