Example 8

The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:

$$2{\text{SO}}_2\text{(g)}+{\text{O}}_2\text{(g)}\rightleftharpoons 2{\text{SO}}_3\text{(g)}$$

A mixture of SO₂ and O₂ was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10⁻² M SO₃, 3.5 × 10⁻³ M O₂, and 3.0 × 10⁻³ M SO₂. Calculate K and K_p at this temperature.

Given: balanced equilibrium equation and composition of equilibrium mixture

Asked for: equilibrium constant

Strategy:

Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain K.

Solution:

Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,

$$K=\frac{{[{\text{SO}}_3]}^2}{{[{\text{SO}}_2]}^2[{\text{O}}_2]}=\frac{{(5.0\times {10}^{-2})}^2}{{(3.0\times {10}^{-3})}^2(3.5\times {10}^{-3})}=7.9\times {10}^4$$

To solve for K_p, we use Equation 16.18, where Δn = 2 − 3 = −1:

$$\begin{array}{cc}\hfill K_{\text{p}}& =K{(RT)}^{\Delta n}\hfill \\ \hfill & =7.9\times {10}^{\text{4}}{[(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K})(800\text{ K})]}^{-1}\hfill \\ \hfill & =1.2\times {10}^3\hfill \end{array}$$

Exercise

Hydrogen gas and iodine react to form hydrogen iodide via the reaction

$${\text{H}}_2\text{(g)}+{\text{I}}_2\text{(g)}\rightleftharpoons \text{2HI(g)}$$

A mixture of H₂ and I₂ was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10⁻² M HI, 6.47 × 10⁻³ M H₂, and 5.94 × 10⁻⁴ M I₂. Calculate K and K_p for this reaction.

Answer: K = 48.8; K_p = 48.8

Example 9

A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl₂. Calculate K at this temperature. The equation for the decomposition of NOCl to NO and Cl₂ is as follows:

$$2\text{NOCl(g)}\rightleftharpoons 2\text{NO(g)}+{\text{Cl}}_2\text{(g)}$$

Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium

Asked for: K

Strategy:

A Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).

B Calculate all possible initial concentrations from the data given and insert them in the table.

C Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.

D Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.

Solution:

A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:

$$K=\frac{{\text{[NO]}}^2[{\text{Cl}}_2]}{{\text{[NOCl]}}^2}$$

To obtain the concentrations of NOCl, NO, and Cl₂ at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.

B Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl]_i = 1.00 mol/2.00 L = 0.500 M. The initial concentrations of NO and Cl₂ are 0 M because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl₂ in a 2.00 L container, so [Cl₂]_f = 0.056 mol/2.00 L = 0.028 M. We insert these values into the following table:

C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl₂, the substance for which initial and final concentrations are known:

Δ[Cl₂] = [0.028 M (final) − 0.00 M (initial)] = +0.028 M

According to the coefficients in the balanced chemical equation, 2 mol of NO are produced for every 1 mol of Cl₂, so the change in the NO concentration is as follows:

$$\Delta \text{[NO]}=\left(\frac{0.028\cancel{\text{mol}{\text{Cl}}_2}}{\text{L}}\right)\left(\frac{2\text{ mol NO}}{1\cancel{\text{mol}{\text{Cl}}_2}}\right)=0.056\text{ M}$$

Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl₂ produced, so the change in the NOCl concentration is as follows:

$$\Delta \text{[NOCl]}=\left(\frac{0.028\cancel{{\text{mol Cl}}_2}}{\text{L}}\right)\left(\frac{-2\text{ mol NOCl}}{1\cancel{{\text{mol Cl}}_2}}\right)=-0.056\text{ M}$$

We insert these values into our table:

D We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl:

[NO]_f = 0.000 M + 0.056 M = 0.056 M [NOCl]_f = 0.500 M + (−0.056 M) = 0.444 M

We can now complete the table:

We can now calculate the equilibrium constant for the reaction:

$$K=\frac{{\text{[NO]}}^2[{\text{Cl}}_2]}{{\text{[NOCl]}}^2}=\frac{{(0.056)}^2(0.028)}{{(0.444)}^2}=4.5\times {10}^{-4}$$

Exercise

The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH₃) by reacting 0.1248 M H₂ and 0.0416 M N₂ at about 500°C. At equilibrium, the mixture contained 0.00272 M NH₃. What is K for the reaction ${\text{N}}_2+3{\text{H}}_2\rightleftharpoons 2{\text{NH}}_3$ at this temperature? What is K_p?

Answer: K = 0.105; K_p = 2.61 × 10⁻⁵

Example 10

The water–gas shift reaction is important in several chemical processes, such as the production of H₂ for fuel cells. This reaction can be written as follows:

$${\text{H}}_2\text{(g)}+{\text{CO}}_2\text{(g)}\rightleftharpoons {\text{H}}_2\text{O(g)}+\text{CO(g)}$$

K = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H₂ and 0.0150 M CO₂ is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?

Given: balanced equilibrium equation, K, and initial concentrations

Asked for: final concentrations

Strategy:

A Construct a table showing what is known and what needs to be calculated. Define x as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of x. From the values in the table, calculate the final concentrations.

B Write the equilibrium equation for the reaction. Substitute appropriate values from the table to obtain x.

C Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain K.

Solution:

A The initial concentrations of the reactants are [H₂]_i = [CO₂]_i = 0.0150 M. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H₂O as x, then Δ[H₂O] = +x. We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x. For example, 1 mol of CO is produced for every 1 mol of H₂O, so the change in the CO concentration can be expressed as Δ[CO] = +x. Similarly, for every 1 mol of H₂O produced, 1 mol each of H₂ and CO₂ are consumed, so the change in the concentration of the reactants is Δ[H₂] = Δ[CO₂] = −x. We enter the values in the following table and calculate the final concentrations.

B We can now use the equilibrium equation and the given K to solve for x:

$$K=\frac{[{\text{H}}_2\text{O}]\text{[CO]}}{[{\text{H}}_2][{\text{CO}}_2]}=\frac{(x)(x)}{(0.0150-x)(0.0150-x)}=\frac{x^2}{{(0.0150-x)}^2}=0.106$$

We could solve this equation with the quadratic formula, but it is far easier to solve for x by recognizing that the left side of the equation is a perfect square; that is,

$$\frac{x^2}{{(0.0150-x)}^2}={\left(\frac{x}{0.0150-x}\right)}^2=0.106$$

(The quadratic formula is presented in Essential Skills 7 in Section 15.7 "Essential Skills".) Taking the square root of the middle and right terms,

$$\begin{array}{cc}\hfill \frac{x}{(0.0150-x)}& ={(0.106)}^{1/2}=0.326\hfill \\ \hfill x& =(0.326)(0.0150)-0.326x\hfill \\ \hfill 1.326x& =0.00489\hfill \\ \hfill x& =0.00369=3.69\times {10}^{-3}\hfill \end{array}$$

C The final concentrations of all species in the reaction mixture are as follows:

$$\begin{array}{cc}\hfill {[{\text{H}}_2]}_{\text{f}}& ={[{\text{H}}_2]}_{\text{i}}+\Delta [{\text{H}}_2]=(0.0150-0.00369)\text{ M}=0.0113\text{ M}\hfill \\ \hfill {[{\text{CO}}_2]}_{\text{f}}& ={[{\text{CO}}_2]}_{\text{i}}+\Delta [{\text{CO}}_2]=(0.0150-0.00369)\text{ M}=0.0113\text{ M}\hfill \\ \hfill {[{\text{H}}_2\text{O}]}_{\text{f}}& ={[{\text{H}}_2\text{O}]}_{\text{i}}+\Delta [{\text{H}}_2\text{O}]=(0+0.00369)\text{ M}=0.00369\text{ M}\hfill \\ \hfill {\text{[CO]}}_{\text{f}}& ={\text{[CO]}}_{\text{i}}+\Delta \text{[CO]}=(0+0.00369)\text{ M}=0.00369\text{ M}\hfill \end{array}$$

We can check our work by inserting the calculated values back into the equilibrium constant expression:

$$K=\frac{[{\text{H}}_2\text{O}]\text{[CO]}}{[{\text{H}}_2][{\text{CO}}_2]}=\frac{{(0.00369)}^2}{{(0.0113)}^2}=0.107$$

To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.

Exercise

Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:

$${\text{H}}_2\text{(g)}+{\text{I}}_2\text{(g)}\rightleftharpoons \text{2HI(g)}$$

K = 54 at 425°C. If 0.172 M H₂ and I₂ are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?

Answer: [HI]_f = 0.270 M; [H₂]_f = [I₂]_f = 0.037 M

Example 11

In the water–gas shift reaction shown in Example 10, a sample containing 0.632 M CO₂ and 0.570 M H₂ is allowed to equilibrate at 700 K. At this temperature, K = 0.106. What is the composition of the reaction mixture at equilibrium?

Given: balanced equilibrium equation, concentrations of reactants, and K

Asked for: composition of reaction mixture at equilibrium

Strategy:

A Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations (x) and the final concentrations.

B Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x.

C Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K.

Solution:

A [CO₂]_i = 0.632 M and [H₂]_i = 0.570 M. Again, x is defined as the change in the concentration of H₂O: Δ[H₂O] = +x. Because 1 mol of CO is produced for every 1 mol of H₂O, the change in the concentration of CO is the same as the change in the concentration of H₂O, so Δ[CO] = +x. Similarly, because 1 mol each of H₂ and CO₂ are consumed for every 1 mol of H₂O produced, Δ[H₂] = Δ[CO₂] = −x. The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.

B We can now use the equilibrium equation and the known K value to solve for x:

$$K=\frac{[{\text{H}}_2\text{O}]\text{[CO]}}{[{\text{H}}_2][{\text{CO}}_2]}=\frac{x^2}{(0.570-x)(0.632-x)}=0.106$$

In contrast to Example 10, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:

x² = 0.106(0.360 − 1.20x + x²)

Collecting terms on one side of the equation,

0.894x² + 0.127x − 0.0382 = 0

This equation can be solved using the quadratic formula:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-0.127\pm \sqrt{{(0.127)}^2-4(0.894)(-0.0382)}}{2(0.894)}=0.148\text{, }-0.290$$

Only the answer with the positive value has any physical significance, so Δ[H₂O] = Δ[CO] = +0.148 M, and Δ[H₂] = Δ[CO₂] = −0.148 M.

C The final concentrations of all species in the reaction mixture are as follows:

$$\begin{array}{cc}\hfill {{\text{[H}}_{\text{2}}\text{]}}_{\text{f}}& ={{\text{[H}}_{\text{2}}\text{]}}_{\text{i}}+\Delta {\text{[H}}_{\text{2}}\text{]}=0.570\text{ M}-0.148\text{ M}=0.422\text{ M}\hfill \\ \hfill {{\text{[CO}}_{\text{2}}\text{]}}_{\text{f}}& ={{\text{[CO}}_{\text{2}}\text{]}}_{\text{i}}+\Delta {\text{[CO}}_{\text{2}}\text{]}=0.632\text{ M}-0.148\text{ M}=0.484\text{ M}\hfill \\ \hfill {{\text{[H}}_{\text{2}}\text{O]}}_{\text{f}}& ={{\text{[H}}_{\text{2}}\text{O]}}_{\text{i}}+\Delta {\text{[H}}_{\text{2}}\text{O]}=0\text{ M}+0.148\text{ M}=0.148\text{ M}\hfill \\ \hfill {\text{[CO]}}_{\text{f}}& ={\text{[CO]}}_{\text{i}}+\Delta \text{[CO]}=0\text{ M}+0.148\text{ M}=0.148\text{ M}\hfill \end{array}$$

We can check our work by substituting these values into the equilibrium constant expression:

$$K=\frac{[{\text{H}}_2\text{O}]\text{[CO]}}{[{\text{H}}_2][{\text{CO}}_2]}=\frac{{(0.148)}^2}{(0.422)(0.484)}=0.107$$

Because K is essentially the same as the value given in the problem, our calculations are confirmed.

Exercise

The exercise in Example 8 showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which K = 54 at 425°C. If a sample containing 0.200 M H₂ and 0.0450 M I₂ is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?

Answer: [HI]_f = 0.0882 M; [H₂]_f = 0.156 M; [I₂]_f = 9.2 × 10⁻⁴ M

Example 12

Atmospheric nitrogen and oxygen react to form nitric oxide:

$${\text{N}}_2\text{(g)}+{\text{O}}_2\text{(g)}\rightleftharpoons 2\text{NO(g)}$$

K_p = 2.0 × 10⁻³¹ at 25°C. What is the partial pressure of NO in equilibrium with N₂ and O₂ in the atmosphere (at 1 atm, $P_{{\text{N}}_2}$ = 0.78 atm and $P_{{\text{O}}_2}$ = 0.21 atm)?

Given: balanced equilibrium equation and values of K_p, $P_{{\text{O}}_2}\text{,}$ and $P_{{\text{N}}_2}$

Asked for: partial pressure of NO

Strategy:

A Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.

B Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration (x).

C Calculate the partial pressure of NO. Check your answer by substituting values into the equilibrium equation and solving for K.

Solution:

A Because we are given K_p and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of O₂ is 0.21 atm and that of N₂ is 0.78 atm. If we define the change in the partial pressure of NO as 2x, then the change in the partial pressure of O₂ and of N₂ is −x because 1 mol each of N₂ and of O₂ is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.

B Substituting these values into the equation for the equilibrium constant,

$$K_{\text{p}}=\frac{{(P_{\text{NO}})}^2}{\left(P_{{\text{N}}_2}\right)\left(P_{{\text{O}}_2}\right)}=\frac{{(2x)}^2}{(0.78-x)(0.21-x)}=2.0\times {10}^{-31}$$

In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x) = 0.78 and (0.21 − x) = 0.21. Substituting these expressions into our original equation,

$$\begin{array}{cc}\hfill \frac{{(2x)}^2}{(0.78)(0.21)}& =2.0\times {10}^{-31}\hfill \\ \hfill \frac{4x^2}{0.16}& =2.0\times {10}^{-31}\hfill \\ \hfill x^2& =\frac{0.33\times {10}^{-31}}{4}\hfill \\ \hfill x& =9.1\times {10}^{-17}\hfill \end{array}$$

C Substituting this value of x into our expressions for the final partial pressures of the substances,

$$\begin{array}{c}\hfill P_{\text{NO}}=2x\text{ atm}=1.8\times {10}^{-\text{16}}\text{ atm}\\ \hfill P_{{\text{N}}_2}=(0.78-x)\text{ atm}=0.78\text{ atm}\\ \hfill P_{{\text{O}}_2}=(0.21-x)\text{ atm}=0.21\text{ atm}\end{array}$$

From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, 2.0 × 10⁻¹⁶ is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or 10⁻³ > K > 10³, then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic NO, an ingredient of smog, does not form from atmospheric concentrations of N₂ and O₂ to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:

$$K_{\text{p}}=\frac{{(P_{\text{NO}})}^2}{\left(P_{{\text{N}}_2}\right)\left(P_{{\text{O}}_2}\right)}=\frac{{(1.8\times {10}^{-16})}^2}{(0.78)(0.21)}=2.0\times {10}^{-31}$$

The final K_p agrees with the value given at the beginning of this example.

Exercise

Under certain conditions, oxygen will react to form ozone, as shown in the following equation:

$$3{\text{O}}_2\text{(g)}\rightleftharpoons 2{\text{O}}_3\text{(g)}$$

K_p = 2.5 × 10⁻⁵⁹ at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere $\left(P_{{\text{O}}_2}=0.21\text{ atm}\right)$?

Answer: 4.8 × 10⁻³¹ atm

Example 13

The chemical equation for the reaction of hydrogen with ethylene (C₂H₄) to give ethane (C₂H₆) is as follows:

$${\text{H}}_2\text{(g)}+{\text{C}}_2{\text{H}}_4\text{(g)}\underset{}{\overset{\text{Ni}}{\rightleftharpoons }}{\text{C}}_2{\text{H}}_6\text{(g)}$$

K = 9.6 × 10¹⁸ at 25°C. If a mixture of 0.200 M H₂ and 0.155 M C₂H₄ is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?

Given: balanced chemical equation, K, and initial concentrations of reactants

Asked for: equilibrium concentrations

Strategy:

A Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.

B Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for x (the change in concentration).

C Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.

Solution:

A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define −x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is +x. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.

B Substituting values into the equilibrium constant expression,

$$K=\frac{[{\text{C}}_2{\text{H}}_6]}{[{\text{H}}_2][{\text{C}}_2{\text{H}}_4]}=\frac{0.155-x}{(0.045+x)x}=9.6\times {10}^{18}$$

Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x) = 0.045 and (0.155 − x) = 0.155] as follows:

$$\begin{array}{ccc}\hfill K& =\frac{0.155}{0.045x}\hfill & =9.6\times {10}^{18}\hfill \\ \hfill & x& =3.6\times {10}^{-19}\hfill \end{array}$$

C The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:

[C₂H₆]_f = (0.155 − x) M = 0.155 [C₂H₄]_f = x M = 3.6 × 10⁻¹⁹ M [H₂]_f = (0.045 + x) M = 0.045 M

We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:

$$K=\frac{[{\text{C}}_2{\text{H}}_6]}{[{\text{H}}_2][{\text{C}}_2{\text{H}}_4]}=\frac{0.155}{(0.045)(3.6\times {10}^{-19})}=9.6\times {10}^{18}$$

This K value agrees with our initial value at the beginning of the example.

Exercise

Hydrogen reacts with chlorine gas to form hydrogen chloride:

$${\text{H}}_2\text{(g)}+{\text{Cl}}_2\text{(g)}\rightleftharpoons 2\text{HCl(g)}$$

K_p = 4.0 × 10³¹ at 47°C. If a mixture of 0.257 M H₂ and 0.392 M Cl₂ is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?

Answer: [H₂]_f = 4.8 × 10⁻³² M; [Cl₂]_f = 0.135 M; [HCl]_f = 0.514 M