1. Test: May 23, 2013 THURSDAY

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3. How fast does aging occur? 3 http://www.ging- strategie.de/Anti%2 0Aging%20Strategie.JPG How fast does the candle burn? http://images.usatoday.com/tec h/_photos/2006/03/28/aprilfig1. jpg

4. So how do we measure the speed of a reaction? Reaction Rate (def.): The speed with which products form from the reactants (or that reactants are used to form products) -usually measured in Molar/ second (M/s) or Molar/ minute (M/min) 4

5. But first, what is necessary for a successful reaction???? COLLISION THEORY OF REACTIONS says: 1. Collision 2. Enough Kinetic Energy 3. Proper orientation (during collision)

6. Collision Theory of Reactions 6 http://staff.um.edu.mt/jgri1/teaching/che2372/notes/10/10_19.gif a)Improper orientation (and maybe not enough kinetic energy) = NO REACTION b)Proper orientation, enough kinetic energy = REACTION

7. Reaction Rates are affected by: 1. Concentration 2. Temperature 3. Catalyst 4. Surface Area (Particle Size) 5. Mixing (Stirring) 7

8. Concentration Effect on Reaction Rates 8 The more concentrated, the higher the reaction rate. Why? http://content.answers.com/main/content/wp/en/thumb/f/ff/525px-Molecular-collisions.jpg

9. Temperature Effect on Reaction Rates 9 The higher the temperature, the faster the rate. Why? HIGHER TEMP = HIGHER KINETIC ENERGY (or speed) = MORE COLLISIONS & MORE ENERGY IN EACH COLLISION (2 factors affected!)

10. Catalyst Effect on Reaction Rates 10 A catalyst LOWERS the activation energy, thus the reaction proceeds faster = MORE PARTICLES HAVE ENOUGH KINETIC ENERGY

11. ACTIVATED COMPLEX or TRANSITION STATE When reactants collide in the proper orientation, they form a high-energy, unstable arrangement of atoms called the ACTIVATED COMPLEX or TRANSITION STATE before forming products. The activated complex or transition state has a structure that is between that of the reactants and products (as bonds are simultaneously being broken/ formed)

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13. Activation Energy Minimum amount of kinetic energy the reactants must have in order to react. AKA the minimum energy needed to create the activated complex or transition state 13

14. Potential Energy vs Time (Exothermic) 14 http://staff.um.edu.mt/jgri1/teaching/che2372/notes/10/theory.html Time Heat of Reaction  H

15. Analogy 15 http://www.800mainstreet.com/7/0007-004-reac_rate2.htm

16. Exo- and Endothermic Reactions Exothermic reaction: reaction gives off energy. reactants products + energy Endothermic reaction: reaction takes in energy. energy + reactants products 16

17. Bonds and Energy When bonds are broken: energy is absorbed. (analogy: stretching a rubber band) When bonds are formed : energy is released. (analogy: releasing a stretched rubber band) 17

18. Average bond energies, kcal/mole C-H98 O-H110 C-C80 C-O78 H-H103 C-N65 O=O116 (2 x 58) C=O187* (2 x 93.5) C=C145 (2 x 72.5) (* as found in CO 2 ) 18

19. Ex: Exothermic H 2 + Cl 2  2 HCl + 183 kJ/mol 19

20. Ex: Endothermic 2HgO + 181.7 kJ  2Hg + O 2 20

21. Net Energy Change  H= Potential Energy of Products -Potential Energy of Reactants  H <0 exothermic  H >0 endothermic 21

22. Energy vs Time (Endothermic) 22 Heat of Reaction  H Activation Energy

23. Particle Size (Surface Area) Mixing The smaller the particle size the faster the rate = MORE SURFACE AREA = MORE COLLISIONS. The more mixing = MORE COLLISIONS 23

24. Inhibitors Substances that negate the effect of the catalyst. 24

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26. Reversible Reactions Occur simultaneously in both directions. 26

27. Example of a Reversible Reaction Heat +N 2 O 4 (g) 2NO 2 (g) 27 Warmer temp http://genchem.chem.wisc.edu/demonstrations/Images/13equil/NO2N2O4.jpg Colder temp

28. Equilibrium A state of balance at which the rates of the forward and reverse reactions are equal. 28

29. Equilibrium Position The relative concentrations of reactants and products at equilibrium. At this point the concentrations don’t change unless a stress is applied to change the equilibrium. 29

30. Le Châtelier’s Principle If a stress is applied to a system in dynamic equilibrium, the system changes to relieve the stress. System stresses: Concentration of reactants or products Temperature Pressure 30

31. Effect of Concentration of Reactants Adding reactant shifts the reaction toward the products. Why? Stress: Increasing reactants Relief: Decreasing reactants Shift: to the right (products) H 2 O (l) + CO 2 (g) H 2 CO 3 (aq) 31

32. Effect of Concentration of Products Adding products shifts the reaction toward the reactants. Why? Stress: Increasing products Relief: Decreasing products Shift: to the left (reactants) H 2 O (l) + CO 2 (g) H 2 CO 3 (aq) 32

33. Effect of Temperature Increasing the temperature causes the equilibrium to shift in the direction that absorbs heat. Stress: Increase in Temp Relief: Decrease in Temp Shift: Towards the left SO 2 (g) + O 2 (g) 2SO 3 (g) + heat 33

34. Effect of Pressure Affects gases only. For unequal number of moles of reactants and products, if pressure is increased, the equilibrium will shift to reduce the number of particles. For equal number of moles of reactants and products, no shift occurs. 2NO 2 (g) N 2 O 4 (g) 34

35. Ex: Effect of Pressure 2NO 2 (g) N 2 O 4 (g) Stress: increasing the pressure Relief: decreasing the pressure Shift: to the right (side of less molecules) 35

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37. Equilibrium Position 37 http://www.chm.davidson.edu/ronutt/che115/K/Sol_1.gif

38. Equilibrium Expression Equilibrium Constant, K eq The concentrations are at the equilibrium position. The K eq is constant for a given temperature regardless of the initial concentrations. 38

39. K eq > 1, products favored in equilibrium K eq < 1, reactants favored in equilibrium K eq is not expressed with units. 39

40. Rules for K eq of Heterogeneous Equilibrium (more than one phase) Only gases and solute concentrations appear in the equilibrium expressions. Pure liquids and solids do not affect the K eq, because they do not change. 40

41. Practice 2 SO 2(g) + O 2(g) ⇄ 2 SO 3(g) K eq = AgCl(s) ⇄ Ag + (aq) +Cl - (aq) K eq = 41

42. Ex. 1: Calculate the K eq, given equilibrium concentrations N 2 O 4 (g) ⇄ 2 NO 2(g) At equilibrium at 10°C, [N 2 O 4 ]= 0.0045 M and [NO 2 ] = 0.030M. Calculate the K eq. Answer: K eq = 0.20 42

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44. Solubility Equilibrium AgCl(s) ↔ Ag + (aq) + Cl - (aq) The concentration of AgCl (s) is constant and AgCl is a strong electrolyte (dissociates almost completely), so K sp = [Ag + ] [Cl - ] K sp is the solubility product constant 44

45. Note The larger the K sp the larger the solubility of the compound. (This comparison can be made when the ion ratios are equal). Ex. Of the following, which is more soluble? AgCl K sp = 1.8x10 -10 AgI K sp = 8.3x10 -17 45

46. Example: Write the K sp for: CuS BaSO 4 SrCO 3 AgI 46

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48. Spontaneous Process A process that occurs without outside intervention. Spontaneous processes can be fast or slow! 48

49. What Factors Determine the Spontaneity of a Reaction? Entropy Enthalpy What ties the two together is: Gibbs Free Energy. 49

50. Entropy Symbol S A measure of molecular randomness or disorder. 50

51. Disorder Chemical processes spontaneously go to a direction of increased entropy. Why? Probability 51

52. Probability of Disorder Is there a higher probability your room will be messy or neat as time goes on? 52

53. System the part of the universe under investigation. 53

54. Entropy of the System Is greater in: Gases than solids. Larger volumes of gases than smaller volumes. Larger number of gas molecules than smaller number of gas molecules. 54

55. Example: Which has more entropy in its system? H 2 O (s) orH 2 O (g) 55

56. Enthalpy (Heat of Reaction) Spontaneity is favored when the process is exothermic (  H<0). 56

57. Example of a spontaneous exothermic reaction: 2SO 2 (g)+ O 2 (g)  2SO 3 (g) + heat  H<0 57

58. So…… : When  H 0 (exothermic) (Greater Disorder) the reaction would be spontaneous. 58

59. Are all spontaneous reactions exothermic and with a greater system disorder? Answer: No. 59

60. What about exothermic and less disorder? Use Gibbs Free Energy Gibbs Free Energy can be used to predict the spontaneity and it ties together the  H and the  S, the two driving forces of reactions.  G=  H-T  S  (all quantities refer to the system) 60

61. Gibbs Free Energy  Energy that can be converted to work.   G<0 for spontaneous processes.   G=0 at equilibrium. 61