13.4  Equilibrium Calculations

Nicole Bouvier-Brown; Saori Shiraki; J. Ryan Hunt; Emily Jarvis

Chapter 13 Fundamental Equilibrium Concepts

13.4 Equilibrium Calculations

Learning Objectives

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

Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches

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:

[latex][/latex]

2 NH₃(g) ⇌ N₂(g) + 3 H₂(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:

[latex][/latex]

[latex]\Delta [N_2] = +x[/latex]

the corresponding changes in the other species concentrations are

[latex][/latex]

[latex]\Delta [H_2] = \Delta [N_2] \: \left( \frac {3 \: mol \: H_2}{1 \: mol \: N_2} \right) = +3x[/latex]

[latex][/latex]

[latex]\Delta [NH_3] = -\Delta [N_2] \: \left( \frac {2 \: mol \: NH_3}{1 \: mol \: N_2} \right) = -2x[/latex]

where the negative sign indicates a decrease in concentration.

Example 13.6 – Determining Relative Changes in Concentration

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

(a)

	C₂H₂(g)	+	2 Br₂(g)	⇌	C₂H₂Br₄(g)
change	+x

(b)

	I₂(aq)	+	I⁻(aq)	⇌	I₃⁻(aq)
change					+x

(c)

	C₃H₈(g)	+	5 O₂(g)	⇌	3 CO₂(g)	+	4 H₂O(g)
change	+x

Solution

(a)

	C₂H₂(g)	+	2 Br₂(g)	⇌	C₂H₂Br₄(g)
change	+x		+2x		−x

(b)

	I₂(aq)	+	I⁻(aq)	⇌	I₃⁻(aq)
change	−x		−x		+x

(c)

	C₃H₈(g)	+	5 O₂(g)	⇌	3 CO₂(g)	+	4 H₂O(g)
change	+x		+5x		−3x		−4x

Check Your Learning

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

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 (for initial, change, and equilibrium) 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.

I₂(aq) + I⁻(aq) ⇌ I₃⁻(aq)

If a solution with the concentrations of I₂ and I⁻ both equal to 1.000 × 10⁻³ M before reaction gives an equilibrium concentration of I₂ of 6.61 × 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:

[latex]K_c = \frac {[I_3^-]}{[I_2] \, [I^-]}[/latex]

[latex][/latex]

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.

	I₂	+	I⁻	⇌	I₃⁻
I (M)	1.000 × 10⁻³		1.000 × 10⁻³		0
C (M)	−x		−x		+x
E (M)	1.000 × 10⁻³ − x		1.000 × 10⁻³ − x		x

At equilibrium the concentration of I₂ is 6.61 × 10⁻⁴ M. Using the equilibrium (E) expression from the ICE table above allows us to solve for “x”:

[latex][/latex]

[latex]1.000 \times 10^{-3} - x = 6.61 \times 10^{-4}[/latex]

[latex][/latex]

[latex]x = 1.000 \times 10^{-3} - 6.61 \times 10^{-4} = 3.39 \times 10^{-4} \: M[/latex]

[latex][/latex]

The ICE table may now be updated with numerical values for all concentration changes (C) and equilibrium concentrations (E):

	I₂	+	I⁻	⇌	I₃⁻
I (M)	1.000 × 10⁻³		1.000 × 10⁻³		0
C (M)	−3.39 × 10⁻⁴		−3.39 × 10⁻⁴		+3.39 × 10⁻⁴
E (M)	6.61 × 10⁻⁴		6.61 × 10⁻⁴		3.39 × 10⁻⁴

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

[latex]K_c = \frac {[I_3^-]}{[I_2] \, [I^-]} = \frac {3.39 \times 10^{-4}}{(6.61 \times 10^{-4})(6.61 \times 10^{-4})} = 776[/latex]

Check Your Learning

Click here for a walkthrough of this problem!

First, we can create an ICE table for the reaction:

	C₂H₅OH	+	CH₃CO₂H	⇌	CH₃CO₂C₂H₅	+	H₂O
I (M)	1		1		0		0
C (M)	-x		-x		+x		+x
E (M)	1 – x		1 – x		x		x

Using the provided equilibrium concentration ([latex]\frac {1}{3}[/latex] M) for the reactants, we can solve for x

[latex]1\ -\ x\ =\ \frac{1}{3}[/latex]

[latex]x\ =\ \frac{2}{3}[/latex]

We can now update the ICE table with numerical values for equilibrium concentrations:

	C₂H₅OH	+	CH₃CO₂H	⇌	CH₃CO₂C₂H₅	+	H₂O
I (M)	1		1		0		0
C (M)	– 2/3		– 2/3		+ 2/3		+ 2/3
E (M)	1 – 2/3 = 1/3		1 – 2/3 = 1/3		2/3		2/3

Finally, we can calculate the value of the equilibrium constant for the reaction:

[latex]K_c=\frac{\left[CH_3CO_2C_2H_5\right]\left[H_2O\right]}{\left[C_2H_5OH\right]\left[CH_3CO_2H\right]}=\frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)}=4[/latex]

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 13.8 – Calculation of a Missing Equilibrium Concentration

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures:

N₂(g) + O₂(g) ⇌ 2 NO(g). K_c= 4.1 × 10⁻⁴ at 2000°C

Calculate the equilibrium concentration of NO(g) in air at 1 atm pressure and 2000°C.

The equilibrium concentrations of N₂ and O₂ at this pressure and temperature are 0.036 M and 0.0089 M, respectively.

Solution

Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:

[latex]K_c = \frac {[NO]^2}{[N_2] \, [O_2]}[/latex]

[latex][/latex]

[latex][NO]^2 = K_c \, [N_2] \, [O_2][/latex]

[latex][/latex]

[latex][NO] = \sqrt {K_c \, [N_2] \, [O_2]}[/latex]

[latex][/latex]

[latex][NO] = \sqrt {(4.1 \times 10^{-4}) \, (0.036) \, (0.0089)} = \sqrt {1.3 \times 10^{-7}} = 3.6 \times 10^{-4}[/latex]

[latex][/latex]

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

[latex][/latex]

To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for K:

[latex]K_c = \frac {[NO]^2}{[N_2] \, [O_2]} = \frac {(3.6 \times 10^{-4})^2}{(0.036) \, (0.0089)} = 4.0 \times 10^{-4}[/latex]

[latex][/latex]

This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.

Check Your Learning

Click here to see a walkthrough for this problem!

The balanced equation for this reaction is

[latex]N_2\ +\ 3\ H_2\ \rightleftharpoons\ 2\ NH_3[/latex]

The equilibrium constant expression is therefore

[latex]K_c=\frac{\left[NH_3\right]^2}{\left[N_2\right]\left[H_2\right]^3}[/latex]

Solving for [NH₃] and substituting in known concentrations, we get

[latex]\left[NH_3\right]=\sqrt{K_c\left[N_2\right]\left[H_2\right]^3}=\sqrt{\left(6.00\ \times10^{-2}\right)\left(4.26\right)\left(2.09\right)^3}=1.53[/latex]

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:

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

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

Example 13.9 – Calculation of Equilibrium Concentrations

Under certain conditions, the equilibrium constant K_c 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 in a mixture that initially contained only PCl₅ at a concentration of 1.00 M?

Solution

Use the stepwise process described earlier.

Step 1. Determine the direction the reaction proceeds.

The balanced equation for the decomposition of PCl₅ is

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

[latex][/latex]

Because only the reactant is present initially Q_c = 0 and the reaction will proceed to the right.

Step 2. Develop an ICE table.

	PCl₅	⇌	PCl₃	+	Cl₂
I (M)	1.00		0		0
C (M)	−x		+x		+x
E (M)	1.00 −x		x		x

[latex][/latex]

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

Substituting the equilibrium concentrations into the equilibrium constant equation gives

[latex]K_c = \frac {[PCl_3] \, [Cl_2]}{[PCl_5]}[/latex]

[latex][/latex]

[latex]0.0211 = \frac {x^2}{1.00 - x}[/latex]

[latex][/latex]

[latex]0.0211 \, (1.00 - x) = x^2[/latex]

[latex][/latex]

[latex]x^2 + 0.0211x - 0.0211 = 0[/latex]

[latex][/latex]

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

[latex]x = \frac {-b \, \pm \, \sqrt {b^2 \, - \, 4ac}}{2a}[/latex]

[latex][/latex]

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

[latex]x = \frac {-0.0211 \, \pm \, \sqrt {(0.0211)^2 \, - \, 4(1)(-0.0211)}}{2(1)}= \frac {-0.0211 \, \pm \, 0.291}{2}[/latex]

[latex][/latex]

The two roots of the quadratic are, therefore,

[latex][/latex]

[latex]x = \frac {-0.0211 \, + \, 0.291}{2} = 0.135[/latex]

and

[latex]x= \frac {-0.0211 \, - \, 0.291}{2} = -0.156[/latex]

[latex][/latex]

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

[PCl₅] = 1.00 − x = 1.00 − 0.135 = 0.87 M

[PCl₃] = x = 0.135 M

[Cl₂] = x = 0.135 M

Step 4. Confirm the calculated equilibrium concentrations.

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

[latex]K_c = \frac {[PCl_3] \, [Cl_2]}{[PCl_5]} = \frac {(0.135) \, (0.135)}{0.87} = 0.021[/latex]

[latex][/latex]

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).

Check Your Learning

Using the ICE table you developed in the quiz above, determine the equilibrium concentrations for CH₃CO₂H, C₂H₅OH, CH₃CO₂C₂H₅, and H₂O.

Click here to see the answers and a walkthrough!

First, we calculate Q to determine which direction the reaction will proceed:

[latex]Q\ =\ \frac{\left[CH_3CO_2C_2H_5\right]\left[H_2O\right]}{\left[CH_3CO_2H\right]\left[C_2H_5OH\right]}=\frac{\left(0.40\right)\left(0.40\right)}{\left(0.15\right)\left(0.15\right)}=7.1[/latex]

Since Q > K, the reaction will proceed to the left (towards the reactants). Develop an ICE table that reflects this:

	C₂H₅OH	+	CH₃CO₂H	⇌	CH₃CO₂C₂H₅	+	H₂O
I (M)	0.15		0.15		0.40		0.40
C (M)	+ x		+ x		– x		– x
E (M)	0.15 + x		0.15 + x		0.40 – x		0.40 – x

Substituting in our equilibrium expressions into K_c, we can solve for x:

[latex]K_c\ =\ \frac{\left[CH_3CO_2C_2H_5\right]\left[H_2O\right]}{\left[CH_3CO_2H\right]\left[C_2H_5OH\right]}[/latex]

[latex]4.0\ =\ \frac{\left(0.40-x\right)\left(0.40\ -\ x\right)}{\left(0.15\ +\ x\right)\left(0.15\ +x\right)}[/latex]

[latex]4.0\ =\ \frac{0.16\ -0.80x\ +x^2}{0.0225\ +0.30x+x^2}[/latex]

[latex]0.09\ +\ 1.2x\ +4.0x^2\ =0.16\ -0.80x+x^2[/latex]

[latex]3.0x^2+2.0x-0.07=0[/latex]

Using the quadratic formula, we find

[latex]x\ =0.033[/latex]

and

[latex]x\ =-0.70[/latex]

We use x = 0.033 because x = -0.70 would yield negative concentrations, which are non-physical. Filling out our ice chart with numerical values, we get

	C₂H₅OH	+	CH₃CO₂H	⇌	CH₃CO₂C₂H₅	+	H₂O
I (M)	0.15		0.15		0.40		0.40
C (M)	+ 0.03		+ 0.03		– 0.03		– 0.03
E (M)	0.18		0.18		0.37		0.37

We can check our answer by substituting the equilibrium concentrations into the K_c expression:

[latex]K_c=\frac{\left(0.37\right)\left(0.37\right)}{\left(0.18\right)\left(0.18\right)}=4.2[/latex]

Check Your Learning

A 1.00-L flask is filled with 1.00 mole 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.

H₂(g) + I₂(g) ⇌ 2 HI(g)

What are the equilibrium concentrations of H₂, I₂, and HI in moles/L?

Click here to see the answers and a walkthrough!

Since there are initially no mols of the product (HI), Q = 0 and the reaction must proceed to the right. We create an ICE table that reflects this:

	H₂	+	I₂	⇌	2 HI
I (M)	1.00		2.00		0
C (M)	−x		–x		+2x
E (M)	1.00 −x		2.00 – x		2x

Substitution of the equilibrium expressions into the K_c expression allows us to solve for x:

[latex]K_c=\frac{\left[HI\right]^2}{\left[H_2\right]\left[I_2\right]}[/latex]

[latex]50.5\ =\frac{\left(2x\right)^2}{\left(1.00-x\right)\left(2.00-x\right)}=\frac{4x^2}{2.00-3.00x+x^2}[/latex]

[latex]101\ -\ 152x\ +\ 50.5x^2=4x^2[/latex]

[latex]46.5x^2-152x+101=0[/latex]

Using the quadratic formula, we find

x = 2.3

and

x = 0.94

We choose x = 0.94 because x = 2.3 yields negative concentrations. Filling out our ice chart with numerical values, we get

	H₂	+	I₂	⇌	2 HI
I (M)	1.00		2.00		0
C (M)	−0.94		-0.94		+2(0.94)
E (M)	0.06		1.06		1.88

We can check our answer by substituting the equilibrium concentrations into the K_c expression:

[latex]K_c=\frac{\left(1.88\right)^2}{\left(0.06\right)\left(1.06\right)}=55.5[/latex]

Although not exact, this value is close to the actual value of K_c (50.5).

Example 13.10 – Calculation of Equilibrium Concentrations Using the Small x Approximation

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

HCN(aq) ⇌ H⁺(aq) + CN⁻(aq) K_c = 4.9 × 10⁻¹⁰

Solution

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

	HCN(aq)	⇌	H⁺(aq)	+	CN⁻(aq)
I (M)	0.15		0		0
C (M)	−x		+x		+x
E (M)	0.15 −x		x		x

Substitute the equilibrium concentration terms into the K_c expression

[latex]K_c = \frac {x^2}{0.15 - x} = 4.9 \times 10^{-10}[/latex]

[latex][/latex]

rearrange to the quadratic form and solve for x

[latex][/latex]

[latex]x^2 + 4.9 \times 10^{-10} x -7.35 \times 10^{-11} = 0[/latex]

[latex][/latex]

[latex]x = 8.6 \times 10^{-6} \: M[/latex]

[latex][/latex]

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

[latex][/latex]

Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), so the initial concentration experiences a negligible change. This allows us to make the “small x approximation”:

[latex][/latex]

if x ≪ 0.15 M, then 0.15 – x ≈ 0.15

[latex][/latex]

If we apply this approximation at the start of an ICE problem with a small K_c , it allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:

[latex]K_c = \frac {x^2}{0.15 \, - \, x} \approx \frac {x^2}{0.15}[/latex]

[latex][/latex]

[latex]4.9 \times 10^{-10} = \frac {x^2}{0.15}[/latex]

[latex][/latex]

[latex]x^2 = (0.15) \, (4.9 \times 10^{-10}) = 7.4 \times 10^{-11}[/latex]

[latex][/latex]

[latex]x = \sqrt {7.4 \times 10^{-11}} = 8.6 \times 10^{-6} \: M[/latex]

[latex][/latex]

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

8.6 × 10⁻⁶ ≪ 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

Click here to see a walkthrough for this problem!

The appropriate ICE table is

	NH₃ (aq)	+	H₂O (l)	⇌	NH₄⁺ (aq)	+	OH^– (aq)
I (M)	0.25		–		0		0
C (M)	– x		–		+ x		+ x
E (M)	0.25 – x		–		x		x

Note that H₂O is a liquid, so it does not affect the equilibrium position.

Because K_c is small and the initial concentration of NH₃ is somewhat large, the small x approximation may be appropriate. Applying the approximation and solving for x yields:

[latex]K_c=\frac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}=\frac{x^2}{0.25\ -\ x}=\frac{x^2}{0.25}[/latex]

[latex]1.8\ \times10^{-5}=\frac{x^2}{0.25}[/latex]

[latex]x\ =\sqrt{\left(1.8\ \times10^{-5}\right)\left(0.25\right)}=0.0021\ M[/latex]

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

0.0021 ≪ 0.25

so the approximation was justified. Filling out our ice chart with numerical values, we get

	NH₃ (aq)	+	H₂O (l)	⇌	NH₄⁺ (aq)	+	OH^– (aq)
I (M)	0.25		–		0		0
C (M)	– 0.0021		–		+ 0.0021		+ 0.0021
E (M)	0.25		–		0.0021		0.0021