1. Thermochemistry Heatand ChemicalChange

2. TEMPERATURE VS. HEAT Temperature is a measure of the average energy of the molecules Heat is the total amount of energy of the molecules

3. Which is at a higher temperature?

4. Which possesses more heat energy?

5. heat energy = specific heat x mass x ∆T Q = sm  T

6. Specific Heat Capacity (s): The amount of heat energy required to raise the temperature of1 gram of a substance by 1 ° C Energy is measured in Joules (J) Symbol=s Unit =J/g °C

7. Specific Heat Capacities : Substance H 2 O(l) J/g ° C 4.18 Al0.890.89 Fe0.450.45 Hg0.140.14 C0.71 ethanol2.44

8. Specific Heat Capacity of Water: 4.184J/gºC You must memorize this number!

9. Specific Heat Capacities : The higher the specific heat capacity, the more energy it takes to change the Temperature of a substance How does water’s high specific heat capacity make it such a valuable resource to living things?

10. You place 10 gram samples of both Al and Fe in a hot oven for a short period of time…. 0.89 J/g  C0.45 J/g  C AlFe Which substance will heat up the fastest?

11. You place 10 gram samples of both Al and Fe in a hot oven for a short period of time…. 0.89 J/g  C0.45 J/g  C AlFe Which substance has to absorb more heat energy to warm up?

12. You now place both the Al and Fe in the same oven and let them sit overnight….. Al Fe 0.89 J/g  C 0.45 J/g  C

13. You now place both the Al and Fe in the same oven and let them sit overnight….. Al Fe 0.89 J/g  C 0.45 J/g  C Which substance will be at the higher temperature at the end of this time?

14. You now place both the Al and Fe in the same oven and let them sit overnight….. Al Fe 0.89 J/g  C 0.45 J/g  C Which substance has absorbed more heat energy to get to this temperature?

15. H 2 O (l) Ethanol 4.18 J/g ° C 2.44 J/g ° C Which substance would be abetter coolant in a car’s radiator? Why?

16. Calculations with Specific Heat Q = Heat (J) s = specific heat (J/g·Cº) m = mass of sample (g)  T = change in temperature (ºC) (T final – T initial ) J = Joules Q = sm  T

17. Calculations with Specific Heat Example 1: How much heat is absorbed when the temperature of a 125 g piece of aluminum increases from 35ºC to 65ºC. The specific heat capacity of Aluminum is 0.90 J/g·ºC. Q = sm  T

18. Calculations with Specific Heat Example 2: The temperature of a 95.4g piece of copper increases from 25.0ºC to 48.0ºC when the copper absorbs 849 J of heat. What is the specific heat of copper? Q = sm  T

19. Calorimetry: The measurement of heat energy transferred during a physical or chemical process

20. Calorimetry: The measurement of heat energy transferred during a physical or chemical process The heat energy that is gained or lost is absorbed or released into water in a calorimeter

21. Calorimetry: The measurement of heat energy transferred during a physical or chemical process The heat energy that is gained or lost is absorbed or released into water in a calorimeter Heat lost by reaction = heat gained by water

22. Calorimetry: Heat lost by reaction = heat gained by water s metal m metal  T metal = s water m water  T water smmmTmsmmmTm = s w m w  T w

23. Calorimetry Experiment

24. Heat lost = Heat Gained Calculations smmmTmsmmmTm = s w m w  T w Example 1: A piece of metal with a mass of 678 g is heated to 99.8°C and dropped into 250.0 mL of water at 26.0°C. The final temperature of the system is 27.1°C. What is the specific heat of the metal?

25. Heat lost = Heat Gained Calculations smmmTmsmmmTm = s w m w  T w Example 2: A piece of metal is heated to 76.8°C and dropped into 75.0 mL of water at 22.0°C.The final temperature of the system is 27.1°C. The specific heat of the metal is 0.98 J/g∙°C. What is the mass of the metal piece?

26. Changes in Heat Energy within a Chemical Reaction Energy can be gained or lost in the form of heat during a chemical reaction When heat energy is gained (or absorbed), a reaction is called endothermic When heat energy is lost (or released), a reaction is called exothermic The change in heat is represented by the symbol  H

27. Surroundings Energy  H= negative Exothermic System Heat Released

28. Exothermic Reactions An exothermic reaction will feel warm or hot to the touch because you feel the heat being released from the reaction!

29. +2H 2O+Heat+2H 2O+Heat CH 4+2O2 CO2CH 4+2O2 CO2 CH 4+2O2CH 4+2O2 E n e r gy Heat CO 2 + 2 H 2 O EXOTHERMIC  H=negative Reactants Products

30. Surroundings  H= positive Endothermic System Energy Heat Absorbed

31. Endothermic Reactions An endothermic reaction will feel cool or cold to the touch because the heat from your hand is being absorbed by the reaction!

32. N 2 + O 2 E n e r gy Heat 2NO2NO N 2 +O 2 +heat  2 N O2 N O ENDOTHERMIC  H = positive Reactants Products

33. Change in Energy (Heat)  H The change in heat energy (  H) in a chemical reaction is called enthalpy  H = negative (-) = exothermic  H = positive (+) = endothermic  H is expressed in kJ (kilojoules)

34. Calculations Using Heat Energy

35. 2 S + 3 O 2  2 SO 3  H = -791.4 kJ How much heat will be released when 3 moles of Sulfur react with excess O 2 in the reaction above? 3 mol S 2 mol S -791.4 kJ = -1187 kJ of heat energy released Is the reaction exothermic or endothermic? How do you know?

36. H 2 + Br 2  2 HBr  H = 72.80 kJ How much heat will be absorbed when 38.2 g of bromine reacts with excess H 2 in the reaction above? 38.2 g Br 2 159.80 g Br 2 1 mol Br 2 72.80 kJ = 17.40 kJ of heat energy absorbed Is the reaction exothermic or endothermic?

37. HESS’S LAW Hess Law Video example

38. HESS’S LAW Sometimes reactions can occur in steps rather than as single reaction Example: 2C + H 2 ---> C 2 H 2 This reaction actually occurs in 3 steps: C 2 H 2 + 5/2 O 2 ---> 2CO 2 + H 2 O C + O 2 ---> CO 2 H 2 + ½ O 2 ---> H 2 O

39. HESS’S LAW Regardless of the number of steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. 2C(s) + H 2 (g) ---> C 2 H 2 (g)  H = ?? C 2 H 2 + 5/2 O 2 ---> 2CO 2 + H 2 O  H = -1299.5 kJ C + O 2 ---> CO 2  H = -393.5 kJ H 2 + ½ O 2 ---> H 2 O  H = -285.8 kJ

40. HESS’S LAW Unfortunately, it is not as simple as just adding up all the Enthalpies. We must manipulate the various steps to the reaction until it matches the final reaction exactly. 2C(s) + H 2 (g) ---> C 2 H 2 (g)  H = ?? C 2 H 2 + 5/2 O 2 ---> 2CO 2 + H 2 O  H = -1299.5 kJ C + O 2 ---> CO 2  H = -393.5 kJ H 2 + ½ O 2 ---> H 2 O  H = -285.8 kJ

41. Let’s do an example together Given the following data: Calculate  H for the reaction 2NO 2  N 2 + 2O 2 N 2 + 2O 2 ---> N 2 O 4  H = 9.6 kJ 2NO 2 ---> N 2 O 4  H = -58.2 kJ

42. Let’s do an example together Given the following data: Calculate  H for the reaction 2S + 2OF 2  SO 2 + SF 4 OF 2 + H 2 O ---> O 2 + 2HF  H = -277 kJ SF 4 + 2H 2 O ---> SO 2 + 4HF  H = 828 kJ S + O 2 ---> SO 2  H = -297 kJ