Chemistry
Equilibrium Constants
Equilibrium constants are numerical values that express the extent of a chemical reaction at equilibrium. They are calculated from the concentrations of reactants and products at equilibrium and are used to predict the direction of a reaction and its equilibrium position. The value of the equilibrium constant provides information about the relative amounts of reactants and products at equilibrium.
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10 Key excerpts on "Equilibrium Constants"
eBook - PDFChemistry
An Industry-Based Introduction with CD-ROM
- John Kenkel, Paul B. Kelter, David S. Hage(Authors)
- 2000(Publication Date)
- CRC Press(Publisher)
Because there is no change in these counts, we can divide the number of students on one side by the number of students on the other side and always obtain the same number. This same concept applied to a chemical equilibrium system results in a number called the equilibrium constant . In a chemical equilibrium system, the “count” is the molar concentration of the chemical species involved. For the purpose of symbolizing this molar concentration (and the equilibrium constant), we now introduce the symbolism of the brackets, “[].” The molar concentration (molarity) of a chemical species is symbolized by enclosing the formula for this species within brackets . For example, [H ] refers to the molar concentration of the hydrogen ion, [NH 3 ] refers to the molar concentration of ammonia, and [Cl ] refers to the molar concentration of the chloride ion. The equilibrium constant for a chemical equilibrium is defined as the molar concentrations of the products of the reaction, raised to the power of their respective balancing coefficients and multiplied together, divided by the molar concentrations of the reactants of the reaction, raised to the power of their balancing coefficients and multiplied together. Thus, for the general reaction (11.11) the equilibrium constant, symbolized as K eq , is defined as in Equation (11.12). (11.12) Knowing the molar concentrations of all the chemicals involved in the equilibrium, we can calculate the number that is given by the equilibrium constant. For Eq. (11.7), K eq is as shown in Equation (11.13). (11.13) and for Eq. (11.8), K eq is shown below, (11.14) and for Eq. (11.9) it is as follows. (11.15) Equations (11.8) and (11.9) represent “ionization” equilibria in which an uncharged chemical species (molecule or formula unit) splits apart into ions. Equation (11.8) represents the ionization of a weak acid, acetic acid, and Eq. (11.9) represents the ionization of a weak base, NH 4 OH.
- Morris Hein, Scott Pattison, Susan Arena, Leo R. Best(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
16.4 Equilibrium Constants Write the general expression for the equilibrium constant and calculate Equilibrium Constants. In a reversible chemical reaction at equilibrium, the concentrations of the reactants and prod- ucts are constant. At equilibrium, the rates of the forward and reverse reactions are equal, and an equilibrium constant expression can be written relating the products to the reactants. For the general reaction aA + bB m cC + dD at a given temperature, the equilibrium constant expression can be written as K eq = [C] c [D] d [A] a [B] b where K eq , the equilibrium constant, is constant at a particular temperature. The quantities in brackets are the concentrations of each substance in moles per liter. The superscript letters a, b, c, and d are the coefficients of the substances in the balanced equation. According to conven- tion, we place the concentrations of the products (the substances on the right side of the equa- tion as written) in the numerator and the concentrations of the reactants in the denominator. KEY TERM equilibrium constant, K eq Note: The exponents (a, b, c and d ) are the same as the coeffi- cients in the balanced equation. E X A M P L E 1 6 . 5 Write equilibrium constant expressions for (a) 3 H 2 (g) + N 2 (g) m 2 NH 3 (g) (b) CO(g) + 2 H 2 (g) m CH 3 OH(g) SOLUTION (a) The only product, NH 3 , has a coefficient of 2. Therefore, the numerator will be [NH 3 ] 2 . Two reactants are present: H 2 , with a coefficient of 3, and N 2 , with a coefficient of 1. The denominator will thus be [H 2 ] 3 [N 2 ]. The equilibrium constant expression is K eq = [NH 3 ] 2 [H 2 ] 3 [N 2 ] (b) For this equation, the numerator is [CH 3 OH] and the denominator is [CO][H 2 ] 2 . The equilibrium constant expression is K eq = [CH 3 OH] [CO][H 2 ] 2 LEARNING OBJECTIVE 374 CHAPTER 16 • Chemical Equilibrium P R A C T I C E 1 6 .
eBook - PDFIntroductory Chemistry
An Active Learning Approach
- Mark Cracolice, Edward Peters, Mark Cracolice(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
720 Chapter 18 Chemical Equilibrium 18.9 The Significance of the Value of K Goal 12 Given an equilibrium equation and the value of the equilibrium constant, identify the direction in which the equilibrium is favored. By definition, an equilibrium constant is a ratio—a fraction. The numerical value of an equilibrium constant may be very large, very small, or anyplace in between. Even though there is no defined intermediate range, equilibria with constants between 0.01 and 100 (10 22 to 10 2 ) will have appreciable quantities of all species present at equilibrium. To see what is meant by “very large” or “very small” K values, consider an equi- librium similar to the hydrogen iodide system studied in Section 18.8. If we substi- tute chlorine for iodine, the equilibrium equation is H 2 sgd 1 Cl 2 sgd (Figure 18.19) m 2 HCl(g). At 25°C, K 5 fHClg 2 fH 2 g fCl 2 g 5 2.4 3 10 33 This is a very large number—10 billion times larger than the number of particles in a mole! The only way an equilibrium constant ratio can become so huge is for the concentration of one or more reacting species to be very close to zero. If the denom- inator of a ratio is nearly zero, the value of the ratio will be very large. A near-zero denominator and large K mean the equilibrium is favored overwhelmingly in the forward direction. By contrast, if the equilibrium constant is very small, it means the concentra- tion of one or more of the species on the right-hand side of the equation is nearly zero. This puts a near-zero number in the numerator of K, and the equilibrium is strongly favored in the reverse direction. You improved your skill at writing equilibrium constant expressions. What did you learn by solving this Active Example? Practice Exercise 18.9 Write the equilibrium constant expressions for the following: a) 2 H 2 O(O) m 2 H 2 (g) 1 O 2 (g) b) CoCl 2 (s) 1 6 H 2 O(g) m CoCl 2 ? 6 H 2 O(s) Figure 18.19 Chlorine gas has a greenish-yellow color.
eBook - PDFChemistry
Structure and Dynamics
- James N. Spencer, George M. Bodner, Lyman H. Rickard(Authors)
- 2011(Publication Date)
- Wiley(Publisher)
Because the reaction proceeds in both directions at the same rate, there is no apparent change in the concentrations of the reactants or the prod- ucts on the macroscopic scale (i.e., the level of objects visible to the naked eye). This model can also be used to predict the direction in which a reaction has to shift to reach equilibrium. If the concentrations of the reactants are too large for the reaction to be at equilibrium, the rate of the forward reaction will be faster than that of the reverse reaction, and some of the reactants will be converted to products until equilibrium is achieved. Conversely, if the concentrations of the reactants are too small, the rate of the reverse reaction will exceed that of the forward reaction, and the reaction will convert some of the excess products back into reactants until the system reaches equilibrium. We can determine the direction in which a reaction has to shift to reach equilibrium by comparing the reaction quotient (Q c ) for the reaction at some moment in time with the equilibrium constant (K c ) for the reaction. The reaction quotient expression is written in much the same way as the equilibrium constant expression. But the concentrations used to calculate Q c describe the system at any moment in time, whereas the concentrations used to calculate K c describe the sys- tem only when it is at equilibrium. To illustrate how the reaction quotient is used, consider the following gas- phase reaction. The equilibrium constant expression for the reaction is written as follows. By analogy, we can write the expression for the reaction quotient as follows. There are three important differences between the equilibrium constant expres- sion and the reaction quotient expression. First, we use brackets, such as [HI], in the equilibrium constant expression to indicate that the reaction is at equilibrium.
eBook - PDF- Juan J. de Pablo, Jay D. Schieber(Authors)
- 2014(Publication Date)
- Cambridge University Press(Publisher)
10 Reaction equilibrium Reactions are an essential component of chemical engineering, and reaction engineering itself is a very broad discipline [43 , 107 ]. However, before we study the time evolution of chemical reactions, it is important to know the final equilibrium state that a system can reach. The equilibrium state of a chemical reaction is determined by thermodynamic principles. In this chapter we examine the thermodynamic principles that govern chemical reactions, and find methods for calculating the final composition of a mixture that results from chemical reactions. The methods consist of two general steps: determination of the equilibrium constant(s) from ther-modynamic properties of the constituent chemical compounds; and determination of the chemical composition of the equilibrium mixture from the initial composition and the equilibrium constant(s). In general, finding this composition requires the use of models with good estimates for the activities (or fugacities) of the constituent species, as shown in the previous two chapters. The equilibrium constant depends only on the temperature and the species present, not on the composition. We also consider how thermodynamic driving forces can influence reaction rates. However, few details are given in this book, since that is a topic best left for other courses and textbooks. Denat-uration of DNA strands is discussed in Section 10.9, where denaturation is applied to polymerase chain reactions (PCR). PCR is a technique that can be used to amplify small amounts of genetic material. Finally, in the last section, Section 10.10, connections are made to statistical mechanics. 10.1 A SIMPLE PICTURE: THE REACTION COORDINATE ......................................................................................................................
eBook - PDFChemistry
The Molecular Nature of Matter
- James E. Brady, Neil D. Jespersen, Alison Hyslop(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
For a given overall chemical composition, the amounts of reactants and products that are present at equilibrium are the same regardless of whether the equilibrium is approached from the direction of pure “reactants,” pure “products,” or any mixture of them. Explain the basics of equilibrium laws The mass action expression is a fraction. The concentrations of the products, raised to powers equal to their coefficients in the chemical equation for a homogeneous equilibrium, are multi- plied together in the numerator. The denominator is con- structed in the same way from the concentrations of the reactants raised to powers equal to their coefficients. The numerical value of the mass action expression is the reaction quotient, Q. At equilibrium, the reaction quotient is equal to the equilibrium constant, K c . If partial pressures of gases are used in the mass action expression, is represented as K P . The magnitude of the equilibrium constant is roughly proportional to the extent to which the reaction proceeds to completion when equilibrium is reached. Equilibrium equations can be manipulated by multiplying the coefficients by a common fac- tor, changing the direction of the reaction, and by adding two or more equilibria. The rules given in the description of the Tools for Problem Solving below apply. Write and convert between equilibrium laws based on molar concentration and gas pressures The values of K P and K c are only equal if the same number of moles of gas are represented on both sides of the chemical equa- tion. When the numbers of moles of gas are different, K P is related to K c by the equation K P = K c (RT) n g . Remember to use R = 0.0821 L atm mol -1 K -1 and T = absolute temperature. Also, be careful to calculate ∆n g as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants in the balanced equation.
- John A. Olmsted, Gregory M. Williams, Robert C. Burk(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
Equilibrium constant expressions have the concentrations of the products in the numerator and the concentrations of the reactants in the denominator, with each concentration raised to the power of its stoichiometric coefficient: 2 2 3 2 2 3 2 4 2 2 3 eq CO eq H O eq 4 3 NH eq O eq eq 2 N eq 2 CH CHO eq eq 2 C H eq O eq 2 2 eq eq 4 eq eq H S eq NH eq (a) ( ) ( ) ( ) ( ) (b) ( ) ( ) (c) ( ) ( ) (d) [Ag ] [SO ] (e) ( ) ( ) K p p p p K p p K p p K K p p + − = = = = = 14.51 This problem describes an equilibrium reaction. We are asked to determine the equilibrium pressures of all the gases. To calculate pressures at equilibrium from initial conditions, set up a concentration table, write the K eq expression, and solve for the pressures. For a gas phase reaction, concentrations must be expressed in bar. Chapter 14 Chemistry, 3ce Student Solutions Manual 276 © John Wiley & Sons Canada, Ltd. 1 2 1 2 2 1 2 1 (3.00 bar)(273 352 K) 6.29 bar 298 K p p p T p T T T + = = = = Reaction: COCl 2 CO + Cl 2 Initial pressure (bar) 6.29 0 0 Change in pressure (bar) – x + x + x Equilibrium pressure (bar) 6.29 – x x x Now substitute into the equilibrium constant expression and solve for x: 2 2 2 2 2 CO eq Cl eq 4 eq COCl eq 2 2 4 3 2 CO eq Cl eq COCl eq ( ) ( ) ( ) 8.3 10 ; assume that 6.29: ( ) 6.29 8.3 (6.29)(8.3 10 ) 5.22 10 6.29 7.2 10 bar ( ) ( ) ( ) 6.29 ( p p x K x p x x x x p p p − −4 − − − = × = = << − × 10 = = × = × = × = = = − 2 2 7.2 10 ) 6.22 bar 7.2 10 is 1.1% of 6.29, so is << 6.29 and the approximation is valid. x − − × = × 14.53 At equilibrium, a molecular picture should show the presence of both reactants and products, in relative amounts that are determined by the value of the equilibrium constant.
eBook - PDF- Morris Hein, Susan Arena, Cary Willard(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
ONLINE LEARNING MODULE Buffers ➥ 16.1 Rates of Reaction • The rate of a reaction is variable and depends on: • Concentration of reacting species • Temperature • Presence of catalysts • Nature of the reactants 16.2 Chemical Equilibrium • In a reversible chemical reaction, the products formed react to produce the original reaction mixture. • The forward reaction is called the reaction to the right. • The reverse reaction is called the reaction to the left. • A system at equilibrium is dynamic. • At chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. 16.3 Le Châtelier’s Principle • If stress is applied to a system at equilibrium, the system will respond in such a way as to relieve the stress and restore equilibrium under a new set of conditions. • When the concentration of a reactant is increased, the equilibrium shifts toward the right. • When the concentration of a product on the right side of the equation is increased, the equilib- rium shifts to the left. • If the concentration of a reactant is decreased, the equilibrium shifts toward the side where the reactant is decreased. • In gaseous reactions: • A decrease in volume causes the equilibrium position to shift toward the side of the reac- tion with the fewest molecules. • If the number of molecules is the same on both sides of a reaction, a decrease in volume has no effect on the equilibrium position. • When heat is added to a reaction, the side of the equation that absorbs heat is favored: • If the reaction is endothermic, the forward reaction increases. • If the reaction is exothermic, the reverse reaction increases. • A catalyst does not shift the equilibrium of a reaction; it only affects the speed at which equilib- rium is reached. • The activation energy for a reaction is the minimum energy required for the reaction to occur: • A catalyst lowers the activation energy for a reaction by providing a different pathway for the reaction.
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
Although some of these problems can be so complicated that a computer is needed to solve them, we can learn the general principles involved by work- ing on simple calculations. Even these, however, require a little applied algebra. This is where the concentration table can be especially helpful. 738 CHAPTER 14 Chemical Equilibrium For example, we can calculate the equilibrium concentrations of each of the four molecules in the reaction CO( g) + H 2 O( g) CO 2 ( g) + H 2 ( g) if we know (a) the equilibrium law (b) the value of the equilibrium constant K c (c) the initial concentrations (d) how much the initial concentrations have changed when equilibrium is reached We can write the equilibrium law from the chemical equation, K c = 4.06 at 500 °C, but it can be looked up in tables if it is not given, and the initial concentrations must be given as part of the question. In this case, let’s say that 0.100 mol of CO and 0.100 mol of H 2 O( g) are placed in a 1.00 liter reaction vessel at 500 °C. All we need to do is determine how much the concentrations change in reaching equilibrium. The concentration table comes in handy here. To build the table, we need quantities to enter into the initial concentrations, changes in concentrations, and equilibrium concentrations rows. Initial Concentrations The initial concentrations of CO and H 2 O are each 0.100 mol / 1.00 L = 0.100 M. Since neither CO 2 nor H 2 is initially placed into the reaction ves- sel, their initial concentrations both are zero. These are shown in red in the table. Changes in Concentration Some CO 2 and H 2 must form for the reaction to reach equilibrium. This also means that some CO and H 2 O must react. But how much? If we knew the answer, we could calculate the equilibrium concentrations. Therefore, the changes in con- centration are our unknown quantities. Let us allow x to be equal to the number of moles per liter of CO that react.
eBook - PDFPhysical Chemistry
Thermodynamics
- Horia Metiu(Author)
- 2006(Publication Date)
- Taylor & Francis(Publisher)
18 CHEMICAL EQUILIBRIUM: THE CONNECTION BETWEEN THE EQUILIBRIUM CONSTANT AND COMPOSITION Introduction §1. The Starting Point. In this chapter, you will learn how the connection between the equilibrium composition and the equilibrium constant is used in the practice of chemistry. Our starting point is the equation K = c i = 1 x ( i ) ν ( i ) (18.1) which is Eq. 17.34. Here { x (1), x (2), . . . , x ( c ) } are the molar fractions of the c compounds participating in the reaction and { ν (1), ν (2), . . . , ν ( c ) } are their stoichiometric coefficients (negative for reactants, positive for products). 359 360 The Connection between K and Composition It seems that if I know K and the initial composition, I have one equation and several unknown quantities { x (1), . . . , x ( c ) } . This is not the case. Eqs 17.6–17.8 give the number of moles of compound i as n ( i ) = n 0 ( i ) + ν ( i ) ξ (18.2) and the molar fraction of compound i as x ( i ) = n 0 ( i ) + ν ( i ) ξ ∑ c j = 1 n 0 ( j ) + ν ( j ) ξ (18.3) Combining Eq. 18.3 with Eq. 18.1 gives K = c i = 1 n 0 ( i ) + ν ( i ) ξ ∑ c j = 1 n 0 ( j ) + ν ( j ) ξ ν ( i ) (18.4) In this equation, the equilibrium constant depends on only one variable: the extent of reaction ξ . The initial number of moles n 0 ( i ) is assumed to be known (chemists doing quantitative work measure the amounts of substances used to perform a reaction). §2. The Use of Eq. 18.4 in Practice. If you know K for a given temperature, pres-sure, and initial composition, you can calculate the equilibrium extent of reaction ξ e by solving Eq. 18.4 for ξ . Once you have ξ e , you can calculate the number of moles and the molar fraction of each reaction participant, at equilibrium, from Eqs 18.2 and 18.3. You can also perform measurements of equilibrium composition at different tem-peratures and pressures, and use the data in Eq. 18.4 to calculate K . Once you know K you can obtain G 0 from G 0 = − RT ln K (18.5) which is Eq.
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