1. Energy The ability to do work or produce heat The ability to do work or produce heat Potential- Stored energy Potential- Stored energy Energy stored in chemical bonds Energy stored in chemical bonds Kinetic- Energy of movement Kinetic- Energy of movement Energy of moving molecules Energy of moving molecules

2. Law of Conservation of Energy Energy can not be created or destroyed, it can only be converted from one form to another. Energy can not be created or destroyed, it can only be converted from one form to another.

3. Chemical Potential Energy Chemical bonds of a compound contain energy Chemical bonds of a compound contain energy That energy can be converted to do work or is converted to heat That energy can be converted to do work or is converted to heat

4. Heat Heat flows from warm objects to cooler objects. Heat flows from warm objects to cooler objects. The two objects will reach equilibrium and record the same temperature The two objects will reach equilibrium and record the same temperature

5. Measuring Heat Calorie- The amount of energy required to raise 1 g of water 1 o C Calorie- The amount of energy required to raise 1 g of water 1 o C 1 food Calorie = 1000 calories 1 food Calorie = 1000 calories Joule- The unit of measurement for heat energy Joule- The unit of measurement for heat energy 1 Joule =.2390 calories 1 Joule =.2390 calories

6. Problem How Joules of heat would be obtained from a 364 nutritional Calorie hamburger? How Joules of heat would be obtained from a 364 nutritional Calorie hamburger? See chart on p. 491 for conversions See chart on p. 491 for conversions

7. Answer 64 Cal x 1000 cal x 1 J = 1523012 J 1 Cal.2390 cal 64 Cal x 1000 cal x 1 J = 1523012 J 1 Cal.2390 cal

8. Specific Heat The amount of energy required to raise 1 gram of any substance 1 o C The amount of energy required to raise 1 gram of any substance 1 o C Specific heat determines how fast or slow that substances gain or lose heat. Specific heat determines how fast or slow that substances gain or lose heat.

9. Specific Heat and Water Water has a high specific heat Water has a high specific heat It gains and loses heat energy very slowly It gains and loses heat energy very slowly This is why bodies of water stay cool long into spring and warm long into fall This is why bodies of water stay cool long into spring and warm long into fall

10. Calculating Heat Change q = c x m x  T q = c x m x  T q = Heat gain or loss- J q = Heat gain or loss- J c = Specific heat - J/(g. o C) c = Specific heat - J/(g. o C) m = mass g m = mass g  T = Change in temperature- o C  T = Change in temperature- o C

11. Calculating Heat Change How much heat is required to raise 2.3 kg of iron 23 o C? How much heat is required to raise 2.3 kg of iron 23 o C? q = c x m x  T q = c x m x  T q =.449 J/(g. o C) x 2300g x 23 o C q =.449 J/(g. o C) x 2300g x 23 o C q = 23752.1 J q = 23752.1 J

12. Calculating Heat Change If there is more than one substance you will need to calculate the heat change separately If there is more than one substance you will need to calculate the heat change separately Ex. Water in an iron pot would require calculations for both iron and water Ex. Water in an iron pot would require calculations for both iron and water

13. Determining Specific Heat 1. Heat the unknown substance to a set temperature 1. Heat the unknown substance to a set temperature 2. Place the substance in a measured mass of water 2. Place the substance in a measured mass of water 3. Measure the temperature change when equilibrium is reached 3. Measure the temperature change when equilibrium is reached

14. Determining Specific Heat 4. Calculate the amount of energy gained by the water. This equals the amount of energy lost by the substance 4. Calculate the amount of energy gained by the water. This equals the amount of energy lost by the substance Use the heat value to solve for the specific heat of the substance Use the heat value to solve for the specific heat of the substance

15. Calorimeter Lab Calorimeter- A device used to measure the amount of heat gained or lost by a substance Calorimeter- A device used to measure the amount of heat gained or lost by a substance See handout for instructions See handout for instructions

16. Thermochemistry The study of heat changes involved in chemical reactions The study of heat changes involved in chemical reactions Reactions are either exothermic (release energy) or endothermic (absorb energy) Reactions are either exothermic (release energy) or endothermic (absorb energy)

17. The Universe A system plus its surroundings A system plus its surroundings The System Surroundings

18. Enthalpy (H) The heat content of a substance at a constant pressure The heat content of a substance at a constant pressure There is no way to measure the total enthalpy of a substance but you can calculate the change in enthalpy for a reaction There is no way to measure the total enthalpy of a substance but you can calculate the change in enthalpy for a reaction

19. Enthalpy Changes To calculate enthalpy change use the following equation To calculate enthalpy change use the following equation  H rxn = H product - H reactants  H rxn = H product - H reactants If  H rxn is positive the reaction is endothermic, if negative the reaction is exothermic If  H rxn is positive the reaction is endothermic, if negative the reaction is exothermic

20. Enthalpy change Exothermic reaction Exothermic reaction EnthalpyEnthalpy 4Fe(s) + 3O 2 (g) Fe 2 O 3 (s)  H= -1625 Product Reactants

21. Enthalpy change Endothermic reaction Endothermic reaction EnthalpyEnthalpy NH 4 + (aq) + NO 3 - (aq) Product NH 4 NO 3 (s) Reactant  H= 27 kJ

22. Heat Gain and Enthalpy q = c x m x  T q = c x m x  T  H = H product - H reactants  H = H product - H reactants q =  H since heat gain or loss is the same as enthalpy change at constant pressure q =  H since heat gain or loss is the same as enthalpy change at constant pressure

23. Writing Thermochemical Equations 1. Write the equation 1. Write the equation 2. Write the  H value to the right of the equation 2. Write the  H value to the right of the equation

24. Writing Thermochemical Equations 3. If the reaction is exothermic the  H value is negative, if endothermic, positive 3. If the reaction is exothermic the  H value is negative, if endothermic, positive

25. Enthalpy of Combustion The amount of energy released when one mole of a substance is burned The amount of energy released when one mole of a substance is burned  H o comb = Heat of combustion  H o comb = Heat of combustion

26. Changes of State  H vap = Heat of vaporization  H vap = Heat of vaporization  H cond = Heat of condensation  H cond = Heat of condensation  H fus = Heat of fusion  H fus = Heat of fusion  H solid = Heat of solid  H solid = Heat of solid

27. Changes of state Changing phase from a solid to a liquid, liquid to gas or vice versa requires a change in energy Changing phase from a solid to a liquid, liquid to gas or vice versa requires a change in energy

28. Changes of state The amount of energy required to convert one mole of a substance from a liquid to a gas or from a liquid to a solid The amount of energy required to convert one mole of a substance from a liquid to a gas or from a liquid to a solid

29. Equations for phase changes H 2 O(l) --> H 2 O(g)  H vap =40.7 kJ H 2 O(l) --> H 2 O(g)  H vap =40.7 kJ H 2 O(g) --> H 2 O(l)  H vap =-40.7 kJ H 2 O(g) --> H 2 O(l)  H vap =-40.7 kJ H 2 O(s) --> H 2 O(l)  H fus = 6.01 kJ H 2 O(s) --> H 2 O(l)  H fus = 6.01 kJ H 2 O(l) --> H 2 O(s)  H vap =-6.01 kJ H 2 O(l) --> H 2 O(s)  H vap =-6.01 kJ

30. Assignment P. 500 14, 15, 16, 17 P. 500 14, 15, 16, 17 p. 504 20-22 p. 504 20-22

31. Lab- Energy Change P. 503 P. 503 Use Microsoft Excel to generate graph Use Microsoft Excel to generate graph

32. Hess’s Law States that if you can add 2 or more equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions = the enthalpy for the final reaction. States that if you can add 2 or more equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions = the enthalpy for the final reaction.

33. Hess’s Law What does this mean? What does this mean? You can predict enthalpy changes for reactions that can not be observed directly. You can predict enthalpy changes for reactions that can not be observed directly. Ex. A reaction that takes several thousand years. Converting carbon to a diamond. Ex. A reaction that takes several thousand years. Converting carbon to a diamond.

34. Applying Hess’s Law What is the enthalpy change for the following reaction? What is the enthalpy change for the following reaction? 2S(s) + 3O 2 (g)  2SO 3 (g) 2S(s) + 3O 2 (g)  2SO 3 (g)

35. Applying Hess’s Law We know the following equations We know the following equations S(s) + O 2 (g)  SO 2 (g) S(s) + O 2 (g)  SO 2 (g)  H=-297 kJ  H=-297 kJ 2SO 3 (g)  2SO 2 (g) + O 2 (g)  H=198 kJ 2SO 3 (g)  2SO 2 (g) + O 2 (g)  H=198 kJ

36. Applying Hess’s Law 1. You need to change the coefficients in the two equations to match the molar amounts in the original equation 1. You need to change the coefficients in the two equations to match the molar amounts in the original equation

37. Applying Hess’s Law 2S(s) + 3O 2 (g)  2SO 3 (g) 2S(s) + 3O 2 (g)  2SO 3 (g) S(s) + O 2 (g)  SO 2 (g) S(s) + O 2 (g)  SO 2 (g)  H=-297 kJ  H=-297 kJ Since there are 2 mol of S in the original equation the second equation must by multiplied by a factor of 2. Since there are 2 mol of S in the original equation the second equation must by multiplied by a factor of 2.

38. Applying Hess’s Law 2 ( S(s) + O 2 (g)  SO 2 (g)  H=-297 kJ ) 2 ( S(s) + O 2 (g)  SO 2 (g)  H=-297 kJ ) 2 S(s) + 2O 2 (g)  2SO 2 (g) 2 S(s) + 2O 2 (g)  2SO 2 (g)  H=2(-297 kJ) = -594 kJ  H=2(-297 kJ) = -594 kJ Now we have the amount of heat required for the reaction of 2 moles of S Now we have the amount of heat required for the reaction of 2 moles of S

39. Applying Hess’s Law Since SO3 is the product in the original reaction we have to reverse the second reaction from 2SO 3 (g)  2SO 2 (g) + O 2 (g)  H=198 kJ to Since SO3 is the product in the original reaction we have to reverse the second reaction from 2SO 3 (g)  2SO 2 (g) + O 2 (g)  H=198 kJ to 2SO 2 (g) + O 2 (g)  2SO 3 (g)  H= -198 kJ 2SO 2 (g) + O 2 (g)  2SO 3 (g)  H= -198 kJ

40. Applying Hess’s Law 2S(s) + 2O 2 (g)  2SO 2 (g)  H= -594 kJ 2S(s) + 2O 2 (g)  2SO 2 (g)  H= -594 kJ 2SO 2 (g) + O 2 (g)  2SO 3 (g)  H= -198 kJ 2SO 2 (g) + O 2 (g)  2SO 3 (g)  H= -198 kJ 2S+2O 2 +2SO 2 + O 2  2SO 2 +2SO 3  H= -792kJ 2S+2O 2 +2SO 2 + O 2  2SO 2 +2SO 3  H= -792kJ 2S + 3O 2  2SO 3  H= -792 kJ 2S + 3O 2  2SO 3  H= -792 kJ So the original reaction evolves 792 kJ So the original reaction evolves 792 kJ

41. Assignment P. 508 # 28 and 29 P. 508 # 28 and 29

42. Enthalpy of formation Elements are free atoms. It takes no heat to form them. They are in the lowest energy state possible. Elements are free atoms. It takes no heat to form them. They are in the lowest energy state possible. Elements are assigned a value of 0 kJ for  H f since there is no energy needed to create them because they are the simplest substances on Earth Elements are assigned a value of 0 kJ for  H f since there is no energy needed to create them because they are the simplest substances on Earth

43. Enthalpy of formation Compounds are formed by combining elements. Energy is required to combine elements, therefore the  H has a value greater than 0 if elements form a compound. Compounds are formed by combining elements. Energy is required to combine elements, therefore the  H has a value greater than 0 if elements form a compound. This is an endothermic reaction. This is an endothermic reaction.

44. Enthalpy of formation  Hrxn =  Hf (product) -  Hf (reactant)  Hrxn =  Hf (product) -  Hf (reactant) Add up all of the  H f products and subtract the sum of the  H f reactants Add up all of the  H f products and subtract the sum of the  H f reactants This equals the net  H of the reaction This equals the net  H of the reaction

45. Reaction Spontaneity Spontaneous Process- a physical or chemical change that occurs with no outside intervention Spontaneous Process- a physical or chemical change that occurs with no outside intervention Most exothermic reactions are spontaneous and most endothermic reactions are not spontaneous. Most exothermic reactions are spontaneous and most endothermic reactions are not spontaneous.

46. Reaction Spontaneity Not all exothermic reactions are spontaneous and not all endothermic reactions are not spontaneous Not all exothermic reactions are spontaneous and not all endothermic reactions are not spontaneous Ice melting is spontaneous but requires energy to occur. This makes in endothermic and spontaneous Ice melting is spontaneous but requires energy to occur. This makes in endothermic and spontaneous

47. Reaction Spontaneity What accounts for an endothermic process being spontaneous? What accounts for an endothermic process being spontaneous? Entropy Entropy

48. Entropy The measure of the randomness and disorder of a system The measure of the randomness and disorder of a system Substances are more likely to exist in a high state of randomness than in a low state of randomness Substances are more likely to exist in a high state of randomness than in a low state of randomness

49. Entropy Substances are more likely to have high entropy than low entropy Substances are more likely to have high entropy than low entropy Spontaneous processes always proceed in such a way that the entropy of the universe increases Spontaneous processes always proceed in such a way that the entropy of the universe increases

50. Entropy Think of a deck of cards Think of a deck of cards What are the odds that well shuffled cards will come out in order? What are the odds that well shuffled cards will come out in order?

51. Entropy Not very high- 1 in 807,000,000,000,000,000,0 00,000,000,000,000,000,00 0,000,000,000,000,000,000, 000,000,000,000,000 Not very high- 1 in 807,000,000,000,000,000,0 00,000,000,000,000,000,00 0,000,000,000,000,000,000, 000,000,000,000,000 Random is favored Random is favored

52. Entropy  S system = S products – S reactants  S system = S products – S reactants If entropy increases in a reaction the entropy is positive If entropy increases in a reaction the entropy is positive If entropy decreases then entropy is negative If entropy decreases then entropy is negative

53. Predicting entropy changes If substances change phase entropy increases as the energy of the substance increases If substances change phase entropy increases as the energy of the substance increases As you go from a solid to a liquid to a gas entropy increases As you go from a solid to a liquid to a gas entropy increases

54. Predicting entropy changes If a gas is dissolved in a solvent entropy decreases If a gas is dissolved in a solvent entropy decreases Entropy increases when the number of gas particles as a product is greater than the number of gas particles as a reactant Entropy increases when the number of gas particles as a product is greater than the number of gas particles as a reactant

55. Predicting entropy changes Entropy increases as solids are dissolved in a solvent Entropy increases as solids are dissolved in a solvent Entropy increases as temperature increases Entropy increases as temperature increases

56. Entropy and Free Energy  S universe > 0  S universe > 0  S universe =  S system +  S surroundings  S universe =  S system +  S surroundings If a reaction increases the entropy of the universe then it will be spontaneous If a reaction increases the entropy of the universe then it will be spontaneous

57. Entropy and Free Energy Gibbs free energy determines if a reaction is spontaneous Gibbs free energy determines if a reaction is spontaneous  G system =  H system - T  S system  G system =  H system - T  S system If  G is negative the reaction is spontaneous, if positive the reaction is not spontaneous If  G is negative the reaction is spontaneous, if positive the reaction is not spontaneous

58. P. 517, 518 Copy chart 16-8, 16-9 Copy chart 16-8, 16-9