13.4 Equilibrium Calculations

Calculation of an Equilibrium Constant

In order to calculate an equilibrium constant, enough information must be available to determine the equilibrium concentrations of all reactants and products. Armed with the concentrations, we can solve the equation for K_c, as it will be the only unknown.

Example 13.2 showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE chart—for Initial, Change, and Equilibrium–will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. The initial concentrations of the reactants and products are provided in the first row of the ICE table (these essentially time-zero concentrations that assume no reaction has taken place). The next row of data contains the changes in concentrations that result when the reaction proceeds toward equilibrium (don’t forget to account for the reaction stoichiometry). The last row contains the concentrations once equilibrium has been reached.

Example 13.6

Calculation of an Equilibrium Constant

Iodine molecules react reversibly with iodide ions to produce triiodide ions.

$${\text{I}}_2(aq)+{\text{I}}^{\text{−}}(aq)\rightleftharpoons {\text{I}}_3{}^{\text{−}}(aq)$$

13.71

If a solution with the concentrations of I₂ and I⁻ both equal to 1.000 $\times$ 10⁻³ M before reaction gives an equilibrium concentration of I₂ of 6.61 $\times$ 10⁻⁴ M, what is the equilibrium constant for the reaction?

Solution

We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −x as the change in concentration of I₂.

Since the equilibrium concentration of I₂ is given, we can solve for x. At equilibrium the concentration of I₂ is 6.61 $\times$ 10⁻⁴ M so that

$$1.000\times {10}^{\mathrm{-3}}-x=6.61\times {10}^{\mathrm{-4}}$$

13.72

$$x=1.000\times {10}^{\mathrm{-3}}-6.61\times {10}^{\mathrm{-4}}$$

13.73

$$=3.39\times {10}^{\mathrm{-4}}M$$

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Now we can fill in the table with the concentrations at equilibrium.

We now calculate the value of the equilibrium constant.

$$K_c=Q_c=\frac{[{\text{I}}_3{}^{\text{−}}]}{[{\text{I}}_2]\left[{\text{I}}^{\text{−}}\right]}$$

13.75

$$=\frac{3.39\times {10}^{\mathrm{-4}}M}{(6.61\times {10}^{\mathrm{-4}}M)(6.61\times {10}^{\mathrm{-4}}M)}=776$$

13.76

Check Your Learning

Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.

$${\text{C}}_2{\text{H}}_5\text{OH}+{\text{CH}}_3{\text{CO}}_2\text{H}\rightleftharpoons {\text{CH}}_3{\text{CO}}_2{\text{C}}_2{\text{H}}_5+{\text{H}}_2\text{O}$$

13.77

When 1 mol each of C₂H₅OH and CH₃CO₂H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when $\frac{1}{3}$ mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)

Answer:

K_c = 4

Calculation of a Missing Equilibrium Concentration

If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.

Calculation of Equilibrium Concentrations from Initial Concentrations

If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium, we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

1. Determine the direction the reaction proceeds to come to equilibrium.
   1. Write a balanced chemical equation for the reaction.
   2. If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate Q_c from the initial concentrations and compare to K_c to determine the direction of change.
2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
   1. Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol x and express the other changes in terms of the smallest change.
   2. Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
3. Solve for the change and the equilibrium concentrations.
   1. Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x, and check any assumptions used to find x.
   2. Calculate the equilibrium concentrations.
4. Check the arithmetic.
   1. Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
      Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.
      In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.

Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products (similar to what was shown in Figure 13.7).

Consider the ionization of 0.150 M HA, a weak acid.

The most obvious way to determine the equilibrium concentrations would be to start with only reactants. This could be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, this is the ICE table and solution.

Setting up and solving the quadratic equation gives

Using the positive (physical) root, the equilibrium concentrations are

A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could be called the “all product” starting point. Assuming all of the HA ionizes gives

Using these as initial concentrations and “y” to represent the concentration of HA at equilibrium, this is the ICE table for this starting point.

Setting up and solving the quadratic equation gives

Retain a few extra significant figures to minimize rounding problems.

Rounding each solution to three significant figures gives

Using the physically significant root (0.140 M) gives the equilibrium concentrations as

Thus, the two approaches give the same results (to three decimal places), and show that both starting points lead to the same equilibrium conditions. The “all reactant” starting point resulted in a relatively small change (x) because the system was close to equilibrium, while the “all product” starting point had a relatively large change (y) that was nearly the size of the initial concentrations. It can be said that a system that starts “close” to equilibrium will require only a ”small” change in conditions (x) to reach equilibrium.

Recall that a small K_c means that very little of the reactants form products and a large K_c means that most of the reactants form products. If the system can be arranged so it starts “close” to equilibrium, then if the change (x) is small compared to any initial concentrations, it can be neglected. Small is usually defined as resulting in an error of less than 5%. The following two examples demonstrate this.

Example 13.9

Approximate Solution Starting Close to Equilibrium

What are the concentrations at equilibrium of a 0.15 M solution of HCN?

$$\text{HCN}(aq)\rightleftharpoons {\text{H}}^{\text{+}}(aq)+{\text{CN}}^{\text{−}}(aq)K_c=4.9\times {10}^{\text{−10}}$$

13.125

Solution

Using “x” to represent the concentration of each product at equilibrium gives this ICE table.

The exact solution may be obtained using the quadratic formula with

$$K_c=\frac{(x)(x)}{0.15-x}$$

13.126

solving

$$x^2+4.9\times {10}^{\text{−10}}-7.35\times {10}^{\text{−11}}=0$$

13.127

$$x=8.56\times {10}^{\text{−6}}M(\text{3 sig. figs.})=8.6\times {10}^{\text{−6}}M(\text{2 sig. figs.})$$

13.128

Thus [H⁺] = [CN^–] = x = 8.6 $\times$ 10^–6 M and [HCN] = 0.15 – x = 0.15 M.

In this case, chemical intuition can provide a simpler solution. From the equilibrium constant and the initial conditions, x must be small compared to 0.15 M. More formally, if $x\ll 0.15,$ then 0.15 – x ≈ 0.15. If this assumption is true, then it simplifies obtaining x

$$K_c=\frac{(x)(x)}{0.15-x}\approx \frac{x^2}{0.15}$$

13.129

$$4.9\times {10}^{\text{−10}}=\frac{x^2}{0.15}$$

13.130

$$x^2=(0.15)(4.9\times {10}^{\text{−10}})=7.4\times {10}^{\text{−11}}$$

13.131

$$x=\sqrt{7.4\times {10}^{\text{−11}}}=8.6\times {10}^{\text{−6}}M$$

13.132

In this example, solving the exact (quadratic) equation and using approximations gave the same result to two significant figures. While most of the time the approximation is a bit different from the exact solution, as long as the error is less than 5%, the approximate solution is considered valid. In this problem, the 5% applies to IF (0.15 – x) ≈ 0.15 M, so if

$$\frac{x}{0.15}\times 100\%=\frac{8.6\times {10}^{\text{−6}}}{0.15}\times 100\%=0.006\%$$

13.133

is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.

Check Your Learning

What are the equilibrium concentrations in a 0.25 M NH₃ solution?

$${\text{NH}}_3(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{NH}}_4{}^{\text{+}}(aq)+{\text{OH}}^{\text{−}}(aq)K_{\text{c}}=1.8\times {10}^{\text{−5}}$$

13.134

Assume that x is much less than 0.25 M and calculate the error in your assumption.

Answer:

$[{\text{OH}}^{\text{−}}]=[{\text{NH}}_4{}^{\text{+}}]=0.0021M;$ [NH₃] = 0.25 M, error = 0.84%

The second example requires that the original information be processed a bit, but it still can be solved using a small x approximation.

Example 13.10

Approximate Solution After Shifting Starting Concentration

Copper(II) ions form a complex ion in the presence of ammonia

$${\text{Cu}}^{\mathrm{2+}}(aq)+{\text{4NH}}_3(aq)\rightleftharpoons \text{Cu}{({\text{NH}}_3)}_4{}^{\mathrm{2+}}(aq)K_{\text{c}}=5.0\times {10}^{13}=\frac{[\text{Cu}{({\text{NH}}_3)}_4{}^{\mathrm{2+}}]}{[{\text{Cu}}^{\mathrm{2+}}(aq)]{[{\text{NH}}_3]}^4}$$

13.135

If 0.010 mol Cu²⁺ is added to 1.00 L of a solution that is 1.00 M NH₃ what are the concentrations when the system comes to equilibrium?

Solution

The initial concentration of copper(II) is 0.010 M. The equilibrium constant is very large so it would be better to start with as much product as possible because “all products” is much closer to equilibrium than “all reactants.” Note that Cu²⁺ is the limiting reactant; if all 0.010 M of it reacts to form product the concentrations would be

$$[{\text{Cu}}^{\mathrm{2+}}]=0.010-0.010=0M$$

13.136

$$[\text{Cu}{({\text{NH}}_3)}_4{}^{\mathrm{2+}}]=0.010M$$

13.137

$$[{\text{NH}}_3]=1.00-4\times 0.010=0.96M$$

13.138

Using these “shifted” values as initial concentrations with x as the free copper(II) ion concentration at equilibrium gives this ICE table.

Since we are starting close to equilibrium, x should be small so that

$$0.96+4x\approx 0.96M$$

13.139

$$0.010-x\approx 0.010M$$

13.140

$$K_c=\frac{(0.010-x)}{x{(0.96-4x)}^4}\approx \frac{(0.010)}{x{(0.96)}^4}=5.0\times {10}^{13}$$

13.141

$$x=\frac{(0.010)}{K_c{(0.96)}^4}=2.4\times {10}^{\text{−16}}M$$

13.142

Select the smallest concentration for the 5% rule.

$$\frac{2.4\times {10}^{\text{−16}}}{0.010}\times 100\%=2\times {10}^{\text{−12}}\%$$

13.143

This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are

$$[{\text{Cu}}^{\mathrm{2+}}]=x=2.4\times {10}^{\text{−16}}M$$

13.144

$$[{\text{NH}}_3]=0.96-4x=0.96M$$

13.145

$$[\text{Cu}{({\text{NH}}_3)}_4{}^{\mathrm{2+}}]=0.010-x=0.010M$$

13.146

By starting with the maximum amount of product, this system was near equilibrium and the change (x) was very small. With only a small change required to get to equilibrium, the equation for x was greatly simplified and gave a valid result well within the 5% error maximum.

Check Your Learning

What are the equilibrium concentrations when 0.25 mol Ni²⁺ is added to 1.00 L of 2.00 M NH₃ solution?

$${\text{Ni}}^{\mathrm{2+}}(aq)+{\text{6NH}}_3(aq)\rightleftharpoons \text{Ni}{({\text{NH}}_3)}_6{}^{\mathrm{2+}}(aq)K_{\text{c}}=5.5\times {10}^8$$

13.147

With such a large equilibrium constant, first form as much product as possible, then assume that only a small amount (x) of the product shifts left. Calculate the error in your assumption.

Answer:

$[\text{Ni}{({\text{NH}}_3)}_6{}^{\mathrm{2+}}]=0.25M,$ [NH₃] = 0.50 M, [Ni²⁺] = 2.9 $\times$ 10^–8 M, error = 1.2 $\times$ 10^–5%

Temperature Dependence of Spontaneity

As was previously demonstrated in the section on entropy in an earlier chapter, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

1. Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
2. Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
3. ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
4. ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in Figure 13.9.

Figure 13.9 There are four possibilities regarding the signs of enthalpy and entropy changes.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:

Such a plot is shown in Figure 13.10. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:

And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.

Figure 13.10 These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

Free Energy and Equilibrium

The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a finite driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium).

In the section on equilibrium, the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.

The free energy change for a process taking place with reactants and products present under nonstandard conditions, ΔG, is related to the standard free energy change, ΔG°, according to this equation:

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. We may use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.13.