1. Guiding Question: What determines whether a reaction takes place and how quickly it happens? Reactions can be looked at from two different perspectives: 1.Rate of Reaction: How fast, how many moles are used up, how many are produced. 2.Mechanism of Rxn: Steps involved in the reaction.

2. Reactions are a process of breaking existing bonds and forming new bonds, and this requires particles to collide. Whether the reaction takes place and how quickly it proceeds first begins with an effective collision.

3. Collision Theory: Effective collisions are defined as: - Collisions with the proper amount of energy - Collisions with the proper orientation or angle

5. Any way we can increase the number of effective collisions, we can make the reaction occur faster. More effective collisions will always lead to a faster reaction rate.

7. Particles must collide with the proper orientation HI + HI --> H 2 + I 2 (H hits H, I hits I)

8. Factors that Affect Rate of Rxn 1.Concentration (abbv [ ] ) The more particles you have in a given area, the more collisions occur. By increasing the number of collision, we increase our chance that the collisions will be effective.

10. 2. Pressure (only applies to gases) When you increase the pressure on a gas, the gas molecules are forced to occupy a smaller volume. This is equivalent to increasing the concentration, it results in an increase in effective collisions.

11. The increase in pressure results in an increase in collisions. Reactions that contain gases proceed faster under HIGH PRESSURE.

12. 3. Temperature (Kelvin) This directly affects the speed of the molecules. The higher the temperature, the faster the molecules are moving and with more energy. The HIGHER the T, the faster the reaction will proceed. Not only will the number of collision increase, but they will be more energetic and more likely to be effective.

13. Temperatures influence on the rate of the reaction is the reason why most chemical reactions require the use of a bunsen burner. The bunsen burner heats up the reactants and gives them enough energy to have effective collisions. Batteries!!!

14. 4. Surface area ( recall solutions) Powdered substances will always react faster than solid substances. By crushing a sample into smaller pieces, we create more surfaces for effective collisions to take place. Vs.

17. This is why we granulate (crush) sugar. If we used a sugar cube, it wouldn’t dissolve as fast. By crushing it, the dissolving process happens faster. If you tried to throw a brick at someone, you would likely miss. However if you broke it into 20 pieces and threw all of those at the same time, you would probably hit your target. This is also why cars that have scratches, rust faster than a car in perfect condition. It creates more exposed area for oxidation (rusting) reaction to take place.

18. 5. Nature (type of molecules) First… we will look at ionic compounds. As a rule, ionic compounds react faster than molecular compounds. Why? Ionic compounds dissociate, or break the bonds that hold them together, resulting in free moving mobile ions. This makes it much easier for these ions to collide, react and form new products.

19. Nature continued… - Rxns are a process of breaking bonds and forming new ones. Ionic compounds have already broken bonds holding them together. Easier to rearrange and “re-bond” with new things = faster rxn rate. - Molecular compounds do not dissociate, they retain the bonds that hold them together. For a rxn to occur, first you must break bonds and then rearrange molecules Finally, re-bond molecules back together. This requires a tremendous amount of energy, so reactions proceed much slower.

20. Rule: Reactions with ionic compounds proceed much faster. Especially when dissolved in water, since the ions have already dissociated. Molecular (Organic) reactions are incredibly slow moving and often require a catalyst to help move them along. Molecules also have a much more complicated structure than an ionic crystal lattice and it requires more energy and time to rearrange the bonds.

21. 6. Catalysts Catalysts are substances that increase the rate of a reaction by providing a shortcut. They are not consumed and not considered part of the reaction, but only used to speed up the rate of reaction.

23. Check Your Understanding: Can you…  Define an effective collision?  Identify the six factors that influence the rate of a reaction?

24. Guiding Question: How is energy exchanged during chemical reactions? Chemical reactions involves breaking bonds and forming new ones. This is always accompanied by energy changes.

25. When chemical reactions take place, bonds are broken and energy is absorbed. As the products are formed and bonds are created, energy is released. If more energy is absorbed than released: This is called an endothermic reaction. If more energy is released than absorbed: This is called an exothermic reaction.

26. Exothermic reactions give off energy… therefore during exothermic reactions, the temperature of the surroundings will increase as the heat flows. Ex: Hot packs

27. Endothermic reactions absorb more energy than they release…. Therefore during endothermic reactions, the temperature of the surroundings will decrease. Ex: Ice cubes in water

28. This heat exchange is known  H: Heat of Reaction. Also known as “enthalpy”  H = H products – H reactants If  H is positive: reactants have less energy than products. Endothermic: +  H If  H is negative: reactants have more energy than products: Exothermic: -  H.

29. Heats of Combustion Heats of Formation Heats of Solution Heats of Neutralization

30. When  H is negative = exothermic. This means heat is released or is a product. Energy will be on the right side of the equation. When  H is positive = endothermic. This means heat is absorbed or is a reactant. Energy will be on the left side of the equation. If the equation is reversed, then the sign is switched!

31. When a reaction is exothermic: Heat is a “product” 4 Al + 3 O 2  2 Al 2 O 3 + 3351 kJ Because heat is located on the products side of the equation, we know this reaction is exothermic. Therefore, the sign of  H would be – 3351 kJ

32. 4 Al + 3 O 2  2 Al 2 O 3 + 3351 kJ This can be interpreted as the formation of 2 moles of Al 2 O 3, released 3351 kJ of energy.  H = -3351 kJ. How could we calculate the  H for the formation of 4 moles of the product? How much energy is released during the consumption of 1.75 moles of O 2 ?

33. When a reaction is endothermic, heat is on the reactants side: N 2 + O 2 + 182.6  2 NO Because energy is on the reactants side, we know that this is an endothermic reaction. Therefore, the of  H would be + 182.6 kJ

34. N 2 + O 2 + 182.6  2 NO This can be interpreted as the formation of 2 moles of NO, requires the absorption of 182.6 kJ of energy. How much energy must be absorbed to produce 4 moles of NO? How much energy is absorbed when 3.5 moles of N 2 are reacted?

35. The moral of the story: Where heat is added to an equation tells us whether it is an exothermic or endothermic reaction. Then, we can say whether it is positive or negative  H.

36. Check Your Understanding: Can you…  Determine if a reaction is endothermic or exothermic?  Use table I to identify exothermic and endothermic reactions?  Use principles of stoichiometry to calculate heat released or absorbed?

37. Guiding Question: How are the energy changes in a reaction monitored over time? The way that energy changes during the course of a reaction is graphed in what is known as a Potential Energy Diagram. Because the energy changes are quite different in exothermic and endothermic reaction, the resulting PE diagrams look much different.

38. Exothermic: H reactants > H products : +  H If you have an exothermic reaction, on the PE diagram, the reactants ARE ALWAYS HIGHER than the products

39. Endothermic: H products > H reactants: -  H If a reaction is endothermic, reactants ARE ALWAYS LOWER than the products

40. Parts of the PE diagram: 1. PE of reactants 2. Energy of Activated Complex 3. Activation Energy for Forward Reaction 4.  H, Heat of Reaction (Heat of Products – Heat of Reactants) (read from graph) 5. PE of products 5

41. Heat of Reactants and Products The amount of energy possessed by the reactants and products (cost vs. what’s in your wallet) To determine: Read from bottom to reactants or products. (Arrow 1 for Reactant, Arrow 5 for products) 5

42. Activation Energy: Minimum energy required for a chemical reaction to proceed. (Energy to get over the hump) To determine: Compare PE of reactants to top of hill (Arrow 3) 5

43. Activated Complex: The highest energy point in a reaction, only temporary… Point where the bonds are rearranged. To determine: Read from bottom to top of highest point (Arrow 2) 5

44. Heat of Reaction:  H (PEp – PEr) Tells us whether a reaction is overall exothermic or endothermic. To determine: Difference between products and reactants. (Arrow 5 – Arrow 1) 5

45. Addition of a catalyst The purpose of a catalyst is to lower the activation energy, so the reaction can proceed faster. It speeds up both the forward and reverse reaction, however it has no effect on the heat of the reaction.

46. Effect of Catalyst on Reaction Rate reactants products Energy activation energy for catalyzed reaction Reaction Progress No catalyst Catalyst lowers the activation energy for the reaction. What is a catalyst? What does it do during a chemical reaction?

47. Ex. CH 4 + 2O 2  CO 2 + 2 H 2 O From Table I H r = -890.4 kJ = exothermic PE Diagram will look like:

48. Ex. N 2 + O 2  2 NO From Table I: H r : + 182.6 kJ = endothermic PE Diagram will look like:

49. For the following rxn: Does the graph represent an endothermic or exothermic reaction? Determine the heat of reaction,  H, for this reaction. Determine the activation energy, Ea for this reaction. What is the energy of the activated complex for this reaction? Determine the reverse activation energy, Ea for this reaction. Endothermic + 50 kJ +200 kJ 250kJ +150 kJ absorbed

50. For the following reaction… Does the graph represent an endothermic or exothermic reaction? Determine the heat of reaction,  H, for this reaction Determine the activation energy, Ea for this reaction. What is the energy of the activated complex for this reaction? Determine the reverse activation energy, Ea for this reaction. Exothermic -20 kJ + 60 kJ 100 kJ + 80 kJ absorbed

51. For the following reaction… Is this reaction exothermic or endothermic? How would the curve look if we added a catalyst?

52. Based on the following reaction, which curve best represents that formation of water? Eq: 2 H2(g) + O2(g) --> 2 H2O(l) + 571.6 kJ Ans: Choice 1, Heat is a product, so it must be exothermic, where Hr > Hp

53. Check Your Understanding: Can you…  Identify exothermic and endothermic PE Diagrams  Label the parts of the PE Diagram  Calculate the Heat of Reaction  Describe how catalysts affect a PE diagram and how it affects the Heat of Reaction

54. Guiding Question: Do reactions stop or can they go on forever? Certain reactions have a definite start and stop point. However, many reactions end up as a balance of both the forward reaction and the reverse reaction. This eventual balance is known as equilibrium. It is known as a dynamic equilibrium because although at the surface it looks like nothing is going on, at the molecular level things are constantly changing.

56. The forward reaction is reactants  products. The reverse reaction is products  reactants vs

57. Equilibrium is defined as: The rate of the forward reaction is EQUAL to the rate of the reverse reaction. The concentrations are constant.

58. For example… Forward Reaction: N 2 + 3 H 2  2 NH 3 + 92 kJ Reverse Reaction: 92 kJ + 2 NH 3  N 2 + 3 H 2 To represent equilibrium, we use a double headed arrow in between the reactions and the products… N 2 + 3 H 2  2 NH 3 + 92 kJ This occurs only when the rates of the opposing reactions are equal

59. Types of Equilibrium: all occur in closed (sealed) sytems 1. Physical Equilibrium Rate of the forward phase change is equal to the rate of the reverse phase change. H 2 O (l)  H 2 O (g) The water is condensing at the same rate that it is vaporizing. The level of the water (concentration) never changes.

60. 2. Solution Equilibrium: Occurs in saturated solutions The rate of dissolving = rate of crystallizing.

61. 3. Chemical Equilibrium Rate of forward = Rate of reverse Equilibrium is dynamic. The system looks like its at a standstill, because the amount of your reactants is constant, as is the amount of your products. However, at a molecular level, things are constantly forming bonds and breaking bonds.

62. Equilibrium can only occur in closed system. This means that the system is sealed and not in contact with the environment. This way nothing gets in or out… especially the heat energy. The concentrations of the reactants and products at equilibrium are constant (not changing). However this does not mean that the concentrations are equal. Equilibrium can be reached when you have 25% product and 75% reactant. So long as the rates are equal.

64. Check Your Understanding: Can you…  Define equilibrium in terms of: rates and concentrations

65. Guiding Question: What happens in equilibrium is disrupted?

66. Once equilibrium is established (this takes time), we can disrupt the system by changing different variables. When a variable is changed, and eq is disrupted, it will auto-correct itself and “shift” equilibrium to fix the problem. This shifting of the equilibrium to compensate for “stress” is known as LeChatelier’s Principle. - when a system is subject to stress, the equilibrium will shift to relieve stress.

67. What kind of “stress” are we talking about? Any factor that will affect the reaction rate is considered a “stress”. Therefore, when we adjust variables like temperature, concentration or pressure (for gases)… the reaction will have to “auto-correct” itself, until equilibrium is re-established.

68. When we say “shift” what do we mean? When we say the reaction will “shift right”, this means that the forward reaction is favored and will happen at a faster rate, and the concentration of the products will increase. When we say the reaction will “shift left”, this means that the reverse reaction is favored and will happen at a faster rate. The concentration of the reactants will increase.

69. 1.Changes in Concentration A + B C+D If we increase the concentration of A, this creates too many molecules on the reactant side. An increase in the number of collisions/ reactions. The system was in equilibrium, but now we have too many reactants. So the eq. shifts to the right to use up extra reactants and produce more product.

70. A + B  C + D + B  C + D +B  A +B  C + D A C + D A We increase the concentration of A Then, we use up the extra A and some of B, and the concentration of C and D increase. Equilibrium shifts right.

71. In other words… + B  C + D By increasing the concentration of A, this results in an increase in the concentration in C and D, and a decrease in B. Equilibrium shifts right. A A

72. What about an increase in the concentration of D? A + B  C + D D By increasing the concentration of D, this mean we have too many molecules on the product side of the equation. In order to balance out, we use up the extra products to produce more of the reactants. Equilibrium shifts left.

73. For the reaction: what if we decrease the concentration of A + B  C + D A A A decrease in the concentration of A, means that instead of an extra reactants, we don’t have enough. So the equilibrium will shift left, to produce more reactants. This results in a decrease in the amount of C and D, and an increase in the amount A and B.

74. For the reaction…what if we decrease the concentration of D? A + B  C + D D When we decrease the concentration of D, we don’t have enough products anymore, so we shift the equilibrium right, to create more products. This decrease the concentration of A and B, and increases the concentration of C and D.

75. Trick for Concentration Shifts A – away and T-towards A: add, shift away T: take out, shift towards

76. The Effect of Pressure: An increase in the pressure will increase the number of collisions. To relieve the effect of an increased pressure, the system will shift in the direction of less molecules/moles of molecules.

77. Ex. Haber Process – know this process -The Haber Process is a way of making ammonia: What will happen if we increase the pressure, by decreasing the volume? N 2 + 3 H 2  2 NH 3 Volume : Pressure: 4 moles of molecules 2 moles of molecules When the pressure is increased, the system will shift in the direction that relieves some of the extra stress, which is the direction with less moles of molecules. In this case, equilibrium shifts RIGHT.

78. Ex. Haber Process – know this process -The Haber Process is a way of making ammonia: What will happen if we decrease the pressure, by increasing the volume? N 2 + 3 H 2  2 NH 3 Volume : Pressure: 4 moles of molecules 2 moles of molecules When the pressure is decreased, the system will has “more room” and will shift in the direction that produced more moles of molecules. In this case, equilibrium shifts LEFT.

79. The Effect of Temperature… -It is helpful to consider heat as a reactant or a product - Exothermic: heat is a product - Endothermic: heat is a reactant When you do this, changing the temperature has the same effect has changing concentration. (A-away, T-towards)

80. For example… N 2 + O 2  2 NO Table I:  H = 182.6 kJ = endothermic So we can rewrite the equation as: 182.6 kJ + N 2 + O 2  2 NO When the temperature increases, its just like increasing a concentration of one of the reactants. When temperature rises, equilibrium shifts RIGHT.

81. For example… N 2 + O 2  2 NO Table I:  H = 182.6 kJ = endothermic So we can rewrite the equation as: 182.6 kJ + N 2 + O 2  2 NO When the temperature decreases, its just like decreasing a concentration of one of the reactants. When temperature lowers, equilibrium shifts left.

82. For example… N 2 + 3 H 2  2 NH 3 On Table I:  H = -91.8kJ So we can rewrite the equation: N 2 + 3 H 2  2 NH 3 + 91.8kJ * by putting it on right products side, we don’t need to have the negative sign*

83. N 2 + 3 H 2  2 NH 3 + 91.8kJ An increase in temperature, increases products, equilibrium will shift LEFT. N 2 + 3 H 2  2 NH 3 + 91.8kJ A decrease in temperature, decreases products, equilibrium will shift RIGHT.

84. The effect of a catalyst… While a catalyst increases the RATE of a reaction, adding a catalyst to the system has NO effect on the equilibrium. There is no change in concentration on either side when a catalyst is added to the system.

85. For the equilibrium system: 2N 2 (g) + O 2 (g) + 66.4kJ  2 N 2 O(g) What are four things that could be done to increase the concentration of NO(g)? Desired Effect: Shift Right 1. Increase concentration of N 2 2. Increase concentration of O 2 3. Increase temperature 4. Increase pressure

86. For the equilibrium system: NaCl (s) + 3.88 kJ  Na +1 (aq) + Cl -1 (aq) What are three things that could be done to increase the concentration of NaCl(s)? Desired Effect: Shift Left 1. Increase concentration of Na +1 2. Increase concentration of Cl -1 3. Decrease temperature

87. Check Your Understanding: Can you…  Determine shifts in equilibrium when factors such as concentration, temperature and pressure are changed?

88. Guiding Question: Will a reaction occur on its own?

89. Determining whether a reaction will occur spontaneously or not depends on the balance between two factors: 1.Enthalpy (heat of the system) 2.Entropy (disorder of the system)

90. Enthalpy is the heat of the reaction, or heat of the system. You know enthalpy as  H. Nature favors reactions that have a DECREASE in ENTHALPY, or have a –  H… In other words, nature favors exothermic reactions.

91. Exothermic reactions have products with less energy and those are more stable which is favored. Most exothermic reactions are spontaneous. Most endothermic reactions are non- spontaneous because they require a constant input of energy

92. Entropy is the degree of disorder or randomness in a sample. It is represented by the symbol  S Mother nature is a wild woman, and will always favor reactions that have an INCREASE in ENTROPY or a +  S. Low EntropyHigh Entropy

93. Rule 1 for Entropy: SOLID < LIQUID < GAS Entropy increases Least entropy  Most entropy

94. Ex. 2 C (s) + 3 H 2 (g)  C 2 H 6 (g) Ex. 4 Al (s) + 3 O 2 (g)  2 Al 2 O 3 (s) There is an INCREASE in entropy, so this reaction is favored. There is an DECREASE in entropy, so this reaction is unfavored.

95. Rule 2: When solids dissolve in a liquid, entropy increases… aqueous ions have more entropy than solid.

97. Putting it all together… In order for a reaction to be spontaneous all the time… you need an EXOTHERMIC reaction with an increase in entropy. Spontaneous = Lazy and Crazy For a reaction to be non-spontaneous all the time, you need an ENDOTHERMIC reaction with a decrease in entropy.

98. Ex. 2 C (s) + 3 H 2 (g)  C 2 H 6 (g) + 84.0kJ This reaction has a –  H.. because heat is given off as a product, it is an EXOTHERMIC reaction. This reaction starts with a solid and a gas… and ends up all gas. Therefore there is a +  S. Since both factors are favored, then this reaction is always spontaneous at all temperatures.

99. Ex. 66.4 kJ + N 2 (g) + 2O 2 (g)  2 NO 2 (g) This reaction has a +  H.. because heat is required as a reactant, it is an ENDOTHERMIC reaction. This reaction starts with a 3 moles of gases… and ends up with 2 moles of gas. Therefore there is a -  S. Since both factors are unfavored, then this reaction is never spontaneous, regardless of temperature.

100. Ex… is this rxn spontaneous? NaOH (s)  Na+(aq) + Cl-(aq) + 44.51 kJ Is it exothermic? Yes Is there an increase in entropy? Yes This reaction is spontaneous

101. Ex… Is this reaction spontaneous? Li+ (aq) + Br-(aq) + 48.83 kJ  LiBr (s) Is it exothermic? No Is there an increase in entropy? No Therefore, the reaction is definitely NOT spontaneous.

102. Check Your Understanding: Can you…  Define entropy and enthalpy  Determine whether entropy increases or decreases in a reaction  Determine whether a reaction will be spontaneous or not