Equilibrium Constants

Klaus Theopold; Paul Flowers; Richard Langley; William R. Robinson

79 Equilibrium Constants

Learning Objectives

By the end of this section, 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
Relate the magnitude of an equilibrium constant to properties of the chemical system

The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Q). For a reversible reaction described by

$m\text{A}+n\text{B}+⇌x\text{C}+y\text{D}$

the reaction quotient is derived directly from the stoichiometry of the balanced equation as

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[\text{C}\right]}^{x}\left[\text{D}{\right]}^{y}}{{\left[\text{A}\right]}^{m}\left[\text{B}{\right]}^{n}}$

where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:

${Q}_{p}=\phantom{\rule{0.2em}{0ex}}\frac{{P}_{\text{C}}^{x}\phantom{\rule{0.2em}{0ex}}{P}_{\text{D}}^{y}}{{P}_{\text{A}}^{m}\phantom{\rule{0.2em}{0ex}}{P}_{\text{B}}^{n}}$

Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q. In most cases, this will introduce only modest errors in calculations involving reaction quotients.

Writing Reaction Quotient Expressions

Write the concentration-based reaction quotient expression for each of the following reactions:

(a) $3{\text{O}}_{2}\left(g\right)⇌2{\text{O}}_{3}\left(g\right)$

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

(c) $4{\text{NH}}_{3}\left(g\right)+7{\text{O}}_{2}\left(g\right)⇌4{\text{NO}}_{2}\left(g\right)+6{\text{H}}_{2}\text{O}\left(g\right)$

Solution

(a) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{O}}_{3}\right]}^{2}}{{\left[{\text{O}}_{2}\right]}^{3}}$

(b) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2}\right]}^{3}}$

(c) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{NO}}_{2}\right]}^{4}\left[{\text{H}}_{2}\text{O}{\right]}^{6}}{{\left[{\text{NH}}_{3}\right]}^{4}\left[{\text{O}}_{2}{\right]}^{7}}$

Check Your Learning

Write the concentration-based reaction quotient expression for each of the following reactions:

(a) $2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2{\text{SO}}_{3}\left(g\right)$

(b) ${\text{C}}_{4}{\text{H}}_{8}\left(g\right)⇌2{\text{C}}_{2}{\text{H}}_{4}\left(g\right)$

(c) $2{\text{C}}_{4}{\text{H}}_{10}\left(g\right)+13{\text{O}}_{2}\left(g\right)⇌8{\text{CO}}_{2}\left(g\right)+10{\text{H}}_{2}\text{O}\left(g\right)$

Answer:

(a) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{SO}}_{3}\right]}^{2}}{{\left[{\text{SO}}_{2}\right]}^{2}\left[{\text{O}}_{2}\right]};$ (b) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{C}}_{2}{\text{H}}_{4}\right]}^{2}}{\left[{\text{C}}_{4}{\text{H}}_{8}\right]};$ (c) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{CO}}_{2}\right]}^{8}\left[{\text{H}}_{2}\text{O}{\right]}^{10}}{{\left[{\text{C}}_{4}{\text{H}}_{10}\right]}^{2}\left[{\text{O}}_{2}{\right]}^{13}}$

Changes in concentrations and Q_c for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.

The numerical value of Q varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:

${\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌{\text{2SO}}_{3}\left(g\right)$

Two different experimental scenarios are depicted in (Figure), one in which this reaction is initiated with a mixture of reactants only, SO₂ and O₂, and another that begins with only product, SO₃. For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:

${Q}_{c}=\frac{\left[{\text{SO}}_{3}{\right]}^{2}}{\left[{\text{SO}}_{2}{\right]}^{2}\phantom{\rule{0.2em}{0ex}}\left[{\text{O}}_{2}\right]}=\frac{{0}^{2}}{\left[{\text{SO}}_{2}{\right]}^{2}\phantom{\rule{0.2em}{0ex}}\left[{\text{O}}_{2}\right]}=0$

As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Q_c), product concentration increases (as does the numerator of Q_c), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Q_c.

If the reaction begins with only product present, the value of Q_c is initially undefined (immeasurably large, or infinite):

${Q}_{c}=\frac{\left[{\text{SO}}_{3}{\right]}^{2}}{\left[{\text{SO}}_{2}{\right]}^{2}\phantom{\rule{0.2em}{0ex}}\left[{\text{O}}_{2}\right]}=\frac{\left[{\text{SO}}_{3}{\right]}^{2}}{0\to \text{∞}}$

In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Q_c decrease with time, the reactant concentrations and the denominator of Q_c increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.

The constant value of Q exhibited by a system at equilibrium is called the equilibrium constant, K:

$K\equiv Q\phantom{\rule{0.2em}{0ex}}\text{at equilibrium}$

Comparison of the data plots in (Figure) shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action: At a given temperature, the reaction quotient for a system at equilibrium is constant.

Evaluating a Reaction Quotient

Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:

$2{\text{NO}}_{2}\left(g\right)⇌{\text{N}}_{2}{\text{O}}_{4}\left(g\right)$

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?

Solution

As for all equilibrium calculations in this text, use the simplified equations for Q and K and disregard any concentration or pressure units, as noted previously in this section.

(a) Before any product is formed, $\left[{\text{NO}}_{2}\right]=\phantom{\rule{0.2em}{0ex}}\frac{0.10\phantom{\rule{0.2em}{0ex}}\text{mol}}{1.0\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}=0.10\phantom{\rule{0.2em}{0ex}}M,$ and [N₂O₄] = 0 M. Thus,

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}=\phantom{\rule{0.2em}{0ex}}\frac{0}{{0.10}^{2}}\phantom{\rule{0.2em}{0ex}}=0$

(b) At equilibrium, ${K}_{c}={Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}=\phantom{\rule{0.2em}{0ex}}\frac{0.042}{{0.016}^{2}}\phantom{\rule{0.2em}{0ex}}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}.$ The equilibrium constant is 1.6 $×$ 10².

Check Your Learning

For the reaction ${\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌{\text{2SO}}_{3}\left(g\right),$ 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?

Answer:

K_c = 4.3

By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.

The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.

To further illustrate this important point, consider the reversible reaction shown below:

$\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.640\phantom{\rule{5em}{0ex}}\text{T}=800\phantom{\rule{0.2em}{0ex}}\text{°C}$

The bar charts in (Figure) represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.

Compositions of three mixtures before (Q_c ≠ K_c) and after (Q_c = K_c) equilibrium is established for the reaction $\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right).$

Predicting the Direction of Reaction

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

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

${K}_{c}=0.64$

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	0.020 M	0.011 M	0.0094 M
[H₂O]_i	0.020 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

Solution

Experiment 1:

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CO}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\text{O}\right]}=\phantom{\rule{0.2em}{0ex}}\frac{\left(0.0040\right)\left(0.0040\right)}{\left(0.020\right)\left(0.020\right)}\phantom{\rule{0.2em}{0ex}}=0.040.$

Q_c < K_c (0.040 < 0.64)

The reaction will proceed in the forward direction.

Experiment 2:

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CO}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\text{O}\right]}=\phantom{\rule{0.2em}{0ex}}\frac{\left(0.037\right)\left(0.046\right)}{\left(0.011\right)\left(0.0011\right)}\phantom{\rule{0.2em}{0ex}}=1.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}$

Q_c > K_c (140 > 0.64)

The reaction will proceed in the reverse direction.

Experiment 3:

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CO}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\text{O}\right]}=\phantom{\rule{0.2em}{0ex}}\frac{\left(0.0015\right)\left(0.0076\right)}{\left(0.0094\right)\left(0.0025\right)}\phantom{\rule{0.2em}{0ex}}=0.48$

Q_c < K_c (0.48 < 0.64)

The reaction will proceed in the forward direction.

Check Your Learning

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

(a) A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:

$2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)⇌2\text{NOCl}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=4.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}$

(b) A 5.0-L flask containing 17 g of NH₃, 14 g of N₂, and 12 g of H₂:

${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.060$

(c) A 2.00-L flask containing 230 g of SO₃(g):

$2{\text{SO}}_{3}\left(g\right)⇌2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.230$

Answer:

(a) Q_c = 6.45 $×$ 10³, forward. (b) Q_c = 0.23, reverse. (c) Q_c = 0, forward.

Homogeneous Equilibria

A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:

$\begin{array}{cccccccc}\hfill {\text{C}}_{2}{\text{H}}_{2}\left(aq\right)+2{\text{Br}}_{2}\left(aq\right)& ⇌\hfill & {\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}\left(aq\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{\left[{\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}\right]}{\left[{\text{C}}_{2}{\text{H}}_{2}\right]\phantom{\rule{0.2em}{0ex}}{\left[{\text{Br}}_{2}\right]}^{2}}\hfill \\ \hfill {\text{I}}_{2}\left(aq\right)+{\text{I}}^{\text{−}}\left(aq\right)& ⇌\hfill & {\text{I}}_{3}{}^{\text{−}}\left(aq\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{\left[{\text{I}}_{3}{}^{\text{−}}\right]}{\left[{\text{I}}_{2}\right]\left[{\text{I}}^{\text{−}}\right]}\hfill \\ \hfill \text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)& ⇌\hfill & {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{\text{−}}\left(aq\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{\text{−}}\right]}{\left[\text{HF}\right]}\hfill \\ \hfill {\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)& ⇌\hfill & {\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{\left[{\text{NH}}_{4}{}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{NH}}_{3}\right]}\hfill \end{array}$

These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.

The equilibria below all involve gas-phase solutions:

$\begin{array}{cccccccc}\hfill {\text{C}}_{2}{\text{H}}_{6}\left(g\right)& ⇌\hfill & {\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{\left[{\text{C}}_{2}{\text{H}}_{4}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[{\text{C}}_{2}{\text{H}}_{6}\right]}\hfill \\ \hfill 3{\text{O}}_{2}\left(g\right)& ⇌\hfill & 2{\text{O}}_{3}\left(g\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{{\left[{\text{O}}_{3}\right]}^{2}}{{\left[{\text{O}}_{2}\right]}^{3}}\hfill \\ \hfill {\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)& ⇌\hfill & 2{\text{NH}}_{3}\left(g\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}^{3}}\hfill \\ \hfill {\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)& ⇌\hfill & 3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{{\left[{\text{CO}}_{2}\right]}^{3}{\left[{\text{H}}_{2}\text{O}\right]}^{4}}{\left[{\text{C}}_{3}{\text{H}}_{8}\right]{\left[{\text{O}}_{2}\right]}^{5}}\hfill \end{array}$

For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (K_c) or partial pressures (K_p) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity:

$PV=nRT$

$\phantom{\rule{0.6em}{0ex}}P=\left(\frac{n}{V}\right)RT$

$\phantom{\rule{0.6em}{0ex}}=MRT$

where P is partial pressure, V is volume, n is molar amount, R is the gas constant, T is temperature, and M is molar concentration.

For the gas-phase reaction $m\text{A}+n\text{B}⇌x\text{C}+y\text{D:}$

${K}_{P}=\phantom{\rule{0.2em}{0ex}}\frac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}$

$=\phantom{\rule{0.2em}{0ex}}\frac{{\left(\left[\text{C]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT\right)}^{x}\left(\left[\text{D]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT{\right)}^{y}}{\left(\left[\text{A]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT{\right)}^{m}{\left(\left[\text{B]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT\right)}^{n}}$

$=\phantom{\rule{0.2em}{0ex}}\frac{\left[\text{C}{\right]}^{x}{\left[\text{D}\right]}^{y}}{\left[\text{A}{\right]}^{m}{\left[\text{B}\right]}^{n}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{{\left(RT\right)}^{x\text{+}y}}{{\left(RT\right)}^{m\text{+}n}}$

$={K}_{c}{\left(RT\right)}^{\left(x\text{+}y\right)-\left(m\text{+}n\right)}$

$={K}_{c}{\left(RT\right)}^{\text{Δ}n}$

And so, the relationship between K_c and K_P is

${K}_{P}={K}_{c}{\left(RT\right)}^{\text{Δ}n}$

where Δn is the difference in the molar amounts of product and reactant gases, in this case:

$\text{Δ}n=\left(x\text{+}y\right)-\left(m\text{+}n\right)$

Calculation of K_P

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

(a) ${\text{C}}_{2}{\text{H}}_{6}\left(g\right)⇌{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)$

(b) $\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)$

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

(d) K_c is equal to 0.28 for the following reaction at 900 °C:

${\text{CS}}_{2}\left(g\right)+4{\text{H}}_{2}\left(g\right)⇌{\text{CH}}_{4}\left(g\right)+2{\text{H}}_{2}\text{S}\left(g\right)$

What is K_P at this temperature?

Solution

(a) Δn = (2) − (1) = 1
K_P = K_c (RT)^Δn = K_c (RT)¹ = K_c (RT)

(b) Δn = (2) − (2) = 0
K_P = K_c (RT)^Δn = K_c (RT)⁰ = K_c

(c) Δn = (2) − (1 + 3) = −2
K_P = K_c (RT)^Δn = K_c (RT)⁻² = $\frac{{K}_{c}}{{\left(RT\right)}^{2}}$

(d) K_P = K_c (RT)^Δn = (0.28)[(0.0821)(1173)]⁻² = 3.0 $×$ 10⁻⁵

Check Your Learning

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

(a) $2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2{\text{SO}}_{3}\left(g\right)$

(b) ${\text{N}}_{2}{\text{O}}_{4}\left(g\right)⇌2{\text{NO}}_{2}\left(g\right)$

(c) ${\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)⇌3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right)$

(d) At 227 °C, the following reaction has K_c = 0.0952:

${\text{CH}}_{3}\text{OH}\left(g\right)⇌\text{CO}\left(g\right)+2{\text{H}}_{2}\left(g\right)$

What would be the value of K_P at this temperature?

Answer:

(a) K_P = K_c (RT)⁻¹; (b) K_P = K_c (RT); (c) K_P = K_c (RT); (d) 160 or 1.6 $×$ 10²

Heterogeneous Equilibria

A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:

$\begin{array}{cccccccc}\hfill {\text{PbCl}}_{2}\left(s\right)& ⇌\hfill & {\text{Pb}}^{2+}\left(aq\right)+2{\text{Cl}}^{\text{−}}\left(aq\right)\hfill & & & \hfill {K}_{c}& =\hfill & \left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{\text{−}}\right]}^{2}\hfill \\ \hfill \text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)& ⇌\hfill & {\text{CaCO}}_{3}\left(s\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{1}{{\text{[CO}}_{2}\right]}\hfill \\ \hfill \text{C}\left(s\right)+2\text{S}\left(g\right)& ⇌\hfill & {\text{CS}}_{2}\left(g\right)\hfill & & & \hfill {K}_{c}& =\hfill & \frac{{\text{[CS}}_{2}\right]}{{\left[\text{S}\right]}^{2}}\hfill \\ \hfill {\text{Br}}_{2}\left(l\right)& ⇌\hfill & {\text{Br}}_{2}\left(g\right)\hfill & & & \hfill {K}_{c}& =\hfill & \left[{\text{Br}}_{2}\left(g\right)\right]\hfill \end{array}$

Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.

Two of the above examples include terms for gaseous species only in their equilibrium constants, and so K_p expressions may also be written:

$\begin{array}{cccccccc}\hfill \text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)& ⇌\hfill & {\text{CaCO}}_{3}\left(s\right)\hfill & & & \hfill {K}_{P}& =\hfill & \frac{1}{{P}_{{\text{CO}}_{2}}}\hfill \\ \hfill \text{C}\left(s\right)+2\text{S}\left(g\right)& ⇌\hfill & {\text{CS}}_{2}\left(g\right)\hfill & & & \hfill {K}_{P}& =\hfill & \frac{{P}_{{\text{CS}}_{2}}}{{\left({P}_{\text{S}}\right)}^{2}}\hfill \end{array}$

Coupled Equilibria

The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.

1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.

$\begin{array}{}\\ \\ \\ \text{A}⇌\text{B}\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c}}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}\hfill \\ \text{B}⇌\text{A}\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c'}}=\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\hfill \end{array}$

${\text{K}}_{\text{c'}}=\frac{1}{{\text{K}}_{\text{c}}}$

2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor:

$\begin{array}{}\\ \\ \\ \text{A}⇌\text{B}\phantom{\rule{6em}{0ex}}{\text{K}}_{\text{c}}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}\hfill \\ \text{xA}⇌\text{xB}\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c'}}=\frac{\left[\text{B}{\right]}^{\text{x}}}{\left[\text{A}{\right]}^{\text{x}}}\hfill \end{array}$

${\text{K}}_{\text{c'}}={\text{K}}_{\text{c}}{}^{\text{x}}$

3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:

$\begin{array}{}\\ \\ \\ \text{A}⇌\text{B}\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c1}}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}\hfill \\ \text{B}⇌\text{C}\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c2}}=\frac{\left[\text{C}\right]}{\left[\text{B}\right]}\hfill \end{array}$

The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:

$\begin{array}{l}\text{A}+\text{B}⇌\text{B}+\text{C}\\ \text{A}+\overline{)\text{B}}⇌\overline{)\text{B}}+\text{C}\\ \text{A}⇌\text{C}\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c'}}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]}\end{array}$

Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:

${\text{K}}_{\text{c1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}_{\text{c2}}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\left[\text{C}\right]}{\left[\text{B}\right]}=\frac{\overline{)\left[\text{B}\right]}\left[\text{C}\right]}{\left[\text{A}\right]\overline{)\left[\text{B}\right]}}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]}={\text{K}}_{\text{c'}}$

${\text{K}}_{\text{c'}}={\text{K}}_{\text{c1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}_{\text{c2}}$

(Figure) demonstrates the use of this strategy in describing coupled equilibrium processes.

Equilibrium Constants for Coupled Reactions

A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:

${\text{2NH}}_{3}\left(g\right)+{\text{3I}}_{2}\left(g\right)⇌\text{N}\left(g{\right)}_{2}+\text{6HI}\left(g\right)$

Use the information below to calculate Kc for this reaction.

$\begin{array}{cccc}{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)\hfill & & & {\text{K}}_{\text{c}1}=0.50\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}400\phantom{\rule{0.2em}{0ex}}°\text{C}\hfill \\ {\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)⇌2\text{HI}\left(g\right)\hfill & & & {\text{K}}_{\text{c}2}=50\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}400\phantom{\rule{0.2em}{0ex}}°\text{C}\hfill \end{array}$

Solution

The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.

Reverse the first coupled reaction equation:

${\text{2NH}}_{3}\left(g\right)⇌{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c}1\text{'}}=\frac{1}{{\text{K}}_{\text{c}1}}=\frac{1}{0.50}=2.0$

Multiply the second coupled reaction by 3:

${\text{3H}}_{2}\left(g\right)+3{\text{I}}_{2}\left(g\right)⇌6\text{HI}\left(g\right)\phantom{\rule{5em}{0ex}}{\text{K}}_{\text{c}2\text{'}}={\text{K}}_{\text{c}2}^{3}={50}^{3}=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}$

Finally, add the two revised equations:

$\begin{array}{c}{\text{2NH}}_{3}\left(g\right)+\overline{)3{\text{H}}_{2}\left(g\right)}+3{\text{I}}_{\text{2}}\left(g\right)⇌{N}_{2}\left(g\right)+\overline{)3{\text{H}}_{2}\left(g\right)}+6\text{HI}\left(g\right)\hfill \\ {\text{2NH}}_{3}\left(g\right)+3{\text{I}}_{\text{2}}\left(g\right)⇌{N}_{2}\left(g\right)+6\text{HI}\left(g\right)\hfill \\ {\text{K}}_{\text{c}}={\text{K}}_{\text{c}1\text{'}}\text{\hspace{0.17em}}{\text{K}}_{\text{c}2\text{'}}=\left(2.0\right)\left(1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\right)=2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\hfill \end{array}$

Check Your Learning

Use the provided information to calculate Kc for the following reaction at 550 °C:

$\begin{array}{cccc}{\text{H}}_{2}\left(g\right)+{\text{CO}}_{2}\left(g\right)⇌\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & {\text{K}}_{\text{c}}=?\hfill \\ \text{CoO}\left(s\right)+\text{CO}\left(g\right)⇌\text{Co}\left(s\right)+{\text{CO}}_{2}\left(g\right)\hfill & & & {\text{K}}_{\text{c}1}=490\hfill \\ \text{CoO}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)⇌\text{Co}\left(s\right)+{\text{H}}_{2}O\left(g\right)\hfill & & & {\text{K}}_{\text{c}1}=67\hfill \end{array}$

Answer:

K_c = 0.14

Key Concepts and Summary

The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, Q. For a reaction at equilibrium, the composition is constant, and Q is called the equilibrium constant, K.

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.

Key Equations

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[\text{C}{\right]}^{x}{\left[\text{D}\right]}^{y}}{\left[\text{A}{\right]}^{m}{\left[\text{B}\right]}^{n}}\phantom{\rule{5em}{0ex}}\text{for the reaction}\phantom{\rule{0.2em}{0ex}}m\text{A}+n\text{B}⇌x\text{C}+y\text{D}$
${Q}_{P}=\phantom{\rule{0.2em}{0ex}}\frac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}\phantom{\rule{5em}{0ex}}\text{for the reaction}\phantom{\rule{0.2em}{0ex}}m\text{A}+n\text{B}⇌x\text{C}+y\text{D}$
P = MRT
K_c = Q_c at equilibrium
K_p = Q_p at equilibrium
K_P = K_c (RT)^Δn

Chemistry End of Chapter Exercises

Instructions

Complete the following exercises by providing answers to each question. There are a total of 25 questions in this section. Please be aware that certain questions may contain subsections.

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.

2. 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).

Answer(s): 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.

3. If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO₂ or with pure N₂O₄?
$2{\text{NO}}_{2}\left(g\right)⇌{\text{N}}_{2}{\text{O}}_{4}\left(g\right)$

4. Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg₂Cl₂, AgCl, PbCl₂, and CuCl.

(a) Write the expression for the equilibrium constant for the reaction represented by the equation $\text{AgCl}\left(s\right)⇌{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{\text{−}}\left(aq\right).$ Is K_c > 1, < 1, or ≈ 1? Explain your answer.

(b) Write the expression for the equilibrium constant for the reaction represented by the equation ${\text{Pb}}^{2+}\left(aq\right)+2{\text{Cl}}^{\text{−}}\left(aq\right)⇌{\text{PbCl}}_{2}\left(s\right).$ Is K_c > 1, < 1, or ≈ 1? Explain your answer.

Answer(s): (a) K_c = [Ag⁺][Cl⁻] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M; (b) ${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{1}{\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{\text{−}}\right]}^{2}}$ > 1 because PbCl₂ is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M).

5. 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.

(a) Write the expression for the equilibrium constant for the reaction represented by the equation ${\text{CaCO}}_{3}\left(s\right)⇌{\text{Ca}}^{2+}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(\text{aq}\right).$ Is K_c > 1, < 1, or ≈ 1? Explain your answer.

(b) Write the expression for the equilibrium constant for the reaction represented by the equation $3{\text{Ba}}^{2+}\left(aq\right)+2{\text{PO}}_{4}{}^{3-}\left(aq\right)⇌{\text{Ba}}_{3}{\left({\text{PO}}_{4}\right)}_{2}\left(s\right).$ Is K_c > 1, < 1, or ≈ 1? Explain your answer.

6. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: $3{\text{C}}_{2}{\text{H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{6}\left(g\right).$ 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.

Answer(s): Since ${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{C}}_{6}{\text{H}}_{6}\right]}{{\left[{\text{C}}_{2}{\text{H}}_{2}\right]}^{3}},$ 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.

7. Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation $\text{KI}\left(aq\right)+{\text{I}}_{2}\left(aq\right)⇌{\text{KI}}_{3}\left(aq\right)$ give the same expression for the reaction quotient. KI₃ is composed of the ions K⁺ and ${\text{I}}_{3}{}^{\text{−}}.$

8. 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?

Answer(s): K_c > 1

9. 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?

10. Write the mathematical expression for the reaction quotient, Q_c, for each of the following reactions:

(a) ${\text{CH}}_{4}\left(g\right)+{\text{Cl}}_{2}\left(g\right)⇌{\text{CH}}_{3}\text{Cl}\left(g\right)+\text{HCl}\left(g\right)$

(b) ${\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2\text{NO}\left(g\right)$

(c) $2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2{\text{SO}}_{3}\left(g\right)$

(d) ${\text{BaSO}}_{3}\left(s\right)⇌\text{BaO}\left(s\right)+{\text{SO}}_{2}\left(g\right)$

(e) ${\text{P}}_{4}\left(g\right)+5{\text{O}}_{2}\left(g\right)⇌{\text{P}}_{4}{\text{O}}_{10}\left(s\right)$

(f) ${\text{Br}}_{2}\left(g\right)⇌2\text{Br}\left(g\right)$

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

(h) ${\text{CuSO}}_{4}\text{·}5{\text{H}}_{2}\text{O}\left(s\right)⇌{\text{CuSO}}_{4}\left(s\right)+5{\text{H}}_{2}\text{O}\left(g\right)$

Answer(s): (a) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CH}}_{3}\text{Cl}\right]\left[\text{HCl}\right]}{\left[{\text{CH}}_{4}\right]\left[{\text{Cl}}_{2}\right]};$ (b) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[\text{NO}\right]}^{2}}{\left[{\text{N}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{O}}_{2}\right]};$ (c) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{SO}}_{3}\right]}^{2}}{{\left[{\text{SO}}_{2}\right]}^{2}\left[{\text{O}}_{2}\right]};$ (d) Q_c = [SO₂]; (e) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{1}{\left[{\text{P}}_{4}\right]{\left[{\text{O}}_{2}\right]}^{5}};$ (f) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[\text{Br}\right]}^{2}}{\left[{\text{Br}}_{2}\right]};$ (g) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CO}}_{2}\right]}{\left[{\text{CH}}_{4}\right]{\left[{\text{O}}_{2}\right]}^{2}};$ (h) Q_c = [H₂O]⁵

11. Write the mathematical expression for the reaction quotient, Q_c, for each of the following reactions:

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

(b) $4{\text{NH}}_{3}\left(g\right)+5{\text{O}}_{2}\left(g\right)⇌4\text{NO}\left(g\right)+6{\text{H}}_{2}\text{O}\left(g\right)$

(c) ${\text{N}}_{2}{\text{O}}_{4}\left(g\right)⇌2{\text{NO}}_{2}\left(g\right)$

(d) ${\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)⇌\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)$

(e) ${\text{NH}}_{4}\text{Cl}\left(s\right)⇌{\text{NH}}_{3}\left(g\right)+\text{HCl}\left(g\right)$

(f) $2\text{Pb}{\left({\text{NO}}_{3}\right)}_{2}\left(s\right)⇌2\text{PbO}\left(s\right)+4{\text{NO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)$

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

(h) ${\text{S}}_{8}\left(g\right)⇌8\text{S}\left(g\right)$

12. 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.

(a) $2{\text{NH}}_{3}\left(g\right)⇌{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=17;$ [NH₃] = 0.20 M, [N₂] = 1.00 M, [H₂] = 1.00 M

(b) $2{\text{NH}}_{3}\left(g\right)⇌{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=6.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4};$ NH₃ = 3.0 atm, N₂ = 2.0 atm, H₂ = 1.0 atm

(c) $2{\text{SO}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⇌2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.230;$ [SO₃] = 0.00 M, [SO₂] = 1.00 M, [O₂] = 1.00 M

(d) $2{\text{SO}}_{3}\left(g\right)⇌2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=16.5;$ SO₃ = 1.00 atm, SO₂ = 1.00 atm, O₂ = 1.00 atm

(e) $2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)⇌2\text{NOCl}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=4.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4};$ [NO] = 1.00 M, [Cl₂] = 1.00 M, [NOCl] = 0 M

(f) ${\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2\text{NO}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=0.050;$ NO = 10.0 atm, N₂ = O₂ = 5 atm

Answer(s): (a) Q_c 25 proceeds left; (b) Q_P 0.22 proceeds right; (c) Q_c undefined proceeds left; (d) Q_P 1.00 proceeds right; (e) Q_P 0 proceeds right; (f) Q_c 4 proceeds left

13. 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.

(a) $2{\text{NH}}_{3}\left(g\right)⇌{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=17;$ [NH₃] = 0.50 M, [N₂] = 0.15 M, [H₂] = 0.12 M

(b) $2{\text{NH}}_{3}\left(g\right)⇌{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=6.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4};$ NH₃ = 2.00 atm, N₂ = 10.00 atm, H₂ = 10.00 atm

(c) $2{\text{SO}}_{3}\left(g\right)⇌2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.230;$ [SO₃] = 2.00 M, [SO₂] = 2.00 M, [O₂] = 2.00 M

(d) $2{\text{SO}}_{3}\left(g\right)⇌2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=6.5\phantom{\rule{0.2em}{0ex}}\text{atm;}$ SO₂ = 1.00 atm, O₂ = 1.130 atm, SO₃ = 0 atm

(e) $2\text{NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)⇌2\text{NOCl}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3};$ NO = 1.00 atm, Cl₂ = 1.00 atm, NOCl = 0 atm

(f) ${\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2\text{NO}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.050;$ [N₂] = 0.100 M, [O₂] = 0.200 M, [NO] = 1.00 M

14. The following reaction has K_P = 4.50 $×$ 10⁻⁵ at 720 K.
${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)$

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 atm

Answer(s): The system will shift toward the reactants to reach equilibrium.

15. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?
${\text{SO}}_{2}{\text{Cl}}_{2}\left(g\right)⇌{\text{SO}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)$

Answer(s): [SO₂Cl₂] = 0.12 M, [Cl₂] = 0.16 M and [SO₂] = 0.050 M. K_c for the reaction is 0.078.

16. Which of the systems described in (Figure) are homogeneous equilibria? Which are heterogeneous equilibria?

(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous

17. Which of the systems described in (Figure) are homogeneous equilibria? Which are heterogeneous equilibria?

18. For which of the reactions in (Figure) does K_c (calculated using concentrations) equal K_P (calculated using pressures)?

Answer(s): This situation occurs in (a) and (b).

19. For which of the reactions in (Figure) does K_c (calculated using concentrations) equal K_P (calculated using pressures)?

20. Convert the values of K_c to values of K_P or the values of K_P to values of K_c.

(a) ${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.50\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}400\phantom{\rule{0.2em}{0ex}}\text{°C}$

(b) ${\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)⇌2\text{HI}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=50.2\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}448\phantom{\rule{0.2em}{0ex}}\text{°C}$

(c) ${\text{Na}}_{2}{\text{SO}}_{4}\text{·}10{\text{H}}_{2}\text{O}\left(s\right)⇌{\text{Na}}_{2}{\text{SO}}_{4}\left(s\right)+10{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=4.08\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-25}\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}25\phantom{\rule{0.2em}{0ex}}\text{°C}$

(d) ${\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=0.122\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}50\phantom{\rule{0.2em}{0ex}}\text{°C}$

Answer(s): (a) K_P = 1.6 $×$ 10⁻⁴; (b) K_P = 50.2; (c) K_c = 5.34 $×$ 10⁻³⁹; (d) K_c = 4.60 $×$ 10⁻³

21. Convert the values of K_c to values of K_P or the values of K_P to values of K_c.

(a) ${\text{Cl}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)⇌2\text{BrCl}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=4.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\text{at}\phantom{\rule{0.2em}{0ex}}25\phantom{\rule{0.2em}{0ex}}\text{°C}$

(b) $2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2{\text{SO}}_{3}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=48.2\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}500\phantom{\rule{0.2em}{0ex}}\text{°C}$

(c) ${\text{CaCl}}_{2}\text{·}6{\text{H}}_{2}\text{O}\left(s\right)⇌{\text{CaCl}}_{2}\left(s\right)+6{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=5.09\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-44}\text{at}\phantom{\rule{0.2em}{0ex}}25\phantom{\rule{0.2em}{0ex}}\text{°C}$

(d) ${\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=0.196\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}60\phantom{\rule{0.2em}{0ex}}\text{°C}$

22. What is the value of the equilibrium constant expression for the change ${\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{2}\text{O}\left(g\right)$ at 30 °C? (See Appendix E.)

Answer(s): ${K}_{P}={P}_{{\text{H}}_{2}\text{O}}=0.042.$

23. Write the expression of the reaction quotient for the ionization of HOCN in water.

24. Write the reaction quotient expression for the ionization of NH₃ in water.

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{NH}}_{4}{}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{HN}}_{3}\right]}$

25. What is the approximate value of the equilibrium constant K_P for the change ${\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}\left(l\right)⇌{\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}\left(g\right)$ at 25 °C. (The equilibrium vapor pressure for this substance is 570 torr at 25 °C.)

Glossary

equilibrium constant (K): value of the reaction quotient for a system at equilibrium; may be expressed using concentrations (K_c) or partial pressures (K_p)

heterogeneous equilibria: equilibria in which reactants and products occupy two or more different phases

homogeneous equilibria: equilibria in which all reactants and products occupy the same phase

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

reaction quotient (Q): mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations (Q_c) or pressures (Q_p)

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Writing Reaction Quotient Expressions

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Evaluating a Reaction Quotient

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Predicting the Direction of Reaction

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Homogeneous Equilibria

Calculation of KP

Solution

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Heterogeneous Equilibria

Coupled Equilibria

Equilibrium Constants for Coupled Reactions

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Calculation of K_P