13.2 The Equilibrium Constant

Learning Objectives

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

Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
Compare values of Q_c and K_c of a system to determine how the reaction will proceed, if at all
Relate the magnitude of an equilibrium constant to properties of the chemical system

Now that we have a symbol [latex]\rightleftharpoons[/latex] to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:

[latex]m\text{A}+n\text{B}+\rightleftharpoons x\text{C}+y\text{D}[/latex]

We can write the reaction quotient (Q) for this equation. When evaluated using concentrations, it is called Q_c. We use brackets to indicate molar concentrations of reactants and products.

[latex]\displaystyle{Q}_{c}=\frac{{\left[\text{C}\right]}^{x}{\left[\text{D}\right]}^{y}}{{\left[\text{A}\right]}^{m}{\left[\text{B}\right]}^{n}}[/latex]

The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction [latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)[/latex] is given by this expression:

[latex]\displaystyle{Q}_{c}=\frac{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}[/latex]

Example 1: Writing Reaction Quotient Expressions

Write the expression for the reaction quotient for each of the following reactions:

[latex]3{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{O}}_{3}\left(g\right)[/latex]
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
[latex]4{\text{NH}}_{3}\left(g\right)+7{\text{O}}_{2}\left(g\right)\rightleftharpoons4{\text{NO}}_{2}\left(g\right)+6{\text{H}}_{2}\text{O}\left(g\right)[/latex]

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Check Your Learning

Write the expression for the reaction quotient for each of the following reactions:

[latex]2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{SO}}_{3}\left(g\right)[/latex]
[latex]{\text{C}}_{4}{\text{H}}_{8}\left(g\right)\rightleftharpoons2{\text{C}}_{2}{\text{H}}_{4}\left(g\right)[/latex]
[latex]2{\text{C}}_{4}{\text{H}}_{10}\left(g\right)+13{\text{O}}_{2}\left(g\right)\rightleftharpoons8{\text{CO}}_{2}\left(g\right)+10{\text{H}}_{2}\text{O}\left(g\right)[/latex]

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The numeric value of Q_c for a given reaction varies; it depends on the concentrations of products and reactants present at the time when Q_c is determined. When pure reactants are mixed, Q_c is initially zero because there are no products present at that point. As the reaction proceeds, the value of Q_c increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease Figure 1. When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.

Three graphs are shown and labeled, “a,” “b,” and “c.” All three graphs have a vertical dotted line running through the middle labeled, “Equilibrium is reached.” The y-axis on graph a is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph a. The first is labeled, “[ S O subscript 2 ];” this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph b is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph b. The first is labeled, “[ S O subscript 2 ];” this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph c is labeled, “Reaction Quotient,” and the x-axis is labeled, “Time.” A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, “k.”

Figure 1. (a) The change in the concentrations of reactants and products is depicted as the 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) reaction approaches equilibrium. (b) The change in concentrations of reactants and products is depicted as the reaction 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) approaches equilibrium. (c) The graph shows the change in the value of the reaction quotient as the reaction approaches equilibrium.

When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as K_c.

That a reaction quotient always assumes the same value at equilibrium can be expressed as:

[latex]\displaystyle{Q}_{c}\text{ at equilibrium}={K}_{c}=\frac{{\left[\text{C}\right]}^{x}{\left[\text{D}\right]}^{y}\ldots }{{\left[\text{A}\right]}^{m}{\left[\text{B}\right]}^{n}\ldots}[/latex]

This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.

Often you will know the concentrations of reactants and products for a particular reaction and want to know whether the system is at equilibrium. If it is not, it is useful to predict how those concentrations will change as the reaction approaches equilibrium.

For example, suppose you are interested in the reaction

\[2 \text{SO}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{SO}_{3} (g)\]

\[ K_{\text{c}} = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = 245 \text{ mol/L} \text{(at 1000 K)} \]

and you have added 0.10 mol of each gas to a container with volume 10.0 L. Is the system at equilibrium? If not, will the concentration of SO₃ be greater than or less than 0.010 mol/L when equilibrium is reached? You can answer these questions by calculating Q and comparing it with K_c. There are three possibilities:

If Q = K_c then the actual concentrations of products (and of reactants) are equal to the equilibrium concentrations and the system is at equilibrium.
If Q < K_c then the actual concentrations of products are less than the equilibrium concentrations; the forward reaction will occur and more products will be formed.
If Q > K_c then the actual concentrations of products are greater than the equilibrium concentrations; the reverse reaction will occur and more reactants will be formed.

For the reaction given above,

\[ Q = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = \frac{(\frac{0.1 mol}{10 L})^{2}}{(\frac{0.1 mol}{10 L})^{2} (\frac{0.1 mol}{10 L})} = 100 \frac{\text{mol}}{\text{L}} \]

(In the expression for Q each actual concentration is enclosed in braces {curly brackets} in order to distinguish it from the equilibrium concentrations, which, in the K_c expression, are enclosed in [square brackets].) In this case Q = 100 mol/L. This is less than K_c, which has the value 245 mol/L. This implies that the concentrations of products are less than the equilibrium concentrations (and the concentrations of reactants are greater than the equilibrium concentrations). Therefore the reaction will proceed in the forward direction, producing more products, until the concentrations reach their equilibrium values.

Example 2: Evaluating a reaction quotient

At 2300 K, the equilibrium constant, K_c, is 1.7 x 10^-3 for the reaction
\[ \text{N}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{NO} (g) \]
A mixture of the three gases at 2300 K has these concentrations, [N₂] = 0.17 M, [O₂] = 0.17 M, and [NO] = 0.034 M.
(a) Is the system at equilibrium?
(b) In which direction must the reaction occur to reach equilibrium?

Solutions

Example 3: Evaluating a Reaction Quotient

Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:

[latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)[/latex]

When 0.10 mol NO₂ is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO₂] = 0.016 M and [N₂O₄] = 0.042 M.

(a) What is the value of the reaction quotient before any reaction occurs?

(b) What is the value of the equilibrium constant for the reaction?

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Check Your Learning

For the reaction [latex]2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{SO}}_{3}\left(g\right)[/latex], the concentrations at equilibrium are [SO₂] = 0.90 M, [O₂] = 0.35 M, and [SO₃] = 1.1 M. What is the value of the equilibrium constant, K_c?

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The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for K_c indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of K_c—much less than 1—indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.

Once a value of K_c is known for a reaction, it can be used to predict directional shifts when compared to the value of Q_c. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure 2 illustrate this. When heated to a consistent temperature, 800 °C, different starting mixtures of CO, H₂O, CO₂, and H₂ react to reach compositions adhering to the same equilibrium (the value of Q_c changes until it equals the value of K_c). This value is 0.640, the equilibrium constant for the reaction under these conditions.

[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){K}_{c}=0.640\text{ T}=800 ^{\circ}\text{C}[/latex]

It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure 2 when the products of the reaction were heated in a closed container. In fact, one technique used to determine whether a reaction is truly at equilibrium is to approach equilibrium starting with reactants in one experiment and starting with products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.

Two sets of bar graphs are shown. The left is labeled, “Before reaction,” and the right is labeled, “At equilibrium.” Both graphs have y-axes labeled, “Concentration ( M ),” and three bars on the x-axes labeled, “Mixture 1,” “Mixture 2,” and “Mixture 3.” The y-axis has a scale beginning at 0.00 and ending at 0.10, with measurement increments of 0.02. The bars on the graphs are color coded and a key is provided with a legend. Red is labeled, “C O;” blue is labeled, “H subscript 2 O;” green is labeled, “C O subscript 2,” and yellow is labeled, “H subscript 2.” The graph on the left shows the red bar for mixture one just above 0.02 and the blue bar near 0.05. For mixture two, the green bar is near 0.05, and the yellow bar is near 0.09. For mixture 3, the red bar is near 0.01, the blue bar is slightly above that with green and yellow topping it off at 0.02. On the right graph, the bar for mixture one shows the red bar slightly above 0.01, the blue bar stacked on it rising slightly above 0.02, the green rising near 0.04, and the yellow bar reaching near 0.05. A label above this bar reads, “Q equals 0.640.” The bar for mixture two shows the red bar slightly above 0.02, the blue bar stacked on it rising near 0.05, the green rising near 0.07, and the yellow bar reaching near 0.10. A label above this bar reads “Q equals 0.640.” The bar for mixture three shows the red bar near 0.01, the blue bar stacked on it rising slightly above 0.01, the green rising near 0.02, and the yellow bar reaching 0.02. A label above this bar reads “Q equals 0.640”.

Figure 2. Concentrations of three mixtures are shown before and after reaching equilibrium at 800 °C for the so-called water gas shift reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g).

Example 4: Predicting the Direction of Reaction

Given below are the starting concentrations of reactants and products for three experiments involving this reaction:

[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]
[latex]{K}_{c}=0.64[/latex]

Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.

Reactants/Products	Experiment 1	Experiment 2	Experiment 3
[CO]_i	0.0203 M	0.011 M	0.0094 M
[H₂O]_i	0.0203 M	0.0011 M	0.0025 M
[CO₂]_i	0.0040 M	0.037 M	0.0015 M
[H₂]_i	0.0040 M	0.046 M	0.0076 M

Show Answer

Check Your Learning

Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.

A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:
[latex]2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightleftharpoons2\text{NOCl}\left(g\right){K}_{c}=4.6\times {10}^{4}[/latex]
A 5.0-L flask containing 17 g of NH₃, 14 g of N₂, and 12 g of H₂:
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right){K}_{c}=0.060[/latex]
A 2.00-L flask containing 230 g of SO₃(g):
[latex]2{\text{SO}}_{3}\left(g\right)\rightleftharpoons2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right){K}_{c}=0.230[/latex]

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In Example 2, it was mentioned that the common practice is to omit units when evaluating reaction quotients and equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are not constant) at high solution concentrations. This may be avoided by computing K_c values using the activities of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:

Activities are dimensionless (unitless) quantities and are in essence “adjusted” concentrations.
For relatively dilute solutions, a substance’s activity and its molar concentration are roughly equal.
Activities for pure condensed phases (solids and liquids) are equal to 1.

As a consequence of this last consideration, Q_c and K_c expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression’s value). Several examples of equilibria yielding such expressions will be encountered below.

Homogeneous Equilibria

A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided below.

[latex]{\text{C}}_{2}{\text{H}}_{2}\left(aq\right)+2{\text{Br}}_{2}\left(aq\right)\rightleftharpoons{\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}\left(aq\right){K}_{c}=\frac{\left[{\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}\right]}{\left[{\text{C}}_{2}{\text{H}}_{2}\right]{\left[{\text{Br}}_{2}\right]}^{2}}[/latex]
[latex]{\text{I}}_{2}\left(aq\right)+{\text{I}}^{-}\left(aq\right)\rightleftharpoons{\text{I}}_{3}{}^{-}\left(aq\right){K}_{c}=\frac{\left[{\text{I}}_{3}{}^{-}\right]}{\left[{\text{I}}_{2}\right]\left[{\text{I}}^{-}\right]}[/latex]
[latex]{\text{Hg}}_{2}{}^{2+}\left(aq\right)+{\text{NO}}_{3}{}^{-}\left(aq\right)+3{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)\rightleftharpoons2{\text{Hg}}^{2+}\left(aq\right)+{\text{HNO}}_{2}\left(aq\right)+4{\text{H}}_{2}\text{O}\left(l\right)[/latex]
[latex]{K}_{c}=\frac{{\left[{\text{Hg}}^{2+}\right]}^{2}\left[{\text{HNO}}_{2}\right]}{\left[{\text{Hg}}_{2}{}^{2+}\right]\left[{\text{NO}}_{3}{}^{-}\right]{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{3}}[/latex]
[latex]\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{\text{-}}\left(aq\right){K}_{c}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{F}}^{-}\right]}{\left[\text{HF}\right]}[/latex]
[latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{NH}}_{4}{}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{c}=\frac{\left[{\text{NH}}_{4}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}[/latex]

In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H₂O(l) is the solvent for these solutions, its concentration does not appear as a term in the K_c expression, as discussed above, even though it may also appear as a reactant or product in the chemical equation.

Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.

[latex]{\text{C}}_{2}{\text{H}}_{6}\left(g\right)\rightleftharpoons{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right){K}_{c}=\frac{\left[{\text{C}}_{2}{\text{H}}_{4}\right]\left[{\text{H}}_{2}\right]}{\left[{\text{C}}_{2}{\text{H}}_{6}\right]}[/latex]
[latex]3{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{O}}_{3}\left(g\right){K}_{c}=\frac{{\left[{\text{O}}_{3}\right]}^{2}}{{\left[{\text{O}}_{2}\right]}^{3}}[/latex]
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right){K}_{c}=\frac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2}\right]}^{3}}[/latex]
[latex]{\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)\rightleftharpoons3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right){K}_{c}=\frac{{\left[{\text{CO}}_{2}\right]}^{3}{\left[{\text{H}}_{2}\text{O}\right]}^{4}}{\left[{\text{C}}_{3}{\text{H}}_{8}\right]{\left[{\text{O}}_{2}\right]}^{5}}[/latex]

Note that the concentration of H₂O(g) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.

Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, [latex]\frac{n}{V}[/latex].

[latex]\begin{array} { }PV&=&nRT \\ P&=&\left(\frac{n}{V}\right)RT\\ &=&MRT\end{array}[/latex]

Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.

Using the partial pressures of the gases, we can write the reaction quotient for the system [latex]{\text{C}}_{2}{\text{H}}_{6}\left(g\right)\rightleftharpoons{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)[/latex] by following the same guidelines for deriving concentration-based expressions:

[latex]\displaystyle{Q}_{P}=\frac{{P}_{{\text{C}}_{2}{\text{H}}_{4}}{P}_{{\text{H}}_{2}}}{{P}_{{\text{C}}_{2}{\text{H}}_{6}}}[/latex]

In this equation we use Q_P to indicate a reaction quotient written with partial pressures: [latex]{P}_{{\text{C}}_{2}{\text{H}}_{6}}[/latex] is the partial pressure of C₂H₆; [latex]{P}_{{\text{H}}_{2}}[/latex], the partial pressure of H₂; and [latex]{P}_{{\text{C}}_{2}{\text{H}}_{6}}[/latex], the partial pressure of C₂H₄. At equilibrium:

[latex]\displaystyle{K}_{P}={Q}_{P}=\frac{{P}_{{\text{C}}_{2}{\text{H}}_{4}}{P}_{{\text{H}}_{2}}}{{P}_{{\text{C}}_{2}{\text{H}}_{6}}}[/latex]

The subscript P in the symbol K_P designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, K_P, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.

Conversion between a value for K_c, an equilibrium constant expressed in terms of concentrations, and a value for K_P, an equilibrium constant expressed in terms of pressures, is straightforward (a K or Q without a subscript could be either concentration or pressure).

The equation relating K_c and K_P is derived as follows. For the gas-phase reaction [latex]m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}[/latex]:

[latex]\begin{array}{ }{K}_{P}&=&\frac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}\\ &=&\frac{{\left(\left[\text{C}\right]\times RT\right)}^{x}{\left(\left[\text{D}\right]\times RT\right)}^{y}}{{\left(\left[\text{A}\right]\times RT\right)}^{m}{\left(\left[\text{B}\right]\times RT\right)}^{n}}\\ &=&\frac{{\left[\text{C}\right]}^{x}{\left[\text{D}\right]}^{y}}{{\left[\text{A}\right]}^{m}{\left[\text{B}\right]}^{n}}\times \frac{{\left(RT\right)}^{x+y}}{{\left(RT\right)}^{m+n}}\\ &=&{K}_{c}{\left(RT\right)}^{\left(x+y\right)-\left(m+n\right)}\\ &=&{K}_{c}{\left(RT\right)}^{\Delta n}\end{array}[/latex]

The relationship between K_c and K_P is

[latex]{K}_{P}={K}_{c}{\left(RT\right)}^{\Delta n}[/latex]

In this equation, Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction [latex]m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}[/latex], we have

[latex]\Delta n=\left(x+y\right)-\left(m+n\right)[/latex]

Example 5: Calculation of K_P

Write the equations for the conversion of K_c to K_P for each of the following reactions:

[latex]{\text{C}}_{2}{\text{H}}_{6}\left(g\right)\rightleftharpoons{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)[/latex]
[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]
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
K_c is equal to 0.28 for the following reaction at 900 °C: [latex]{\text{CS}}_{2}\left(g\right)+4{\text{H}}_{2}\left(g\right)\rightleftharpoons{\text{CH}}_{4}\left(g\right)+2{\text{H}}_{2}\text{S}\left(g\right)[/latex]
What is K_P at this temperature?

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Check Your Learning

Write the equations for the conversion of K_c to K_P for each of the following reactions, which occur in the gas phase:

[latex]2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{SO}}_{3}\left(g\right)[/latex]
[latex]{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right)[/latex]
[latex]{\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)\rightleftharpoons3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right)[/latex]
At 227 °C, the following reaction has K_c = 0.0952: [latex]{\text{CH}}_{3}\text{OH}\left(g\right)\rightleftharpoons\text{CO}\left(g\right)+2{\text{H}}_{2}\left(g\right)[/latex]
What would be the value of K_P at this temperature?

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Heterogeneous Equilibria

A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).

Some heterogeneous equilibria involve chemical changes; for example:

[latex]{\text{PbCl}}_{2}\left(s\right)\rightleftharpoons{\text{Pb}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right){K}_{c}=\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^{2}[/latex]
[latex]\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)\rightleftharpoons{\text{CaCO}}_{3}\left(s\right){K}_{c}=\frac{1}{\left[{\text{CO}}_{2}\right]}[/latex]
[latex]\text{C}\left(s\right)+2\text{S}\left(g\right)\rightleftharpoons{\text{CS}}_{2}\left(g\right){K}_{c}=\frac{\left[{\text{CS}}_{2}\right]}{{\left[\text{S}\right]}^{2}}[/latex]

Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:

[latex]{\text{Br}}_{2}\left(l\right)\rightleftharpoons{\text{Br}}_{2}\left(g\right){K}_{c}=\left[{\text{Br}}_{2}\right][/latex]

We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:

[latex]\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)\rightleftharpoons{\text{CaCO}}_{3}\left(s\right){K}_{P}=\frac{1}{{P}_{{\text{CO}}_{2}}}[/latex]
[latex]\text{C}\left(s\right)+2\text{S}\left(g\right)\rightleftharpoons{\text{CS}}_{2}\left(g\right){K}_{P}=\frac{{P}_{C{S}_{2}}}{{\left({P}_{S}\right)}^{2}}[/latex]

Key Concepts and Summary

For any reaction that is at equilibrium, the reaction quotient Q is equal to the equilibrium constant K for the reaction. If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant. At equilibrium, the values of the concentrations of the reactants and products are constant. Their particular values may vary depending on conditions, but the value of the reaction quotient will always equal K (K_c when using concentrations or K_P when using partial pressures).

A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases. We can decide whether a reaction is at equilibrium by comparing the reaction quotient with the equilibrium constant for the reaction.

Key Equations

[latex]Q=\frac{{\left[\text{C}\right]}^{\text{x}}{\left[\text{D}\right]}^{\text{y}}}{{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}}\text{ where }m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}[/latex]
[latex]{Q}_{P}=\frac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}\text{ where }m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}[/latex]
P = MRT
K_P = K_c(RT)^Δn

Exercises

Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
Explain why an equilibrium between Br₂(l) and Br₂(g) would not be established if the container were not a closed vessel shown in Figure 4 in Section 13.1 Chemical Equilibrium.
If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO₂ or with pure N₂O₄? [latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)[/latex]
Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg₂Cl₂, AgCl, PbCl₂, and CuCl.
1. Write the expression for the equilibrium constant for the reaction represented by the equation [latex]\text{AgCl}\left(s\right)\rightleftharpoons{\text{Ag}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\text{.}[/latex] Is K_c > 1, <1, or ≈1? Explain your answer.
2. Write the expression for the equilibrium constant for the reaction represented by the equation [latex]{\text{Pb}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\rightleftharpoons{\text{PbCl}}_{2}\left(s\right)\text{.}[/latex] Is K_c > 1, <1, or ≈1? Explain your answer.
Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble.
1. Write the expression for the equilibrium constant for the reaction represented by the equation [latex]{\text{CaCO}}_{3}\left(s\right)\rightleftharpoons{\text{Ca}}^{2+}\left(aq\right)+{\text{CO}}_{3}{}^{-}\left(aq\right)\text{.}[/latex] Is K_c > 1, <1, or ≈1? Explain your answer.
2. Write the expression for the equilibrium constant for the reaction represented by the equation [latex]3{\text{Ba}}^{2+}\left(aq\right)+2{\text{PO}}_{4}{}^{3-}\left(aq\right)\rightleftharpoons{\text{Ba}}_{3}{\left({\text{PO}}_{4}\right)}_{2}\left(s\right)\text{.}[/latex] Is K_c > 1, <1, or ≈1? Explain your answer.
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: [latex]3{\text{C}}_{2}{\text{H}}_{2}\left(g\right)\rightarrow{\text{C}}_{6}{\text{H}}_{6}\left(g\right)\text{.}[/latex] Which value of K_c would make this reaction most useful commercially? K_c ≈ 0.01, K_c ≈ 1, or K_c ≈ 10. Explain your answer.
Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation [latex]\text{KI}\left(aq\right)+{\text{I}}_{2}\left(aq\right)\rightleftharpoons{\text{KI}}_{3}\left(aq\right)[/latex] give the same expression for the reaction quotient. KI₃ is composed of the ions K⁺ and [latex]{\text{I}}_{3}{}^{-}[/latex].
The reaction quotient for the complete form of the equation is [latex]{Q}_{c}=\frac{\left[{\text{KI}}_{3}\right]}{\left[\text{KI}\right]\left[{\text{I}}_{2}\right]}\text{.}[/latex]
For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is K_c > 1, <1, or ≈1 for a titration reaction?
For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is K_c > 1, <1, or ≈1 for a useful precipitation reaction?
Write the mathematical expression for the reaction quotient, Q_c, for each of the following reactions:
1. [latex]{\text{CH}}_{4}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightleftharpoons{\text{CH}}_{3}\text{Cl}\left(g\right)+\text{HCl}\left(g\right)[/latex]
2. [latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)[/latex]
3. [latex]2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{SO}}_{3}\left(g\right)[/latex]
4. [latex]{\text{BaSO}}_{3}\left(s\right)\rightleftharpoons\text{BaO}\left(s\right)+{\text{SO}}_{2}\left(g\right)[/latex]
5. [latex]{\text{P}}_{4}\left(g\right)+5{\text{O}}_{2}\left(g\right)\rightleftharpoons{\text{P}}_{4}{\text{O}}_{10}\left(s\right)[/latex]
6. [latex]{\text{Br}}_{2}\left(g\right)\rightleftharpoons2\text{Br}\left(g\right)[/latex]
7. [latex]{\text{CH}}_{4}\left(g\right)+2{\text{O}}_{2}\left(g\right)\rightleftharpoons{\text{CO}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(l\right)[/latex]
8. [latex]{\text{CuSO}}_{4}\cdot 5{\text{H}}_{2}\text{O}\left(s\right)\rightleftharpoons{\text{CuSO}}_{4}\left(s\right)+5{\text{H}}_{2}\text{O}\left(g\right)[/latex]
Write the mathematical expression for the reaction quotient, Q_c, for each of the following reactions:
1. [latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
2. [latex]4{\text{NH}}_{3}\left(g\right)+5{\text{O}}_{2}\left(g\right)\rightleftharpoons4\text{NO}\left(g\right)+6{\text{H}}_{2}\text{O}\left(g\right)[/latex]
3. [latex]{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right)[/latex]
4. [latex]{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\rightleftharpoons\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)[/latex]
5. [latex]{\text{NH}}_{4}\text{Cl}\left(s\right)\rightleftharpoons{\text{NH}}_{3}\left(g\right)+\text{HCl}\left(g\right)[/latex]
6. [latex]2\text{Pb}{\left({\text{NO}}_{3}\right)}_{2}\left(s\right)\rightleftharpoons2\text{PbO}\left(s\right)+4{\text{NO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)[/latex]
7. [latex]2{\text{H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{H}}_{2}\text{O}\left(l\right)[/latex]
8. [latex]{\text{S}}_{8}\left(g\right)\rightleftharpoons8\text{S}\left(g\right)[/latex]
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
1. [latex]2{\text{NH}}_{3}\left(g\right)\rightleftharpoons{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right){K}_{c}=17\text{;}[/latex] [NH₃] = 0.20 M, [N₂] = 1.00 M, [H₂] = 1.00 M
2. [latex]2{\text{NH}}_{3}\left(g\right)\rightleftharpoons{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right){K}_{P}=6.8\times {10}^{4}\text{;}[/latex] initial pressures: NH₃ = 3.0 atm, N₂ = 2.0 atm, H₂ = 1.0 atm
3. [latex]2{\text{SO}}_{3}\left(g\right)\rightleftharpoons2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\text{(}g\text{)}{K}_{c}=0.230\text{;}[/latex] [SO₃] = 0.00 M, [SO₂] = 1.00 M, [O₂] = 1.00 M
4. [latex]2{\text{SO}}_{3}\left(g\right)\rightleftharpoons2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right){K}_{P}=16.5\text{;}[/latex] initial pressures: SO₃ = 1.00 atm, SO₂ = 1.00 atm, O₂ = 1.00 atm
5. [latex]2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightleftharpoons2\text{NOCl}\left(g\right){K}_{c}=4.6\times {10}^{4}\text{;}[/latex] [NO] = 1.00 M, [Cl₂] = 1.00 M, [NOCl] = 0 M
6. [latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right){K}_{P}=0.050\text{;}[/latex] initial pressures: NO = 10.0 atm, N₂ = O₂ = 5 atm
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
1. [latex]2{\text{NH}}_{3}\left(g\right)\rightleftharpoons{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right){K}_{c}=17\text{;}[/latex] [NH₃] = 0.50 M, [N₂] = 0.15 M, [H₂] = 0.12 M
2. [latex]2{\text{NH}}_{3}\left(g\right)\rightleftharpoons{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right){K}_{P}=6.8\times {10}^{4}\text{;}[/latex] initial pressures: NH₃ = 2.00 atm, N₂ = 10.00 atm, H₂ = 10.00 atm
3. [latex]2{\text{SO}}_{3}\left(g\right)\rightleftharpoons2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right){K}_{c}=0.230\text{;}[/latex] [SO₃] = 2.00 M, [SO₂] = 2.00 M, [O₂] = 2.00 M
4. [latex]2{\text{SO}}_{3}\left(g\right)\rightleftharpoons2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right){K}_{P}=6.5\text{atm;}[/latex] initial pressures: SO₂ = 1.00 atm, O₂ = 1.130 atm, SO₃ = 0 atm
5. [latex]2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightleftharpoons2\text{NOCl}\left(g\right){K}_{P}=2.5\times {10}^{3}\text{;}[/latex] initial pressures: NO = 1.00 atm, Cl₂ = 1.00 atm, NOCl = 0 atm
6. [latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right){K}_{c}=0.050\text{;}[/latex] [N₂] = 0.100 M, [O₂] = 0.200 M, [NO] = 1.00 M
The following reaction has K_P = 4.50 × 10⁻⁵ at 720 K: [latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P(NH₃) = 93 atm, P(N₂) = 48 atm, and P(H₂) = 52
Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?
Which of the systems described in question 10 give homogeneous equilibria? Which give heterogeneous equilibria?
Which of the systems described in question 11 give homogeneous equilibria? Which give heterogeneous equilibria?
For which of the reactions in question 10 does K_c (calculated using concentrations) equal K_P (calculated using pressures)?
For which of the reactions in question 11 does K_c (calculated using concentrations) equal K_P (calculated using pressures)?
Convert the values of K_c to values of K_P or the values of K_P to values of K_c.
1. [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]
2. [latex]{\text{H}}_{2}+{\text{I}}_{2}\rightleftharpoons2\text{HI}{K}_{c}=50.2\text{ at }448^{\circ}\text{C}[/latex]
3. [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]
4. [latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right){K}_{P}=0.122\text{ at }50^{\circ}\text{C}[/latex]
Convert the values of K_c to values of K_P or the values of K_P to values of K_c.
1. [latex]{\text{Cl}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)\rightleftharpoons2\text{BrCl}\left(g\right){K}_{c}=4.7\times {10}^{-2}\text{ at }25^{\circ}\text{C}[/latex]
2. [latex]2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{SO}}_{3}\left(g\right){K}_{P}=48.2\text{ at }500^{\circ}\text{C}[/latex]
3. [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]
4. [latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right){K}_{P}=0.196\text{ at }60^{\circ}\text{C}[/latex]
What is the value of the equilibrium constant expression for the change [latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right)[/latex] at 30 °C?
Write the expression of the reaction quotient for the ionization of HOCN in water.
Write the reaction quotient expression for the ionization of NH₃ in water.
What is the approximate value of the equilibrium constant K_P for the change [latex]{\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}\left(l\right)\rightleftharpoons{\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}\left(g\right)[/latex] at 25 °C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)

Show Selected Answers

2. Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle, because the system is not closed; one of the components of the equilibrium, the Br₂ vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.

4. The expressions for the equilibrium constant are as follows:

K_c = [Ag+][Cl−] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M;
[latex]{K}_{c}=\frac{1}{\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^{2}}[/latex] > 1 because PbCl₂ is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M).

6. Since [latex]{K}_{c}=\frac{\left[{\text{C}}_{6}{\text{H}}_{6}\right]}{{\left[{\text{C}}_{2}{\text{H}}_{2}\right]}^{3}}[/latex], a value of K_c ≈ 10 means that C₆H₆ predominates over C₂H₂. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.

8. K_c > 1; the product must be formed in overwhelmingly large proportions.

10. The expressions are as follows:

[latex]{Q}_{c}=\frac{\left[{\text{CH}}_{3}\text{Cl}\right]\left[\text{HCl}\right]}{\left[{\text{CH}}_{4}\right]\left[{\text{Cl}}_{2}\right]}[/latex]
[latex]{Q}_{c}=\frac{{\left[\text{NO}\right]}^{2}}{\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}[/latex]
[latex]{Q}_{c}=\frac{{\left[{\text{SO}}_{3}\right]}^{2}}{{\left[{\text{SO}}_{2}\right]}^{2}\left[{\text{O}}_{2}\right]}[/latex]
Q_c = [SO₂]
[latex]{Q}_{c}=\frac{1}{\left[{\text{P}}_{4}\right]{\left[{\text{O}}_{2}\right]}^{5}}[/latex]
[latex]{Q}_{c}=\frac{{\left[\text{Br}\right]}^{2}}{\left[{\text{Br}}_{2}\right]}[/latex]
[latex]{Q}_{c}=\frac{\left[{\text{CO}}_{2}\right]}{\left[{\text{CH}}_{4}\right]{\left[{\text{O}}_{2}\right]}^{2}}[/latex]
(h) Q_c = [H₂O]⁵

12. The reaction quotients and directions are as follows:

[latex]{Q}_{c}=\frac{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2}\right]}^{3}}{{\left[{\text{NH}}_{3}\right]}^{2}}=\frac{\left(1.00\right){\left(1.00\right)}^{3}}{{\left(0.20\right)}^{2}}=25[/latex] Q_c < K_c, proceeds left;
[latex]{Q}_{P}=\frac{{P}_{{\text{N}}_{2}}{\left({P}_{{\text{H}}_{2}}\right)}^{3}}{{\left({P}_{{\text{NH}}_{3}}\right)}^{2}}=\frac{\left(2.0\right){\left(1.0\right)}^{3}}{{\left(3.0\right)}^{2}}=0.22[/latex] Q_P < K_P, proceeds right;
[latex]{Q}_{c}=\frac{{\left[{\text{SO}}_{2}\right]}^{2}\left[{\text{O}}_{2}\right]}{{\left[{\text{SO}}_{3}\right]}^{2}}=\frac{{\left(1.00\right)}^{2}\left(1.00\right)}{\left(0\right)}=\text{undefined}[/latex] Q_c > K_c, proceeds left;
[latex]{Q}_{P}=\frac{{\left({P}_{{\text{SO}}_{2}}\right)}^{2}{P}_{{\text{O}}_{2}}}{{\left({P}_{{\text{SO}}_{3}}\right)}^{2}}=\frac{{\left(1.00\right)}^{2}\left(1.00\right)}{{\left(1.00\right)}^{2}}=1.00[/latex] Q_P > K_P, proceeds right;
[latex]{Q}_{P}=\frac{{\left({P}_{\text{NOCl}}\right)}^{2}}{{\left({P}_{\text{NO}}\right)}^{2}{P}_{{\text{Cl}}_{2}}}=\frac{{\left(0\right)}^{2}}{{\left(1.00\right)}^{2}\left(1.00\right)}=0[/latex] Q_P < K_P, proceeds right;
[latex]{Q}_{c}=\frac{{\left[\text{NO}\right]}^{2}}{\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}=\frac{{\left(10.0\right)}^{2}}{\left(5.00\right)\left(5.00\right)}=4[/latex] Q_c > K_c, proceeds left

14. The reaction quotient expression for this problem is [latex]{Q}_{p}=\frac{{\left({P}_{{\text{NH}}_{3}}\right)}^{2}}{\left({P}_{{\text{N}}_{2}}\right){\left({P}_{{\text{H}}_{2}}\right)}^{3}}\text{.}[/latex] Plugging in the given values of partial pressures gives [latex]\frac{{\left(93\right)}^{2}}{\left(\left(48\right)\times {\left(52\right)}^{3}\right)}[/latex], so Q_p = 1.3 × 10⁻³. Since this value is larger than K_P (4.50 × 10⁻⁵), the system will shift toward the reactants to reach equilibrium.

16. The answers are as follows:

homogenous
homogenous
homogenous
heterogeneous
heterogeneous
homogenous
heterogeneous
heterogeneous

18. When the number of gaseous components are the same on both sides of the equilibrium expression, K_c will equal K_P. This situation occurs in (a) and (b).

20. [latex]{K}_{P}={K}_{c}{\left(RT\right)}^{\Delta n}[/latex], where Δn is the sum of gaseous products minus the sum of gaseous reactants.

Δn = (2) − (1 + 3) = −2, K_P = 0.50[0.08206 × 673.15]⁻² = 1.6 × 10⁻⁴
Δn = (2) − (1 + 1) = 0, K_P = K_c(RT)⁰ = K_c = 50.2
Δn = (10) − (0) = 10, K_c = K_P(RT)^−Δn, K_c = 4.08 × 10⁻²⁵[0.08206 × 298.15]⁻¹⁰ = 5.31 × 10⁻³⁹
Δn = (1) − (0) = 1, K_c = 0.122(0.08206 × 323.15)⁻¹ = 4.60 × 10⁻³

22. The equilibrium expression for this transformation is [latex]{K}_{P}={P}_{{\text{H}}_{2}\text{O}}\text{.}[/latex] The vapor pressure of H₂O at 30 ºC is 31.8 torr. Converting to atmospheres gives [latex]3.18\cancel{\text{torr}}\times \frac{1\text{atm}}{760\cancel{\text{torr}}}=0.042\text{atm.}[/latex] Therefore, [latex]{K}_{P}={\text{P}}_{{\text{H}}_{2}\text{O}}=0.042\text{.}[/latex]

24. [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{NH}}_{4}{}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\text{.}[/latex] Because the concentration of water is a constant, the term [H₂O] is normally incorporated into the reaction quotient as well as the final equilibrium constant. [latex]{Q}_{c}=\frac{\left[{\text{NH}}_{4}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{HN}}_{3}\right]}[/latex]

Glossary

equilibrium constant (K): value of the reaction quotient for a system at equilibrium

heterogeneous equilibria: equilibria between reactants and products in different phases

homogeneous equilibria: equilibria within a single phase

K_c: equilibrium constant for reactions based on concentrations of reactants and products

K_P: equilibrium constant for gas-phase reactions based on partial pressures of reactants and products

law of mass action: when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant

reaction quotient (Q): ratio of the product of molar concentrations (or pressures) of the products to that of the reactants, each concentration (or pressure) being raised to the power equal to the coeffi

Chapter 13: Chemical Equilibrium

Learning Objectives

Example 1: Writing Reaction Quotient Expressions

Check Your Learning

Example 2: Evaluating a reaction quotient

Example 3: Evaluating a Reaction Quotient

Check Your Learning

Example 4: Predicting the Direction of Reaction

Experiment 1

Experiment 2

Experiment 3

Check Your Learning

Homogeneous Equilibria

Example 5: Calculation of K_P

Check Your Learning

Heterogeneous Equilibria

Key Concepts and Summary

Key Equations

Exercises

Glossary

Candela Citations

Chapter 13: Chemical Equilibrium