7.2: Equilibrium Constants

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Learning Objectives

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

Derive reaction quotients from chemical equations
Calculate values of reaction quotients and equilibrium constants
Relate the magnitude of an equilibrium constant to properties of the chemical system
Describe the ways in which an equilibrium system can be stressed
Predict the response of a stressed equilibrium using Le Chatelier’s principle

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

\[ m \mathrm{A}+n \mathrm{B} \rightleftharpoons x \mathrm{C} + y \mathrm{D} \]

the reaction quotient is derived directly from the stoichiometry of the balanced equation where the subscript c denotes the use of molar concentrations in the expression as

\[ Q_c=\frac{[\mathrm{C}]^x[\mathrm{D}]^y}{[\mathrm{A}]^m[\mathrm{B}]^n} \]

Example

Writing Reaction Quotient Expressions

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

(a) $3 O_{2} (g) ⇌ 2 O_{3} (g)$

(b) $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

Solution

(a) $ Q_c=\frac{[\mathrm{O_3}]^2}{[\mathrm{O_2}]^3} $

(b) $ Q_c=\frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}] [\mathrm{H_2}]^3} $

Check Your Learning

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

(a) $2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$

(b) $C_{4} H_{8} (g) ⇌ 2 C_{2} H_{4} (g)$

Answer:

(a) $Q_c=\frac{[\mathrm{SO_3}]^2 }{[\mathrm{SO_2}]^2 [\mathrm{O_2}]} $

(b) $ Q_c=\frac{[\mathrm{C_2H_4}]^2 }{[\mathrm{C_4H_8}]} $

Four graphs are shown and labeled, “a,” “b,” “c,” and “d.” All four 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.” The y-axis on graph d is labeled, “Reaction Quotient,” and the x-axis is labeled, “Time.” A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, “k.”

Figure $\PageIndex{1}$ 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:

{2SO}_{2} (g) + O_{2} (g) ⇌ {2SO}_{3} (g)

Two different experimental scenarios are depicted in Figure $\PageIndex{1}$ , 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:

\[ Q_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}=\frac{0^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}= 0 \]

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):

\[ Q_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}=\frac{[\mathrm{SO}_3]}{0}=\infty \]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:

\[ K \equiv q ~ at ~ equlibrium \] $equilibrium$ Comparison of the data plots in Figure $\PageIndex{1}$ 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.

Example

Evaluating a Reaction Quotient

Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:

2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g)

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 reaction quotent when the system reacheconstant 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, $[{NO}_{2}] = \frac{0.10 mol}{1.0 L} = 0.10 M,$ and [N₂O₄] = 0 M. Thus,

Q_{c} = \frac{[N_{2} O_{4}]}{{[{NO}_{2}]}^{2}} = \frac{0}{{0.10}^{2}} = 0

(b) At equilibrium, $K_{c} = Q_{c} = \frac{[N_{2} O_{4}]}{{[{NO}_{2}]}^{2}} = \frac{0.042}{{0.016}^{2}} = 1.6 \times 10^{2} .$ The reaction quotient is 1.6 $\times$ 10².

Check Your Learning

For the reaction

{2SO}_{2} (g) + O_{2} (g) ⇌ {2SO}_{3} (g),

the concentrations at equilibrium are [SO₂] = 0.90 M, [O₂] = 0.35 M, and [SO₃] = 1.1 M. What is the value of the reaction quotient at equlibrium?

Answer:

Q_c = 4.3

By its definition, the reaction quotient at equilibrium is called the equilibrium constant. 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 equilibrium constant will reach equilibrium when most of the reactant has been converted to product, whereas a small equilibrium constant indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of the equilibrium constant 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 the equilibrium constant 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, K_c is the symbol for the equilibrium constant:

CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g) K_{c} = 0.640 T = 800 ° C

The bar charts in Figure $\PageIndex{2}$ 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.

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, labeled “0.0243,” and the blue bar near 0.05, labeled “0.0243.” For mixture two, the green bar is near 0.05, labeled “0.0468,” and the yellow bar is near 0.09, labeled “0.0468.” For mixture 3, the red bar is near 0.01, labeled “0.0330,” the blue bar is slightly above that, labeled “0.190,” with green and yellow topping it off at 0.02. Green is labeled “0.00175” and yellow is labeled “0.00160.” On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled “0.0135,” the blue bar stacked on it rising slightly above 0.02, labeled “0.0135,” the green rising near 0.04, labeled “0.0108,” and the yellow bar reaching near 0.05, labeled “0.0108.” A label above this bar reads, “Q equals 0.640.” The bar for mixture two shows the red bar slightly above 0.02, labeled “0.0260,” the blue bar stacked on it rising near 0.05, labeled “0.0260,” the green rising near 0.07, labeled “0.0208,” and the yellow bar reaching near 0.10, labeled “0.0208.” A label above this bar reads “Q equals 0.640.” The bar for mixture three shows the red bar near 0.01, labeled “0.0231,” the blue bar stacked on it rising slightly above 0.01, labeled “0.00909,” the green rising near 0.02, labeled “0.0115,” and the yellow bar reaching 0.02, labeled “0.0117.” A label above this bar reads “Q equals 0.640”.

Figure $\PageIndex{2}$ Compositions of three mixtures before (Q_c ≠ K_c) and after (Q_c = K_c) equilibrium is established for the reaction

CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g) .

Example

Predicting the Direction of Reaction

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

CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g)

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} = \frac{[{CO}_{2}] [H_{2}]}{[CO] [H_{2} O]} = \frac{(0.0040) (0.0040)}{(0.020) (0.020)} = 0.040 .

Q_c < K_c (0.040 < 0.64)

The reaction will proceed in the forward direction.

Experiment 2:

Q_{c} = \frac{[{CO}_{2}] [H_{2}]}{[CO] [H_{2} O]} = \frac{(0.037) (0.046)}{(0.011) (0.0011)} = 1.4 \times 10^{2}

Q_c > K_c (140 > 0.64)

The reaction will proceed in the reverse direction.

Experiment 3:

Q_{c} = \frac{[{CO}_{2}] [H_{2}]}{[CO] [H_{2} O]} = \frac{(0.0015) (0.0076)}{(0.0094) (0.0025)} = 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 NO (g) + {Cl}_{2} (g) ⇌ 2 NOCl (g) K_{c} = 4.6 \times 10^{4}

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

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g) K_{c} = 0.060

2 {SO}_{3} (g) ⇌ 2 {SO}_{2} (g) + O_{2} (g) K_{c} = 0.230

Answer:

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

Predicting the Direction of a Reversible Reaction

As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient ($Q$) is equal to the equilibrium constant (K). We next address what happens when a system at equilibrium is disturbed so that $Q$ is no longer equal to $K$. If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of $Q$ will no longer equal the value of $K$. To re-establish equilibrium, the system will either shift toward the products (if $Q < K$) or the reactants (if $Q > K$) until $Q$ returns to the same value as $K$.

This process is described by Le Chatelier's principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in $Q$; the reaction will shift to re-establish $Q = K$.

Le Chatelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of Q and K for the system to predict the changes.

A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium.

The stress on the system in Figure $\PageIndex{3}$ is the reduction of the equilibrium concentration of SCN⁻ (lowering the concentration of one of the reactants would cause Q to be larger than K). As a consequence, Le Chatelier's principle leads us to predict that the concentration of Fe(SCN)²⁺ should decrease, increasing the concentration of SCN⁻ part way back to its original concentration, and increasing the concentration of Fe³⁺ above its initial equilibrium concentration.

Three capped test tubes held vertically in clamps are shown in pictures labeled, “a,” “b,” and “c.” The test tube in picture a is half filled with a clear, orange liquid. The test tube in picture b is half filled with a dark, burgundy liquid. The test tube in picture c is half filled with a slightly cloudy, orange liquid. — **Figure $\PageIndex{1}$:** (a) The test tube contains 0.1 M Fe³⁺. (b) Thiocyanate ion has been added to solution in (a), forming the red Fe(SCN)²⁺ ion. $\ce{Fe}^{3+}(aq)+\ce{SCN^-}(aq)\rightleftharpoons \ce{Fe(SCN)^{2+}}(aq)$. *(c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN− as the white solid AgSCN.* $\ce{Ag^+}(aq)+\ce{SCN^-}(aq)\rightleftharpoons \ce{AgSCN}(s)$. The decrease in the SCN− concentration shifts the first equilibrium in the solution to the left, decreasing the concentration (and lightening color) of the Fe(SCN)²⁺. (credit: modification of work by Mark Ott).

The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:

\[\ce{H}_{2(g)}+\ce{I}_{2(g)} \rightleftharpoons \ce{2HI}_{(g)} \label{13.4.1a} \]

\[K_c=\mathrm{50.0 \; at\; 400°C} \label{13.4.1b} \]

The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with $\mathrm{[H_2] = [I_2]} = 0.221\; M$ and $\ce{[HI]} = 1.563 \;M$ is at equilibrium; for this mixture, $Q_c = K_c = 50.0$. If $\ce{H_2}$ is introduced into the system so quickly that its concentration doubles before it begins to react (new $\ce{[H_2]} = 0.442\; M$), the reaction will shift so that a new equilibrium is reached, at which

$\ce{[H_2]} = 0.374\; M$,
$\ce{[I_2]} = 0.153\; M$, and
$\ce{[HI]} = 1.692\; M$.

This gives:

\[Q_c=\mathrm{\dfrac{[HI]^2}{[H_2][I_2]}}=\dfrac{(1.692)^2}{(0.374)(0.153)}=50.0=K_c \label{13.4.2} \]

We have stressed this system by introducing additional $\ce{H_2}$. The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess $\ce{H_2}$, reducing the amount of uncombined $\ce{I_2}$, and forming additional $\ce{HI}$.

Effect of Change in Pressure on Equilibrium

Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for $K_c$) or partial pressure (for $K_P$). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.

As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.

Consider what happens when we increase the pressure on a system in which $\ce{NO}$, $\ce{O_2}$, and $\ce{NO_2}$ are at equilibrium:

\[\ce{2NO (g) + O2(g) \rightleftharpoons 2NO2(g)} \label{13.4.3} \]

The formation of additional amounts of $\ce{NO2}$ decreases the total number of molecules in the system because each time two molecules of $\ce{NO_2}$ form, a total of three molecules of $\ce{NO}$ and $\ce{O_2}$ are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of $\ce{NO_2}$ into $\ce{NO}$ and $\ce{O_2}$, which tends to restore the pressure.

Now consider this reaction:

\[\ce{N2 (g) + O2 (g) \rightleftharpoons 2NO (g)} \label{13.4.4} \]

Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.

Effect of Change in Temperature on Equilibrium

Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle.

When hydrogen reacts with gaseous iodine, heat is evolved.

\[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g) } \;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{13.4.5} \]

Because this reaction is exothermic, we can write it with heat as a product.

\[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} + \text{heat} \label{13.4.6} \]

Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of $\ce{H2}$ and $\ce{I2}$ and a reduction in the concentration of $\ce{HI}$. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide.

When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the $\ce{HI}$ system increases the equilibrium constant: At the new equilibrium the concentration of $\ce{HI}$ has increased and the concentrations of $\ce{H2}$ and $\ce{I2}$ decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.

Temperature affects the equilibrium between $\ce{NO_2}$ and $\ce{N_2O_4}$ in this reaction

\[\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{13.4.7} \]

The positive ΔH value tells us that the reaction is endothermic and could be written

\[\text{heat}+\ce{N2O4(g) \rightleftharpoons 2NO2(g)} \label{13.4.8} \]

At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown $\ce{NO_2}$ molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless $\ce{N_2O_4}$ increases, and the concentration of brown $\ce{NO_2}$ decreases, causing the brown color to fade.

Catalysts Do Not Affect Equilibrium

As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.

The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation

\[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \label{13.4.9} \]

A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year. Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.

Fritz Haber

Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures.

For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate.

\[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \nonumber \]

Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008. The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen ($\ce{N_2}$) is nutritionally unavailable due the tremendous stability of the nitrogen-nitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation).

In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.”¹ Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.

Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.

Summary

Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side.

Table $\PageIndex{1}$: Effects of Disturbances of Equilibrium and K
Disturbance	Observed Change as Equilibrium is Restored	Direction of Shift	Effect on K
reactant added	added reactant is partially consumed	toward products	none
product added	added product is partially consumed	toward reactants	none
decrease in volume/increase in gas pressure	pressure decreases	toward side with fewer moles of gas	none
increase in volume/decrease in gas pressure	pressure increases	toward side with more moles of gas	none
temperature increase	heat is absorbed	toward products for endothermic, toward reactants for exothermic	changes
temperature decrease	heat is given off	toward reactants for endothermic, toward products for exothermic	changes

Footnotes

Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764.

Glossary

equilibrium: state of a reversible reaction in which the forward and reverse processes occur at equal rates

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

Le Châtelier’s principle: an equilibrium subjected to stress will shift in a way to counter the stress and re-establish equilibrium

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)

reversible reaction: chemical reaction that can proceed in both the forward and reverse directions under given conditions

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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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Fritz Haber