Chemistry for Majors: Atoms First

Module 13: Fundamental Equilibrium Concepts

Equilibrium Calculations

Learning Objectives

By the end of this module, you will be able to:

Write equations representing changes in concentration and pressure for chemical species in equilibrium systems
Use algebra to perform various types of equilibrium calculations

We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.

Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.

On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:

2{\text{NH}}_{3}\left(g\right){\rightleftharpoons}{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)

If a sample of ammonia decomposes in a closed system and the concentration of N₂ increases by 0.11 M, the change in the N₂ concentration, Δ[N₂], the final concentration minus the initial concentration, is 0.11 M. The change is positive because the concentration of N₂ increases.

The change in the H₂ concentration, Δ[H₂], is also positive—the concentration of H₂ increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H₂ is three times the change in the concentration of N₂, because for each mole of N₂ produced, 3 moles of H₂ are produced.

\Delta\left[{\text{H}}_{2}\right]=3\times \Delta\left[{\text{N}}_{2}\right]

=3\times \left(0.11M\right)=0.33M

The change in concentration of NH₃, Δ[NH₃], is twice that of Δ[N₂]; the equation indicates that 2 moles of NH₃ must decompose for each mole of N₂ formed. However, the change in the NH₃ concentration is negative because the concentration of ammonia decreases as it decomposes.

\Delta\left[{\text{NH}}_{3}\right]=-2\times \Delta\left[{\text{N}}_{2}\right]=-2\times \left(0.11M\right)=-0.22M

We can relate these relationships directly to the coefficients in the equation

\begin{array}{ccccc}{ }2\text{NH}_3\left(g\right)&\rightleftharpoons&\text{N}_2(g)&+&3\text{H}_2\left(g\right)\\ \Delta\left[\text{NH}_3\right]=-2\times\Delta\left[\text{N}_2\right]&{ }&\Delta\left[\text{N}_2\right]=0.11M&{ }&\Delta\left[\text{H}_2\right]=3\times\Delta\left[\text{N}_2\right]\end{array}

Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.

If we did not know the magnitude of the change in the concentration of N₂, we could represent it by the symbol x.

\Delta\left[{\text{N}}_{2}\right]=x

The changes in the other concentrations would then be represented as:

\Delta\left[{\text{H}}_{2}\right]=3\times \Delta\left[{\text{N}}_{2}\right]=3x

\Delta\left[{\text{NH}}_{3}\right]=-2\times \Delta\left[{\text{N}}_{2}\right]=-2x

The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.

\begin{array}{ccccc}{ }2\text{NH}_3\left(g\right)&\rightleftharpoons&\text{N}_2(g)&+&3\text{H}_2\left(g\right)\\ -2x&{ }&x&{ }&3x\end{array}

The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.

Example 1: Determining Relative Changes in Concentration

Complete the changes in concentrations for each of the following reactions.

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

Show Answer

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

Check Your Learning

Complete the changes in concentrations for each of the following reactions:

$\begin{array}{ccccc}2{\text{SO}}_{2}\text{(}g\text{)}&+\qquad & {\text{O}}_{2}\text{(}g\text{)}\qquad & \rightleftharpoons\qquad & 2{\text{SO}}_{3}\text{(}g\text{)}\qquad \\ & & x & \qquad \end{array}$
$\begin{array}{ccc}{\text{C}}_{4}{\text{H}}_{8}\text{(}g\text{)}\qquad & \rightleftharpoons\qquad & 2{\text{C}}_{2}{\text{H}}_{4}\text{(}g\text{)}\qquad \\ & & -2x\qquad \end{array}$
$\begin{array}{ccccccc}4{\text{NH}}_{3}\text{(}g\text{)}&+\qquad & 7{\text{H}}_{2}\text{O}\text{(}g\text{)}\qquad & \rightleftharpoons\qquad & 4{\text{NO}}_{2}\text{(}g\text{)}&+\qquad & 6{\text{H}}_{2}\text{O}\text{(}g\text{)}\end{array}$

Show Answer

$\begin{array}{ccccc}2{\text{SO}}_{2}\text{(}g\text{)}&+\qquad & {\text{O}}_{2}\text{(}g\text{)}\qquad & \rightleftharpoons\qquad & 2{\text{SO}}_{3}\text{(}g\text{)}\qquad \\ 2x & & x & -2x & \end{array}$
$\begin{array}{ccc}{\text{C}}_{4}{\text{H}}_{8}\text{(}g\text{)}\qquad & \rightleftharpoons\qquad & 2{\text{C}}_{2}{\text{H}}_{4}\text{(}g\text{)}\qquad \\ x & & -2x\qquad \end{array}$
There are two possible solutions for this question:

$\begin{array}{ccccccc}4{\text{NH}}_{3}\text{(}g\text{)}&+\qquad & 7{\text{H}}_{2}\text{O}\text{(}g\text{)}\qquad & \rightleftharpoons\qquad & 4{\text{NO}}_{2}\text{(}g\text{)}&+\qquad & 6{\text{H}}_{2}\text{O}\text{(}g\text{)}\\ 4x& &7x& &-4x& &-6x \end{array}$

or

$\begin{array}{ccccccc}4{\text{NH}}_{3}\text{(}g\text{)}&+\qquad & 7{\text{H}}_{2}\text{O}\text{(}g\text{)}\qquad & \rightleftharpoons\qquad & 4{\text{NO}}_{2}\text{(}g\text{)}&+\qquad & 6{\text{H}}_{2}\text{O}\text{(}g\text{)}\\ -4x& &-7x& &4x& &6x \end{array}$

Exercises

A reaction is represented by this equation:
$\text{A}\left(aq\right)+2\text{B}\left(aq\right)\rightleftharpoons2\text{C}\left(aq\right){K}_{c}=1\times {10}^{3}$
1. Write the mathematical expression for the equilibrium constant.
2. Using concentrations ≤1 M, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.
A reaction is represented by this equation:
$2\text{W}\left(aq\right)\rightleftharpoons\text{X}\left(aq\right)+2\text{Y}\left(aq\right){K}_{c}=5\times {10}^{-4}$
1. Write the mathematical expression for the equilibrium constant.
2. Using concentrations of ≤1 M, make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.

Show Answer to Question 1

{K}_{c}=\frac{{\left[\text{C}\right]}^{2}}{\left[\text{A}\right]{\left[\text{B}\right]}^{2}}\text{.}

There are many different sets of equilibrium concentrations; two are [A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.

Calculations Involving Equilibrium Concentrations

Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Q_c (i.e., the law of mass action) to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Q_c = K_c (at equilibrium) in all of these situations and that there are only three basic types of calculations:

Calculation of an equilibrium constant. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated.
Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated.

A similar list could be generated using Q_P, K_P, and partial pressure. We will look at solving each of these cases in sequence.

Calculation of an Equilibrium Constant

Since the law of mass action is the only equation we have to describe the relationship between K_c and the concentrations of reactants and products, any problem that requires us to solve for K_c must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for K_c, as it will be the only unknown.

Example 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. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.

Example 2: Calculation of an Equilibrium Constant

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

{\text{I}}_{2}\left(aq\right)+{\text{I}}^{-}\left(aq\right)\rightleftharpoons{\text{I}}_{3}{}^{-}\left(aq\right)

If a solution with the concentrations of I₂ and 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 Answer

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.

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, “I subscript 2 plus sign I superscript negative sign equilibrium arrow I subscript 3 superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.000 times 10 to the negative third power, negative x, [ I subscript 2 ] subscript i minus x. The second column has the following: 1.000 times 10 to the negative third power, negative x, [ I superscript negative sign ] subscript i minus x. The third column has the following: 0, positive x, [ I superscript negative sign ] subscript i plus x.

Since the equilibrium concentration of I₂ is given, we can solve for x. At equilibrium the concentration of 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

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{\left[{\text{I}}_{3}{}^{-}\right]}{\left[{I}_{2}\right]\left[{\text{I}}^{-}\right]}

=\frac{3.39\times {10}^{-4}M}{\left(6.61\times {10}^{-4}M\right)\left(6.61\times {10}^{-4}M\right)}=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 not a solvent in this reaction.)

Show Answer

K_c = 4

Exercises

What is the value of the equilibrium constant at 500 °C for the formation of NH₃ according to the following equation?
${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)$

An equilibrium mixture of NH₃(g), H₂(g), and N₂(g) at 500 °C was found to contain 1.35 M H₂, 1.15 M N₂, and 4.12 × 10^-1M NH₃.
Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.

${\text{CH}}_{4}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)\rightleftharpoons3{\text{H}}_{2}\left(g\right)+\text{CO}\left(g\right)$

What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH₄, 0.126 M; H₂O, 0.242 M; CO, 0.126 M; H₂ 1.15 M, at a temperature of 760 °C?
A 0.72-mol sample of PCl₅ is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl₃(g) and 0.40 mol of Cl₂(g). Calculate the value of the equilibrium constant for the decomposition of PCl₅ to PCl₃ and Cl₂ at this temperature.
At 1 atm and 25 °C, NO₂ with an initial concentration of 1.00 M is 3.3 × 10^-3% decomposed into NO and O₂. Calculate the value of the equilibrium constant for the reaction

$2{\text{NO}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)+{\text{O}}_{2}\left(g\right)$
Calculate the value of the equilibrium constant K_P for the reaction
$2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightleftharpoons2\text{NOCl}\left(g\right)$
from these equilibrium pressures: NO, 0.050 atm; Cl₂, 0.30 atm; NOCl, 1.2 atm.
When heated, iodine vapor dissociates according to this equation:
${\text{I}}_{2}\left(g\right)\rightleftharpoons2\text{I}\left(g\right)$

At 1274 K, a sample exhibits a partial pressure of I₂ of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, K_P, for the decomposition at 1274 K.
A sample of ammonium chloride was heated in a closed container:
${\text{NH}}_{4}\text{Cl}\left(s\right)\rightleftharpoons{\text{NH}}_{3}\left(g\right)+\text{HCl}\left(g\right)$

At equilibrium, the pressure of NH₃(g) was found to be 1.75 atm. What is the value of the equilibrium constant K_P for the decomposition at this temperature?
At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant K_P for the transformation at 60 °C?

${\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right)$

Show Selected Answer

1. The reaction may be written as

{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)

The equilibrium constant for the reaction is

{K}_{c}=\frac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2}\right]}^{3}}=\frac{{\left(4.12\times {10}^{-1}\right)}^{2}}{\left(1.15\right){\left(1.35\right)}^{3}}=\frac{{\left(0.170\right)}^{2}}{\left(1.15\right){\left(2.46\right)}^{3}}0.0600=6.00\times10^{-2}

3. The decomposition of PCl₅ to PCl₃ and Cl₂ is given as

\begin{array}{l}{\text{PCl}}_{5}\left(g\right)\rightleftharpoons{\text{PCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\qquad \\{K}_{c}=\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}\qquad \end{array}

Let x = change in [PCl₅].

	[PCl₅]	[PCl₃]	[Cl₂]
Initial concentration (M)	0.72	0	0
Change (M)	−x	x	x
Equilibrium concentration (M)	0.72 − x = 0.32	0 + x = 0.40	0 + x = 0.40

{K}_{c}=\frac{\left(0.40\right)\left(0.40\right)}{\left(0.32\right)}=0.50

5. The equilibrium equation is

{K}_{P}=\frac{{\left[\text{NOCl}\right]}^{2}}{{\left[\text{NO}\right]}^{2}\left[{\text{Cl}}_{2}\right]}=\frac{{\left(1.2\right)}^{2}}{{\left(0.050\right)}^{2}\left(0.30\right)}=\frac{1.44}{\left(2.5\times {10}^{-3}\right)\left(0.30\right)}=1.9\times {10}^{3}

7. Because the decomposition must generate the same pressure of HCl as NH₃, 1.75 atm of HCl must be present.

{K}_{p}={P}_{{\text{NH}}_{3}}{P}_{\text{HCl}}=\left(1.75\text{atm}\right)\left(1.75\text{atm}\right)=3.06

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.

Example 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,

{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)

, is 4.1 × 10⁻⁴. Find the concentration of NO(g) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N₂] = 0.036 mol/L and [O₂] 0.0089 mol/L.

Show Answer

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

\begin{array}{rll}{ }{K}_{c}={Q}_{c}&=&\frac{{\left[\text{NO}\right]}^{2}}{\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}\\{\left[\text{NO}\right]}^{2}&=&{K}_{c}\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]\\\left[\text{NO}\right]&=&\sqrt{{K}_{c}\left[{\text{N}}_{2}\right]\left[{\text{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}{ }{Q}_{c}&=&\frac{{\left[\text{NO}\right]}^{2}}{\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}\\&=&\frac{{\left(3.6\times {10}^{-4}\right)}^{2}}{\left(0.036\right)\left(0.0089\right)}\\&=&4.0\times {10}^{-4}={K}_{c}\end{array}

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

Check Your Learning

The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10⁻². Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.

Show Answer

1.53 mol/L

Exercises

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

${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right){K}_{c}=0.50\text{ at }400^{\circ}\text{C}$

Calculate the equilibrium molar concentration of NH₃.
Carbon reacts with water vapor at elevated temperatures.

$\text{C}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\rightleftharpoons\text{CO}\left(g\right)+{\text{H}}_{2}\left(g\right){K}_{c}=0.2\text{ at }1000^{\circ}\text{C}$

What is the concentration of CO in an equilibrium mixture with [H₂O] = 0.500 M at 1000 °C?
Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.

$\text{CoO}\left(s\right)+\text{CO}\left(g\right)\rightleftharpoons\text{Co}\left(s\right)+{\text{CO}}_{2}\left(g\right){K}_{c}=4.90\times {10}^{2}\text{ at }550^{\circ}\text{C}$

What concentration of CO remains in an equilibrium mixture with [CO₂] = 0.100 M?
A student solved the following problem and found [N₂O₄] = 0.16 M at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N₂O₄ in a mixture formed from a sample of NO₂ with a concentration of 0.10 M?

$2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right){K}_{c}=160$
A student solved the following problem and found the equilibrium concentrations to be [SO₂] = 0.590 M, [O₂] = 0.0450 M, and [SO₃] = 0.260 M. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:

$2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{SO}}_{3}\left(g\right){K}_{c}=4.32$

What are the equilibrium concentrations of all species in a mixture that was prepared with [SO₃] = 0.500 M, [SO₂] = 0 M, and [O₂] = 0.350 M

Show Selected Answers

1. Write the equilibrium constant expression and solve for [NH₃].

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

[NH₃]² = 1.2 × (0.24)³ × 0.50 = 0.0083

[NH₃] = 9.1 × 10^-2M

3. Write the equilibrium constant expression and solve for [CO].

\begin{array}{l}{K}_{c}=\frac{\left[{\text{CO}}_{2}\right]}{\left[\text{CO}\right]}=\frac{0.100}{\left[\text{CO}\right]}=4.90\times {10}^{2}\qquad \\ \left[\text{CO}\right]=\frac{0.100}{4.90\times {10}^{2}}=2.0\times {0}^{-4}M\qquad \end{array}

5. Calculate Q based on the calculated concentrations and see if it is equal to K_c. Because Q does equal 4.32, the system must be at equilibrium.

Calculation of Changes in Concentration

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.

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.
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).
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.
Check the arithmetic.
1. Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
2. 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.
3. 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.

Example 4: Calculation of Concentration Changes as a Reaction Goes to Equilibrium

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

Show Answer

Use the stepwise process described above.

Step 1: Determine the direction the reaction proceeds.

The balanced equation for the decomposition of PCl₅ is

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

Because we have no products initially, Q_c = 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.

Let us represent the increase in concentration of PCl₃ by the symbol x. The other changes may be written in terms of x by considering the coefficients in the chemical equation.

\begin{array}{llll}{\text{PCl}}_{5}\left(g\right)\qquad & \rightleftharpoons\qquad & {\text{PCl}}_{3}\left(g\right)+\qquad & {\text{Cl}}_{2}\left(g\right)\qquad \\ -x\qquad & & x\qquad & x\qquad \end{array}

The changes in concentration and the expressions for the equilibrium concentrations are:

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

Substituting the equilibrium concentrations into the equilibrium constant equation gives

\begin{array}{rcl}{K}_{c}&=&\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}=0.0211\\{}&=&\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}\end{array}

This equation contains only one variable, x, the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.

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

Hence

x=\frac{-0.0211+0.291}{2}=0.135

x=\frac{-0.0211-0.291}{2}=-0.156

.

Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M.

The equilibrium concentrations are

$\left[{\text{PCl}}_{5}\right]=1.00-0.135=0.87M$
$\left[{\text{PCl}}_{3}\right]=x=0.135M$
$\left[{\text{Cl}}_{2}\right]=x=0.135M$

Step 4: Check the arithmetic.

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

{K}_{c}=\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}=\frac{\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 K_c given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.

Check Your Learning

Acetic acid, CH₃CO₂H, reacts with ethanol, C₂H₅OH, to form water and ethyl acetate, CH₃CO₂C₂H₅.

{\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{C}}_{2}{\text{H}}_{5}\text{OH}\rightleftharpoons{\text{CH}}_{3}{\text{CO}}_{2}{\text{C}}_{2}{\text{H}}_{5}+{\text{H}}_{2}\text{O}

The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH₃CO₂H, 0.15 M in C₂H₅OH, 0.40 M in CH₃CO₂C₂H₅, and 0.40 M in H₂O are mixed in enough dioxane to make 1.0 L of solution?

Show Answer

[CH₃CO₂H] = 0.36 M, [C₂H₅OH] = 0.36 M, [CH₃CO₂C₂H₅] = 0.17 M, [H₂O] = 0.17 M

Check Your Learning

A 1.00-L flask is filled with 1.00 moles of H₂ and 2.00 moles of I₂. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H₂, I₂, and HI in moles/L?

{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\rightleftharpoons2\text{HI}\left(g\right)

Show Answer

[H₂] = 0.06 M, [I₂] = 1.06 M, [HI] = 1.88 M

Exercises

Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
1. $\begin{array}{llll}2{\text{SO}}_{3}\left(g\right)\qquad & \rightleftharpoons\qquad & 2{\text{SO}}_{2}\left(g\right)+\qquad & {\text{O}}_{2}\left(g\right)\qquad \\ \text{ }\qquad & & \text{ }\qquad & +x\qquad \\ \text{ }\qquad & & \text{ }\qquad & 0.125M\qquad \end{array}$
2. $\begin{array}{lllll}4{\text{NH}}_{3}\left(g\right)\qquad & +3{\text{O}}_{2}\left(g\right)\qquad & \rightleftharpoons\qquad & 2{\text{N}}_{2}\left(g\right)+\qquad & 6{\text{H}}_{2}\text{O}\left(g\right)\qquad \\ \text{ }\qquad & 3x\qquad & & \text{ }\qquad & \text{ }\qquad \\ \text{ }\qquad & 0.24M\qquad & & \text{ }\qquad & \text{ }\qquad \end{array}$
3. Change in pressure:
  
  $\begin{array}{llll}2{\text{CH}}_{4}\left(g\right)\qquad & \rightleftharpoons\qquad & {\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\qquad & 3{\text{H}}_{2}\left(g\right)\qquad \\ \text{ }\qquad & & x\qquad & \text{ }\qquad \\ \text{ }\qquad & & 25\text{torr}\qquad & \text{ }\qquad \end{array}$
4. Change in pressure:
  
  $\begin{array}{lllll}{\text{CH}}_{4}\left(g\right)+\qquad & {\text{H}}_{2}\text{O}\left(g\right)\qquad & \rightleftharpoons\qquad & \text{CO}\left(g\right)+\qquad & 3{\text{H}}_{2}\left(g\right)\qquad \\ \text{ }\qquad & x\qquad & & \text{ }\qquad & \text{ }\qquad \\ \text{ }\qquad & 5\text{atm}\qquad & & \text{ }\qquad & \text{ }\qquad \end{array}$
5. $\begin{array}{llll}{\text{NH}}_{4}\text{Cl}\left(s\right)\qquad & \rightleftharpoons\qquad & {\text{NH}}_{3}\left(g\right)+\qquad & \text{HCl}\left(g\right)\qquad \\ & & x\qquad & \text{ }\qquad \\ & & \qquad 1.03\times {10}^{-4}M\qquad & \text{ }\qquad \end{array}$
6. change in pressure:
  
  $\begin{array}{cccc}\text{Ni}\left(s\right)+\qquad & 4\text{CO}\left(g\right)\qquad & \rightleftharpoons\qquad & \text{Ni}{\left(\text{CO}\right)}_{4}\left(g\right)\qquad \\ & 4x\qquad & & \text{ }\qquad \\ & \qquad 0.40\text{atm}\qquad & & \text{ }\qquad \end{array}$
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
1. $\begin{array}{cccc}2{\text{H}}_{2}\left(g\right)+\qquad & {\text{O}}_{2}\left(g\right)\qquad & \rightleftharpoons\qquad & 2{\text{H}}_{2}\text{O}\left(g\right)\qquad \\ \text{ }\qquad & \text{ }\qquad & & +2x\qquad \\ \text{ }\qquad & \text{ }\qquad & & 1.50M\qquad \end{array}$
2. $\begin{array}{ccccc}{\text{CS}}_{2}\left(g\right)+\qquad & 4{\text{H}}_{2}\left(g\right)\qquad & \rightleftharpoons\qquad & {\text{CH}}_{4}\left(g\right)+\qquad & 2{\text{H}}_{2}\text{S}\left(g\right)\qquad \\ x\qquad & \text{ }\qquad & & \text{ }\qquad & \text{ }\qquad \\ 0.020M\qquad & \text{ }\qquad & & \text{ }\qquad & \text{ }\qquad \end{array}$
3. Change in pressure:
  
  $\begin{array}{cccc}{\text{H}}_{2}\left(g\right)+\qquad & {\text{Cl}}_{2}\left(g\right)\qquad & \rightleftharpoons\qquad & 2\text{HCl}\left(g\right)\qquad \\ x\qquad & \text{ }\qquad & & \text{ }\qquad \\ 1.50\text{atm}\qquad & \text{ }\qquad & & \text{ }\qquad \end{array}$
4. Change in pressure:
  
  $\begin{array}{ccccc}2{\text{NH}}_{3}\left(g\right)\qquad & +2{\text{O}}_{2}\left(g\right)\qquad & \rightleftharpoons\qquad & {\text{N}}_{2}\text{O}\left(g\right)+\qquad & 3{\text{H}}_{2}\text{O}\left(g\right)\qquad \\ \text{ }\qquad & \text{ }\qquad & & \text{ }\qquad & x\qquad \\ \text{ }\qquad & \text{ }\qquad & & \text{ }\qquad & 60.6\text{torr}\qquad \end{array}$
5. $\begin{array}{cccc}{\text{NH}}_{4}\text{HS}\left(s\right)\qquad & \rightleftharpoons\qquad & {\text{NH}}_{3}\left(g\right)+\qquad & {\text{H}}_{2}\text{S}\left(g\right)\qquad \\ & & x\qquad & \text{ }\qquad \\ & & 9.8\times {10}^{-6}M\qquad & \text{ }\qquad \end{array}$
6. Change in pressure:
  
  $\begin{array}{cccc}\text{Fe}\left(s\right)+\qquad & 5\text{CO}\left(g\right)\qquad & \rightleftharpoons\qquad & \text{Fe}{\left(\text{CO}\right)}_{4}\left(g\right)\qquad \\ & \text{ }\qquad & & x\qquad \\ & \text{ }\qquad & & 0.012\text{atm}\qquad \end{array}$
Why are there no changes specified for Ni in question 11 , part (f)? What property of Ni does change?
Why are there no changes specified for NH₄HS in question 12, part (e)? What property of NH₄HS does change?

Show Selected Answers

1. The changes in concentrations (or pressure, if requested) are as follows:

$\begin{array}{llll}2{\text{SO}}_{3}\left(g\right)\qquad & \rightleftharpoons\qquad & 2{\text{SO}}_{2}\left(g\right)+\qquad & {\text{O}}_{2}\left(g\right)\qquad \\ \text{ }\qquad -2x & &2x\qquad & +x\qquad \\ -0.250M\qquad & & 0.250M\qquad & 0.125M\qquad \end{array}$
$\begin{array}{lllll}4{\text{NH}}_{3}\left(g\right)\qquad & +3{\text{O}}_{2}\left(g\right)\qquad & \rightleftharpoons\qquad & 2{\text{N}}_{2}\left(g\right)+\qquad & 6{\text{H}}_{2}\text{O}\left(g\right)\qquad \\ 4x\qquad & 3x\qquad & & -2x\qquad & -6x\qquad \\ 0.32M\qquad & 0.24M\qquad & &-0.16M\qquad & -0.48M\qquad \end{array}$
Change in pressure:

$\begin{array}{llll}2{\text{CH}}_{4}\left(g\right)\qquad & \rightleftharpoons\qquad & {\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\qquad & 3{\text{H}}_{2}\left(g\right)\qquad \\ -2x\qquad & & x\qquad & 3x\qquad \\ -50\text{ torr}\qquad & & 25\text{ torr}\qquad & 75\text{ torr}\qquad \end{array}$
Change in pressure:

$\begin{array}{lllll}{\text{CH}}_{4}\left(g\right)+\qquad & {\text{H}}_{2}\text{O}\left(g\right)\qquad & \rightleftharpoons\qquad & \text{CO}\left(g\right)+\qquad & 3{\text{H}}_{2}\left(g\right)\qquad \\ x\qquad & x\qquad & & -x\qquad & -3x\qquad \\ 5\text{ atm}\qquad & 5\text{atm}\qquad & & -5\text{ atm}\qquad & -15\text{ atm}\qquad \end{array}$
$\begin{array}{llll}{\text{NH}}_{4}\text{Cl}\left(s\right)\qquad & \rightleftharpoons\qquad & {\text{NH}}_{3}\left(g\right)+\qquad & \text{HCl}\left(g\right)\qquad \\ & & x\qquad & x\qquad \\ & & \qquad 1.03\times {10}^{-4}M\qquad & 1.03\times10^{-4}M\qquad \end{array}$
change in pressure:

$\begin{array}{cccc}\text{Ni}\left(s\right)+\qquad & 4\text{CO}\left(g\right)\qquad & \rightleftharpoons\qquad & \text{Ni}{\left(\text{CO}\right)}_{4}\left(g\right)\qquad \\ & 4x\qquad & & x\qquad \\ & \qquad 0.40\text{atm}\qquad & & 0.1\text{ atm}\qquad \end{array}$

3. Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.

Exercises

Assume that the change in concentration of 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 N₂O₄ with chloroform as the solvent.
  
  ${\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right){K}_{c}=1.07\times {10}^{-5}$
  in chloroform
2. Show that the change is small enough to be neglected.
Assume that the change in concentration of COCl₂ is small enough to be neglected in the following problem.
1. Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl₂ with an initial concentration of 0.3166 M.
  
  ${\text{COCl}}_{2}\left(g\right)\rightleftharpoons\text{CO}\left(g\right)+{\text{Cl}}_{2}\left(g\right){K}_{c}=2.2\times {10}^{-10}$
2. Show that the change is small enough to be neglected.
Assume that the change in pressure of H₂S is small enough to be neglected in the following problem.
1. Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H₂S with an initial pressure of 0.824 atm.
  
  $2{\text{H}}_{2}\text{S}\left(g\right)\rightleftharpoons2{\text{H}}_{2}\left(g\right)+{\text{S}}_{2}\left(g\right){K}_{P}=2.2\times {10}^{-6}$
2. Show that the change is small enough to be neglected.
What are all concentrations after a mixture that contains [H₂O] = 1.00 M and [Cl₂O] = 1.00 M comes to equilibrium at 25 °C?

${\text{H}}_{2}\text{O}\left(g\right)+{\text{Cl}}_{2}\text{O}\left(g\right)\rightleftharpoons2\text{HOCl}\left(g\right){K}_{c}=0.0900$
What are the concentrations of PCl₅, PCl₃, and Cl₂ in an equilibrium mixture produced by the decomposition of a sample of pure PCl₅ with [PCl₅] = 2.00 M?

${\text{PCl}}_{5}\left(g\right)\rightleftharpoons{\text{PCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right){K}_{c}=0.0211$

Show Selected Answers

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

	[NO₂]	[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[{\text{NO}}_{2}\right]}^{2}}{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}=\frac{{\left(2x\right)}^{2}}{\left(0.129-x\right)}=\frac{{\left(2x\right)}^{2}}{0.129}\qquad \\ {x}^{2}=\frac{0.129\times 1.07\times {10}^{-5}}{4}=3.45\times {10}^{-7}\qquad \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 N₂O₄ is far less than the 5% maximum allowed.

3. (a) Write the balanced equation, and then set up a table with initial pressures and the changed pressures using x as the change in pressure. The simplest way to find the coefficients for the x values is to use the coefficient in the balanced equation.

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, “2 H subscript 2 S ( g ) equilibrium arrow 2 H subscript 2 ( g ) plus S subscript 2 ( g ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.824, negative 2 x, 0.824 minus 2 x. The second column has the following: 0, negative 2 x, positive 2 x. The third column has the following: 0, negative 2 x, positive 2 x.

{K}_{P}=2.2\times {10}^{-6}=\frac{\left({P}_{{\text{S}}_{2}}\right){\left({P}_{{\text{H}}_{2}}\right)}^{2}}{{\left({P}_{{\text{H}}_{2}\text{S}}\right)}^{2}}=\frac{\left[x\right]{\left[2x\right]}^{2}}{{\left[0.824-2x\right]}^{2}}

Since the value of K_p is much smaller than 0.824, the 2x term in 0.824 − 2x is deemed negligible and, therefore, can be dropped.

\begin{array}{l}{ }2.2\times {10}^{-6}=\frac{\left[x\right]{\left[2x\right]}^{2}}{{\left[0.824\right]}^{2}}\qquad \\ 2.2\times {10}^{-6}=\frac{4{\text{X}}^{3}}{\left[0.679\right]}\qquad \end{array}

1.494 × 10⁻⁶ = 4x³
3.73 × 10⁻⁷ = x³
7.20 × 10⁻³ = x

Final equilibrium pressures:

[H₂S] = 0.824 − 2x = 0.824 − 2(7.20 × 10⁻³) = 0.824 - 0.0144 = 0.810 atm
[H₂] = 2x = 2(7.2 × 10⁻³) = 0.014 atm
[S₂] = [x] = 0.0072 atm

(b) The 2x is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2x is

\frac{0.014}{0.824}\times 100\%=1.7\%

, which is less than allowed by the “5% test.” The error is, indeed, negligible.

5. As all species are gases and are in M concentration units, a simple K_c equilibrium can be solved using the balanced equation:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, “P C l subscript 5 ( g ) equilibrium arrow P C l subscript 3 ( g ) plus C l subscript 2 ( g ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 2.00, negative x, 2.00 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.

{K}_{c}=0.0211=\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}=\frac{\left[x\right]\left[x\right]}{\left[2.00-x\right]}

As the value of K_c is substantial when compared with 2.00 M, the x terms in 2.00 − x cannot be disregarded. Thus, x must be solved by using the quadratic formula.

0.0211=\frac{{x}^{2}}{\left[2.00-x\right]}=0.0422-0.0211x={x}^{2}

Begin by arranging the terms in the form of the quadratic equation:

ax² + bx + c = 0
x² + 0.0211x − 0.0422 = 0

Next, solve for x using the quadratic formula.

x=\frac{\text{-}b\pm\sqrt{{b}^{2}-4ac}}{2a}=\frac{-0.0211\pm\sqrt{{\left(0.0211\right)}^{2}-4\left(1\right)\left(-0.0422\right)}}{2\left(1\right)}

=\frac{-0.0211\pm\sqrt{0.0004452+0.1688}}{2}=\frac{-0.0211\pm 0.4114}{2}

= 0.195 M or −0.216 M

The process of dissociation renders only positive quantities. Thus, x must be a positive value when factored into the solution so as to guarantee a realistic result. The final equilibrium concentrations are: [PCl₃] = 2.00 − x = 2.00 − 0.195 = 1.80 M; [PC₃] = [Cl₂] = x = 0.195 M; [PCl₃] = [Cl₂] = x = 0.195 M.

Key Concepts and Summary

The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemical equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to approach three basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we can solve for the equilibrium constant; when given the equilibrium constant and some of the concentrations involved, we can solve for the missing concentrations; and when given the equilibrium constant and the initial concentrations, we can solve for the concentrations at equilibrium.

Exercises

Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H₂ and 1.25 mol of I₂ in a 5.00-L flask at 448 °C.

${\text{H}}_{2}+{\text{I}}_{2}\rightleftharpoons2\text{HI}{K}_{c}=50.2\text{ at }448^{\circ}\text{C}$
What is the pressure of BrCl in an equilibrium mixture of Cl₂, Br₂, and BrCl if the pressure of Cl₂ in the mixture is 0.115 atm and the pressure of Br₂ in the mixture is 0.450 atm?

${\text{Cl}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)\rightleftharpoons2\text{BrCl}\left(g\right){K}_{P}=4.7\times {10}^{-2}$
What is the pressure of CO₂ in a mixture at equilibrium that contains 0.50 atm H₂, 2.0 atm of H₂O, and 1.0 atm of CO at 990 °C?

${\text{H}}_{2}\left(g\right)+{\text{CO}}_{2}\left(g\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right)+\text{CO}\left(g\right){K}_{P}=1.6\text{ at }990^{\circ}\text{C}$
Sodium sulfate 10-hydrate, Na₂SO₄
$\cdot$
10H₂O, dehydrates according to the equation

${\text{Na}}_{2}{\text{SO}}_{4}\cdot 10{\text{H}}_{2}\text{O}\left(s\right)\rightleftharpoons{\text{Na}}_{2}{\text{SO}}_{4}\left(s\right)+10{\text{H}}_{2}\text{O}\left(g\right){K}_{P}=4.08\times {10}^{-25}\text{ at }25^{\circ}\text{C}$

What is the pressure of water vapor at equilibrium with a mixture of Na₂SO₄
$\cdot$
10H₂O and NaSO₄?
Calcium chloride 6-hydrate, CaCl₂
$\cdot$
6H₂O, dehydrates according to the equation

${\text{CaCl}}_{2}\cdot 6{\text{H}}_{2}\text{O}\left(s\right)\rightleftharpoons{\text{CaCl}}_{2}\left(s\right)+6{\text{H}}_{2}\text{O}\left(g\right){K}_{P}=5.09\times {10}^{-44}\text{ at }25^{\circ}\text{C}$

What is the pressure of water vapor at equilibrium with a mixture of CaCl₂
$\cdot$
6H₂O and CaCl₂?
Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl₂ produced when a sample of NOCl with a pressure of 0.500 atm comes to equilibrium according to this reaction:

$2\text{NOCl}\left(g\right)\rightleftharpoons 2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right){K}_{P}=4.0\times {10}^{-4}$
Calculate the equilibrium concentrations of NO, O₂, and NO₂ in a mixture at 250 °C that results from the reaction of 0.20 M NO and 0.10 M O₂. (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium.)

$2\text{NO}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right){K}_{c}=2.3\times {10}^{5}\text{ at }250^{\circ}\text{C}$
Calculate the equilibrium concentrations that result when 0.25 M O₂ and 1.0 M HCl react and come to equilibrium. (Hint: K_c is large; assume the reaction goes to completion, then comes back to equilibrium.)

$4\text{HCl}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{Cl}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(g\right){K}_{c}=3.1\times {10}^{13}$
One of the important reactions in the formation of smog is represented by the equation

${\text{O}}_{3}\left(g\right)+\text{NO}\left(g\right)\rightleftharpoons{\text{NO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right){K}_{P}=6.0\times {10}^{34}$

What is the pressure of O₃ remaining after a mixture of O₃ with a pressure of 1.2 × 10^-8 atm and NO with a pressure of 1.2 × 10^-8 atm comes to equilibrium? (Hint: K_P is large; assume the reaction goes to completion then comes back to equilibrium.)
Calculate the pressures of NO, Cl₂, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl₂. (Hint: K_P is small; assume the reverse reaction goes to completion then comes back to equilibrium.)
Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H₂ and 63.5 g of iodine at 448 °C.

${\text{H}}_{2}+{\text{I}}_{2}\rightleftharpoons2\text{HI}{K}_{c}=50.2\text{ at }448^{\circ}\text{C}$
Butane exists as two isomers, n-butane and isobutane.

K_P = 2.5 at 25 °CWhat is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?
What is the minimum mass of CaCO₃ required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (K_c) is 0.050 for the decomposition reaction of CaCO₃ at that temperature?

${\text{CaCO}}_{3}\left(s\right)\rightleftharpoons\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)$
The equilibrium constant (K_c) for this reaction is 1.60 at 990 °C:
${\text{H}}_{2}\left(g\right)+{\text{CO}}_{2}\left(g\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right)+\text{CO}\left(g\right)$

Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H₂, 2.00 mol of CO₂, 0.750 mol of H₂O, and 1.00 mol of CO to a 5.00-L container at 990 °C.

Show Selected Answers

2. Write the equilibrium constant expression and solve for P_BrCl.

\begin{array}{l}{K}_{P}=\frac{{\left({P}_{\text{BrCl}}\right)}^{2}}{{P}_{{\text{Cl}}_{2}}{P}_{{\text{Br}}_{2}}}=\frac{{\left({P}_{\text{BrCl}}\right)}^{2}}{\left(0.115\right)\left(0.450\right)}=4.7\times {10}^{-2}\\ {\left({P}_{\text{BrCl}}\right)}^{2}=0.115\times 0.450\times 4.7\times {10}^{-2}=2.43\times {10}^{-3}\text{atm}\end{array}

P_BrCl = 4.9 × 10^-2 atm

4. Because two of the substances involved in the equilibrium are solids, their activities are 1, and their pressures are constant and do not appear in the equilibrium expression.

\begin{array}{l}{K}_{p}=4.08\times {10}^{-25}={\left({P}_{{\text{H}}_{2}\text{O}}\right)}^{10}\\ {P}_{{\text{H}}_{2}\text{O}}=\sqrt[10]{4.08\times {10}^{-25}}=3.64\times {10}^{-3}\text{atm}\end{array}

7. As the equilibrium constant for the reaction is large at ≈ 10⁵, first assume that the reaction is complete—that is, that one or both reactants are completely consumed.

This table has two main columns and four rows. The first cell in the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, “2 N O ( g ) plus O subscript 2 ( g ) right-facing arrow 2 N O subscript 2 ( g ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, 2 x, 0. The second column has the following: 0.10, x, 0. The third column has the following: 0, 2 x, 0.20.

As only NO₂ exists at this stage, the reaction will establish equilibrium:

As this equilibrium is the reverse of that originally given, the new K_c will be the inverse of the given K_c—that is,

\frac{1}{{K}_{c}}

\begin{array}{ll}{K}_{\text{new}}\qquad & =\frac{1}{{K}_{c}}=\frac{{\left[\text{NO}\right]}^{2}\left[{\text{O}}_{2}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}\qquad \\ \qquad & =\frac{1}{2.3\times {10}^{5}}=4.35\times {10}^{-6}=\frac{{\left[2x\right]}^{2}\left[x\right]}{{\left[0.20-2x\right]}^{2}}\qquad \end{array}

Because the new equilibrium constant is small compared with the 0.2 M term, the 2x can be dropped and the equilibrium solved.

4.35\times {10}^{-6}=\frac{\left(4{x}^{2}\right)x}{\left[0.04\right]}

4x³ = 1.74 × 10⁻⁷

x³ = 4.35 × 10⁻⁸

x = 3.52 × 10⁻³

Final equilibrium concentrations:

[NO₂] = 0.20 − 2x = 0.20 − 2(0.00352) = 0.19 M
[NO] = 2x = 2(0.00352) = 0.0070 M
[O₂] = x = 0.0035 M

2x is small compared to 0.20 M because

\left(\frac{2x}{0.20}\right)\times 100\%=\left(\frac{0.00352}{0.20}\right)\times 100\%=1.7\%

which is less than the maximum 5% allowed.

9. Assume that the reaction goes to completion, where one or both reactants are completely used up:

This table has two main columns and four rows. The first cell in the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, “O subscript 3 ( g ) plus N O ( g ) right-facing arrow N O subscript 2 ( g ) plus O subscript 2 ( g ).” Under the second column is a subgroup of four columns and three rows. The first column has the following: 1.2 times ten to the negative 8, negative x, 0. The second column has the following: 1.2 times ten to the negative 8, negative x, 0. The third column has the following: 0, positive x, 1.2 times ten to the negative 8. The fourth column as the following: 0, positive x, 1.2 times ten to the negative 8.

Now assume the equilibrium will be established by the reaction moving toward the reactants:

{K}_{P}=\frac{{P}_{{\text{NO}}_{2}}{P}_{{\text{O}}_{2}}}{{P}_{{\text{O}}_{3}}{P}_{\text{NO}}}=\frac{\left(1.2\times {10}^{-8}-x\right)\left(1.2\times {10}^{-8}-x\right)}{\left(x\right)\left(x\right)}

As K_P is much larger than 1.2 × 10⁻⁸, the X terms in the expression 1.2 × 10⁻⁸ − x are relatively negligible and, therefore, can be dropped.

\begin{array}{l}{ }6.0\times {10}^{34}=\frac{\left(1.2\times {10}^{-8}\right)\left(1.2\times {10}^{-8}\right)}{\left(x\right)\left(x\right)}\qquad \\ {x}^{2}=\frac{\left(1.44\times {10}^{-16}\right)}{6.0\times {10}^{34}}=2.4\times {10}^{-51}\qquad \end{array}

x = 4.9 × 10⁻²⁶ atm

{P}_{{\text{O}}_{3}}=x=4.9\times {10}^{-26}\text{atm}

Clearly x is much smaller than 1.2 × 10⁻⁴ so the assumption is valid.

11. The number of moles of I₂ is

\begin{array}{l}{ }\text{mol}=\frac{63.5\text{g}}{253.809\text{g}{\text{mol}}^{-1}}=0.250\text{mol}{\text{I}}_{2}\qquad \\ {K}_{c}=\frac{{\left[\text{HI}\right]}^{2}}{\left[{\text{H}}_{2}\right]\left[{\text{I}}_{2}\right]}\qquad \end{array}

The unit for each concentration term is moles per liter. If the volume were known for this exercise, the number of moles in each term should be divided by this volume. However, there are two terms in the numerator and two terms in the denominator, so these volumes cancel one another. Consequently, for any expression with the same number of numerator terms as denominator terms, the number for moles can be used in place of moles per liter. In this exercise, the volume is not needed even though it is given.

(mol HI)² = K × mol H₂ × mol I₂

= 50.2 × 1.25 mol × 0.250 mol

= 15.7 mol²

mol HI =

\sqrt{15.7{\text{mol}}^{2}}=3.96\text{mol}

Mass(HI) = 3.96 mol × 127.9124 g/mol = 507 g

13. At equilibrium the concentration of CO₂ = K_c = 0.50 M. The number of moles CO₂ in the system is then mol CO₂ = 6.5 L × 0.50 mol/L = 3.3 mol. The minimum moles of CaCO₂ required is then just more than:

3.3\text{mol}{\text{CO}}_{2}\times \frac{1\text{mol}{\text{CaCO}}_{3}}{1\text{mol}{\text{CO}}_{2}}=3.3\text{mol}{\text{CaCO}}_{3}

3.3\text{mol}{\text{CO}}_{2}\times \frac{1\text{mol}{\text{CaCO}}_{3}}{100.1\text{g}{\text{CaCO}}_{3}}=330\text{g}

Exercises

At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of N₂O₄ and NO₂ are
${\text{P}}_{{\text{N}}_{2}{\text{O}}_{4}}=0.70\text{atm}$
and
${\text{P}}_{{\text{NO}}_{2}}=0.30\text{atm.}$
1. Predict how the pressures of NO₂ and N₂O₄ will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?
2. Calculate the partial pressures of NO₂ and N₂O₄ when they are at equilibrium at 9.0 atm and 25 °C.
In a 3.0-L vessel, the following equilibrium partial pressures are measured: N₂, 190 torr; H₂, 317 torr; NH₃, 1.00 × 10³ torr:
${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)$