Chapter 14: Fundamental Equilibrium Concepts

# 14.4 Equilibrium Calculations

### Learning Outcomes

• Calculate equilibrium constants from experimental equilibrium concentration or pressure measurements
• Calculate equilibrium concentrations from the equilibrium constant and all but one of the reactant or product equilibrium concentrations or pressures
• Calculate equilibrium concentrations from the equilibrium constant and initial equilibrium concentrations or pressures

Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology—for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.

Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:

$\ce{2NH3}(g)\rightleftharpoons\ce{N2}(g)+\ce{3H2}(g)$

As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH3 only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount x:

$\Delta[\ce{N2}] = +x$

the corresponding changes in the other species concentrations are

$\Delta[\ce{H}_2] =$$\Delta[\ce{N2}]\left(\dfrac{\ce{3mol} \ce{H2}}{1\text{mol}\ce{N2}}\right) = +3x$

$\Delta[\ce{NH3}] =$$\Delta[\ce{N2}]\left(\dfrac{\text{2mol} \ce{NH3}}{1\text{mol}\ce{N2}}\right) = -2x$

where the negative sign indicates a decrease in concentration.

### Example 14.4.1: Determining Relative Changes in Concentration

Derive the missing terms representing concentration changes for each of the following reactions.

1. $\begin{array}{ccccc}{\ce{C}}_{2}{\ce{H}}_{2}\text{(}g\text{)}&+\hfill & 2{\ce{Br}}_{2}\text{(}g\text{)}\hfill & \rightleftharpoons& {\ce{C}}_{2}{\ce{H}}_{2}{\ce{Br}}_{4}\text{(}g\text{)}\hfill \\ x\end{array}$
2. $\begin{array}{ccccc}{\ce{I}}_{2}\text{(}aq\text{)}&+\hfill & {\ce{I}}^{-}\text{(}aq\text{)}\hfill & \rightleftharpoons\hfill & {\ce{I}}_{3}{}^{-}\text{(}aq\text{)}\hfill \\ & & & & x \end{array}$
3. $\begin{array}{cccccc}{\ce{C}}_{3}{\ce{H}}_{8}\text{(}g\text{)}&+\hfill & 5{\ce{O}}_{2}\text{(}g\text{)}\hfill & \rightleftharpoons\hfill & 3{\ce{CO}}_{2}\text{(}g\text{)}+\hfill & 4{\ce{H}}_{2}\ce{O}\text{(}g\text{)}\hfill \\ x \end{array}$
Show Solution
1. $\begin{array}{ccccc}{\ce{C}}_{2}{\ce{H}}_{2}\text{(}g\text{)}&+\hfill & 2{\ce{Br}}_{2}\text{(}g\text{)}\hfill & \rightleftharpoons\hfill & {\ce{C}}_{2}{\ce{H}}_{2}{\ce{Br}}_{4}\text{(}g\text{)}\hfill \\ x\hfill& & 2x\hfill & & -x\hfill \end{array}$
2. $\begin{array}{ccccc}{\text{I}}_{2}\text{(}aq\text{)}&+\hfill & {\ce{I}}^{-}\text{(}aq\text{)}\hfill & \rightleftharpoons\hfill & {\ce{I}}_{3}{}^{-}\text{(}aq\text{)}\hfill \\ -x\hfill& & -x\hfill & & x\hfill \end{array}$
3. $\begin{array}{cccccc}{\ce{C}}_{3}{\ce{H}}_{8}\text{(}g\text{)}&+\hfill & 5{\ce{O}}_{2}\text{(}g\text{)}\hfill & \rightleftharpoons\hfill & 3{\ce{CO}}_{2}\text{(}g\text{)}+\hfill & 4{\ce{H}}_{2}\ce{O}\text{(}g\text{)}\hfill \\ x\hfill & & 5x\hfill & & -3x\hfill & -4x\hfill \end{array}$

## Calculations 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 14.4.2. 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 14.4.2: Calculation of an Equilibrium Constant

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

$\ce{I2}(aq)+\ce{I-}(aq)\rightleftharpoons\ce{I3-}(aq)$

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

Show Solution

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

${K}_{c}=\dfrac{\left[{\ce{I}}_{3}{}^{-}\right]}{\left[{I}_{2}\right]\left[{\ce{I}}^{-}\right]}$

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 $\ce{I2}$ is 6.61 × 10−4M so that

$1.000\times {10}^{-3}-x=6.61\times {10}^{-4}$
$x=1.000\times {10}^{-3}-6.61\times {10}^{-4}$
$=3.39\times {10}^{-4}M$

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:

${K}_{c}={Q}_{c}=\dfrac{\left[{\ce{I}}_{3}{}^{-}\right]}{\left[{I}_{2}\right]\left[{\ce{I}}^{-}\right]}$
$=\dfrac{3.39\times {10}^{-4}M}{\left(6.61\times {10}^{-4}M\right)\left(6.61\times {10}^{-4}M\right)}=776$

### 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.

### Example 14.4.3: Calculation of a Missing Equilibrium Concentration

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, $\ce{N2}(g)+\ce{O2}(g)\rightleftharpoons\ce{2NO}(g)$, is 4.1 × 10−4. Find the concentration of $\ce{NO}$(g) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N2] = 0.036 mol/L and $\ce{O2}$ 0.0089 mol/L.

Show Solution

We are given all of the equilibrium concentrations except that of $\ce{NO}$. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.

$\begin{array}{rll}{ }{K}_{c}&=&\dfrac{{\left[\ce{NO}\right]}^{2}}{\left[{\ce{N}}_{2}\right]\left[{\ce{O}}_{2}\right]}\\{\left[\ce{NO}\right]}^{2}&=&{K}_{c}\left[{\ce{N}}_{2}\right]\left[{\ce{O}}_{2}\right]\\\left[\ce{NO}\right]&=&\sqrt{{K}_{c}\left[{\ce{N}}_{2}\right]\left[{\ce{O}}_{2}\right]}\\ &=&\sqrt{\left(4.1\times {10}^{-4}\right)\left(0.036\right)\left(0.0089\right)}\\ &=&\sqrt{1.31\times {10}^{-7}}\\ &=&3.6\times {10}^{-4}\end{array}$

Thus [NO] is 3.6 × 10−4 mol/L at equilibrium under these conditions.

We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.

$\begin{array}{ }{K}_{c}&=&\dfrac{{\left[\ce{NO}\right]}^{2}}{\left[{\ce{N}}_{2}\right]\left[{\ce{O}}_{2}\right]}\\&=&\dfrac{{\left(3.6\times {10}^{-4}\right)}^{2}}{\left(0.036\right)\left(0.0089\right)}\\&=&4.0\times {10}^{-4}\end{array}$

The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.

### Calculation of Changes in Concentration

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 14.4.4: Calculation of Concentration Changes as a Reaction Goes to Equilibrium

Under certain conditions, the equilibrium constant for the decomposition of $\ce{PCl5}$(g) into $\ce{PCl3}$(g) and $\ce{Cl2}$(g) is 0.0211. What are the equilibrium concentrations of $\ce{PCl5}$, $\ce{PCl3}$, and $\ce{Cl2}$ if the initial concentration of $\ce{PCl5}$ was 1.00 M?

Show Solution

Use the stepwise process described above.

#### Step 1: Determine the direction the reaction proceeds.

The balanced equation for the decomposition of $\ce{PCl5}$ is

${\ce{PCl}}_{5}\left(g\right)\rightleftharpoons{\ce{PCl}}_{3}\left(g\right)+{\ce{Cl}}_{2}\left(g\right)$

Because we have no products initially, Qc = 0 and the reaction will proceed to the right.

#### Step 2: Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.

Develop an ICE table.

#### Step 3: Solve for the change and the equilibrium concentrations.

Solve for the change and the equilibrium concentrations.

Substituting the equilibrium concentrations into the equilibrium constant equation gives

$\begin{array}{rcl}{K}_{c}&=&\dfrac{\left[{\ce{PCl}}_{3}\right]\left[{\ce{Cl}}_{2}\right]}{\left[{\ce{PCl}}_{5}\right]}=0.0211\\{}&=&\dfrac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}\\0.0211&=&\dfrac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}\\0.0211\left(1.00-x\right)&=&{x}^{2}\\{x}^{2}+0.0211 x-0.0211&=&0\end{array}$

Essential Mathematics shows us an equation of the form ax2 + bx + c = 0 can be rearranged to solve for x:

$x=\dfrac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}$

In this case, $a = 1, b = 0.0211$, and $c = −0.0211$. Substituting the appropriate values for a, b, and c yields:

$\begin{array}{rcl}{x}&=&\dfrac{-0.0211\pm\sqrt{\left(0.0211\right)^{2}-4\left(1\right)\left(-0.0211\right)}}{2\left(1\right)}\\{}&=&\dfrac{-0.0211\pm\sqrt{\left(4.45\times{10}^{-4}\right)+\left(8.44\times{10}^{-2}\right)}}{2}\\{}&=&\dfrac{-0.0211\pm0.291}{2}\end{array}$

The two roots of the quadratic are, therefore,

$x=\dfrac{-0.0211+0.291}{2}=0.135$ or $x=\dfrac{-0.0211-0.291}{2}=-0.156$.

For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M.

The equilibrium concentrations are

• $[\ce{PCl5}]=1.00-0.135=0.87M$
• $[\ce{PCl3}]=x=0.135M$
• $[\ce{Cl2}]=x=0.135M$

#### Step 4: Check the arithmetic.

Confirm the calculated equilibrium concentrations.

Substitution into the expression for Kc (to check the calculation) gives

${K}_{c}=\dfrac{\left[\ce{PCl3}\right]\left[\ce{Cl2}\right]}{\left[\ce{PCl5}\right]}=\dfrac{\left(0.135\right)\left(0.135\right)}{0.87}=0.021$

The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.

### Example 14.4.5: Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption

What are the concentrations at equilibrium of a 0.15 M solution of $\ce{HCN}$?

$\ce{HCN}(aq) \rightleftharpoons \ce{H+}(aq) + \ce{CN-} (aq)$ $K_c = 4.9 \text{x} 10^{-10}$

Show Solution

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

Substitute the equilibrium concentration terms into the Kc expression

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

rearrange to the quadratic form and solve for x

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

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

Thus $\ce{H+}$ = $\ce{CN-}$ = x = 8.6 × 10–6 M and $\ce{HCN}$ = 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:

if $x \ll$ 0.15 M, then (0.15 – x) ≈ 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:

Kc = $\dfrac{(x)(x)}{0.15-x}$ ≈ $\dfrac{x^2}{0.15}$

4.9 × 10-10 = $\dfrac{x^2}{0.15}$

x2 = (0.15)(4.9 × 10-10) = 7.4 × 10-11

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

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

8.6 × 10-6 $\ll$ 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.

### Key Concepts and Summary

Calculating values for equilibrium constants and/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations.

### Try It

1. Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.

$\ce{CH4}(g)+\ce{H2O}(g)\rightleftharpoons\ce{3H2}(g)+\ce{CO}(g)$

What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: $\ce{CH4}$, 0.126 M; $\ce{H2O}$, 0.242 M; $\ce{CO}$, 0.126 M; $\ce{H2}$ 1.15 M, at a temperature of 760 °C?

2. Analysis of the gases in a sealed reaction vessel containing $\ce{NH3}$, $\ce{N2}$, and $\ce{H2}$ at equilibrium at 400 °C established the concentration of $\ce{N2}$ to be 1.2 M and the concentration of $\ce{H2}$ to be 0.24 M.

$\ce{N2}\left(g\right)+\ce{3H2}(g)\rightleftharpoons\ce{2NH3}(g){K}_{c}=0.50\text{ at }400^{\circ}\text{C}$

Calculate the equilibrium molar concentration of $\ce{NH3}$.

3. Assume that the change in concentration of $\ce{N2O4}$ is small enough to be neglected in the following problem.

1. Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of $\ce{N2O4}$ with chloroform as the solvent.
$\ce{N2O4}\left(g\right)\rightleftharpoons\ce{2NO2}(g){K}_{c}=1.07\times {10}^{-5}$ in chloroform
2. Show that the change is small enough to be neglected.
Show Selected Solutions

2. Write the equilibrium constant expression and solve for $\ce{NH3}$.

${K}_{c}=\frac{{\left[{\ce{NH}}_{3}\right]}^{2}}{\left[{\ce{N}}_{2}\right]{\left[{\ce{H}}_{2}\right]}^{3}}=\frac{{\left[{\ce{NH}}_{3}\right]}^{2}}{\left[1.2\right]{\left[0.24\right]}^{3}}=0.50$

$\ce{NH3}$2 = 1.2 × (0.24)3 × 0.50 = 0.0083

$\ce{NH3}$ = 9.1 × 10-2M

3. (a) Write the starting conditions, change, and equilibrium constant in tabular form.

$\ce{NO2}$ $\ce{N2O4}$
Initial concentration (M) 0 0.129
Change (M) +2x x
Equilibrium concentration (M) 2x 0.129 − x

Since K is very small, ignore x in comparison with 0.129 M. The equilibrium expression is

$\begin{array}{l} {K}_{c}=1.07\times {10}^{-5}=\frac{{\left[{\ce{NO}}_{2}\right]}^{2}}{\left[{\ce{N}}_{2}{\ce{O}}_{4}\right]}=\frac{{\left(2x\right)}^{2}}{\left(0.129-x\right)}=\frac{{\left(2x\right)}^{2}}{0.129}\hfill \\ {x}^{2}=\frac{0.129\times 1.07\times {10}^{-5}}{4}=3.45\times {10}^{-7}\hfill \end{array}$

x = 5.87 × 10−4

The concentrations are:

• [NO2] = 2x = 5.87 × 10-4 = 1.17 × 10-3M
• [N2O4] = 0.129 – x = 0.129 – 5.87 × 10-4 = 0.128 M

(b) Percent error $=\frac{5.87\times {10}^{-4}}{0.129}\times 100\%=0.455\%$. The change in concentration of $\ce{N2O4}$ is far less than the 5% maximum allowed.