1. Warm up u P 4 + N 2 O  P 4 O 6 + N 2 u Balance the equation. u What is the Limiting/Excess reactant for 12 mol P 4 and 14 mole N 2 O

2. Warm up u True or false: all chemical reactions release energy.

3. Thermochemistry Heat and Chemical Change

4. 4 Section 11.1 The Flow of Energy - Heat u OBJECTIVES: Explain the relationship between energy and heat.

5. 5 Section 11.1 The Flow of Energy - Heat u OBJECTIVES: Distinguish between heat capacity and specific heat.

6. 6 Energy and Heat u Thermochemistry - concerned with heat changes that occur during chemical reactions u Energy - capacity for doing work or supplying heat

7. 7 Energy and Heat u Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.

8. Exothermic and Endothermic Processes u In studying heat changes, think of defining these two parts: the system the surroundings

9. Exothermic and Endothermic Processes u The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed. All the energy is accounted for as work, stored energy, or heat.

10. Exothermic and Endothermic Processes u Heat flowing into a system from it’s surroundings: defined as positive q has a positive value called endothermic –system gains heat as the surroundings cool down

11. Exothermic and Endothermic Processes u Heat flowing out of a system into it’s surroundings: defined as negative q has a negative value called exothermic –system loses heat as the surroundings heat up

12. u GUMMY BEAR SACRIFICE

14. 14 Exothemic and Endothermic u Every reaction has an energy change associated with it u Exothermic reactions release energy, usually in the form of heat. u Endothermic reactions absorb energy u Energy is stored in bonds between atoms

15. Warm up u Define exothermic and give one CHEMICAL reaction example. u Define endothermic and give one CHEMICAL reaction example.

16. 16 Heat Capacity and Specific Heat u A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 o C. Used except when referring to food a Calorie, written with a capital C, always refers to the energy in food 1 Calorie = 1 kilocalorie = 1000 cal.

17. 17 Heat Capacity and Specific Heat u Joule-- the SI unit of heat and energy 4.184 J = 1 cal u Specific Heat Capacity - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 o C (abbreviated “C”)

18. 18 Heat Capacity and Specific Heat u For water, C = 4.18 J/(g o C), and also C = 1.00 cal/(g o C) u Thus, for water: it takes a long time to heat up, and it takes a long time to cool off! u Water is used as a coolant!

20. 20 Heat Capacity and Specific Heat u To calculate, use the formula: u q = mass (g) x  T x C u heat abbreviated as “q” u  T = change in temperature u C = Specific Heat u Units are either J/(g o C) or cal/(g o C)

21. q = mass (g) x  T x C u You have 250ml of water at 25 o C. In order to start your experiment you need the water to be exactly the same as a human core temperature, 37 o C. The specific heat of water is 1cal/(g o C). u How much heat energy(in calories) do you need to add to your sample?

22. q = mass (g) x  T x C u You are the scientist assigned to explore an asteroid, caught by NASA, to look for exotic materials. You have decided to use specific heat to identify your sample. u Mass = 100g u q = 100 cal u  T = 0.00 to 32.46 o C u What is the rock of the asteroid made of?

23. q = mass (g) x  T x C u A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron. Pg. T18

24. Warm up u Imagine how the graph of an endothermic and an exothermic reaction would look. u In your lab journal, make a line graph for each with energy on the y-axis and time on the x-axis.

25. 25 Section 11.2 Measuring and Expressing Heat Changes u OBJECTIVES: Construct equations that show the heat changes for chemical and physical processes.

26. 26 Section 11.2 Measuring and Expressing Heat Changes u OBJECTIVES: Calculate heat changes in chemical and physical processes.

27. Calorimetry u Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes.

28. Calorimetry u For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system

29. Calorimetry u Changes in enthalpy =  H u q =  H These terms will be used interchangeably u Thus, q =  H = m x C x  T u  H is negative for an exothermic reaction u  H is positive for an endothermic reaction

30. 30 C + O 2  CO 2 Energy ReactantsProducts  C + O 2 C O 2 395kJ + 395 kJ

31. 31 In terms of bonds C O O C O O Breaking this bond will require energy. C O O O O C Making these bonds gives you energy. In this case making the bonds gives you more energy than breaking them.

32. 32 Exothermic u The products are lower in energy than the reactants u Releases energy

33. 33 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ CaCO 3 + 176 kJ  CaO + CO 2

34. Graph these diagrams! 2H 2(g) + O 2(g) ---> 2H 2 O (g) + heat 2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O ΔH comb = -5320 kJ N 2(g) + 2 O 2(g) ----> 2 NO 2(g) ΔH = +67.6 kJ NH 3 (g) + HCl(g) ---> NH 4 Cl(s) ΔH = -176 kJ E a = 50kJ 4 NO(g) + 6 H 2 O(l) ---> 4 NH 3 (g) + 5 O 2 (g) ΔH = +1170 kJ E a = 456kJ

35. Graph these diagrams!

37. 37 Chemistry Happens in MOLES u An equation that includes energy is called a thermochemical equation  CH 4 + 2O 2  CO 2 + 2H 2 O + 802.2 kJ u 1 mole of CH 4 releases 802.2 kJ of energy. u When you make 802.2 kJ you also make 2 moles of water

38. 38 Thermochemical Equations u A heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 o C

39. 39 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u If 10. 3 grams of CH 4 are burned completely, how much heat will be produced? 10. 3 g CH 4 16.05 g CH 4 1 mol CH 4 802.2 kJ = 514 kJ

40. 40 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u How many liters of O 2 at STP would be required to produce 23 kJ of heat? u How many grams of water would be produced with 506 kJ of heat?

41. Summary, so far...

42. 42 Enthalpy u The heat content a substance has at a given temperature and pressure u Can’t be measured directly because there is no set starting point u The reactants start with a heat content u The products end up with a heat content u So we can measure how much enthalpy changes

43. 43 Enthalpy u Symbol is H  Change in enthalpy is  H (delta H) u If heat is released, the heat content of the products is lower  H is negative (exothermic) u If heat is absorbed, the heat content of the products is higher  H is positive (endothermic)

44. 44 Energy ReactantsProducts  Change is down  H is <0

45. 45 Energy ReactantsProducts  Change is up  H is > 0

46. 46 Heat of Reaction u The heat that is released or absorbed in a chemical reaction  Equivalent to  H  C + O 2 (g)  CO 2 (g) + 393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ u In thermochemical equation, it is important to indicate the physical state  H 2 (g) + 1/2O 2 (g)  H 2 O(g)  H = -241.8 kJ  H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -285.8 kJ

47. 47 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance u Note Table 11.4, page 305

48. 48 Section 11.3 Heat in Changes of State u OBJECTIVES: Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.

49. 49 Section 11.3 Heat in Changes of State u OBJECTIVES: Calculate heat changes that occur during melting, freezing, boiling, and condensing.

50. 50 Heats of Fusion and Solidification u Molar Heat of Fusion (  H fus ) - the heat absorbed by one mole of a substance in melting from a solid to a liquid u Molar Heat of Solidification (  H solid ) - heat lost when one mole of liquid solidifies

51. 51 Heats of Vaporization and Condensation u Molar Heat of Vaporization (  H vap ) - the amount of heat necessary to vaporize one mole of a given liquid. u Table 11.5, page 308

52. 52 Heats of Vaporization and Condensation u Molar Heat of Condensation (  H cond ) - amount of heat released when one mole of vapor condenses u  H vap = -  H cond

53. 53 Heats of Vaporization and Condensation u Note Figure 11.5, page 310 u The large values for  H vap and  H cond are the reason hot vapors such as steam is very dangerous You can receive a scalding burn from steam when the heat of condensation is released!

54. 54 Heats of Vaporization and Condensation u H 2 0 (g)  H 2 0 (l)  H cond = - 40.7kJ/mol u Sample Problem 11-5, page 311

55. 55 Heat of Solution u Heat changes can also occur when a solute dissolves in a solvent. u Molar Heat of Solution (  H soln ) - heat change caused by dissolution of one mole of substance

56. 56 Section 11.4 Calculating Heat Changes u OBJECTIVES: Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.

57. 57 Section 11.4 Calculating Heat Changes u OBJECTIVES: Calculate heat changes using standard heats of formation.

58. 58 Hess’s Law u If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Called Hess’s law of heat summation u Example shown on page 314 for graphite and diamonds

59. 59 Why Does It Work? u If you turn an equation around, you change the sign:  If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ  then, H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ u also, u If you multiply the equation by a number, you multiply the heat by that number:  2 H 2 O(g)  H 2 (g) + O 2 (g)  H =+571.0 kJ

60. 60 Standard Heats of Formation  The  H for a reaction that produces 1 mol of a compound from its elements at standard conditions u Standard conditions: 25°C and 1 atm. u Symbol is u The standard heat of formation of an element = 0 u This includes the diatomics

61. 61 What good are they? u Table 11.6, page 316 has standard heats of formation u The heat of a reaction can be calculated by: subtracting the heats of formation of the reactants from the products HoHo = (Products) -(Reactants)

62. 62 Examples  CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = - 74.86 kJ/molO 2 (g) = 0 kJ/molCO 2 (g) = - 393.5 kJ/molH 2 O(g) = - 241.8 kJ/mol u  H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]  H= - 802.4 kJ