13.4 Equilibrium Calculations

Calculation of an Equilibrium Constant

The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example 13.2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.

Example 13.7

Calculation of an Equilibrium Constant

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

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

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

To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:

$$\begin{array}{ccc}{K}_C\hfill & =\hfill & [{\text{I}}_3{}^{\text{−}}]\hfill \\ \hfill & \hfill & [{\text{I}}_2][{\text{I}}^{\text{−}}]\hfill \end{array}$$

Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.

At equilibrium the concentration of I₂ is 6.61 $\times$ 10⁻⁴ M so that

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

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

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

The ICE table may now be updated with numerical values for all its concentrations:

Finally, substitute the equilibrium concentrations into the K expression and solve:

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

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

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.

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

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 a solute in this reaction.)

Answer:

K_c = 4

Calculation of a Missing Equilibrium Concentration

When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.

Calculation of Equilibrium Concentrations from Initial Concentrations

Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:

1. Identify the direction in which the reaction will proceed to reach equilibrium.
2. Develop an ICE table.
3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.
4. Confirm the calculated equilibrium concentrations.

The last two example exercises of this chapter demonstrate the application of this strategy.

Example 13.10

Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption

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

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

Solution

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

Substitute the equilibrium concentration terms into the K_c expression

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

Rearrange to the quadratic form and solve for x

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

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

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

Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change:

$$\text{if}x\ll 0.15\text{M},\text{then}(0.15-x)\approx 0.15$$

This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:

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

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

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

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

The value of x calculated is, indeed, much less than the initial concentration

$$8.6\times {10}^{\mathrm{-6}}\ll 0.15$$

and so the approximation was justified. If this simplified approach were to yield a value for x that did not justify the approximation, the calculation would need to be repeated without making the approximation.

Check Your Learning

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

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

Answer:

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

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 13.8.

Figure 13.8 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 13.9. 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:

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 13.9 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 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 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 (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change, 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. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.13.

For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 13.1.

To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure 13.10). If a system consists of reactants and products in nonequilibrium amounts (Q ≠ K), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Figure 13.10 These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.