Chemistry
Equilibrium Concentrations
Equilibrium concentrations refer to the relative amounts of reactants and products in a chemical reaction when the rates of the forward and reverse reactions are equal. At equilibrium, the concentrations of the reactants and products remain constant over time. These concentrations are determined by the equilibrium constant, which is a measure of the extent to which a reaction proceeds.
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10 Key excerpts on "Equilibrium Concentrations"
eBook - PDF- John A. Olmsted, Gregory M. Williams, Robert C. Burk(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
As this occurs, concentrations of starting materials decrease, and concentrations of products increase. These concentration changes are related in two ways, as described in Chapter 1. To review, the concentration of each reagent at equilibrium is its initial concentration plus the change that has occurred: [ Equilibrium concentration ] = [ Initial concentration ] + [ Change in concentration during reaction ] [A] eq = [A] i + Δ[A] (14.5) Furthermore, stoichiometric ratios link the changes in concentration of the various reagents: [ Change in concentration of B ] = [ Stoichiometric ratio of B to A ][ Change in concentration of A ] Δ[B ] = v B ___ v A Δ[A] (14.6) As we show in upcoming examples, some changes in con- centration are positive whereas others are negative, depend- ing on whether a reagent is produced or consumed. These relationships provide complete stoichiometric information regarding the equilibrium. Just as amounts tables are useful in doing stoichiometric calculations, a concentration table that provides initial concentrations, changes in concentrations, and equilibrium concentra- tions is an excellent way to organize Step 5 of the problem- solving procedure for equilibrium problems. Figure 14.11 diagrams how to complete a concentration table: Equation 14.5 applies to every column, but Equation 14.6 applies only across the “change” row. Initial Conditions and Concentration Tables When we do equilibrium calculations, we are usually interested in the concentrations of spe- cies present at equilibrium. In many cases, however, we have information about what we call initial concentrations, before any net change has occurred. Initial concentrations are the concentrations that would be present if it were possible to mix all the reactants but block the reactions that lead to equilibrium. These concentrations are easy to calculate from the initial conditions, but they seldom exist in reality because substances begin to react as soon as they are mixed.
- Morris Hein, Scott Pattison, Susan Arena, Leo R. Best(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
A double arrow is used in the equation to indicate that the reaction is reversible. Any system at equilibrium represents a dynamic state in which two or more opposing processes are taking place at the same time and at the same rate. A chemical equilibrium is a dynamic system in which two or more opposing chemical reactions are going on at the same time and at the same rate. When the rate of the forward reaction is exactly equal to the rate of the reverse reaction, a condition of chemical equilibrium exists (see purple line in KEY TERMS reversible chemical reaction equilibrium chemical equilibrium NO 2 Figure 16.2 Reversible reaction of NO 2 and N 2 O 4 . More of the dark brown molecules are visible in the heated container on the right than in the room-temperature tube on the left. LEARNING OBJECTIVE N 2 O 4 Richard Megna/Fundamental Photographs Richard Megna/Fundamental Photographs 366 CHAPTER 16 • Chemical Equilibrium P R A C T I C E 1 6 . 2 What symbolism is used in a chemical equation to indicate that a chemical reaction is reversible? Figure 16.1). The concentrations of the products and the reactants are not changing, and the system appears to be at a standstill because the products are reacting at the same rate at which they are being formed. Chemical equilibrium: rate of forward reaction = rate of reverse reaction A saturated salt solution is in a condition of equilibrium: NaCl(s) m Na + (aq) + Cl - (aq) At equilibrium, salt crystals are continuously dissolving, and Na + and Cl - ions are continu- ously crystallizing. Both processes are occurring at the same rate. The ionization of weak electrolytes is another chemical equilibrium system: HC 2 H 3 O 2 (aq) + H 2 O(l) m H 3 O + (aq) + C 2 H 3 O 2 - (aq) In this reaction, the equilibrium is established in a 1 M solution when the forward reaction has gone about 1%—that is, when only 1% of the acetic acid molecules in solution have ionized.
eBook - PDF- Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
Chapter 5 GENERAL CONCEPTS OF CHEMICAL EQUILIBRIUM “The worst form of inequality is to try to make unequal things equal.” —Aristotle KEY THINGS TO LEARN FROM THIS CHAPTER The equilibrium constant (key Equations: 5.12, 5.15) Calculation of Equilibrium Concentrations Using Excel Goal Seek to solve one-variable equations The systematic approach to equilibrium calculations: mass balance and charge balance equations Activity and activity coefficients (key Equation: 5.19) Thermodynamic equilibrium constants (key Equation: 5.23) Even though in a chemical reaction the reactants may almost quantitatively react to form the products, reactions never go in only one direction. In fact, reactions reach an equilibrium in which the rates of reactions in both directions are equal. In this chapter we review the equilibrium concept and the equilibrium constant and describe gen- eral approaches for calculations using equilibrium constants. We discuss the activity of ionic species along with the calculation of activity coefficients. These values are required for calculations using thermodynamic equilibrium constants, that is, for the diverse ion effect, described at the end of the chapter. They are also used in potentio- metric calculations (Chapter 20). 5.1 Chemical Reactions: The Rate Concept In 1863 Guldberg and Waage described what we now call the law of mass action, which states that the rate of a chemical reaction is proportional to the “active masses” of the reacting substances present at any time. The active masses may be concentrations or pressures. Guldberg and Waage derived an equilibrium constant by defining equilib- rium as the condition when the rates of the forward and reverse reactions are equal.
eBook - PDFChemistry
An Industry-Based Introduction with CD-ROM
- John Kenkel, Paul B. Kelter, David S. Hage(Authors)
- 2000(Publication Date)
- CRC Press(Publisher)
Temperature and pressure are parameters that, when varied, create a “stress” on the equilibrium. Stresses on equilibrium systems cause changes in the system. The nature of these changes are described by the principle of Le Châtelier. Le Châtelier’s Principle can be expressed as follows: when a stress is placed on a system at equilibrium, the equilibrium shifts to partially relieve the stress. An equilibrium shift is defined as a spontaneous and momentary increase in the rate of an equilibrium reaction in one direction or the other, causing a temporary situation in which the reaction is not at equilibrium since the rates of the two opposing reactions are no longer equal. This phenomenon is tem-porary, since equilibrium is re-established at some later moment. When the equilibrium is reestablished, the concentrations of the reactants and products are not what they were previously because the temporary rate imbalance caused a surge toward one side and away from the other. The equilibrium constant may or may not change during this process depending on what caused the shift. This will be discussed below in conjunction with the various stresses that can be placed on a chemical equilibrium and cause it to shift. 11.7 Effect of Concentration Change The change in concentration of a chemical participating in a chemical equilibrium is a stress on the equilibrium system. Thus, such a change would cause the equilibrium to shift, as stated by Le Châtelier’s principle. Relieving the stress in this case means that if the concentration increases, the equilibrium shifts so as to partially decrease it. If the concentration decreases, the equilibrium shifts so as to partially increase it. An increase in concen-tration may be caused simply by opening the reaction vessel and adding more of the chemical.
eBook - PDFChemistry
An Atoms First Approach
- Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste, , Steven Zumdahl, Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. For Review Key Terms chemical equilibrium Section 12.2 law of mass action equilibrium expression equilibrium constant equilibrium position Section 12.4 homogeneous equilibria heterogeneous equilibria Section 12.5 reaction quotient (Q) Section 12.7 Le Châtelier’s principle Chemical Equilibrium ❯ When a reaction takes place in a closed system, it reaches a condition where the concentra- tions of the reactants and products remain constant over time ❯ Dynamic state: reactants and products are interconverted continually ❯ Forward rate 5 reverse rate ❯ The law of mass action: for the reaction jA 1 kB m mC 1 nD K 5 fCg m fDg n fAg j fBg k 5 equilibrium constant ❯ A pure liquid or solid is never included in the equilibrium expression ❯ For a gas-phase reaction, the reactants and products can be described in terms of their partial pressures and the equilibrium constant is called K p : K p 5 KsRTd Dn where Dn is the sum of the coefficients of the gaseous products minus the sum of the coef- ficients of the gaseous reactants Equilibrium Position ❯ A set of reactant and product concentrations that satisfies the equilibrium constant expression ❯ There is one value of K for a given system at a given temperature ❯ There are an infinite number of equilibrium positions at a given temperature depending on the initial concentrations ❯ A small value of K means the equilibrium lies to the left; a large value of K means the equi- librium lies to the right ❯ The size of K has no relationship to the speed at which equilibrium is achieved ❯ Q, the reaction quotient, applies the law of mass action to initial concentrations rather than Equilibrium Concentrations ❯ If Q .
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 - 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.
eBook - PDF- Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
Video: Goal Seek Equilibrium This is because no judgments on what can or cannot be neglected is needed. A video illustrating the use of Goal Seek to solve an equilibrium problem is on the text website. Example 6.3 A and B react as follows: A + 2B 2C K = [C] 2 [A][B] 2 Assume 0.10 mol of A is reacted with 0.20 mol of B in a volume of 1000 mL; K = 1.0 × 10 10 . What are the Equilibrium Concentrations of A, B, and C? Solution We have stoichiometrically equal amounts of A and B, so both are virtually all reacted, with trace amounts remaining. Let x represent the equilibrium concentration of A. At equilibrium, we have A x + 2B 2x 2C 0.20 − 2x ≈ 0.20 For each mole of A that either reacts (or is produced), we produce (or remove) two moles of C, and consume (or produce) two moles of B. Substituting into the equilibrium constant expression, (0.20) 2 (x)(2x) 2 = 1.0 × 10 10 0.040 4x 3 = 1.0 × 10 10 x = [A] = 3 4.0 × 10 −2 4.0 × 10 10 = 3 1.0 × 10 −12 = 1.0 × 10 −4 M B = 2x = 2.0 × 10 −4 M (analytically detectable, but not appreciable compared to the starting}?> concentration) DISSOCIATION EQUILIBRIA Calculations involving dissociating species are not much different from the example just given for chemical reactions. Example 6.4 Calculate the Equilibrium Concentrations of A and B in a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 × 10 −6 . Solution AB A + B K eq = [A][B] [AB] 202 CHAPTER 6 GENERAL CONCEPTS OF CHEMICAL EQUILIBRIUM Both [A] and [B] are unknown and equal. Let x represent their equilibrium con- centrations. The concentration of AB at equilibrium is equal to its initial analytical concentration minus x. AB 0.10 − x A x + B x The value of K eq is quite small, so we are probably justified in neglecting x compared In a dissociation, neglect x compared to the initial concentration C if C ≥ 100K eq in a dissociation. to 0.10. Otherwise, we will have to use a quadratic equation.
eBook - PDF- Morris Hein, Susan Arena, Cary Willard(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
• 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 KEY TERMS: Le Châtelier’s principle catalyst activation energy • 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 equilibrium 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 reaction 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 equilibrium 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. 16.4 Equilibrium Constants KEY TERM: equilibrium constant, K eq • For the general reaction aA + bB ⎯ ⇀ ↽ ⎯ cC + dD K eq = [C] c [D] d _ [A] a [B] b Chapter 16 Review
- 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.
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