Chemistry
Calculating Equilibrium Constant
The equilibrium constant is a measure of the extent to which a chemical reaction reaches equilibrium. It is calculated by taking the concentration of the products raised to their coefficients in the balanced chemical equation and dividing by the concentration of the reactants raised to their coefficients. The equilibrium constant provides insight into the position of the equilibrium and the relative concentrations of reactants and products.
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12 Key excerpts on "Calculating Equilibrium Constant"
eBook - ePubChemistry
Concepts and Problems, A Self-Teaching Guide
- Richard Post, Chad Snyder, Clifford C. Houk(Authors)
- 2020(Publication Date)
- Jossey-Bass(Publisher)
equilibrium constant is the ratio of the concentration of the products divided by the concentration of the reactants at equilibrium and at a specified temperature. The equilibrium constant (product concentrations divided by reactant concentrations) is valid only at a specified temperature after the reaction has gone to (completion, equilibrium) __________Answer: equilibriumBelow is a reversible reaction and the expression for the equilibrium constant for this reversible reaction.The symbol Keqrepresents the equilibrium constant and the brackets [] represent the concentration (usually in moles per liter) of each product and reactant. Look at the placement of each reactant and product in the equilibrium constant expression. In the equilibrium constant expression for a reversible reaction, the (products, reactants) ____________ are located in the numerator or upper part of the fraction and the (products, reactants) ________________ are located in the denominator or lower part of the fraction.Answer: products; reactantsThe standard equation, then, for Keqis as follows.Write the equilibrium constant expression for the following reversible reaction.Keq= ___________________Answer:(Since there are two products, they should be placed in the upper part of the fraction. The one reactant belongs in the lower part of the fraction.)
eBook - PDF- Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
Chapter Six GENERAL CONCEPTS OF CHEMICAL EQUILIBRIUM “The worst form of inequality is to try to make unequal things equal.” — Aristotle Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ● The equilibrium constant (key equations: 6.12, 6.15), pp. 194 ● Calculation of equilibrium concentrations, p. 195 ● Using Excel Goal Seek to solve one-variable equations, p. 197 ● The systematic approach to equilibrium calculations: mass balance and charge balance equations, p. 204 ● Activity and activity coefficients (key equation: 6.19), p. 211 ● Thermodynamic equilibrium constants (key equation: 6.23), p. 217 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 general 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 potentiometric calculations (Chapter 13). 6.1 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 equilibrium as the condition when the rates of the forward and reverse reactions are equal.
eBook - ePubRapid Review of Chemistry for the Life Sciences and Engineering
With Select Applications
- Armen S. Casparian, Gergely Sirokman, Ann Omollo(Authors)
- 2021(Publication Date)
- CRC Press(Publisher)
4 Chemical EquilibriumDOI: 10.1201/9781003092759-44.1 Basic Concept
Reaction equations describe substances, called reactants, which when put together react and produce other different substances, called products. It may appear as if only products remain after the reaction is finished. In reality, many reactions do not go to completion, even if the reactants are present in stoichiometric ratios or amounts. Rather, they reach a condition known as equilibrium, denoted by double, reversible arrows in the reaction equation. Equilibrium means that there is a balance between the reactant side and the product side, or simply between the reactants and the products, and that the reaction is reversible. The chemistry of many air pollutants falls and many bodily functions under the heading of equilibrium reactions. The equilibrium condition is dynamic, not static, allowing microscopic changes in reactant and product concentrations to take place, such that no net change in reactant or product concentrations occurs, provided that no external stresses are applied. At any given time, all species in the reaction equation—reactants and products—are present at equilibrium in varying amounts. The relationship among these varying amounts can be described by a mathematical formula known as the equilibrium constant expression or simply the equilibrium expression. The equilibrium expression is set equal to an equilibrium constant symbolized by Kc .An equilibrium reaction can be generally represented as follows:a A + b B ⇋ g G + h H(4.1)The Kc expression can then be expressed as follows:(4.2)K c=[ G ]g[ H ]h[ A ]a[ B ]bwhere a, b, g, and h represent the stoichiometric coefficients in the balanced reaction, and the brackets [ ] indicate molar concentrations. The simplest interpretation of Kc is that it is a measure of the extent to which a reaction goes toward completion, i.e., a reaction where the product side is favored. The meaning of Kc
eBook - PDF- William R. Robinson, Edward J. Neth, Paul Flowers, Klaus Theopold, Richard Langley(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
It may help if we keep in mind that Q c = K c (at equilibrium) in all of these situations and that there are only three basic types of calculations: 1. Calculation of an equilibrium constant. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated. 2. Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated. 3. Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated. A similar list could be generated using Q P , K P , and partial pressure. We will look at solving each of these cases in sequence. Calculation of an Equilibrium Constant Since the law of mass action is the only equation we have to describe the relationship between K c and the concentrations of reactants and products, any problem that requires us to solve for K c must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for K c , as it will be the only unknown. Example 13.6 showed us how to determine the equilibrium constant of a reaction if we know the concentrations Chapter 13 | Fundamental Equilibrium Concepts 697 of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE chart—for Initial, Change, and Equilibrium–will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question.
eBook - ePub- Patrick E. McMahon, Rosemary McMahon, Bohdan Khomtchouk(Authors)
- 2019(Publication Date)
- CRC Press(Publisher)
constant, K (or K (subscript) to indicate a specific type of reaction). One conceptual view of equilibrium is described through the general rate expressions for the forward and reverse reactions (a more exact derivation can be developed using the techniques described in Chapter 24): k (forward) [A] x [B] y = k (reverse) [C] x [D] y ; this equation can be rewritten as: k (forward) k (reverse) = [C] x [D] y [A] x [B] y where k (forward) k (reverse) is related to the equilibrium constant K. An equation that expresses the correct equilibrium concentration ratio equal to the equilibrium constant (either the symbol or its actual numerical value) is termed the equilibrium expression: K (forward reaction) = [C] c [D ] d / [A] a [B] b The equilibrium constant, K, has a specific numerical value determined by the concentrations of each molecule in the balanced equation under specific. conditions. Example: Use the general definitions to write the complete equilibrium expression for the following reversible reaction; the symbol, K, is used since no numerical value is given. CH 4 + 2 Cl 2 → CH 2 Cl 2 + 2 HCl ← K = [C H 2 C l 2 ] [HCl ] 2 [C H 4 ] [C l 2 ] 2 Example: Assume that under certain conditions, the concentrations at equilibrium of each species in the previous equation are: [CH 2 Cl 2 ] = 0.850 M; [HCl] = 0.145 M [CH 4 ] = 0.0522 M; [Cl 2 ] = 0.166 M. Calculate the numerical value of K. K= [CH 2 Cl 2 ] [HCl] 2 [CH 4 ] [Cl 2 ] 2 = (0.850 M) (0.145 M) 2 (0.0522 M) (0.166 M) 2 K = 1 2. 4 The expression for the reverse reaction could be stated as: K (r e v e r s e r e a c t i o n) = [A] a [B ] b [C] c [D ] d The symmetry of the expressions for the forward and reverse reactions shows
eBook - ePub- Jeffrey Gaffney, Nancy Marley(Authors)
- 2017(Publication Date)
- Elsevier(Publisher)
7.2, when the reactants of a reversible reaction are first combined, the forward reaction occurs rapidly increasing the concentrations of the products and decreasing the concentrations of the reactants. As the reaction proceeds and the concentration of products increases, the forward reaction slows and the reverse reaction begins to occur. After some time, an equilibrium is reached where the concentrations of reactants and products do not change with time and the forward and reverse reaction rates are equal. This is known as a dynamic equilibrium because, although forward and reverse reactions still occur, the rate of the forward reaction equals the rate of the reverse reaction. So, the ratio of concentrations of the products to the concentrations of the reactants remains constant, although the reaction is constantly proceeding in both directions. Fig. 7.2 A reversible chemical reaction reaches dynamic equilibrium when the rate of the forward and reverse reactions are equal (left) and the concentrations of reactants and products do not change with time (right). 7.2 The Equilibrium Constant Since the concentrations of reactants and products of a reversible reaction at equilibrium are constant, an expression can be written for an equilibrium constant (K eq) that describes the concentrations of the products and reactants at equilibrium. For the generic reversible reaction (3) at equilibrium, the equilibrium constant expression is; K eq = C c D d A a B b (4) As with the acid ionization constants described in Section 5.2, the equilibrium constant for a reversible reaction is expressed as the ratio of the equilibrium concentrations of the products in the numerator to the equilibrium concentrations of the reactants in the denominator. The superscript letters are the stoichiometric coefficients of the reactants and products in the balanced chemical 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 - 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.
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.
- Morris Hein, Susan Arena, Cary Willard(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
253 CHAPTER 16 CHEMICAL EQUILIBRIUM SOLUTIONS TO REVIEW QUESTIONS 1. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. 2. The rate of a reaction increases when the concentration of one of the reactants increases. The increase in concentration causes the number of collisions between the reactants to increase. The rate of a reaction, being proportional to the frequency of such collisions, as a result, will increase. 3. An increase in temperature causes the rate of reaction to increase, because it increases the velocity of the molecules. Faster moving molecules increase the number and effectiveness of the collisions between molecules resulting in an increase in the rate of the reaction. 4. At 25°C both tubes would appear the same. 5. The reaction is endothermic because the increased temperature increases the concentration of product (NO 2 ) present at equilibrium. 6. If pure HI is placed in a vessel at 700 K, some of it will decompose. Since the reaction is reversible ( ) + 2 2 H I 2 HI , HI molecules will react to produce H 2 and I 2 . 7. Acids stronger than acetic acid are: benzoic, cyanic, formic, hydrofluoric, and nitrous acids (all equilibrium constants are greater than the equilibrium constant for acetic acid). Acids weaker than acetic acid are: carbolic, hydrocyanic, and hypochlorous acids (all have equilibrium constants smaller than the equilibrium constant for acetic acid). All have one ionizable hydrogen atom. 8. In an endothermic process heat is absorbed or used by the system so it should be placed on the reactant side of a chemical equation. In an exothermic process heat is given off by the system so it belongs on the product side of a chemical equation. 9. A catalyst speeds up the rate of a reaction by lowering the activation energy. A catalyst is not used up in the reaction. 10. It is very important to specify the temperature because equilibrium constants change with changing temperatures.
eBook - PDF- Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
Adjusting the pH is a common way of shifting the equilibrium. The common ion effect can be used to make analytical reactions more favorable or quantitative. The adjustment of acidity, for example, is frequently used to shift equi- libria. Titrations with potassium dichromate, for example, are favored in acid solution, since protons are consumed in the reaction. Titrations with iodine, a weak oxidizing agent, are usually done in slightly alkaline solution to shift the equilibrium toward completion of the reaction, for example, in titrating arsenic(III): H 3 AsO 3 + I 2 + H 2 O ⇌ H 3 AsO 4 + 2I - + 2H + 5.13 Systematic Approach to Equilibrium Calculations—How to Solve Any Equilibrium Problem Now that some familiarity has been gained with equilibrium problems, we will con- sider a systematic approach for calculating equilibrium concentrations that will work with all equilibria, no matter how complex. It consists of identifying the unknown con- centrations involved and writing a set of simultaneous equations equal to the number of unknowns. Simplifying assumptions are made with respect to relative concentrations Pdf _Folio:1 87 188 CHAPTER 5 GENERAL CONCEPTS OF CHEMICAL EQUILIBRIUM of species (not unlike the approach we have already taken) to shorten the solving of the equations. This approach involves writing expressions for mass balance of species and one for charge balance of species as part of our equations. We will first describe how to arrive at these expressions. Mass Balance Equations The principle of mass balance is based on the law of mass conservation, and it states that the number of atoms of an element remains constant in chemical reactions because no atoms are produced or destroyed. The principle is expressed mathematically by equating the concentrations, usually in molarities. The equations for all the pertinent chemical equilibria are written, from which appropriate relations between species con- centrations are written.
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