Equilibrium Basic Concepts

1. Equilibrium Basic Concepts • Reversiblereactions do not go to completion. • They can occur in either direction • Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. • A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate. • Chemical equilibria are dynamic equilibria. • Molecules are continually reacting, even though the overall composition of the reaction mixture does not change. • Double arrow (⇋) indicates that both the forward and reverse reactions are occurring simultaneously and is read “is in equilibrium with” • At equilibrium, the composition of the system no longer changes with time • Composition of an equilibrium mixture is independent of the direction from which equilibrium is approached

2. Equilibrium Basic Concepts • One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction. • Equilibrium state is achieved when the rate of the forward reaction equals the rate of the reverse reaction: Ratef = Rater • 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. The rates of the forward and reverse reactions can be represented as: • • The ratio of the rate constants yields a new constant, the equilibrium constant (K), a unitless quantity and is defined as K = kf /kr. • Equilibrium constants are uniless because they actually involve a thermodynamic quantity called activity. Activities are directly related to molarity • • This implies that equilibrium mixture is determined by the magnitudes of the rate constants for the forward and reverse reactions.

3. Equilibrium Basic Concepts Product Favored, ∆G˚ negative Systems can reach equilibrium when reactants have NOT converted completely to products. In the case ∆Grxn is < ∆Gorxn state with both reactants and products present is MORE STABLE than complete conversion.

4. Developing an Equilibrium Constant Expression • The ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced equation) is always a constant under a given set of conditions: [C]c[D]d [A]a[B]b is called the equilibrium equation. • This relationship is known as the law of mass action where – Kc is the equilibrium constant for the reaction – Right side of the equation is called the equilibrium constant expression • Relationship is true for any pair of opposing reactions regardless of the mechanism of the reaction or of the number of steps in the mechanism. Kc =

5. The Equilibrium Constant Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.

6. One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction. Equil []’s 0.028 M 0.172 M 0.086 M The Equilibrium Constant

7. The Equilibrium Constant Products favored 103 > K > 10-3 Reactants favored • – Values of K greater than 103 indicate a strong tendency for reactants to form products, so equilibrium lies to the right, favoring the formation of products (kf>>kr) • – Values of K less than 10–3 indicate that the ratio of products to reactants at equilibrium is very small; reactants do not tend to form products readily, and equilibrium lies to the left, favoring the formation of reactants (kf<<kr) • – Values of K between 103 and 10–3 are not very large or small, so there is no strong tendency to form either products or reactants; at equilibrium, there are significant amounts of both products and reactants (kf  kr)

8. The Equilibrium Constant • Equilibrium can be approached from either direction in a chemical reaction, so the equilibrium constant expression and the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. • When a reaction is written in the reverse direction, cC + dD ⇋ aA + bB K and the equilibrium constant expression are inverted: K´= [A]a [B]b [C]c [D]d so K´ = 1/K

9. The value of Kc depends upon how the balanced equation is written. From the prior examples we have: PCl5 PCl3 + Cl2 Variation of Kc with the Form of the Balanced Equation Equil []’s 0.028 M 0.172 M 0.086 M Kc = [PCl3][Cl2]/[PCl5] = 0.53 PCl3 + Cl2 PCl5 Equil. []’s 0.172 M 0.086 M 0.028 M Kc’= [PCl5]/[PCl3][Cl2] = 1.89 2PCl5 2PCl3 + 2Cl2 Equil []’s 0.028 M 0.172 M 0.086 M Kc ’’= [PCl3] 2[Cl2] 2/[PCl5] 2 = 0.28 2PCl3 + 2Cl2 2PCl5 Equil. []’s 0.172 M 0.086 M 0.028 M Kc ‘“= [PCl5]2/[PCl3]2[Cl2]2 = 3.56

10. The Equilibrium Constant • At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were introduced in an evacuated 1.00-liter container. After equilibrium was established the equilibrium concentrations were determined to be 0.20 mole of NH3, 0.70 moles N2, and 0.60 moles H2. Calculate Kc for the reaction.

11. Partial Pressure and Kc • For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations. • For gases, the pressure is proportional to the concentration. • We can see this by looking at the ideal gas law. • PV = nRT • P = nRT/V • n/V = M • P= MRT and M = P/RT = (PC)c(PD)d (PA)a (PB)b {[C](RT)}c{[D](RT)}d {[A](RT)}a{[B](RT)}b [C]c(RT)c[D]d(RT)d [A]a(RT)a[B]b(RT)b [C]c[D]d (RT)c(RT)d [A]a[B]b (RT)a(RT)b Kp = x = = (nprod-nreact ) = Kc{(RT)(c+d)-(a+b)} = Kc(RT) Dn = Kc(RT)

12. Partial Pressures and the Equilibrium Constant • Consider this system at equilibrium at 5000C. 2SO2(g) + O2(g) 2SO3(g) • Kp is unitless. • Partial pressures are expressed in atmospheres or mmHg, so the molar concentration of a gas and its partial pressure do not have the same numerical value but are related by the ideal gas constant R and the temperature Kp = Kc(RT)n Kc = Kp(RT)-n where K is the equilibrium constant expressed in units of concentration and n is the difference between the number of moles of gaseous products and gaseous reactants; temperature is expressed in Kelvins. • If n = 0, Kp = Kc

13. Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel, (a) How many moles of I2 remain unreacted at equilibrium? Relationship Between Kp and Kc (b) What are the equilibrium partial pressures of H2, I2 and HI? (c) What is the total pressure in the reaction vessel?

14. Relationship Between Kp and Kc • Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?

15. Homogeneous and Heterogeneous Equilibria • Homogeneous equilibrium – When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid – Concentrations of the reactants and products can vary over a wide range • Heterogeneous equilibrium – A system whose reactants, products, or both are in more than one phase – An example is the reaction of a gas with a solid or liquid • Molar concentrations of pure liquids and solids do not vary with temperature, so they are treated as constants, this simplifies their equilibrium constant expressions

16. Heterogeneous equilibria have more than one phase present. For example, a gas and a solid or a liquid and a gas. How does the equilibrium constant differ for heterogeneous equilibria? Pure solids and liquids have activities of unity. Solvents in very dilute solutions have activities that are essentially unity. The Kc and Kp for the reaction shown above are: Heterogeneous Equlibria

17. Heterogeneous Equlibria

18. Heterogeneous Equlibria • What are Kc and Kp for this reaction?

19. Heterogeneous Equlibria • What are Kc and Kp for this reaction?

20. Equilibrium Constant Expressions for the Sums of Reactions An aside: Similar to Hess’ Law we can add several reactions together to get to an overall reaction not listed in a table. To determine K for the sum of the reactions simply multiply all the values for each intermediate step to get the K for the overall reaction.

21. Solving Equilibrium Problems • Two fundamental kinds of equilibrium problems 1. Those in which the concentrations of the reactants and products at equilibrium are given and the equilibrium constant for the reaction needs to be calculated 2. Those in which the equilibrium constant and the initial concentrations of reactants are known and the concentration of one or more substances at equilibrium needs to be calculated

22. Calculating an Equilibrium Constant from Equilibrium Concentrations • can be calculated when equilibrium concentrations, or partial pressures are substituted into the equilibrium constant expression for the reaction. • When equilibrium concentrations are not given the equilibrium concentrations can be obtained from the initial concentrations of the reactants and the balanced equation for the reaction, as long as the equilibrium concentration of one of the substances is known. • Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.

23. Uses of the Equilibrium Constant, Kc • The equilibrium constant, Kc, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

24. Uses of the Equilibrium Constant, Kc • The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

25. The Reaction Quotient • The mass action expression or reaction quotient has the symbol Q. • Q has the same form as Kc • The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values. • Why do we need another “equilibrium constant” that does not use equilibrium concentrations? • Q will help us predict how the equilibrium will respond to an applied stress. • Q may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and products, regardless of whether the system is at equilibrium • To make this prediction we compare Q with Kc.

26. The Reaction Quotient (Q) • Comparing the magnitudes of Q and K allows the determination of whether a reaction mixture is already at equilibrium and, if it is not, how to predict whether its composition will change with time (whether the reaction will proceed to the right or to the left) 1. If Q = K, the system is at equilibrium, no further change in the composition of the system will occur unless the conditions are changed 2. If Q < K, then the ratio of the concentrations of products to the concentration of reactants is less than the ratio at equilibrium; reaction will proceed to the right, forming products at the expense of reactants 3. If Q > K, then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium; reaction will proceed to the left, forming reactants at the expense of products

27. The Reaction Quotient • The equilibrium constant for the following reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

28. Le Châtelier’s Principle • When a system at equilibrium is perturbed in some way, the effects of the perturbation can be predicted qualitatively using Le Châtelier’sprinciple. •Thisprinciple states that if a stress is applied to a system at equilibrium, the composition of the system will change to relieve the applied stress. • Stress occurs when any change in the system affects the magnitude of Q or K. • •Three types of stresses can change the composition of an equilibrium mixture: • A change in the concentrations (or partial pressures) of the components by the addition or removal of reactants or products • A change in the total pressure or volume • A change in the temperature of the system

29. Disturbing a System at Equlibrium: Predictions • Changes in Concentration of Reactants and/or Products • Also true for changes in pressure for reactions involving gases. • • An equilibrium is disturbed by adding or removing a reactant or product • 1. Stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance • a. Add reactant—reaction shifts right toward product • b. Add product—reaction shifts left toward reactant • 2. Stress of removing reactant or product is relieved by reaction in the direction that replenishes the removed substance • a. Remove reactant—reaction shifts left • b. Remove product—reaction shifts right

30. Disturbing a System at Equlibrium: Predictions • Changes in Volume (and pressure for reactions involving gases) • Increase in pressure (due to decrease in volume) results in a reaction in the direction of a fewer number of moles of gas • Decrease in pressure (due to increase in volume) results in a reaction in the direction of a greater number of moles of gas • Decrease volume—molarity increases • If reactant side has more moles of gas • Increase in denominator is greater than increase in numerator and Qc< Kc • To return to equilibrium, Qcmust increase; the numerator of the Qc expression must increase and denominator must decrease—it shifts toward fewer moles of gas (reactants to products) 3. If product side has more moles of gas • Increase in numerator is greater than increase in denominator and Qc > Kc • b. To return to equilibrium, Qcmust decrease; the denominator of the Qcexpression must decrease and the numerator must increase—it shifts toward fewer moles of gas (products to reactants) • Predict what will happen if the volume of this system at equilibrium is changed by changing the pressure at constant temperature:

31. Disturbing a System at Equlibrium: Predictions • Changing the Reaction Temperature • Changes in temperature can change the value of K without affecting Q (Q K) • • Predictions on the response of a system to a change in requires knowledge regarding Hrxn. Remember: • 1. Exothermic (heat is lost by the system, H<0): reactants ⇋ products + heat • 2. Endothermic (heat is gained by the system, H>0): reactants + heat ⇋ products • • Le Châtelier’s principle predicts • 1. that an exothermic reaction will shift to the left (toward reactants) if the temperature of the system is increased (heat is added); • 2. that an endothermic reaction will shift to the right (toward the products) if the temperature of the system is increased; • 3. that if Hrxn= 0, then a change in temperature has no affect on composition. • Increasing the temperature increases the magnitude of the equilibrium constant for an endothermic reaction • Increasing the temperature decreases the equilibrium constant for an exothermic reaction

32. Introduction of a Catalyst Catalysts decrease the activation energy of both the forward and reverse reaction equally. Catalysts do not affect the position of equilibrium. The concentrations of the products and reactants will be the same whether a catalyst is introduced or not. Equilibrium will be established faster with a catalyst. Disturbing a System at Equlibrium: Predictions

33. Disturbing a System at Equlibrium: Predictions • How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?

34. Disturbing a System at Equlibrium: Predictions • How will an increase in temperature affect each of the following reactions?

35. A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2. Calculate the equilibrium constant. Disturbing a System at Equlibrium: Predictions

36. Disturbing a System at Equlibrium: Predictions • An additional 0.80 mole of Cl2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established.

37. Controlling the Products of Reactions • One of the primary goals of modern chemistry is to control the identity and quantity of the products of chemical reactions. • Two approaches 1. To get a high yield of a desired compound, make the rate of the desired reaction much faster than the rate of any of the other possible reactions that might occur in the system; altering reaction conditions to control reaction rates, thereby obtaining a single product or set of products is called kinetic control. 2. Thermodynamic control—consists of adjusting conditions so that at equilibrium only the desired products are present in significant quantities.

38. Disturbing a System at Equlibrium: Predictions • Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?

39. An Application of Equilibrium: The Haber Process • The Haber process is used for the commercial production of ammonia. • This is an enormous industrial process in the US and many other countries. • Ammonia is the starting material for fertilizer production. Fritz Haber 1868-1934 Nobel Prize, 1918 Carl Bosch 1874-1940 Nobel Prize, 1931

40. Relationship Between Gorxnand the Equilibrium Constant • The relationship for K at conditions other than thermodynamic standard state conditionsis derived from this equation. When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

41. Acids and bases Autoionization reaction of liquid water pH, pOH, and pKw conjugate acid-base pairs acid or base strength and the magnitude ofKa, Kb, pKa, and pKb leveling effect To be able to predict whether reactants or products are favored in an acid-base equilibrium usemolecular structure and acid and base strengths use Ka and Kb values to calculate the percent ionization and pH of a solution of an acid or a base calculate the pH at any point in an acid-base titration common ion effects and the position of an acid-base equilibrium how a buffer works and how to use the Henderson-Hasselbalch equation to calculate the pH of a buffer

42. Acids and bases • There are three classes of strong electrolytes. • Strong Water Soluble Acids • Remember the list of strong acids from Chapter 4. • Strong Water Soluble Bases • The entire list of these bases was also introduced in Chapter 4. • Most Water Soluble Salts • The solubility guidelines from Chapter 4 will help you remember these salts. • Weak acids and bases ionize or dissociate partially, much less than 100%, and is often less than 10%.

43. Table of Common Ions Common Negative Ions (Anions)

44. Table of Common Ions Common Positive Ions (Cations)

45. Water Solubility of Ionic Compounds If one ion from the “Soluble Compound” list is present in a compound, the compound is water soluble.

46. Acids and bases • Most salts of strong or weak electrolytes can dissolve in water to produce a neutral, basic, or acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (A–) or the conjugate acid of a weak base as the cation (BH+), or possibly both. •Salts that contain small, highly charged metal ions produce acidic solutions in H2O. • The most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. •The reaction of a salt with water to produce an acidic or basic solution is called a hydrolysis reaction, which is just an acid-base reaction in which the acid is a cation or the base is an anion.

47. Acids and bases • Two species that differ by only a proton constitute a conjugate acid-base pair. • 1. Conjugate base has one less proton than its acid; A– is the conjugate base of HA • 2. Conjugate acid has one more proton than its base; BH+ is the conjugate acid of B • 3. Conjugates are weaker than strong parents and stronger than weak parents. • 4. All acid-base reactions involve two conjugate acid-basepairs. • HCl (aq)+ H2O (l) H3O+(aq)+ Cl–(aq) parent acid parent base conjugate acid conjugate base Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion, H3O+. Substances that can behave as both an acid and a base are said to be amphoteric.

48. Acids and bases can be defined in different ways: • 1. Arrhenius definition: An acid is a substance that dissociates in water to produce H+ ions (protons), and a base is a substance that dissociates in water to produce OH– ions (hydroxide); an acid-base reaction involves the reaction of a proton with the hydroxide ion to form water. • – Three limitations • 1. Definition applied only to substances in aqueous solutions. • 2. Definition restricted to substances that produce H+ and OH– ions • 3. Definition does not explain why some compounds containing hydrogen such as CH4 dissolve in water and do not give acidic solutions • 2. Brønsted–Lowry definition: An acid is any substance that can donate a proton, and a base is any substance that can accept a proton; acid-base reactions involve two conjugate acid-base pairs and the transfer of a proton from one substance (the acid) to another (the base). Not restricted to aqueous solutions, expanding to include other solvent systems and acid-base reactions for gases and solids. Not restricted to bases that only produce OH– ions . Acids still restricted to substances that produce H+ ions. Limitation 3 not dealt with. • 3. Lewis definition: A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor .

49. K2 KH K1 CO2+3H2O H2CO3+2H2O HCO3-+H3O++H2O CO32-+2H3O+ Acids and bases CO2 + H2O H2CO3 H2CO3 + H2O HCO3- + H+ HCO3- + H2O CO32- + H+ Baking Soda NaHCO3 Soda Pop CO2 Atmosphere .8317 5.61×10−11 2.5×10−4 CO2 + CaCO3 + H2O 2HCO3- + Ca2- Biological Calcification (not a reversible reaction) unneeded critter's shells fall to ocean floor

50. You should know the strong acids & bases