Chapter 15

1. Chapter 15 Chemical Equilibrium (and Kinetics)

2. Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. (See sec’s 2 & 12) Equilibrium applies to the extent of a reversible reaction (sec’s 3 - 11), the concentration of product that has appeared after an unlimited time, or once no further change occurs. At equilibrium: rateforward= ratereverse A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other.

3. Kinetics: Reaction Rates Some chemical reactions proceed rapidly: like the precipitation reactions studied in Chp 7, where the products form practically the instant the two solutions are mixed. Other reactions proceed slowly: like the rusting of an iron nail left outside. Rate of reaction is measured in the amount of reactant that changes into product in a given period of time, in moles of reactant used per second. Chemists and engineers study ways of controlling reaction rates.

4. Factors Affecting Reaction Rate Reaction rate (kinetics) is affected by the concentrations of the reactants, the temperature, the pressure, agitation, and the presence of catalysts. Concentration: high concentration of reactants can make it faster, but high concentration of products can slow it. Temperature: usually increases because particles are moving faster and colliding more. Pressure: affects rate only if a gas is present. Agitation: temporary increase in rate because collisions increase. Catalyst: made to increase rates via various mechanisms.

5. Factors Affecting Reaction Rate:Reactant Concentration The higher the concentration of reactant molecules, the faster the reaction will generally go. - increases the frequency of reactant molecule collisions (see next 2 slides) Since reactants are consumed as the reaction proceeds, the speed of a reaction generally slows over time.

6. Factors Affecting Reaction Rate: Concentration & Collision Theory In order for a reaction to take place, the reacting molecules must collide with each other. Once molecules collide they may react together or they may not, depending on two factors - Whether the collision has enough energy to start to break the bonds holding reactant molecules together; and Whether the reacting molecules collide in the proper orientation for new bonds to form.

7. Factors Affecting Reaction Rate: Concentration & Collision Theory Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions. The higher the frequency of effective collisions the faster the reaction rate. There is a minimum energy needed for a collision to be effective – we call this the activation energy. The lower the activation energy “barrier” the faster the reaction will be.

8. Factors Affecting Reaction Rate:Temperature Increasing temperature increases the reaction rate: - for each 10°C rise in temperature, the speed of the reaction generally doubles. Increasing the temperature increases the number of molecules in the sample with enough energy so that their collisions can overcome the activation energy. Increasing the temperature also increases the frequency of collisions – so the rate increases because the frequency of effective collisions increases.

9. Activation Energy The energy barrier that prevents any collision between molecules from being an effective collision is called the activation energy. The larger the activation energy of a reaction, the slower it will be at a given temperature.

10. Exothermic Reaction Activation Energy, large Activation Energy, small Reactants Relative Potential Energy DHreaction Products Progress of Reaction

11. Endothermic Reaction Activation Energy Products Relative Potential Energy DHreaction Reactants Progress of Reaction

12. High temperatures lead to more molecules with enough energy to overcome the activation energy, and more frequent reactant collisions, therefore a faster reaction rate Low temperatures lead to fewer molecules with enough energy to overcome the activation energy, and less frequent reactant collisions, therefore a slower reaction rate Effect of Temperature on Rate

13. Factors Affecting Reaction Rate: Catalysts A catalyst is a substance that increases the rate of a reaction, but is not consumed in the reaction. Catalysts lower the activation energy of a reaction. Catalysts work by providing a easier pathway for the reaction.

14. Catalysts’ Effect onActivation Energy

15. Nature’s Catalysts: Enzymes Enzymes are protein molecules produced by living organisms that catalyze chemical reactions. The enzyme molecules have an active site that organic molecules bind to. When the organic molecule is bound to the active site, certain bonds are weakened. This allows a particular chemical change to occur easier and quicker, i.e. the activation energy is lowered.

16. Example of an Enzyme: Sucrase

17. Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up. However, if the products are allowed to accumulate; they will start reacting together to form the original reactants. This is called the reverse reaction.

18. Reaction Dynamics The forward reaction slows down as the amounts of reactants decreases. At the same time the reverse reaction speeds up as the concentration of the products increases. Eventually the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. Dynamic equilibrium is reached when the rates of two opposite processes are the same.

19. Introduction to Equilibrium At equilibrium, the [reactants] and [products] remain constant over time! (Remember, [ ] mean concentration of a substance in Molarity.) Equilibrium is approachable from any direction! Q = reaction quotient Keq = equilibrium constant At equilibrium Q = Keq

20. Introduction to Equilibrium Q = Reaction Quotient = [Products]m/[Reactants]n Given a generic reaction: aA + bB  cC + dD Q = [C]c[D]d [A]a[B]b If at equilibrium, Q = Keq Example: given the reaction N2O4(g) 2 NO2(g), we would write the Keq expression: Keq = [NO2]2/[N2O4]

21. [NO2]2 [NO2]eq2 [N2O4] [N2O4]eq 0.193 1 0.1000 0.0000 0.0000 3.57x10-3 9.83x10-3 ∞ 2 0.0000 0.1000 9.24x10-4 3 0.0500 0.0500 0.0500 2.04x10-3 0.146 4 0.0750 0.0250 0.00833 2.75x10-3 0.170 Table: Initial and Equilibrium Concentration Ratios for the N2O4-NO2 System at 200.0C (473 K) Ratio(Q) Value of Keq Initial Equilibrium Experiment [N2O4] [NO2] [N2O4]eq [NO2]eq 10.4 10.4 10.4 _____

22. The change in Q during the N2O4-NO2 reaction. The initial part is covered by Kinetics; the Q = K part is covered by Equilibrium!

23. Reaction Progress Reaction Progress reactants products Equilibrium: no net change reactants products Reaction direction and the relative sizes of Q and Keq. Figure Keq = Kc for many reactions

24. small K large K intermediate K The range of equilibrium constants

25. WRITE AN EQUILIBRIUM EXPRESSION FOR: CaCO3(s) CaO(s) + CO2(g) Keq = [CaO][CO2]/[CaCO3] But solids don’t change in concentration (moles/liter). Keq = CCaO[CO2]/CCaCO3 Rearrange to Keq' = [CO2] NO solids or solvents are written into the equilibrium constant expression or reaction quotient.

26. The reaction quotient for a heterogeneous system. solids do not change their concentrations

27. PRACTICE WRITING Keq (Kc) FOROTHER REACTION SYSTEMS: • C(s) + H2O(g) CO(g) + H2(g) Keq = [CO][H2]/[H2O] • HNO2(aq) + H2O(l) H3O+(aq) + NO2-(aq) Ka = [H3O+][NO2-]/[HNO2] (Use subscript a for acid) • CaCO3(s) Ca2+(aq) + CO32-(aq) Ksp = [Ca2+][CO32-] (Use subscript sp for insol cmpd) • 2 N2O5(g) 4 NO2(g) + O2(g) Keq = [NO2]4[O2]/[N2O5]2 • NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb = [NH4+][OH-]/[NH3] (Use subscript b for base)

28. Calculating Keq The value of the equilibrium constant may be determined by measuring the concentrations of all the reactants and products in the mixture after the reaction reaches equilibrium, then substituting in the expression for Keq. Even though you may have different amounts of reactants and products in the equilibrium mixture, the value of Keq will always be the same: - the value of Keq depends only on the temperature - the value of Keq does not depend on the amounts of reactants or products you start with

29. Calculating the value of Keq Example: A mixture of CH4, C2H2 and H2 is allowed to come to equilibrium at 1700°C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? 2 CH4(g)  C2H2(g) + 3 H2(g) Write the expression: Keq = [C2H2][H2]3/[CH4]2 Fill in the data: Keq = (0.0451)(0.112)3/(0.0203)2 = 0.154

30. Using Keq to find the concentration at equilibrium In an equilibrium mixture the concentrations of H2 and I2 are both 0.020 M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340°C Write the expression: Keq = [HI]2/[H2][I2] Put in the known data: 69 = [HI]2/(0.020)2 Rearrange and solve: [HI] = 0.166 M or 0.17 M

31. Using Keq and initial concentrations to find equilibrium concentrations We can find equilibrium concentrations if we know initial conditions and the value of Keq. Find the equilibrium concentrations for both the reactant and product for the reaction below, given that you start with 2.00 moles of the reactant in a 1.00 L container. Initial concentration of the reactant is therefore 2.00 M and the product is 0. A. Set up an ICE table: N2O4(g) 2 NO2(g) Keq = 0.20 Initial [ ] 2.00M 0.0M (Note Q = 0) Change -x +2x Equil [ ] (2.00-x) (2x) Keq = [NO2]2/[N2O4] B. Determine changes – using regular old stoichiometry.

32. Using Keq and initial concentrations to find equilibrium concentrations C. Put everything in the Keq expression: Keq = [NO2]2/[N2O4] = 0.20 Keq =(2x)2/(2.00-x)=0.20M Rearrange and solve for x, using the quadratic equation. x= 0.29, but you’re not done. Plug x into equstions: [N2O4] = 2.00 - 0.29 = 1.71M [NO2] = 2x = 0.58 M D. Plug these concentrations back into Keq and solve, if close to 0.20M, then they are correct.

33. Solubility & Solubility Product, Ksp Even “insoluble” salts dissolve somewhat in water insoluble = less than 0.1 g per 100 g H2O. Equilibrium constant for this process is called the solubility product constant, Ksp. Given a salt, AnYm, Ksp = [A+]n[Y-]m If there is undissolved solid in equilibrium with the solution, the solution is saturated. Larger Ksp = more soluble salt.

34. Example - Determine the Kspof PbBr2 if its solubility is 1.44 x 10-2 M PbBr2(s) Pb2+(aq) + 2 Br–(aq) I -- 0 0 E -- 0.0144 0.0288 Ksp = [Pb2+][Br–]2 = (0.0144)(0.0288)2 = 1.19 x 10-5

35. LeChatelier’s Principle When a chemical system at equilibrium is subjected to a stress, the system will return to equilibrium by shifting to reduce the stress.

36. Principle 1: The first stress we look at is a change in concentration of reactant or product: If the concentration increases, the system reacts to consume some of it. If the concentration decreases, the system reacts to produce some of it.

37. Principle 1: Concentration The effect of added Cl2 on the PCl3-Cl2-PCl5 system. PCl3(g) + Cl2(g)  PCl5(g) After adding more reactant chlorine, we see both reactants decreased and the product increased, to reestablish equilibrium.

38. To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2; 2H2S(g) + O2(g) 2S(s) + 2 H2O(g) Sample Problem Predicting the Effect of a Change in Concentration on the Equilibrium Position What happens to (a) [H2O] if O2 is added? (b) [H2S] if O2 is added? (c) [O2] if H2S is removed? (d) [H2S] if sulfur is added?

39. PRINCIPLE 2: EFFECT OF INCREASE OR DECREASE IN PRESSURE Decreasing the size of the container increases the concentration of all the gases in the container increases their partial pressures. If their partial pressures increase, then the total pressure in the container will increase. According to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure. The way to reduce the pressure is to reduce the number of molecules in the container. When the volume decreases, the equilibrium shifts to the side with fewer molecules.

40. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side. Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. PRINCIPLE 2: EFFECT OF INCREASE OR DECREASE IN PRESSURE

41. PRINCIPLE 2: INCREASE OR DECREASE PRESSURE TO MAKE MORE CO? C(s) + CO2(g) 2 CO(g) Remember, the system will adjust to restore pressure: If P decreases, system shifts to side with more moles of gas. If P increases, system shifts to side with fewer moles of gas. We would want to decrease the partial pressure of CO.

42. PROBLEM: How would you change the pressure (volume) of each of the following reactions to increase the yield of the products. (a) CaCO3(s) CaO(s) + CO2(g) (b) S(s) + 3F2(g) SF6(g) (c) Cl2(g) + I2(g) 2ICl(g) Sample Problem Predicting the Effect of a Change in Pressure on the Equilibrium Position SOLUTION: (a) CO2 is the only gas present. To increase its yield, we should increase the volume, decrease the pressure. (b) There are more moles of gaseous reactants than products, so we should decrease the volume, increase the pressure, to shift the reaction to the right. (c) There are an equal number of moles of gases on both sides of the reaction, therefore a change in volume/pressure will have no effect.

43. Principle 3: The Effect of Temperature Changes on Equilibrium for Exothermic Reactions For an exothermic reaction, heat is a “product.” Increasing the temperature is like adding heat. According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat. The concentrations of C and D will decrease and the concentrations of A and B will increase. The value of Keqwill decrease. How will decreasing the temperature affect the system? aA + bB cC + dD + Heat

44. Principle 3: The Effect of Temperature Changes on Equilibrium for Endothermic Reactions For an endothermic reaction, heat is a “reactant.” Increasing the temperature is like adding heat. According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat. The concentrations of C and D will increase and the concentrations of A and B will decrease. The value of Keqwill increase. How will decreasing the temperature affect the system? Heat+ aA + bB cC + dD

45. Practice - Predict the Effect on the Equilibrium when the Temperature is Reduced • 2 CO2(g)Û 2 CO(g) + O2(g) endothermic • BaSO4(s)Û Ba2+(aq) + SO42-(aq) endothermic • CH4(g) + 2 O2(g)Û CO2(g) + 2 H2O(l) exothermic

46. PROBLEM: How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions? (a) CaO(s) + H2O(l) Ca(OH)2(aq) DH0 = -82kJ (b) CaCO3(s) CaO(s) + CO2(g) DH0 = 178kJ (c) SO2(g) S(s) + O2(g) DH0 = 297kJ PLAN: Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on Kc. (a) CaO(s) + H2O(l) Ca(OH)2(aq) heat (b) CaCO3(s) + heat CaO(s) + CO2(g) (c) SO2(g) + heat S(s) + O2(g) Sample Problem Predicting the Effect of a Change in Temperature on the Equilibrium Position SOLUTION: An increase in temperature will shift the reaction to the left, decrease [Ca(OH)2], and decrease Kc. The reaction will shift right resulting in an increase in [CO2] and increase in Kc. The reaction will shift right resulting in an decrease in [SO2] and increase in Kc.

48. ACID STRENGTH

49. Calculating equilibrium for weak acids The ICE table and the quadratic equation (if needed): to find equilibrium concentration, we set up a table and use some algebra. Given 0.100 M acetic acid, what is the H3O+ concentration at equilibrium? What is the pH? Ka is 1.8 x 10-5. Ka = [H3O+][CH3COO-]/[CH3COOH] CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq) Initial: 0.100 M --------- 0.0 M 0.0 M Change: -x +x +x Equilibrium: 0.1 – x x x 1.8 x 10-5 = (x)(x)/(0.1-x) = x2/(0.1-x) Solve for the value of x, which will also be [H3O+], by rearranging and simplifying the equation. The pH will then be found.