1. 1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 6 Lecture Outline Prepared by Andrea D. Leonard University of Louisiana at Lafayette

2. 6.1Energy 2 Energy is the capacity to do work. Potential energy is stored energy. The law of conservation of energy states that the total energy in a system does not change. Energy cannot be created or destroyed. Kinetic energy is the energy of motion.

3. 6.1 Energy 3 Chemical bonds store potential energy. Reactions that form products having lower potential energy than the reactants are favored. A compound with lower potential energy is more stable than a compound with higher potential energy.

4. 6.1Energy A. The Units of Energy 4 A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1 o C. A joule (J) is another unit of energy. 1 cal = 4.184 J Both joules and calories can be reported in the larger units kilojoules (kJ) and kilocalories (kcal). 1,000 J = 1 kJ 1,000 cal = 1 kcal 1 kcal = 4.184 kJ

5. 6.2Energy Changes in Reactions 5 When molecules come together and react, bonds are broken in the reactants and new bonds are formed in the products. Bond breaking always requires an input of energy. Bond formation always releases energy. Cl To cleave this bond, 58 kcal/mol must be added. To form this bond, 58 kcal/mol is released.

6. 6.2Energy Changes in Reactions A. Bond Dissociation Energy 6  H is the energy absorbed or released in a reaction; it is called the heat of reaction or the enthalpy change. When energy is absorbed, the reaction is said to be endothermic and  H is positive (+). When energy is released, the reaction is said to be exothermic and  H is negative (−). Cl To form this bond,  H = −58 kcal/mol. To cleave this bond,  H = +58 kcal/mol.

7. 6.2Energy Changes in Reactions A. Bond Dissociation Energy 7 The bond dissociation energy is the  H for breaking a covalent bond by equally dividing the e − between the two atoms. Bond dissociation energies are positive values, because bond breaking is endothermic (  H > 0). Bond formation always has negative values, because bond formation is exothermic (  H < 0). HH H + H HH  H = +104 kcal/mol  H = −104 kcal/mol

8. 6.2 Energy Changes in Reactions A. Bond Dissociation Energy 8 The stronger the bond, the higher its bond dissociation energy. In comparing bonds formed from elements in the same group, bond dissociation energies generally decrease going down the column.

9. 6.2Energy Changes in Reactions B. Calculations Involving  H Values 9  H indicates the relative strength of the bonds broken and formed in a reaction. More energy is released in forming bonds than is needed to break the bonds. When  H is negative: The bonds formed in the products are stronger than the bonds broken in the reactants. CH 4 (g) + 2 O 2 (g)CO 2 (g) + 2 H 2 O(l)  H = −213 kcal/mol Heat is released

10. 6.2 Energy Changes in Reactions B. Calculations Involving  H Values 10 More energy is needed to break bonds than is released in the formation of new bonds. When  H is positive: The bonds broken in the reactants are stronger than the bonds formed in the products. 6 CO 2 (g) + 6 H 2 O(l)C 6 H 12 O 6 (aq) + 6 O 2 (g) ΔH = +678 kcal/mol Heat is absorbed

11. 6.2 Energy Changes in Reactions B. Calculations Involving  H Values 11

12. 6.3Energy Diagrams 12 For a reaction to occur, two molecules must collide with enough kinetic energy to break bonds. The orientation of the two molecules must be correct as well.

13. 6.3Energy Diagrams 13 E a, the energy of activation, is the difference in energy between the reactants and the transition state.

14. 6.3Energy Diagrams 14 The E a is the minimum amount of energy that the reactants must possess for a reaction to occur. E a is called the energy barrier and the height of the barrier determines the reaction rate. When the E a is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow. When the E a is low, many molecules have enough energy to cross the energy barrier, and the reaction is fast.

15. 6.3Energy Diagrams 15 The difference in energy between the reactants and the products is the  H. If  H is negative, the reaction is exothermic:

16. 6.3Energy Diagrams 16 If  H is positive, the reaction is endothermic:

17. 6.4Reaction Rates A. How Concentration and Temperature Affect Reaction Rate 17 Increasing the concentration of the reactants: Increases the number of collisions Increases the reaction rate Increasing the temperature of the reaction: Increases the kinetic energy of the molecules Increases the reaction rate

18. 6.4Reaction Rates B. Catalysts 18 A catalyst is a substance that speeds up the rate of a reaction. A catalyst is recovered unchanged in a reaction, and does not appear in the product. Catalysts accelerate a reaction by lowering E a without affecting  H.

19. 6.4Reaction Rates B. Catalysts 19 The uncatalyzed reaction (higher E a ) is slower. The catalyzed reaction (lower E a ) is faster.   H is the same for both reactions.

20. 6.4Reaction Rates C. Focus on the Human Body: Biological Catalysts 20 Enzymes (usually protein molecules) are biological catalysts held together in a very specific three- dimensional shape. The enzyme lactase converts the carbohydrate lactose into the two sugars glucose and galactose. People who lack adequate amounts of lactase suffer from abdominal cramping and diarrhea because they cannot digest lactose when it is ingested. The active site binds a reactant, which then under- goes a very specific reaction with an enhanced rate.

21. 6.5Equilibrium 21 A reversible reaction can occur in either direction. CO(g) + H 2 O(g)CO 2 (g) + H 2 (g) The forward reaction proceeds to the right. The forward reaction proceeds to the right. The reverse reaction proceeds to the left. The reverse reaction proceeds to the left. The system is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. The net concentrations of reactants and products do not change at equilibrium.

22. 6.5Equilibrium A. The Equilibrium Constant 22 The relationship between the concentration of the products and the concentration of the reactants is the equilibrium constant, K. Brackets, [ ], are used to symbolize concentration in moles per liter (mol/L). For the reaction: a A + b Bc C + d D equilibrium constant =K = [C] c [D] d [A] a [B] b = [products] [reactants]

23. 6.5Equilibrium A. The Equilibrium Constant 23 For the following balanced chemical equation: N 2 (g) + O 2 (g) 2 NO(g) equilibrium constant =K= [N 2 ] [O 2 ] [NO] 2 The coefficient becomes the exponent.

24. 6.5Equilibrium B. The Magnitude of the Equilibrium Constant 24 When K is much greater than 1, When K is much less than 1, [products] [reactants] The numerator is larger. [products] [reactants] The denominator is larger. equilibrium lies to the right and favors the products. equilibrium lies to the left and favors the reactants.

25. 6.5Equilibrium B. The Magnitude of the Equilibrium Constant 25 When K is around 1 (0.01 < K < 100), [products] [reactants] Both are similar in magnitude. both reactants and products are present in similar amounts.

26. 6.5Equilibrium B. The Magnitude of the Equilibrium Constant 26 For the reaction: 2 H 2 (g) + O 2 (g) 2 H 2 O(g)K = 2.9 x 10 82 The product is favored because K > 1. The equilibrium lies to the right.

27. 6.5Equilibrium B. The Magnitude of the Equilibrium Constant 27

28. 6.5Equilibrium C. Calculating the Equilibrium Constant 28 HOW TO Calculate the Equilibrium Constant for a Reaction Example Calculate K for the reaction between the general reactants A 2 and B 2. The equilibrium concentrations are as follows: [A 2 ] = 0.25 M[B 2 ] = 0.25 M[AB] = 0.50 M A 2 + B 2 2 AB

29. 6.5Equilibrium C. Calculating the Equilibrium Constant 29 HOW TO Calculate the Equilibrium Constant for a Reaction A 2 + B 2 2 AB Step [1] Write the expression for the equilibrium constant from the balanced equation. [AB] 2 [A 2 ][B 2 ] K =

30. 6.5Equilibrium C. Calculating the Equilibrium Constant 30 HOW TO Calculate the Equilibrium Constant for a Reaction Step [2] Substitute the given concentrations in the equilibrium expression and calculate K. [AB] 2 [A 2 ][B 2 ] K == [0.50] 2 [0.25][0.25] = 0.25 0.0625 = 4.0 The unit of the answer is always mol/L (or M), which is usually omitted.

31. 6.6Le Châtelier’s Principle 31 If a chemical system at equilibrium is disturbed or stressed, the system will react in a direction that counteracts the disturbance or relieves the stress. Some of the possible disturbances: 1) Concentration changes 2) Temperature changes 3) Pressure changes

32. 6.6Le Châtelier’s Principle A. Concentration Changes 32 2 CO(g) + O 2 (g)2 CO 2 (g) What happens if [CO(g)] is increased? The concentration of O 2 (g) will decrease. The concentration of CO 2 (g) will increase.

33. 6.6Le Châtelier’s Principle A. Concentration Changes 33 2 CO(g) + O 2 (g)2 CO 2 (g) What happens if [CO 2 (g)] is increased? The concentration of CO(g) will increase. The concentration of O 2 (g) will increase.

34. 6.6Le Châtelier’s Principle A. Concentration Changes 34 What happens if a product is removed? The concentration of ethanol will decrease. The concentration of the other product (C 2 H 4 ) will increase.

35. 6.6Le Châtelier’s Principle B. Temperature Changes 35 When the temperature is increased, the reaction that absorbs heat is favored. An endothermic reaction absorbs heat, so increasing the temperature favors the forward reaction.

36. 6.6Le Châtelier’s Principle B. Temperature Changes 36 Conversely, when the temperature is decreased, the reaction that adds heat is favored. An exothermic reaction releases heat, so increasing the temperature favors the reverse reaction.

37. 6.6Le Châtelier’s Principle C. Pressure Changes 37 When pressure increases, equilibrium shifts in the direction that decreases the number of moles in order to decrease pressure.

38. 6.6Le Châtelier’s Principle C. Pressure Changes 38 When pressure decreases, equilibrium shifts in the direction that increases the number of moles in order to increase pressure.

39. 6.6 Le Châtelier’s Principle Summary 39