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Ch. 18 Chemical Equilibrium

Ch. 18 Chemical Equilibrium. Chapter 18. 18.1 The Nature of Chemical Equilibrium. List two everyday processes that can easily be reversed and two that cannot. The freezing of water and the melting of ice can be reversed The cooking of an egg or the lighting of a match cannot be reversed.

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Ch. 18 Chemical Equilibrium

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  1. Ch. 18 Chemical Equilibrium

  2. Chapter 18 18.1 The Nature of Chemical Equilibrium • List two everyday processes that can easily be reversed and two that cannot. • The freezing of water and the melting of ice can be reversed • The cooking of an egg or the lighting of a match cannot be reversed. • Reactions that “go to completion” refer to a reaction in which all the reactants are converted to products

  3. Chapter 18 Reversible Reactions • But theoretically, every reaction can proceed in two directions, forward and reverse. • Essentially all chemical reactions are considered to be reversible under suitable conditions. • A chemical reaction in which the products can react to re-form the reactants is called areversible reaction.

  4. Chapter 18 • Chemical equilibrium is established when the rate of its forward reaction equals the rate of its reverse reaction • When this happens the amounts of products (concentrations) and reactants remain constant. • Both reactions continue, but there is no net change in their concentration

  5. Chapter 18 • products of the forward reaction favored, lies to the right • products of the reverse reaction favored,lies to the left • Neither reaction is favored

  6. Chapter 18 • After equilibrium is reached, the individual concentrations of A, B, C, and D undergo no further change if conditions remain the same. • i.e. : the ratio of their concentrations remains constant. • The equilibrium constant is designated by the letter K. • The expression for K =

  7. Chapter 18 • The Equilibrium Constant Expression, K, is the ratio of the product of the concentrations of substances formed at equilibrium to the product of the concentrations of reacting substances. Each concentration is raised to a power equal to the coefficient of that substance in the chemical equation. • K is dependent on temperature and always determined experimentally • Only the concentrations of substances that can actually change are included in K. • So pure solids and liquids are omitted because their concentrations cannot change.

  8. Chapter 18 What would be true if K = 1? • Concentration of products = concentration of reactants What would be true if K is greater than 1? • Large K values indicate that the products are favored (forward reaction) What would be true if K is less than 1? • Small K values indicate the reactants are favored (reverse reaction)

  9. What does knowing the value of Keq help us predict? • Whether the forward of the reverse reaction is favored • Therefore, tell us if the reactants or products will be in higher concentrations when equilibrium is established *Does not tell us how quickly the reaction reaches the equilibrium

  10. Examples of Equilibrium Constant Expressions • Make sure you always balance an equation before writing an equilibrium constant expression

  11. Examples of Equilibrium Constant Expressions

  12. Chapter 18 Sample Problem A An equilibrium mixture of N2, O2 , and NO gases at 1500 K is determined to consist of 6.4  10–3 mol/L of N2, 1.7  10–3 mol/L of O2, and 1.1  10–5 mol/L of NO. What is the equilibrium constant for the system at this temperature?

  13. Section 1 The Nature of Chemical Equilibrium Chapter 18 The Equilibrium Expression, continued Sample Problem A Solution Given:[N2] = 6.4  10–3 mol/L [O2] = 1.7  10–3 mol/L [NO] = 1.1  10–5 mol/L Unknown:K Solution: The balanced chemical equation is The chemical equilibrium expression is

  14. Chapter 18 Section 18.2 Le Châtelier’s principle • Imagine children playing on a seesaw. • Five boys are sitting on one side and five girls on the other, and the seesaw is just balanced. • Then, one girl gets off, and the system is no longer at equilibrium. • One way to get the seesaw in balance again is for one of the boys to move toward the girls’side.

  15. Chapter 18 Section 2 Shifting Equilibrium • Le Châtelier’s principle states that if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that tends to relieve the stress. • This principle is true for all dynamic equilibria, chemical as well as physical. • Changes in pressure, concentration, and temperature illustrate Le Châtelier’s principle because they alter the equilibrium position and thereby change the relative amounts of reactants and products

  16. Chapter 18 Predicting the Direction of Shift: Pressure • Change in pressure affects only equilibrium systems in which gases are involved. • An increase in pressure is an applied stress. • The system can reduce the total pressure by reducing the number of molecules. • Therefore, increase in pressure favors the reaction that produces fewer gas molecules

  17. Chapter 18 • Ex: Haber process for the synthesis of ammonia • When pressure is applied, the equilibrium will shift to the right, and produce more NH3. 4 molecules of gas 2 molecules of gas • A reduction in the total number of molecules leads to a decrease in pressure. Note: changes in pressure do not affect the value of the equilibrium constant

  18. Chapter 18 Changes in Concentration • An increase in the concentration of a reactant is a stress on the equilibrium system. • To relieve the stress, some of the added A reacts with B to form products C and D. • Increase in concentration of one substance shifts the equilibrium to the opposite side (i.e. opposite side favored)

  19. Chapter 18 • Note: changes in concentration have no effect on the value of the equilibrium constant. • Such changes have an equal effect on the numerator and the denominator of the chemical equilibrium expression.

  20. Chapter 18 Changes in Temperature • Reversible reactions are exothermic in one direction and endothermic in the other. • The addition of energy in the form of heat shifts the equilibrium so that energy is absorbed. • An increase in temperature favors the endothermic reaction • A decreased in temperature (removal of energy) favors the exothermic reaction

  21. Chapter 18 • A rise in temperature increases the rate of any reaction. • In an equilibrium system, the rates of the opposing reactions are raised unequally. **The value of the equilibrium constant for a given system is affected by the temperature.

  22. Section 2 Shifting Equilibrium Chapter 18 • Haber Process: exothermic • A rise in temperature increases the rate of any reaction, but the rates of the opposing reactions are raised unequally. • The value of the equilibrium constant for a given system is affected by the temperature. • A high temperature favors the decomposition of ammonia, the endothermic reaction. • At low temperatures, the forward reaction is too slow to be commercially useful.

  23. Temperature Changes Affect an Equilibrium System

  24. Chapter 18 Reactions That Go to Completion • Some reactions involving compounds formed by the chemical interaction of ions in solutions appear to go to completion in the sense that the ions are almost completely removed from solution.

  25. Chapter 18 3 Examples of Reactions that go to Completion 1. Formation of a Gas H2CO3(aq) H2O(l) + CO2(g) • Reaction goes practically to completion because one of the products, CO2, escapes as a gas if the container is open to the air.

  26. Chapter 18 2. Formation of a Precipitate • If chemically equivalent amounts of the two solutes are mixed, almost all of the Ag+ ions and Cl− ions combine and separate from the solution as a precipitate of AgCl. • AgCl is only very sparingly soluble in water. • The reaction thus effectively goes to completion because an essentially insoluble product is formed.

  27. Chapter 18 3. Formation of a Slightly Ionized Product • Neutralization reactions between H3O+ ions from aqueous acids and OH− ions from aqueous bases result in the formation of water molecules, which are only slightly ionized. • Hydronium ions and hydroxide ions are almost entirely removed from the solution. • The reaction effectively runs to completion because the product is only slightly ionized.

  28. Chapter 18 Common-Ion Effect • The phenomenon in which the addition of an ion common to two solutes brings about precipitation or reduced ionization of a reactant • Ex: hydrogen chloride gas is bubbled into a saturated solution of sodium chloride.

  29. Chapter 18 Common-Ion Effect, continued • As the hydrogen chloride dissolves in sufficient quantity, it increases the concentration of Cl− ions in the solution, which is a stress on the equilibrium system. • The system can compensate by forming some solid NaCl. The NaCl precipitates out, relieving the stress of added chloride (reactions shifts to the left) • The new equilibrium has a greater concentration of Cl− ions but a decreased concentration of Na+ ions.

  30. Particle Model for the Common-Ion Effect Section 2 Shifting Equilibrium Chapter 18

  31. Chapter 18 Solubility Product • Soluble ionic compounds dissolve in water until they are at equilibrium with their ions • The solubility product constant, Ksp, of a substance is the product of the molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation.

  32. Section 4 Solubility Equilibrium Chapter 18 Solubility Product, continued • For dissolution of the ionic compound AaBb (A cation, B anion), the solubility-production expression is AaBb (s) → aA + bB • The equilibrium expression is written without including the solid species (or water) • The numerical value of Ksphelps us determine how soluble a compound is in water Ksp = [A]a [B]b

  33. Section 4 Solubility Equilibrium Chapter 18 Solubility Product, continued • Compounds with large Ksp values are more soluble • Compounds with smaller Ksp values are less soluble • The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent. • It has an infinite number of possible values at a given temperature and is dependent on other conditions, such as the presence of a common ion.

  34. Section 4 Solubility Equilibrium Chapter 18 Solubility Product, continued • For a saturated solution of CaF2, the equilibrium equation is • The expression for the solubility product constant is Ksp = [Ca] [F]2 • The solubility of CaF2 is 8.6  10−3 g/100 g of water at 25°C. Expressed in moles per liter this concentration becomes 1.1  10−3 mol/L.

  35. Chapter 18 Section 3 Equilibria of Acids, Bases, and Salts Ionization Constant of Acids & Bases • When we write the equiblium constant expression for the ionization of an acid/base we call them • Ka:acid ionization constant. • Kb: base ionization constant. • The acid ionization constant, Ka , is constant for a specified temperature • Weak acids, like CH3COOH, remainly largely unionized as will have small Ka values. • Strong acids, like HNO3, ionize completely and will have larger Ka values.

  36. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 Ionization Constant of Water, Kw • The self-ionization of water is an equilibrium reaction. • Equilibrium is established with a very low concentration of H3O+ and OH− ions. Kw=[H3O+][OH–] = 1.0  10-14

  37. Determining Ksp for Reactions at Chemical Equilibrium Section 4 Solubility Equilibrium Chapter 18

  38. Section 4 Solubility Equilibrium Chapter 18 Solubility Product, continued • CaF2 dissociates to yield twice as many F− ions as Ca2+ ions. [Ca2+] = 1.1  10−3 mol/L [F− ] = 2.2  10−3 mol/L Ksp = 5.3  10-9 • Calculations of Kspordinarily should be limited to two significant figures.

  39. Solubility Product Constants at 25°C Section 4 Solubility Equilibrium Chapter 18

  40. Section 4 Solubility Equilibrium Chapter 18 Solubility Product, continued Sample Problem B Calculate the solubility product constant, Ksp ,for copper(I) chloride, CuCl, given that the solubility of this compound at 25°C is 1.08  10–2 g/100. g H2O.

  41. Section 4 Solubility Equilibrium Chapter 18 Solubility Product, continued Sample Problem B Solution Given:solubility of CuCl = 1.08  10−2 g CuCl/100. g H2O Unknown:Ksp Solution: Ksp=[Cu+][Cl–] [Cu+] = [Cl–] = solubility in mol/L

  42. Section 4 Solubility Equilibrium Chapter 18 Solubility Product, continued Sample Problem B Solution, continued 1.09  10-3 mol/L CuCl [Cu+] = [Cl–]=1.09  10-3 mol/L Ksp= (1.09  10-3)(1.09  10-3) = 1.19  10-6

  43. Section 4 Solubility Equilibrium Chapter 18 Calculating Solubilities • The solubility product constant can be used to determine the solubility of a sparingly soluble salt. • How many moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C? • The molar solubility of BaCO3 is 7.1  10−5 mol/L.

  44. Section 4 Solubility Equilibrium Chapter 18 Calculating Solubilities, continued Sample Problem C Calculate the solubility of silver bromide, AgBr, in mol/L, using the Kspvalue for this compound.

  45. Section 4 Solubility Equilibrium Chapter 18 Calculating Solubilities, continued Sample Problem C Solution Given:Ksp = 5.0 10−13 Unknown:solubility of AgBr Solution: [Ag+] = [Br−], so let [Ag+] = x and [Br−] = x

  46. Section 4 Solubility Equilibrium Chapter 18 Limitations on the Use of Ksp • The solubility product principle can be very useful when applied to solutions of sparingly soluble substances. • It cannot be applied very successfully to solutions of moderately soluble or very soluble substances. • The positive and negative ions attract each other, and this attraction becomes appreciable when the ions are close together. • Sometimes it is necessary to consider two equilibria simultaneously.

  47. Equilibrium Calculations Section 4 Solubility Equilibrium Chapter 18

  48. Multiple Choice 1. A chemical reaction is in equilibrium when A. forward and reverse reactions have ceased. B. the equilibrium constant equals 1. C. forward and reverse reaction rates are equal. D. No reactants remain. Standardized Test Preparation Chapter 18

  49. 1. A chemical reaction is in equilibrium when A. forward and reverse reactions have ceased. B. the equilibrium constant equals 1. C. forward and reverse reaction rates are equal. D. No reactants remain. Standardized Test Preparation Chapter 18 Multiple Choice

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