1. Chapter 7: Hot & Cold Packs

2. Introductory Activity How many things can you think of in everday life that either give off heat or absorb heat? Which of these things are physical processes? Which are chemical processes?

3. Hot & Cold Packs This chapter will introduce the chemistry needed to understand how Hot & Cold Packs work  Section 7.1: Endothermic & Exothermic  Section 7.2: Calorimetry and heat capacity  Section 7.3: Changes in State  Section 7.4: Heat of a Chemical Reaction  Section 7.5: Hess’s Law

4. Hot/Cold Packs Transfer of energy Use System & Surroundings between Materials ability to absorb energy without noticeable temperature change Effect on temperature depends on Physical change Is determined with Calorimetry Can be done in Chemical change

5. Section 7.1—Endothermic and Exothermic Why are hot packs “hot” and cold packs feel “cold”

6. Endothermic & Exothermic When the system absorbs energy from the surroundings, it’s an endothermic process When the system releases energy to the surroundings, it’s an exothermic process

7. System & Surroundings It’s very important to define the system & surroundings correctly to use the endo- and exothermic definitions! Most people define the system too broadly  They include everything in the beaker or container as the system However, the system is only the molecules undergoing the change

8. System & Surroundings The system is only made of the molecules undergoing the change The water molecules & the container… Your hand and the air… Even the thermometer… They are all the surroundings Note that water is made up of water molecules—not a solid chunk of water…but for this picture, it’s best to represent water as one thing since it’s the surroundings and focus on the molecules reacting as the system.

9. Exothermic & You You touch the beaker and it feels hot Energy is being transferred TO YOU You are the surroundings When energy moves from system to surroundings, it’s exothermic

10. Exothermic & the Thermometer The temperature (measured by the thermometer) is related to the average kinetic energy of the molecules in the container The majority of the molecules in a solution are water If the temperature is increasing, the energy of the water molecules is increasing Since water is the surrounding (it’s not actually reacting), energy is being transferred to the surroundings Exothermic shows an increase in the temperature within the container

11. Endothermic The opposite is also true If the container feels cold to you, energy is being transferred FROM YOU (the surroundings) into the system— endothermic If the thermometer goes down, energy is being transferred FROM the water molecules (surroundings) into the system-- endothermic

12. Let’s Practice Example: Identify the system and surroundings when you hold an ice cube while it melts. Is this endo- or exothermic?

13. Let’s Practice Example: Identify the system and surroundings when you hold an ice cube while it melts. Is this endo- or exothermic? System: Water molecules in the form of ice Surroundings: You and the air It feels cold to you…so energy is leaving you (surroundings) When energy goes from surroundings to system it’s endothermic

14. Section 7.2—Calorimetry & Heat Capacity Why do some things get hot more quickly than others?

15. Temperature Temperature – proportional to the average kinetic energy of the molecules Energy due to motion (Related to how fast the molecules are moving) As temperature increases Molecules move faster

16. Heat & Enthalpy Heat (q)– The flow of energy from higher temperature particles to lower temperature particles Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy” Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume

17. Energy Units The most common energy units are Joules (J) and calories (cal) 4.18 J1.00 cal 1000 J 1000 cal 1 kJ 1 Cal (food calorie) = = = Energy Equivalents These equivalents can be used in dimensional analysis to convert units

18. Heat Capacity Specific Heat Capacity (C p ) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C

19. Heat Capacity High Heat CapacityLow Heat Capacity Takes a large amount of energy to noticeably change temp Small amount of energy can noticeably change temperature Heats up slowly Cools down slowly Maintains temp better with small condition changes Heats up quickly Cools down quickly Quickly readjusts to new conditions A pool takes a long time to warm up and remains fairly warm over night. The air warms quickly on a sunny day, but cools quickly at night A cast-iron pan stays hot for a long time after removing from oven. Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly

20. What things affect temperature change? Heat Capacity of substance  The higher the heat capacity, the slower the temperature change Mass of sample  The larger the mass, the more molecules there are to absorb energy, so the slower the temperature change Energy added or removed Mass of sample Specific heat capacity of substance Change in temperature

21. Positive & Negative  T Change in temperature (  T) is always T 2 – T 1 (final temperature – initial temperature)  If temperature increases,  T will be positive A substance goes from 15°C to 25°C. 25°C - 15°C = 10°C This is an increase of 10°C  If temperature decreases,  T will be negative A substance goes from 50°C to 35°C 35°C – 50°C = -15°C This is a decrease of 15°C

22. Positive & Negative  H Energy must be put in for temperature to increase  A “+”  T will have a “+”  H Energy must be removed for temperature to decrease  A “-”  T will have a “-”  H

23. Example Example: If 285 J is added to 45 g of water at 25°C, what is the final temperature? Cp water = 4.18 J/g°C

24. Let’s Practice #1 Example: How many joules must be removed from 25 g of water at 75°C to drop the temperature to 30°? Cp water = 4.18 J/g°C

25. Let’s Practice #2 Example: If the specific heat capacity of aluminum is 0.900 J/g°C, what is the final temperature if 437 J is added to a 30.0 g sample at 15°C

26. Calorimetry

27. 1 st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes This is also referred to as the Law of Conservation of Energy Conservation of Energy If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

28. Calorimetry Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system  H surroundings = -  H system Because of the Law of Conservation of Energy, The energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system. (m×Cp×  T) surroundings = - (m×Cp×  T) system Don’t forget the “-” sign on one side Make sure to keep all information about surroundings together and all information about system together—you can’t mix and match!

29. Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature Two objects at different temperatures So you know that T 2 for the system is the same as T 2 for the surroundings!

30. An example of Calorimetry Example: A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal?

31. Metal: m = 23.8 g T 1 = 100.0°C T 2 = 32.5°C C p = ? Water: m = 50.0 g T 1 = 24°C T 2 = 32.5°C C p = 4.18 J/g°C C p = 1.04 J/g°C An example of Calorimetry Example: A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal?

32. Let’s Practice #3 Example: A 10.0 g of aluminum (specific heat capacity is 0.900 J/g°C) at 95.0°C is placed in a container of 100.0 g of water (specific heat capacity is 4.18 J/g°C) at 25.0°. What’s the final temperature?

33. Section 7.3—Changes in State What’s happening when a frozen ice pack melts?

34. The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature) Change in State To melt or boil, intermolecular forces must be broken Breaking intermolecular forces requires energy A sample with solid & liquid will not rise above the melting point until all the solid is gone. The same is true for a sample of liquid & gas

35. Melting When going from a solid to a liquid, some of the intermolecular forces are broken The Enthalpy of Fusion (H fus ) is the amount of energy needed to melt 1 gram of a substance  The enthalpy of fusion of water is 80.87 cal/g or 334 J/g All samples of a substance melt at the same temperature, but the more you have the longer it takes to melt (requires more energy). Energy needed to melt Mass of the sample Energy needed to melt 1 g

36. Example Example: Find the enthalpy of fusion of a substance if it takes 5175 J to melt 10.5 g of the substance.

37. Vaporization When going from a liquid to a gas, all of the rest of the intermolecular forces are broken The Enthalpy of Vaporization (H vap ) is the amount of energy needed to boil 1 gram of a substance  The H vap of water is 547.2 cal/g or 2287 J/g All samples of a substance boil at the same temperature, but the more you have the longer it takes to boil (requires more energy). Energy needed to boil Mass of the sample Energy needed to boil 1 g

38. Example Example: If the enthalpy of vaporization of water is 547.2 cal/g, how many calories are needed to boil 25.0 g of water?

39. Solid Liquid Gas Melting Vaporizing or Evaporating Condensing Freezing Increasing molecular motion (temperature) Changes in State go in Both Directions

40. Going the other way The energy needed to melt 1 gram (H fus ) is the same as the energy released when 1 gram freezes.  If it takes 547 J to melt a sample, then 547 J would be released when the sample freezes.  H will = -547 J The energy needed to boil 1 gram (H vap ) is the same as the energy released when 1 gram is condensed.  If it takes 2798 J to boil a sample, then 2798 J will be released when a sample is condensed.  H will = -2798 J

41. Example Example: How much energy is released with 157.5 g of water is condensed? H vap water = 547.2 cal/g

42. Heating Curves Melting & Freezing Point Boiling & Condensing Point Heating curves show how the temperature changes as energy is added to the sample

43. Going Up & Down +H+H -H-H Moving up the curve requires energy, while moving down releases energy

44. States of Matter on the Curve Gas Only Energy added increases temp Liquid & gas Energy added breaks remaining IMF’s Liquid Only Energy added increases temp Solid & Liquid Energy added breaks IMF’s Solid Only Energy added increases temp

45. Different Heat Capacities Gas Only Cp = 0.48 cal/g°C Liquid Only Cp = 1.00 cal/g°C Solid Only Cp = 0.51 cal/g°C The solid, liquid and gas states absorb water differently—use the correct Cp!

46. Changing States Liquid & gas H vap = 547.2 cal/g Solid & Liquid H fus = 80.87 cal/g

47. Adding steps together If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice to zero before it could melt. Then you’d melt the ice Then you’d warm that water from 0°C to your final 75° You can calculate the enthalpy needed for each step and then add them together

48. Example: How many calories are needed to change 15.0 g of ice at -12.0°C to steam at 137.0°C? Example Useful information: Cp ice = 0.51 cal/g°C Cp water = 1.00 cal/g°C Cp steam = 0.48 cal/g°C H fus = 80.87 cal/g H vap = 547.2 cal/g

49. Example: How many needed to change 40.5 g of water at 25°C to steam at 142°C? Let’s Practice Useful information: Cp ice = 0.51 cal/g°C Cp water = 1.00 cal/g°C Cp steam = 0.48 cal/g°C H fus = 80.87 cal/g H vap = 547.2 cal/g

50. Section 7.4—Energy of a Chemical Reaction What’s happening in those hot/cold packs that contain chemical reactions?

51. Enthalpy of Reaction Enthalpy of Reaction (  H rxn ) – Net energy change during a chemical reaction +  H rxn means energy is being added to the system—endothermic -  H rxn means energy is being released from the system—exothermic

52. Enthalpy of Formation Enthalpy of Formation (H f ) – Energy change when 1 mole of a compound is formed from elemental states Heat of formation equations: H 2 (g) + ½ O 2 (g)  H 2 O (g) C (s) + O 2 (g)  CO 2 (g) A table with Enthalpy of Formation values can be found in the Appendix of your text Be sure to look up the correct state of matter: H 2 O (g) and H 2 O (l) have different H f values!

53. The overall enthalpy of reaction is the opposite of H f for the reactants and the H f for the products Reactants are broken apart and Products are formed. Breaking apart reactants is the opposite of Enthalpy of Formation. Forming products is the Enthalpy of Formation.  H rxn = sum of H f of all products – the sum of H f reactants Enthalpy of Formation & Enthalpy of Reaction This is not the way a reaction occurs—reactants break apart and then rearrange…remember Collision Theory from Chpt 2! But for when discussing overall energy changes, this manner of thinking is acceptable.

54. Example Example: Find the  Hrxn for: CH 4 (g) + 2 O 2 (g)  2 H 2 O (g) + CO 2 (g) H f (kJ/mole) CH 4 (g)-74.81 O 2 (g)0 H 2 O (g)-241.8 CO 2 (g)-393.5

55. Let’s Practice #1 H f (kJ/mole) CH 3 OH (l)-238.7 O 2 (g)0 H 2 O (l)-285.8 CO 2 (g)-393.5 Example: Find the  Hrxn for: 2 CH 3 OH (l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O (l)

56. Enthalpy & Stoichiometry The Enthalpy of Reaction can be used along with the molar ratio in the balanced chemical equation This allows Enthalpy of Reaction to be used in stoichiometry equalities

57. Example: If 1275 kJ is released, how many grams of B 2 O 3 is produced? B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (s) + 2 H 2 O (g)  H = -2035 kJ Example

58. Let’s Practice #2 If you need to produce 47.8 g B 2 O 3, how many kilojoules will be released? B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (s) + 2 H 2 O (g)  H = -2035 kJ

59. Section 7.5—Hess’s Law How can we find the enthalpy of a reaction using step-wise reactions?

60. Hess’s Law Hess’ Law – The sum of the energy changes during a series of reactions is equal to the sum of the reaction. In other words…if you go from Reactant A to Product Z all in one step, you will have the same total energy change as someone that went from A to Z in 7 step—the energy from each of their 7 steps would add up to your 1 step energy change.

61. Example 1 Label each step-wise equations with letters (“a”, “b”, “c”) if not already done for you. Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc

62. For the first reactant in the overall reaction, find the step-wise reaction that has the same chemical and the same state of matter. It doesn’t have to be on the reactants side of the step-wise equation If it is on the correct “side” write it as is. Write it’s label beside it, too (“a”, “b”) If it’s on the wrong “side”, flip the equation. If you flip it, write it’s label as “-a” or “-b”. 2 Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Example

63. Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one. 3 Example

64. Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one. 3 -b 2HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Example

65. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Use any un-used step-wise equations to get rid of unwanted things. Putting them on opposite sides will allow them to cancel. 4 -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Example

66. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation. Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation. 5 2 2 6 2 4 Example

67. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation. Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation. 5 2 2 6 2 4 2 4 22 Example

68. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation. Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation. 5 2 2 6 2 4 2 4 22 5 2 N 2 (g) + 5 O 2 (g)  2 N 2 O 5 (g) Example

69. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) 2 2 6 2 4 2 4 22 5 2 N 2 (g) + 5 O 2 (g)  2 N 2 O 5 (g) Use the step-wise “labels” as a math expression for solving for  H rxn. 6 2 × (-74.1 kJ) 2 × -(76.6 kJ) 1 × -(-571.6 kJ) 270.2 kJ Example

70. What did you learn about hot/cold packs?

71. Hot/Cold Packs Transfer of energy Use System & Surroundings between Materials ability to absorb energy without noticeable temperature change Effect on temperature depends on Physical change Is determined with Calorimetry Can be done in Chemical change