15.2: The Equilibrium Constant
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Learning Objectives
- To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
- To write an equilibrium constant expression for any reaction.
Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the decomposition reaction of N2O4 to NO2. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:
forward rate=kf[N2O4]
and
reverse rate=kr[NO2]2
At equilibrium, the forward rate equals the reverse rate (definition of equilibrium):
kf[N2O4]=kr[NO2]2
so
kfkr=[NO2]2[N2O4]
The ratio of the rate constants gives us a new constant, the equilibrium constant (K), which is defined as follows:
K=kfkr
Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions.
The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.
Table 15.2.1. lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation ???. At equilibrium the magnitude of the quantity [NO2]2/[N2O4] is essentially the same for all five experiments. In fact, no matter what the initial concentrations of NO2 and N2O4 are, at equilibrium the quantity [NO2]2/[N2O4] will always be 6.53±0.03×10−3 at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations.
Initial Concentrations | Concentrations at Equilibrium | ||||
---|---|---|---|---|---|
Experiment | [N2O4] (M) | [NO2] (M) | [N2O4] (M) | [NO2] (M) | K=[NO2]2/[N2O4] |
1 | 0.0500 | 0.0000 | 0.0417 | 0.0165 | 6.54×10−3 |
2 | 0.0000 | 0.1000 | 0.0417 | 0.0165 | 6.54×10−3 |
3 | 0.0750 | 0.0000 | 0.0647 | 0.0206 | 6.56×10−3 |
4 | 0.0000 | 0.0750 | 0.0304 | 0.0141 | 6.54×10−3 |
5 | 0.0250 | 0.0750 | 0.0532 | 0.0186 | 6.50×10−3 |
Developing an Equilibrium Constant Expression
In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form
aA+bB⇌cC+dD
where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass action and can be stated as follows:
K=[C]c[D]d[A]a[B]b
where K is the equilibrium constant for the reaction. Equation ??? is called the equilibrium equation, and the right side of Equation ??? is called the equilibrium constant expression. The relationship shown in Equation ??? is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism.
The equilibrium constant can vary over a wide range of values. The values of K shown in Table 15.2.2, for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than 103 indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between H2 and Cl2 to produce HCl, which has an equilibrium constant of 1.6×1033 at 300 K. Because H2 is a good reductant and Cl2 is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of K less than 10−3 indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants.
Reaction | Temperature (K) | Equilibrium Constant (K) |
---|---|---|
*Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures. | ||
S(s)+O2(g)⇌SO2(g) | 300 | 4.4×1053 |
2H2(g)+O2(g)⇌2H2O(g) | 500 | 2.4×1047 |
\(H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2 | HCl_{(g)}\)300 | 1.6×1033 |
H2(g)+Br2(g)⇌2HBr(g) | 300 | 4.1×1018 |
2NO(g)+O2(g)⇌2NO2(g) | 300 | 4.2×1013 |
3H2(g)+N2(g)⇌2NH3(g) | 300 | 2.7×108 |
H2(g)+D2(g)⇌2HD(g) | 100 | 1.92 |
H2(g)+I2(g)⇌2HI(g) | 300 | 2.9×10−1 |
I2(g)⇌2I(g) | 800 | 4.6×10−7 |
Br2(g)⇌2Br(g) | 1000 | 4.0×10−7 |
Cl2(g)⇌2Cl(g) | 1000 | 1.8×10−9 |
F2(g)⇌2F(g) | 500 | 7.4×10−13 |
You will also notice in Table 15.2.2 that equilibrium constants have no units, even though Equation ??? suggests that the units of concentration might not always cancel because the exponents may vary. In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in Equation ???, the units of concentration cancel, which makes K unitless as well:
[A]measured[A]standardstate=MM=molLmolL
In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M.
Many reactions have equilibrium constants between 1000 and 0.001 (103≥K≥10−3), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form HD:
H2(g)+D2(g)⇌2HD(g)
The equilibrium constant expression for this reaction is
K=[HD]2[H2][D2]
with K varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of H2, D2, and HD contains significant concentrations of both product and reactants.
Figure 15.2.3 summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants ⇌ products. Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equations ??? and ???), when kf>>kr, K is a large number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when kf<<kr, K is a very small number, and the reaction produces almost no products as written. Systems for which kf≈kr have significant concentrations of both reactants and products at equilibrium.
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A large value of the equilibrium constant K means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium.
Example 15.2.1
Write the equilibrium constant expression for each reaction.
- N2(g)+3H2(g)⇌2NH3(g)
- CO(g)+12O2(g)⇌CO2(g)
- 2CO2(g)⇌2CO(g)+O2(g)
Given: balanced chemical equations
Asked for: equilibrium constant expressions
Strategy:
Refer to Equation ???. Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator.
Solution:
The only product is ammonia, which has a coefficient of 2. For the reactants, N2 has a coefficient of 1 and H2 has a coefficient of 3. The equilibrium constant expression is as follows:
[NH3]2[N2][H2]3
The only product is carbon dioxide, which has a coefficient of 1. The reactants are CO, with a coefficient of 1, and O2, with a coefficient of 12. Thus the equilibrium constant expression is as follows:
[CO2][CO][O2]1/2
This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for O2. The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2:
Exercise 15.2.1
Write the equilibrium constant expression for each reaction.
- N2O(g)⇌N2(g)+12O2(g)
- 2C8H18(g)+25O2(g)⇌16CO2(g)+18H2O(g)
- H2(g)+I2(g)⇌2HI(g)
Answer:
- K=[N2][O2]1/2[N2O]
- K=[CO2]16[H2O]18[C8H18]2[O2]25
- K=[HI]2[H2][I2]
Example 15.2.2
Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants.
- H2(g)+I2(g)⇌2HI(g) K(700K)=54
- 2CO2(g)⇌2CO(g)+O2(g) K(1200K)=3.1×10−18
- PCl5(g)⇌PCl3(g)+Cl2(g) K(613K)=97
- 2O3(g)⇌3O2(g) K(298K)=5.9×1055
Given: systems and values of K
Asked for: composition of systems at equilibrium
Strategy:
Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both.
Solution:
- Only system 4 has K>>103, so at equilibrium it will consist of essentially only products.
- System 2 has K<<10−3, so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
- Both systems 1 and 3 have equilibrium constants in the range 103≥K≥10−3, indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.
Exercise 15.2.2
Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:
3H2(g)+N2(g)⇌2NH3(g)
Values of the equilibrium constant at various temperatures were reported as
- K25°C=3.3×108,
- K177°C=2.6×103, and
- K327°C=4.1.
At which temperature would you expect to find the highest proportion of H2 and N2 in the equilibrium mixture?
Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia?
Answer:
- 327°C, where K is smallest
- 25°C
Variations in the Form of the Equilibrium Constant Expression
Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation ??? in reverse, we obtain the following:
cC+dD⇌aA+bB
The corresponding equilibrium constant K′ is as follows:
K′=[A]a[B]6b[C]c[D]d
This expression is the inverse of the expression for the original equilibrium constant, so K′=1/K. That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction N2O4 \rightleftharpoons 2NO_2\) is as follows:
K=[NO2]2[N2O4]
but for the opposite reaction, 2NO2⇌N2O4, the equilibrium constant K′ is given by the inverse expression:
K′=[N2O4][NO2]2
Consider another example, the formation of water: 2H2(g)+O2(g)⇌2H2O(g). Because H2 is a good reductant and O2 is a good oxidant, this reaction has a very large equilibrium constant (K=2.4×1047 at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form O2 and H2, is very small: K′=1/K=1/(2.4×1047)=4.2×10−48. As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into H2 and O2.
The equilibrium constant for a reaction written in reverse is the inverse of the equilibrium constant for the reaction as written originally.
Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction
2NO2⇌N2O4
as
NO2⇌12N2O4
with the equilibrium constant K″ is as follows:
K′′=[N2O4]1/2[NO2]
The values for K′ (Equation ???) and K″ are related as follows:
K′′=(K′)1/2=√K′
In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by n, then the new equilibrium constant is the original equilibrium constant raised to the nth power.
Example 15.2.3: The Haber Process
At 745 K, K is 0.118 for the following reaction:
N2(g)+3H2(g)⇌2NH3(g)
What is the equilibrium constant for each related reaction at 745 K?
- 2NH3(g)⇌N2(g)+3H2(g)
- 12N2(g)+32H2(g)⇌NH3(g)
Given: balanced equilibrium equation, K at a given temperature, and equations of related reactions
Asked for: values of K for related reactions
Strategy:
Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate K for each reaction.
Solution:
The equilibrium constant expression for the given reaction of N2(g) with H2(g) to produce NH3(g) at 745 K is as follows:
K=[NH3]2[N2][H2]3=0.118
This reaction is the reverse of the one given, so its equilibrium constant expression is as follows:
K′=1K=[N2][H2]3[NH3]2=10.118=8.47
In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows:
Exercise
At 527°C, the equilibrium constant for the reaction
2SO2(g)+O2(g)⇌2SO3(g)
is 7.9×104. Calculate the equilibrium constant for the following reaction at the same temperature:
Answer
3.6×10−3
Equilibrium Constant Expressions for Systems that Contain Gases
For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol Kp is used to denote equilibrium constants calculated from partial pressures. For the general reaction aA+bB⇌cC+dD, in which all the components are gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation):
Kp=(PC)c(PD)d(PA)a(PB)b
Thus Kp for the decomposition of N2O4 (Equation 15.1) is as follows:
Kp=(PNO2)2PN2O4
Like K, Kp is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity.The “effective pressure” is called the fugacity, just as activity is the effective concentration.
Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of K and Kp are usually different. They are, however, related by the ideal gas constant (R) and the absolute temperature (T):
Kp=K(RT)Δn
where K is the equilibrium constant expressed in units of concentration and Δn is the difference between the numbers of moles of gaseous products and gaseous reactants (np−nr). The temperature is expressed as the absolute temperature in Kelvin. According to Equation ???, Kp=K only if the moles of gaseous products and gaseous reactants are the same (i.e., Δn=0). For the decomposition of N2O4, there are 2 mol of gaseous product and 1 mol of gaseous reactant, so Δn=1. Thus, for this reaction,
Kp=K(RT)1=KRT
Example 15.2.4: The Haber Process (again)
The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows:
N2(g)+3H2(g)⇌2NH3(g)
What is Kp for this reaction at the same temperature?
Given: equilibrium equation, equilibrium constant, and temperature
Asked for: Kp
Strategy:
Use the coefficients in the balanced chemical equation to calculate Δn. Then use Equation ??? to calculate K from Kp.
Solution:
This reaction has 2 mol of gaseous product and 4 mol of gaseous reactants, so Δn=(2−4)=−2. We know K, and T=745K. Thus, from Equation ???, we have the following:
Kp=K(RT)−2=K(RT)2=0.118{[0.08206(L⋅atm)/(mol⋅K)][745K]}2=3.16×10−5
Because Kp is a unitless quantity, the answer is Kp=3.16×10−5.
Exercise 15.2.4
Calculate Kp for the reaction
2SO2(g)+O2(g)⇌2SO3(g)
at 527°C, if K=7.9×104 at this temperature.
Answer
Kp=1.2×103
Equilibrium Constant Expressions for the Sums of Reactions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of N2 with O2 to give NO2. This reaction is an important source of the NO2 that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (step 1), N2 reacts with O2 at the high temperatures inside an internal combustion engine to give NO. The released NO then reacts with additional O2 to give NO2 (step 2). The equilibrium constant for each reaction at 100°C is also given.
N2(g)+O2(g)⇌2NO(g) K1=2.0×10−25
2NO(g)+O2(g)⇌2NO2(g)K2=6.4×109
Summing reactions (step 1) and (step 2) gives the overall reaction of N2 with O2:
N2(g)+2O2(g)⇌2NO2(g) K3=?
The equilibrium constant expressions for the reactions are as follows:
K1=[NO]2[N2][O2] K2=[NO2]2[NO]2[O2] K3=[NO2]2[N2][O2]2
What is the relationship between K1, K2, and K3, all at 100°C? The expression for K1 has [NO]2 in the numerator, the expression for K2 has [NO]2 in the denominator, and [NO]2 does not appear in the expression for K3. Multiplying K1 by K2 and canceling the [NO]2 terms,
K1K2=[NO]2[N2][O2]×[NO2]2[NO]2[O2]=[NO2]2[N2][O2]2=K3
Thus the product of the equilibrium constant expressions for K1 and K2 is the same as the equilibrium constant expression for K3:
K3=K1K2=(2.0×10−25)(6.4×109)=1.3×10−15
The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, ΔH for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.
To determine K for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
Example 15.2.6
The following reactions occur at 1200°C:
- CO(g)+3H2(g)⇌CH4(g)+H2O(g) K1=9.17×10−2
- CH4(g)+2H2S(g)⇌CS2(g)+4H2(g) K2=3.3×104
Calculate the equilibrium constant for the following reaction at the same temperature.
- CO(g)+2H2S(g)⇌CS2(g)+H2O(g)+H2(g) K3=?
Given: two balanced equilibrium equations, values of K, and an equilibrium equation for the overall reaction
Asked for: equilibrium constant for the overall reaction
Strategy:
Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of K for that equation. Calculate K for the overall equation by multiplying the equilibrium constants for the individual equations.
Solution:
The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:
CO(g)+3H2(g)⇌CH4(g)+H2O(g)
CH4(g)+2H2S(g)⇌CS2(g)+3H2(g)+H2(g)
CO(g)+3H2(g)⇌CS2(g)+H2O(g)+H2(g)
The values for K1 and K2 are given, so it is straightforward to calculate K3:
K3=K1K2=(9.17×10−2)(3.3×104)=3.03×103
Exercise 15.2.6
In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.
- 18S8(s)+O2(g)⇌SO2(g) K1=4.4×1053
- SO2(g)+12O2(g)⇌SO3(g) K2=2.6×1012
- 18S8(s)+32O2(g)⇌SO3(g) K3=?
Answer: K3=1.1×1066
Summary
The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same K. For a system at equilibrium, the law of mass action relates K to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, K and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures (Kp) is related to K by the ideal gas constant (R), the temperature (T), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.
- The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants.
- For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.
- Definition of equilibrium constant in terms of forward and reverse rate constants: K=kfkr
- Equilibrium constant expression (law of mass action): K=[C]c[D]d[A]a[B]b
- Equilibrium constant expression for reactions involving gases using partial pressures: Kp=(PC)c(PD)d(PA)a(PB)b
- Relationship between Kp and K: Kp=K(RT)Δn