1. This continues our discussion of kinetics (Chapter 13) from the previous lecture. We will also start Chapter 14 in this lecture.

2. The Effect of Temperature on Rate The rate constant (k) is only constant with constant T As T increases, k increases Guideline: an increase of 10˚C will double the rate constant and, thus, the reaction rate.

3. Effect of Temperature on Reaction Rate Svante Arrhenius investigated this relationship and showed that: T is the temperature in Kelvin R is the gas constant in energy units, 8.314 J/mol K E a is the activation energy A is a factor called the frequency factor

4. How the Arrhenius equation is used Measure the rate constant at different temperatures Plot ln k versus 1/T to get ln A and E a Measure at two temperatures and use the following to get Ea

5. For the reaction CH 4 + 2 S 2 --> CS 2 + 2 H 2 S at 550˚C the rate constant was measured to be 1.1 M -1 s -1 and at 625 ˚C the rate constant was measured to be 6.4 M -1 s -1. Calculate the activation energy for this reaction. What is the activation energy and what role does it play in the temperature dependence of a reaction?

7. The Arrhenius Equation: The Exponential Factor The exponential factor in the Arrhenius equation is a number between 0 and 1 It represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier That extra energy comes from the kinetic energy of the molecules that get transferred during collisions

9. Determine the activation energy and frequency factor for the reaction O 3(g)  O 2(g) + O (g) given the following data: Temp, Kk, M -1 ∙s -1 Temp, Kk, M -1 ∙s -1 6003.37 x 10 3 13007.83 x 10 7 7004.83 x 10 4 14001.45 x 10 8 8003.58 x 10 5 15002.46 x 10 8 9001.70 x 10 6 16003.93 x 10 8 10005.90 x 10 6 17005.93 x 10 8 11001.63 x 10 7 18008.55 x 10 8 12003.81 x 10 7 19001.19 x 10 9

10. Effective Collisions Orientation Effect

11. Reaction Mechanisms Reactions often do not occur in the single step indicated by the chemical equation. Reaction Mechanism: collection of elementary steps that make up the overall reaction Elementary Step: simple interaction involving one or more molecules. Overall reaction: The sum of all of the individual steps must equal the chemical reaction equation Reaction intermediate: compound produced in one step that is consumed in another step. Doesn’t appear in the overall reaction.

12. Molecularity The number of reactant particles in an elementary step is called its molecularity A unimolecular step involves 1 reactant particle A bimolecular step involves 2 reactant particles they may be the same kind of particle A termolecular step involves 3 reactant particles these are exceedingly rare in elementary steps

13. Rate Laws for Elementary Steps each step in the mechanism is like its own little reaction – with its own activation energy and own rate law the rate law for an overall reaction must be determined experimentally but the rate law of an elementary step can be deduced from the equation of the step H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g) 1) H 2 (g) + ICl(g)  HCl(g) + HI(g) Rate = k 1 [H 2 ][ICl] 2) HI(g) + ICl(g)  HCl(g) + I 2 (g) Rate = k 2 [HI][ICl]

14. Rate Laws of Elementary Steps

15. Rate Determining Step In most mechanisms, one step occurs slower than the other steps Product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction The slowest step in the mechanism is called the rate determining step the slowest step has the largest activation energy The rate law of the rate determining step determines the rate law of the overall reaction

16. Another Reaction Mechanism NO 2(g) + CO (g)  NO (g) + CO 2(g) Rate obs = k[NO 2 ] 2 1) NO 2(g) + NO 2(g)  NO 3(g) + NO (g) Rate = k 1 [NO 2 ] 2 slow 2) NO 3(g) + CO (g)  NO 2(g) + CO 2(g) Rate = k 2 [NO 3 ][CO] fast The first step in this mechanism is the rate determining step. The first step is slower than the second step because its activation energy is larger. The rate law of the first step is the same as the rate law of the overall reaction.

17. Validating a Mechanism in order to validate (not prove) a mechanism, two conditions must be met: 1. the elementary steps must sum to the overall reaction 2. the rate law predicted by the mechanism must be consistent with the experimentally observed rate law

18. Mechanisms with a Fast, Reversible Initial Step

19. Catalysts Catalysts: substances that affect the rate of a reaction without being consumed Provide an alternative mechanism for the reaction with a lower activation energy Consumed in an early mechanism step, then produced in a later step mechanism without catalyst O 3(g) + O (g)  2 O 2(g) V. Slow mechanism with catalyst Cl (g) + O 3(g)  O 2(g) + ClO (g) Fast ClO (g) + O (g)  O 2(g) + Cl (g) Slow

20. Energy Profile of Catalyzed Reaction polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals

21. Catalysts homogeneous catalysts are in the same phase as the reactant particles Cl (g) in the destruction of O 3(g) heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system

22. Enzymes because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate protein molecules that catalyze biological reactions are called enzymes enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction

23. Chapter 14: Chemical Equilibrium For most of the rest of the course we’ll be talking about Chemical Equilibrium Chapter 14: Introduction Chapter 15: Acids and Bases Chapter 16: Aqueous Ionic Equilibria Chapter 17: Free Energy and Thermodynamics Chapter 18: Electrochemistry

24. Dynamic Equilibrium Eventually the forward reaction and the reverse reaction reach the same rate--this condition is called--dynamic equilibrium Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant because the chemicals are being consumed and made at the same rate

25. H 2 (g) + I 2 (g)  2 HI(g) Concentration  Time  Equilibrium Established Since the [HI] at equilibrium is larger than the [H 2 ] or [I 2 ], we say the position of equilibrium favors products As the reaction proceeds, the [H 2 ] and [I 2 ] decrease and the [HI] increases Since the reactant concentrations are decreasing, the forward reaction rate slows down And since the product concentration is increasing, the reverse reaction rate speeds up Once equilibrium is established, the concentrations no longer change At equilibrium, the forward reaction rate is the same as the reverse reaction rate

26. Equilibrium Constant There is a relationship between concentrations of reactants and products at equilibrium called the Law of Mass Action The concentrations of reactants and products at equilibrium are related to a constant K called the equilibrium constant. We treat K as a unitless constant

27. Writing Equilibrium Constant Expressions For the reaction aA (aq) + bB (aq)  cC (aq) + dD (aq) the equilibrium constant expression is: For the reaction 2 N 2 O 5  4 NO 2 + O 2 the equilibrium constant expression is:

28. What Does the Value of K eq Imply? When K eq >> 1, at equilibrium there will be many more product molecules present than reactant molecules the position of equilibrium favors products When K eq << 1, at equilibrium there will be many more reactant molecules present than product molecules the position of equilibrium favors reactants

29. Relationships between K and Chemical Equations When the reaction is written backwards, the equilibrium constant is inverted. When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor. When you add chemical equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations

30. Equilibrium Constants for Reactions Involving Gases The concentration of a gas in a mixture is proportional to its partial pressure. The equilibrium constant can be expressed as the ratio of the partial pressures of the gases.

31. K c and K p In K p, the partial pressures are always in atm K p and K c are not necessarily equal because of the difference in units K p = K c when  n = 0 the relationship between them is:  n is the difference between the number of moles of reactants and moles of products

32. Heterogeneous Equilibria Pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction its amount can change, but the amount of it in solution doesn’t because it isn’t in solution Therefore, solids and liquids are not included in the equilibrium constant expression For the reaction aA(s) + bB(aq)  cC(l) + dD(aq) the equilibrium constant expression is:

33. Calculating Equilibrium Constants from Measured Equilibrium Concentrations The most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium The equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same

34. Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g)  2HI(g) @ 445°C Initial Equilibrium Equilibrium Constant [H 2 ][I 2 ][HI][H 2 ][I 2 ][HI] 0.50 0.00.11 0.78 0.0 0.500.055 0.39 0.50 0.165 1.17 1.00.50.00.530.0330.934

35. One mole each of ethanol (C 2 H 5 OH) and acetic acid (CH 3 CO 2 H) are dissolved in water at 100˚C and a total volume of 250 mL. At equilibrium, 0.25 mol of acetic acid is consumed in producing ethyl acetate (CH 3 CO 2 C 2 H 5 ) and water. Calculate K c.