1. Chapter 17 Thermochemistry Section 17.1 The Flow of Energy

2. u Thermochemistry - studies energy changes in reactions u q is representing the quantity of heat u Heat flows from high temperature to low temperature 2

3. The Universe u Can be divided into 2 pieces u System- the part you are investigating u Surroundings- the rest of the universe u Law of conservation of energy- Energy can’t be created or destroyed The energy of the universe is constant Energy change of System + Energy change of surroundings = 0 3

4. Exothermic u System releases energy u Heat flows out u Surroundings get hotter u q is negative System Surroundings Energy 4

5. Endothermic u System absorbs energy u Heat flows in u Surroundings get cooler u q is positive System Surroundings Energy 5

6. 6 Units of Energy u Energy is measured in Joules or calories u calorie is amount of heat to change 1 g of water by 1  C u Food Calories are kilocalories 1 Calorie = 1000 calories u 1 cal = 4.184 J u Enthalpy (H) – heat content at constant pressure  H  H = q u If temperature goes up exothermic

7. 7 Thermochemistry u Every reaction has an energy change associated with it u Energy is stored in bonds between atoms u Making bonds gives energy u Breaking bonds takes energy

8. 8 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

9. 9 Exothermic u The products are lower in energy than the reactants u Releases energy u Often releases heat

10. 10 C + O 2  CO 2 Energy ReactantsProducts  C + O 2 CO 2 -395kJ + 395 kJ

11. 11 Endothermic u The products are higher in energy than the reactants u Absorbs energy u Absorbs heat

12. 12 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 +176 kJ CaCO 3 + 176 kJ  CaO + CO 2

13. Section 17.2 Measuring and Expressing Enthalpy Changes 13

14. 14 Chemistry Happens in u An equation that includes energy is called a thermochemical equation  CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u Energy is a product in this example u 1 mole of CH 4 makes 802.2 kJ of energy. u When you make 802.2 kJ you make 2 moles of water

15. 15 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? x = 515 kJ 1 mol 802.2 kJ 0.642 mol x

16. 16 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?

17. 17 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u How many grams of water would be produced with 506 kJ of heat?

18. Section 17.3 Heat in Change of State 18

19. 19 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 say what state  H 2 (g) + ½ O 2 (g)  H 2 O(g)  H = - 241.8 kJ  H 2 (g) + ½ O 2 (g)  H 2 O(l)  H = - 285.8 kJ

20. 20 Energy ReactantsProducts Change is down  H is <0 + heat

21. 21 Energy ReactantsProducts Change is up  H is > 0 Reactants + heat

22. Choose all that apply... C(s) + 2 S(g)  CS 2 ( 1 )  H = 89.3 kJ Which of the following are true? A) This reaction is exothermic B) It could also be written C(s) + 2 S(g) + 89.3 kJ  CS 2 (l) C) The products have higher energy than the reactants 22

23. 23 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance at 25  C and 1 atm  C 2 H 4 + 3 O 2  2 CO 2 + 2 H 2 O  C 2 H 6 + O 2  CO 2 + H 2 O  2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O  C 2 H 6 + (7/2) O 2  2 CO 2 + 3 H 2 O u Always exothermic

24. Heat and phase change endothermic u Melting and vaporizing are endothermic Breaking things apart exothermic u Freezing and condensing are exothermic Forming connections 24

25. Heat of Solution u  H soln - heat change when one mole of solute is dissolved. u q =  H soln x n u Sometimes endothermic Ammonium nitrate for cold packs u Sometimes exothermic Acids and bases 25

26. 26 Standard Heat 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 is 0 u This includes the diatomics

27. 27 What good are they? u There are tables (pg. 530) of heats of formations u For most compounds it is negative Because you are making bonds Making bonds is exothermic u The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products

28. 28 Example  CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = -74.86 kJO 2 (g) = 0 kJCO 2 (g) = - 393.5 kJH 2 O(g) = - 241.8 kJ   H = [-393.5 kJ + 2 (-241.8 kJ)] - [-74.86 kJ + 2 (0 kJ )]   H = - 802.2 kJ

29. 29 Examples  2 SO 3 (g)  2SO 2 (g) + O 2 (g)

30. 30 Why Does It Work?  If H 2 (g) + 1/2 O 2 (g)  H 2 O(l)  H= - 285.5 kJ  then H 2 O(l)  H 2 (g) + 1/2 O 2 (g)  H = + 285.5 kJ u If you turn an equation around, you change the sign  2 H 2 O(l)  H 2 (g) + O 2 (g)  H = + 571.0 kJ u If you multiply the equation by a number, you multiply the heat by that number. Twice the moles, twice the heat

31. 31 Why does it work? u You make the products, so you need their heats of formation u You “unmake” the reactants so you have to subtract their heats.

32. Energy ReactantsProducts  reactants products elements 32

33. Energy ReactantsProducts  reactants products elements 33

34. Section 17.4 Calculating Heats of Reaction 34

35. Any two processes that both start with the same reactants in the same state and finish with the same products in the same state will have the same enthalpy change. Hess’s law states that the overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process. 35

36. When phosphorus is burned in excess chlorine 4 mol of phosphorus pentachloride, PCl 5, is synthesized. P 4 (s) + 10Cl 2 (g)  4PCl 5 (g)  H = -1596 kJ Phosphorus pentachloride may also be prepared in a two-step process. Step 1: P 4 (s) + 6Cl 2 (g)  4PCl 3 (g)  H = -1224 kJ Step 2: PCl 3 (g) + Cl 2 (g)  PCl 5 (g)  H = -93 kJ 36

37. The second reaction must take place four times for each occurrence of the first reaction in the two-step process. This two-step process is more accurately described by the following equations. P 4 (s) + 6Cl 2 (g)  4PCl 3 (g)  H = -1224 kJ 4PCl 3 (g) + 4Cl 2 (g)  4PCl 5 (g)  H = 4(-93 kJ) = -372 kJ 37

38. So, the total change in enthalpy by the two-step process is as follows: (-1224 kJ) + (-372 kJ) = -1596 kJ This enthalpy change,  H, for the two-step process is the same as the enthalpy change for the direct route of the formation of PCl 5. This example is in agreement with Hess’s law. 38

39. The enthalpy of the formation of CO, when CO 2 and solid carbon are reactants, is found using the equations below. 2C(s) + O 2 (g)  2CO(g)  H = -221 kJ C(s) + O 2 (g)  CO 2 (g)  H = -393 kJ You cannot simply add these equations because CO 2 would not be a reactant. 39

40. Adding the two equations gives the equation for the formation of CO by using CO 2 and C. C(s) + CO 2 (g)  2CO(g)  H = ?? kJ 2C(s) + O 2 (g)  2CO(g)  H = _ 221 kJ CO 2 (g)  C(s) + O 2 (g)  H = 393 kJ 2C(s) + O 2 (g) + CO 2 (g)  2CO(g) + C(s) + O 2 (g)  H = 172 kJ Oxygen and carbon that appear on both sides of the equation can be cancelled. C(s) + CO 2 (g)  2CO(g)  H = 172 kJ 40

41. The enthalpy change in forming 1 mol of a substance is called the standard enthalpy of formation of the substance,  H f ° The values of the standard enthalpies of formation for elements are 0. The following equation is used to determine the enthalpy change of a chemical reaction from the standard enthalpies of formation.  H reaction =  H products -  H reactants 41