13.5 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:

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

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.

[latex]\Delta\left[{\text{H}}_{2}\right]=3\times \Delta\left[{\text{N}}_{2}\right][/latex]
[latex]=3\times \left(0.11M\right)=0.33M[/latex]

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.

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

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

[latex]\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}[/latex]

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.

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

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

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

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

[latex]\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}[/latex]

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.

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

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Check Your Learning

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

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

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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.

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

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?

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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.

[latex]{\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}[/latex]

When 1 mol each of C₂H₅OH and CH₃CO₂H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when [latex]\frac{1}{3}[/latex] mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)

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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, [latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)[/latex], 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.

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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.

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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?

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Use the stepwise process described above.

Determine the direction the reaction proceeds.The balanced equation for the decomposition of PCl₅ is[latex]{\text{PCl}}_{5}\left(g\right)\rightleftharpoons{\text{PCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)[/latex]Because we have no products initially, Q_c = 0 and the reaction will proceed to the right.
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.[latex]\begin{array}{llll}{\text{PCl}}_{5}\left(g\right)\hfill & \rightleftharpoons\hfill & {\text{PCl}}_{3}\left(g\right)+\hfill & {\text{Cl}}_{2}\left(g\right)\hfill \\ -x\hfill & & x\hfill & x\hfill \end{array}[/latex]The changes in concentration and the expressions for the equilibrium concentrations are:
Solve for the change and the equilibrium concentrations.Substituting the equilibrium concentrations into the equilibrium constant equation gives[latex]{K}_{c}=\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}=0.0211[/latex]
[latex]=\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}[/latex]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.[latex]0.0211=\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}[/latex]
[latex]0.0211\left(1.00-x\right)={x}^{2}[/latex]
[latex]{x}^{2}+0.0211 x-0.0211=0[/latex]

Essential Mathematics shows us 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]

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{{\left(0.0211\right)}^{2}-4\left(1\right)\left(-0.0211\right)}}{2\left(1\right)}[/latex]
[latex]=\frac{-0.0211\pm\sqrt{\left(4.45\times {10}^{-4}\right)+\left(8.44\times {10}^{-2}\right)}}{2}[/latex]
[latex]=\frac{-0.0211\pm0.291}{2}[/latex]

Hence

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

or

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

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

[latex]\left[{\text{PCl}}_{5}\right]=1.00-0.135=0.87M[/latex]
[latex]\left[{\text{PCl}}_{3}\right]=x=0.135M[/latex]
[latex]\left[{\text{Cl}}_{2}\right]=x=0.135M[/latex]
Check the arithmetic.Substitution into the expression for K_c (to check the calculation) gives[latex]{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[/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). Thus, the calculated equilibrium concentrations check.

Check Your Learning

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

[latex]{\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}[/latex]

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?

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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?
[latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\rightleftharpoons2\text{HI}\left(g\right)[/latex]

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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

A reaction is represented by this equation: [latex]\text{A}\left(aq\right)+2\text{B}\left(aq\right)\rightleftharpoons2\text{C}\left(aq\right){K}_{c}=1\times {10}^{3}[/latex]
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: [latex]2\text{W}\left(aq\right)\rightleftharpoons\text{X}\left(aq\right)+2\text{Y}\left(aq\right){K}_{c}=5\times {10}^{-4}[/latex]
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.
What is the value of the equilibrium constant at 500 °C for the formation of NH₃ according to the following equation?
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
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.
[latex]{\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)[/latex]
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
[latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)+{\text{O}}_{2}\left(g\right)[/latex]
Calculate the value of the equilibrium constant K_P for the reaction [latex]2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightleftharpoons2\text{NOCl}\left(g\right)[/latex] 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: [latex]{\text{I}}_{2}\left(g\right)\rightleftharpoons2\text{I}\left(g\right)[/latex]
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: [latex]{\text{NH}}_{4}\text{Cl}\left(s\right)\rightleftharpoons{\text{NH}}_{3}\left(g\right)+\text{HCl}\left(g\right)[/latex]
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?
[latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right)[/latex]
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
1. [latex]\begin{array}{llll}2{\text{SO}}_{3}\left(g\right)\hfill & \rightleftharpoons\hfill & 2{\text{SO}}_{2}\left(g\right)+\hfill & {\text{O}}_{2}\left(g\right)\hfill \\ \text{ }\hfill & & \text{ }\hfill & +x\hfill \\ \text{ }\hfill & & \text{ }\hfill & 0.125M\hfill \end{array}[/latex]
2. [latex]\begin{array}{lllll}4{\text{NH}}_{3}\left(g\right)\hfill & +3{\text{O}}_{2}\left(g\right)\hfill & \rightleftharpoons\hfill & 2{\text{N}}_{2}\left(g\right)+\hfill & 6{\text{H}}_{2}\text{O}\left(g\right)\hfill \\ \text{ }\hfill & 3x\hfill & & \text{ }\hfill & \text{ }\hfill \\ \text{ }\hfill & 0.24M\hfill & & \text{ }\hfill & \text{ }\hfill \end{array}[/latex]
3. Change in pressure:
  [latex]\begin{array}{llll}2{\text{CH}}_{4}\left(g\right)\hfill & \rightleftharpoons\hfill & {\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\hfill & 3{\text{H}}_{2}\left(g\right)\hfill \\ \text{ }\hfill & & x\hfill & \text{ }\hfill \\ \text{ }\hfill & & 25\text{torr}\hfill & \text{ }\hfill \end{array}[/latex]
4. Change in pressure:
  [latex]\begin{array}{lllll}{\text{CH}}_{4}\left(g\right)+\hfill & {\text{H}}_{2}\text{O}\left(g\right)\hfill & \rightleftharpoons\hfill & \text{CO}\left(g\right)+\hfill & 3{\text{H}}_{2}\left(g\right)\hfill \\ \text{ }\hfill & x\hfill & & \text{ }\hfill & \text{ }\hfill \\ \text{ }\hfill & 5\text{atm}\hfill & & \text{ }\hfill & \text{ }\hfill \end{array}[/latex]
5. [latex]\begin{array}{llll}{\text{NH}}_{4}\text{Cl}\left(s\right)\hfill & \rightleftharpoons\hfill & {\text{NH}}_{3}\left(g\right)+\hfill & \text{HCl}\left(g\right)\hfill \\ & & x\hfill & \text{ }\hfill \\ & & \hfill 1.03\times {10}^{-4}M\hfill & \text{ }\hfill \end{array}[/latex]
6. change in pressure:
  [latex]\begin{array}{cccc}\text{Ni}\left(s\right)+\hfill & 4\text{CO}\left(g\right)\hfill & \rightleftharpoons\hfill & \text{Ni}{\left(\text{CO}\right)}_{4}\left(g\right)\hfill \\ & 4x\hfill & & \text{ }\hfill \\ & \hfill 0.40\text{atm}\hfill & & \text{ }\hfill \end{array}[/latex]
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
1. [latex]\begin{array}{cccc}2{\text{H}}_{2}\left(g\right)+\hfill & {\text{O}}_{2}\left(g\right)\hfill & \rightleftharpoons\hfill & 2{\text{H}}_{2}\text{O}\left(g\right)\hfill \\ \text{ }\hfill & \text{ }\hfill & & +2x\hfill \\ \text{ }\hfill & \text{ }\hfill & & 1.50M\hfill \end{array}[/latex]
2. [latex]\begin{array}{ccccc}{\text{CS}}_{2}\left(g\right)+\hfill & 4{\text{H}}_{2}\left(g\right)\hfill & \rightleftharpoons\hfill & {\text{CH}}_{4}\left(g\right)+\hfill & 2{\text{H}}_{2}\text{S}\left(g\right)\hfill \\ x\hfill & \text{ }\hfill & & \text{ }\hfill & \text{ }\hfill \\ 0.020M\hfill & \text{ }\hfill & & \text{ }\hfill & \text{ }\hfill \end{array}[/latex]
3. Change in pressure:
  [latex]\begin{array}{cccc}{\text{H}}_{2}\left(g\right)+\hfill & {\text{Cl}}_{2}\left(g\right)\hfill & \rightleftharpoons\hfill & 2\text{HCl}\left(g\right)\hfill \\ x\hfill & \text{ }\hfill & & \text{ }\hfill \\ 1.50\text{atm}\hfill & \text{ }\hfill & & \text{ }\hfill \end{array}[/latex]
4. Change in pressure:
  [latex]\begin{array}{ccccc}2{\text{NH}}_{3}\left(g\right)\hfill & +2{\text{O}}_{2}\left(g\right)\hfill & \rightleftharpoons\hfill & {\text{N}}_{2}\text{O}\left(g\right)+\hfill & 3{\text{H}}_{2}\text{O}\left(g\right)\hfill \\ \text{ }\hfill & \text{ }\hfill & & \text{ }\hfill & x\hfill \\ \text{ }\hfill & \text{ }\hfill & & \text{ }\hfill & 60.6\text{torr}\hfill \end{array}[/latex]
5. [latex]\begin{array}{cccc}{\text{NH}}_{4}\text{HS}\left(s\right)\hfill & \rightleftharpoons\hfill & {\text{NH}}_{3}\left(g\right)+\hfill & {\text{H}}_{2}\text{S}\left(g\right)\hfill \\ & & x\hfill & \text{ }\hfill \\ & & 9.8\times {10}^{-6}M\hfill & \text{ }\hfill \end{array}[/latex]
6. Change in pressure:
  [latex]\begin{array}{cccc}\text{Fe}\left(s\right)+\hfill & 5\text{CO}\left(g\right)\hfill & \rightleftharpoons\hfill & \text{Fe}{\left(\text{CO}\right)}_{4}\left(g\right)\hfill \\ & \text{ }\hfill & & x\hfill \\ & \text{ }\hfill & & 0.012\text{atm}\hfill \end{array}[/latex]
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?
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.
[latex]{\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}[/latex]
Calculate the equilibrium molar concentration of NH₃.
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.
[latex]{\text{H}}_{2}+{\text{I}}_{2}\rightleftharpoons2\text{HI}{K}_{c}=50.2\text{ at }448^{\circ}\text{C}[/latex]
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?
[latex]{\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}[/latex]
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?
[latex]{\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}[/latex]
Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
[latex]\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}[/latex]
What concentration of CO remains in an equilibrium mixture with [CO₂] = 0.100 M?
Carbon reacts with water vapor at elevated temperatures.
[latex]\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}[/latex]
What is the concentration of CO in an equilibrium mixture with [H₂O] = 0.500 M at 1000 °C?
Sodium sulfate 10-hydrate, Na₂SO₄ [latex]\cdot[/latex] 10H₂O, dehydrates according to the equation
[latex]{\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}[/latex]
What is the pressure of water vapor at equilibrium with a mixture of Na₂SO₄ [latex]\cdot[/latex] 10H₂O and NaSO₄?
Calcium chloride 6-hydrate, CaCl₂ [latex]\cdot[/latex] 6H₂O, dehydrates according to the equation
[latex]{\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}[/latex]
What is the pressure of water vapor at equilibrium with a mixture of CaCl₂ [latex]\cdot[/latex] 6H₂O and CaCl₂?
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:
[latex]2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{SO}}_{3}\left(g\right){K}_{c}=4.32[/latex]
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
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?
[latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right){K}_{c}=160[/latex]
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.
  [latex]{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right){K}_{c}=1.07\times {10}^{-5}[/latex] 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.
  [latex]{\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}[/latex]
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.
  [latex]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}[/latex]
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?
[latex]{\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[/latex]
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?
[latex]{\text{PCl}}_{5}\left(g\right)\rightleftharpoons{\text{PCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right){K}_{c}=0.0211[/latex]
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:
[latex]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}[/latex]
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.)
[latex]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}[/latex]
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.)
[latex]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}[/latex]
One of the important reactions in the formation of smog is represented by the equation
[latex]{\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}[/latex]
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.
[latex]{\text{H}}_{2}+{\text{I}}_{2}\rightleftharpoons2\text{HI}{K}_{c}=50.2\text{ at }448^{\circ}\text{C}[/latex]
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?
[latex]{\text{CaCO}}_{3}\left(s\right)\rightleftharpoons\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)[/latex]
The equilibrium constant (K_c) for this reaction is 1.60 at 990 °C: [latex]{\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)[/latex]
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.
At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of N₂O₄ and NO₂ are [latex]{\text{P}}_{{\text{N}}_{2}{\text{O}}_{4}}=0.70\text{atm}[/latex] and [latex]{\text{P}}_{{\text{NO}}_{2}}=0.30\text{atm.}[/latex]
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: [latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
1. How will the partial pressures of H₂, N₂, and NH₃ change if H₂ is removed from the system? Will they increase, decrease, or remain the same?
2. Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.
The equilibrium constant (K_c) for this reaction is 5.0 at a given temperature: [latex]\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)\rightleftharpoons{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)[/latex]
1. On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H₂ in a liter. How many moles of CO₂ were there in the equilibrium mixture?
2. Maintaining the same temperature, additional H₂ was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H₂ in a liter. How many moles of CO₂ were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.
Antimony pentachloride decomposes according to this equation: [latex]{\text{SbCl}}_{5}\left(g\right)\rightleftharpoons{\text{SbCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)[/latex]
An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl₅, 9.14 g of SbCl₃, and 2.84 g of Cl₂. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?
Consider the reaction between H₂ and O₂ at 1000 K
[latex]2H_{2}(g)+O_{2}(g)\rightleftharpoons{2H_{2}O(g)}[/latex] [latex]K_{P}=\frac{(P_{H_{2}O})^{2}}{(P_{O_{2}})(P_{H_{2}})^{3}}=1.33\times{10^{20}}[/latex]
If 0.500 atm of H2 and 0.500 atm of O2 are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?
An equilibrium is established according to the following equation
[latex]{\text{Hg}}_{2}{}^{2+}\left(aq\right)+{\text{NO}}_{3}{}^{-}\left(aq\right)+3{\text{H}}^{+}\left(aq\right)\rightleftharpoons2{\text{Hg}}^{2+}\left(aq\right)+{\text{HNO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right){K}_{c}=4.6[/latex]
What will happen in a solution that is 0.20 M each in [latex]{\text{Hg}}_{2}{}^{2+}[/latex], [latex]{\text{NO}}_{3}{}^{-}[/latex], H⁺, Hg²⁺, and HNO₂?
1. [latex]{\text{Hg}}_{2}{}^{2+}[/latex] will be oxidized and [latex]{\text{NO}}_{3}{}^{-}[/latex] reduced.
2. [latex]{\text{Hg}}_{2}{}^{2+}[/latex] will be reduced and [latex]{\text{NO}}_{3}{}^{-}[/latex] oxidized.
3. Hg²⁺ will be oxidized and HNO₂ reduced.
4. Hg²⁺ will be reduced and HNO₂ oxidized.
5. There will be no change, because all reactants and products have an activity of 1.
Consider the equilibrium: [latex]4{\text{NO}}_{2}\left(g\right)+6{\text{H}}_{2}\text{O}\left(g\right)\rightleftharpoons4{\text{NH}}_{3}\left(g\right)+7{\text{O}}_{2}\left(g\right)[/latex]
1. What is the expression for the equilibrium constant (K_c) of the reaction?
2. How must the concentration of NH₃ change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
3. If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO₂?
4. If the change in the pressure of NO₂ is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O₂ change?
The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO₂), is partially regulated by the concentration of H₃O⁺ and dissolved CO₂ in the blood. Although the equilibrium is complicated, it can be summarized as
[latex]{\text{HbO}}_{2}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CO}}_{2}\left(g\right)\rightleftharpoons{\text{CO}}_{2}-\text{Hb}-{\text{H}}^{+}+{\text{O}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)[/latex]
1. Write the equilibrium constant expression for this reaction.
2. Explain why the production of lactic acid and CO₂ in a muscle during exertion stimulates release of O₂ from the oxyhemoglobin in the blood passing through the muscle.
The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.
[latex]{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightarrow{\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}\left(aq\right)[/latex]
Rate = k[C₁₂H₂₂O₁₁]
In neutral solution, k = 2.1 × 10^-11/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C₆H₁₂O₆, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 10⁵ at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1).
The density of trifluoroacetic acid vapor was determined at 118.1 °C and 468.5 torr, and found to be 2.784 g/L. Calculate K_c for the association of the acid.
Liquid N₂O₃ is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO₂. At 25 °C, a value of K_P = 1.91 has been established for this decomposition. If 0.236 moles of N₂O₃ are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N₂O₃(g), NO₂(g), and NO(g).
A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N₂, 1.00 M; H₂, 0.50 M; and NH₃, 0.25 M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M?
A 0.010 M solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010 M solution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.
1. Which acid has the larger equilibrium constant for ionization
  HA [latex]\left[\text{HA}\left(aq\right)\rightleftharpoons{\text{A}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\right][/latex] or HB [latex]\left[\text{HB}\left(aq\right)\rightleftharpoons{\text{H}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)\right][/latex] ?
2. What are the equilibrium constants for the ionization of these acids?
  (Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A^– or B^–), and the hydrogen ion (H⁺). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)

Show Selected Answers

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

3. The reaction may be written as

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

The equilibrium constant for the reaction is

[latex]{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}}[/latex]

= 0.0600 = 6.00 × 10^-2

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

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

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

[latex]{K}_{c}=\frac{\left(0.40\right)\left(0.40\right)}{\left(0.32\right)}=0.50[/latex]

7. The equilibrium equation is

[latex]{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}[/latex]

9. Because the decomposition must generate the same pressure of HCl as NH₃, 1.75 atm of HCl must be present. [latex]{K}_{p}={P}_{{\text{NH}}_{3}}{P}_{\text{HCl}}=\left(1.75\text{atm}\right)\left(1.75\text{atm}\right)=3.06[/latex]

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

[latex]\begin{array}{llll}2{\text{SO}}_{3}\left(g\right)\hfill & \rightleftharpoons\hfill & 2{\text{SO}}_{2}\left(g\right)+\hfill & {\text{O}}_{2}\left(g\right)\hfill \\ \text{ }\hfill -2x & &2x\hfill & +x\hfill \\ -0.250M\hfill & & 0.250M\hfill & 0.125M\hfill \end{array}[/latex]
[latex]\begin{array}{lllll}4{\text{NH}}_{3}\left(g\right)\hfill & +3{\text{O}}_{2}\left(g\right)\hfill & \rightleftharpoons\hfill & 2{\text{N}}_{2}\left(g\right)+\hfill & 6{\text{H}}_{2}\text{O}\left(g\right)\hfill \\ 4x\hfill & 3x\hfill & & -2x\hfill & -6x\hfill \\ 0.32M\hfill & 0.24M\hfill & &-0.16M\hfill & -0.48M\hfill \end{array}[/latex]
Change in pressure:
[latex]\begin{array}{llll}2{\text{CH}}_{4}\left(g\right)\hfill & \rightleftharpoons\hfill & {\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\hfill & 3{\text{H}}_{2}\left(g\right)\hfill \\ -2x\hfill & & x\hfill & 3x\hfill \\ -50\text{ torr}\hfill & & 25\text{ torr}\hfill & 75\text{ torr}\hfill \end{array}[/latex]
Change in pressure:
[latex]\begin{array}{lllll}{\text{CH}}_{4}\left(g\right)+\hfill & {\text{H}}_{2}\text{O}\left(g\right)\hfill & \rightleftharpoons\hfill & \text{CO}\left(g\right)+\hfill & 3{\text{H}}_{2}\left(g\right)\hfill \\ x\hfill & x\hfill & & -x\hfill & -3x\hfill \\ 5\text{ atm}\hfill & 5\text{atm}\hfill & & -5\text{ atm}\hfill & -15\text{ atm}\hfill \end{array}[/latex]
[latex]\begin{array}{llll}{\text{NH}}_{4}\text{Cl}\left(s\right)\hfill & \rightleftharpoons\hfill & {\text{NH}}_{3}\left(g\right)+\hfill & \text{HCl}\left(g\right)\hfill \\ & & x\hfill & x\hfill \\ & & \hfill 1.03\times {10}^{-4}M\hfill & 1.03\times10^{-4}M\hfill \end{array}[/latex]
change in pressure:
[latex]\begin{array}{cccc}\text{Ni}\left(s\right)+\hfill & 4\text{CO}\left(g\right)\hfill & \rightleftharpoons\hfill & \text{Ni}{\left(\text{CO}\right)}_{4}\left(g\right)\hfill \\ & 4x\hfill & & x\hfill \\ & \hfill 0.40\text{atm}\hfill & & 0.1\text{ atm}\hfill \end{array}[/latex]

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

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

[latex]{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[/latex]

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

[NH₃] = 9.1 × 10^-2M

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

[latex]\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}[/latex]

P_BrCl = 4.9 × 10^-2 atm

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

[latex]\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}\hfill \\ \left[\text{CO}\right]=\frac{0.100}{4.90\times {10}^{2}}=2.0\times {0}^{-4}M\hfill \end{array}[/latex]

21. 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.

[latex]\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}[/latex]

23. 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.

25. (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

[latex]\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}\hfill \\ {x}^{2}=\frac{0.129\times 1.07\times {10}^{-5}}{4}=3.45\times {10}^{-7}\hfill \end{array}[/latex]

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 [latex]=\frac{5.87\times {10}^{-4}}{0.129}\times 100\%=0.455\%[/latex]. The change in concentration of N₂O₄ is far less than the 5% maximum allowed.

27. (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.

[latex]{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}}[/latex]

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.

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

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 [latex]\frac{0.014}{0.824}\times 100\%=1.7\%[/latex], which is less than allowed by the “5% test.” The error is, indeed, negligible.

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

[latex]{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]}[/latex]

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.

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

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.

[latex]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)}[/latex]

[latex]=\frac{-0.0211\pm\sqrt{0.0004452+0.1688}}{2}=\frac{-0.0211\pm 0.4114}{2}[/latex]

= 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.

31. 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.

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, [latex]\frac{1}{{K}_{c}}[/latex].

[latex]\begin{array}{ll}{K}_{\text{new}}\hfill & =\frac{1}{{K}_{c}}=\frac{{\left[\text{NO}\right]}^{2}\left[{\text{O}}_{2}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}\hfill \\ \hfill & =\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}}\hfill \end{array}[/latex]

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

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

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

[latex]\left(\frac{2x}{0.20}\right)\times 100\%=\left(\frac{0.00352}{0.20}\right)\times 100\%=1.7\%[/latex] which is less than the maximum 5% allowed.

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

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

[latex]{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)}[/latex]

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.

[latex]\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)}\hfill \\ {x}^{2}=\frac{\left(1.44\times {10}^{-16}\right)}{6.0\times {10}^{34}}=2.4\times {10}^{-51}\hfill \end{array}[/latex]

x = 4.9 × 10⁻²⁶ atm

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

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

35. The number of moles of I₂ is

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

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 = [latex]\sqrt{15.7{\text{mol}}^{2}}=3.96\text{mol}[/latex]

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

37. 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:

[latex]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}[/latex]

[latex]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}[/latex]

39. (a) The reaction is [latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\text{.}[/latex] At equilibrium, [latex]{K}_{P}=\frac{\left({P}_{{\text{N}}_{2}{\text{O}}_{4}}\right)}{{\left({P}_{{\text{NO}}_{2}}\right)}^{2}}\text{.}[/latex] The value of K_P must remain the same when the pressure increases to 9.0 atm. Both gases must increase in pressure.

(b) [latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\text{.}[/latex] At equilibrium, [latex]{K}_{P}=\frac{{P}_{{\text{N}}_{2}{\text{O}}_{4}}}{{\left({P}_{{\text{NO}}_{2}}\right)}^{2}}=\frac{0.70}{{\left(0.30\right)}^{2}}=\frac{0.70}{0.09}=7.78.[/latex] For a total pressure of 9.0 atm, the pressure of N₂O₄ is 9.0 – x; that of NO is x.

[latex]{K}_{P}=\frac{9.0-x}{{x}^{2}}=7.78[/latex]

7.78x² + x − 9.0 = 0

Use the quadratic expression, where

[latex]x=\frac{\text{-}b\pm \sqrt{{b}^{2}-4ac}}{2a}=\frac{-1\pm \sqrt{1+4\left(7.78\right)\left(-9.0\right)}}{2\left(7.78\right)}.[/latex]

[latex]=\frac{-1\pm 16.765}{15.56}=1.013[/latex]

For the plus sign (the negative sign gives a negative pressure which is impossible)

[latex]x=\frac{15.765}{15.56}=1.013[/latex]

The other answer is extraneous. The pressures are [latex]{P}_{{\text{N}}_{2}{\text{O}}_{4}}[/latex] = 8.0 atm and [latex]{P}_{{\text{NO}}_{2}}[/latex] = 1.0 atm

41. (a) For the above reaction, [latex]{K}_{c}=\frac{\left[{\text{CO}}_{2}\right]\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\left[{\text{H}}_{2}\text{O}\right]}=5.0.[/latex] The concentrations at equilibrium are 0.20 M CO, 0.30 M H₂O, and 0.90 M H₂. Substitution gives [latex]K=5.0=\frac{\left[{\text{CO}}_{2}\right]\left[0.90\right]}{\left[0.20\right]\left[0.30\right]};[/latex] [latex]\left[{\text{CO}}_{2}\right]=\frac{5.0\left(0.20\right)\left(0.30\right)}{0.90}=0.33M;[/latex] Amount of CO₂ = 0.33 mol × 1 = 0.33 mol.

(b) At the particular temperature of reaction, K_c remains constant at 5.0. The new concentrations are 0.40 M CO, 0.30 M H₂O, and 1.2 M H₂.

[latex]\frac{\left[{\text{CO}}_{2}\right]\left(1.2\right)}{\left(0.40\right)\left(0.30\right)}=5.0[/latex]

[latex]{\left[\text{CO}\right]}^{2}=0.50M[/latex]

Amount of CO₂ = 0.50 mol × 1 = 0.50 mol. Added H₂ forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H₂ is added.

43. P_H₂ = 8.64 × 10⁻¹¹atm

P_O₂ = 0.250atm

P_H₂O = 0.500 atm

45. (a) [latex]{K}_{c}=\frac{{\left[{\text{NH}}_{3}\right]}^{4}{\left[{\text{O}}_{2}\right]}^{7}}{{\left[{\text{NO}}_{2}\right]}^{4}{\left[{\text{H}}_{2}\text{O}\right]}^{6}}\text{.}[/latex] (b) Because [NH₃] is in the numerator of K_c, [NH₃] must increase for Q_c to reach K_c. (c) That decrease in pressure would decrease [NO₂] because the NO₂ must convert to NH₃ to reduce the effects of the expansion and the consequent relative decrease in products. (d) The relative pressures are controlled by the stoichiometry of the reaction.

[latex]{P}_{{\text{O}}_{2}}=\frac{7}{4}{P}_{{\text{NO}}_{2}}=\frac{7}{4}\left(28\text{torr}\right)=49\text{torr}[/latex]

47. [fructose] = 0.15 M

49. Write the balanced equilibrium expression. With all of the species as gases, it is a straightforward K_P problem to solve. However, all species must be converted to pressures, from other units related to concentration. For N₂O₃, with 0.236 mol in 1.52 L at 25 ºC: PV = nRT

[latex]\begin{array}{lll}P& =& \frac{n}{V}RT\\ & =& \frac{0.236\cancel{\text{mol}}}{1.52\cancel{\text{L}}}\times \frac{\left(0.08206\cancel{\text{L}}\text{atm}\right)\left(298.15\cancel{\text{K}}\right)}{\cancel{\text{mol}}\cancel{\text{K}}}\\ & =& 3.80\text{atm}\end{array}[/latex]

Write the balanced equation and the equilibrium changes:

[latex]{K}_{P}=\frac{\left({P}_{\text{NO}}\right)\left({P}_{{\text{NO}}_{2}}\right)}{\left({P}_{{\text{N}}_{2}{\text{O}}_{3}}\right)}[/latex]

[latex]1.91\text{atm}=\frac{\left(x\right)\left(x\right)}{\left(3.80-x\right)}[/latex]

Because 3.80 is substantial when compared with the equilibrium constant, the value of X must be considered. A quadratic equation must be solved:

[latex]1.91=\frac{{x}^{2}}{\left(3.80-x\right)}[/latex]

7.258 − 1.91x = x²

0 = x² +1.91x – 7.258

0 = ax² + bx + c

[latex]\begin{array}{lll}x& =& \frac{\text{-}b\pm \sqrt{{b}^{2}-4ac}}{2a}\\ & =& \frac{-1.91\pm \sqrt{{\left(1.91\right)}^{2}-4\left(1\right)\left(-7.258\right)}}{2\left(1\right)}\\ & =& \frac{-1.91\pm \sqrt{3.648+29.032}}{2}\\ & =& \frac{-1.91\pm \sqrt{32.680}}{2}\\ & =& \frac{-1.91\pm 5.717}{2}\end{array}[/latex]

= 1.90 atm or -3.81 atm

As negative pressure is not possible in this case, we use the positive value only. The final pressures are: [latex]{P}_{{\text{N}}_{2}{\text{O}}_{3}}[/latex] = 3.80 − x = 3.80 − 1.90 = 1.90 atm and [latex]{P}_{\text{NO}}={P}_{{\text{NO}}_{2}}={P}_{{\text{NO}}_{2}}x=1.90\text{atm}[/latex]

51. (a) Recall that osmotic pressure is a colligative property, depending on the total number of particles present in the system. The osmotic pressure equation is π = iMRT. Given that the two acids face identical reaction conditions, the osmotic pressure (π) will increase only with a corresponding (>1) in the degree of ionization (i). Because both acids produce two ions per molecule dissociated, the number of particles in solution depends on the size of their respective equilibrium constants (K_c(HA) and K_c(HB)):

[latex]\text{HA}\rightleftharpoons{\text{H}}^{+}+{\text{A}}^{-}\text{HB}\rightleftharpoons{\text{H}}^{+}+{\text{B}}^{-}[/latex]

Because HB has a greater osmotic pressure (0.345 atm) than HA (0.293 atm), HB must produce more particles. Therefore, HB ionizes to a greater degree and has the larger K_c.

(b) Determine the value of i for each acid. For HA,

[latex]\pi =iMRT=0.293\text{atm}=i\left(0.010M\right)\left(\frac{0.08206\text{L atm}}{\text{mol K}}\right)\left(298.15\text{K}\right)[/latex]

i_HY = 1.2

For HB,

[latex]\pi =iMRT=0.345\text{atm}=i\left(0.010M\right)\left(\frac{0.08206\text{L atm}}{\text{mol K}}\right)\left(298.15\text{K}\right)[/latex]

i_HZ = 1.4

The i values mean that for every mole of originally undissociated species, i moles particles are produced in solution. The dissociations can be expressed in tabular form. For HA,

0.012 M = xM + xM + (0.010 − x) M

x = 0.002 M

For HB, the only difference from the table for HA is the value of i_HB. x = 0.004 M.

Now calculate the values of K_c(HA) and K_c(HB).

For HA,

[latex]{K}_{c}\left(\text{HA}\right)=\frac{\left[x\right]\left[x\right]}{\left[0.010-x\right]}=\frac{\left(0.002\right)\left(0.002\right)}{0.008}=5\times {10}^{-4}[/latex]

For HB,

[latex]{K}_{c}\left(\text{HB}\right)=\frac{\left[x\right]\left[x\right]}{\left[0.010-x\right]}=\frac{\left(0.004\right)\left(0.004\right)}{0.006}=3\times {10}^{-3}[/latex]

Chapter 13: Chemical Equilibrium

Learning Objectives

Example 1: Determining Relative Changes in Concentration

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Calculations Involving Equilibrium Concentrations

Calculation of an Equilibrium Constant

Example 2: Calculation of an Equilibrium Constant

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Calculation of a Missing Equilibrium Concentration

Example 3: Calculation of a Missing Equilibrium Concentration

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Calculation of Changes in Concentration

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

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Key Concepts and Summary

Exercises

Candela Citations