1. Chapter 6 Thermochemistry

2. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 2  Capacity to do work or to produce heat.  Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed.  The total energy content of the universe is constant. Energy

3. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 3  Potential energy – energy due to position or composition.  Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Energy

4. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 4  Heat involves the transfer of energy between two objects due to a temperature difference.  Work – force acting over a distance.  Energy is a state function; work and heat are not  State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Energy

5. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 5  System – part of the universe on which we wish to focus attention.  Surroundings – include everything else in the universe. Chemical Energy

6. Section 6.1 The Nature of Energy open mass & energyExchange: closed energy isolated nothing 6.2

7. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 7  Endothermic Reaction:  Heat flow is into a system.  Absorb energy from the surroundings.  Exothermic Reaction:  Energy flows out of the system.  Energy gained by the surroundings must be equal to the energy lost by the system. Chemical Energy

8. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 8 Is the freezing of water an endothermic or exothermic process? Explain. CONCEPT CHECK!

9. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 9 Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. a)Your hand gets cold when you touch ice. b)The ice gets warmer when you touch it. c)Water boils in a kettle being heated on a stove. d)Water vapor condenses on a cold pipe. e)Ice cream melts. Exo Endo Exo Endo CONCEPT CHECK!

10. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 10 For each of the following, define a system and its surroundings and give the direction of energy transfer. a)Methane is burning in a Bunsen burner in a laboratory. b)Water drops, sitting on your skin after swimming, evaporate. CONCEPT CHECK!

11. Section 6.1 The Nature of Energy Thermodynamics  The study of energy and its interconversions is called thermodynamics.  Law of conservation of energy is often called the first law of thermodynamics. Copyright © Cengage Learning. All rights reserved 11

12. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 12 Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.  To change the internal energy of a system: ΔE = q + w q represents heat w represents work Internal Energy

13. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 13  Thermodynamic quantities consist of two parts:  Number gives the magnitude of the change.  Sign indicates the direction of the flow. Internal Energy

14. Section 6.1 The Nature of Energy Internal Energy  Sign reflects the system’s point of view.  Endothermic Process:  q is positive  Exothermic Process:  q is negative Copyright © Cengage Learning. All rights reserved 14

15. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 15  Sign reflects the system’s point of view.  System does work on surroundings:  w is negative  Surroundings do work on the system:  w is positive Internal Energy

16. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 16  For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: w = –PΔV  To convert between L·atm and Joules, use 1 L·atm = 101.3 J. Work

17. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 17 Which of the following performs more work? a)A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work. EXERCISE!

18. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 18 1.Calculate the change in internal energy of the system when it undergoes an exothermic process releasing 140 J of heat and a gas is evolved which does 85 J of work during expansion. 2.A balloon is heated by adding 240 J of heat. It expands doing 135 J of work. Calculate the change in internal energy of the balloon. 3.50.0 g of Fe(s) is cooled from 100  C to 90  C losing 225 J of heat. Calculate the change in internal energy of the Fe(s)

19. Section 6.2 Enthalpy and Calorimetry Change in Enthalpy  State function  ΔH = q at constant pressure  ΔH = H products – H reactants Copyright © Cengage Learning. All rights reserved 19

20. Section 6.2 Enthalpy and Calorimetry Consider the combustion of propane: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. –252 kJ Copyright © Cengage Learning. All rights reserved 20 EXERCISE!

21. Section 6.2 Enthalpy and Calorimetry Calorimetry  Science of measuring heat  Specific heat capacity:  The energy required to raise the temperature of one gram of a substance by one degree Celsius.  Molar heat capacity:  The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright © Cengage Learning. All rights reserved 21

22. Section 6.2 Enthalpy and Calorimetry Calorimetry  Heat change = s × m × ΔT s = specific heat capacity (J/°C·g) m = mass of system in units of g ΔT = change in temperature (°C) Copyright © Cengage Learning. All rights reserved 22

23. Section 6.2 Enthalpy and Calorimetry 1.The molar heat capacity of Al is 24.4J/mol- o C. Calculate its specific heat capacity. Answer: 0.904 J/g- o C 2.The specific heat capacity of Hg is 0.14J/g- o C. Calculate the amount of heat required to raise the temperature of 10.0g of Hg from 25.0  C to 30.0  C. Answer: 7.0J 3.The specific heat of copper is 0.385 J/g°C. Calculate its molar heat capacity. Answer: 24.5 J/mol- o C 4.If 100.0 J of heat is supplied to 10.0g of Copper, what is the temperature change of the metal? The specific heat of copper is 0.385 J/g°C. Answer: 26.0 °C

24. Section 6.2 Enthalpy and Calorimetry A Coffee–Cup Calorimeter Made of Two Styrofoam Cups Copyright © Cengage Learning. All rights reserved 24 q rxn = - q solution =-m x s x  t

25. Section 6.2 Enthalpy and Calorimetry When 50.0 mL of 0.100M AgNO 3 (aq) and 50.0 mL of 0.100 M HCl(aq) are mixed in a coffee cup calorimeter, the temperature increased from 22.30  C to 23.11  C. AgNO 3 (aq) + HCl(aq)  HNO 3 (aq) + AgCl(s) Calculate the enthalpy change for the reaction per mol HCl. The density of the solution is 1.03 g/mL. Its specific heat = 4.18J/g  C.

26. Section 6.2 Enthalpy and Calorimetry 1.When 4.25 g of NH 4 NO 3 is dissolved in 60.0g of water in a coffe-cup calorimeter, the temperature drops from 22.0  C to 16.9  C. Calculate the enthalpy change,  H in kJ/mol NH 4 NO 3. The specific heat of the solution = 4.184 J/g  C.

27. Section 6.2 Enthalpy and Calorimetry Constant-Volume Calorimetry No heat enters or leaves! 6.5 Reaction at Constant V  H ~ q rxn  H = q rxn =  E q rxn = - q calorimeter q calorimeter = C calorimeter x  t

28. Section 6.2 Enthalpy and Calorimetry When 4.25 g of NH 4 NO 3 is dissolved in 60.0g of water in a coffe-cup calorimeter, the temperature drops from 22.0  C to 16.9  C. Calculate the enthalpy change,  H in kJ/mol NH 4 NO 3. The specific heat of the solution = 4.184 J/g  C.

29. Section 6.2 Enthalpy and Calorimetry When a hot object and a cold object are brought in contact, q cold = -q hot q cold = heat change of the cold object q hot = heat change of the hot object Heat flows from the hot to the cold till they are both at the same final temperature, equilibrium temperature, T q cold = m cold x s cold x (T-T i,cold ) q hot = m hot x s hot x (T-T i,hot )

30. Section 6.2 Enthalpy and Calorimetry A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Copyright © Cengage Learning. All rights reserved 30 CONCEPT CHECK!

31. Section 6.2 Enthalpy and Calorimetry A 100.0 g sample of water at 90.°C is added to a 500.0 g sample of water at 10.°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 23°C Copyright © Cengage Learning. All rights reserved 31 CONCEPT CHECK!

32. Section 6.2 Enthalpy and Calorimetry You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (s H2O = 4.18 J/°C·g and s Fe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18°C Copyright © Cengage Learning. All rights reserved 32 CONCEPT CHECK!

33. Section 6.3 Hess’s Law  In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Copyright © Cengage Learning. All rights reserved 33

34. Section 6.3 Hess’s Law  This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH 2 and ΔH 3. N 2 (g) + O 2 (g) → 2NO(g) ΔH 2 = 180 kJ 2NO(g) + O 2 (g) → 2NO 2 (g) ΔH 3 = – 112 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 2 + ΔH 3 = 68 kJ ΔH 1 = ΔH 2 + ΔH 3 = 68 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 1 = 68 kJ Copyright © Cengage Learning. All rights reserved 34

35. Section 6.3 Hess’s Law The Principle of Hess’s Law Copyright © Cengage Learning. All rights reserved 35

36. Section 6.3 Hess’s Law Characteristics of Enthalpy Changes  If a reaction is reversed, the sign of ΔH is also reversed.  The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer. Copyright © Cengage Learning. All rights reserved 36

37. Section 6.3 Hess’s Law Example  Consider the following data:  Calculate ΔH for the reaction Copyright © Cengage Learning. All rights reserved 37

38. Section 6.3 Hess’s Law Problem-Solving Strategy  Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal.  Reverse any reactions as needed to give the required reactants and products.  Multiply reactions to give the correct numbers of reactants and products. Copyright © Cengage Learning. All rights reserved 38

39. Section 6.3 Hess’s Law Example  Reverse the two reactions:  Desired reaction: Copyright © Cengage Learning. All rights reserved 39

40. Section 6.3 Hess’s Law Example  Multiply reactions to give the correct numbers of reactants and products: 4( ) 4( ) 3( ) 3( )  Desired reaction:

41. Section 6.3 Hess’s Law Example  Final reactions:  Desired reaction: ΔH = +1268 kJ Copyright © Cengage Learning. All rights reserved 41

42. Section 6.4 Standard Enthalpies of Formation Standard Enthalpy of Formation (ΔH f °)  Change in enthalpy that accompanies the formation of one mole of a substance from its elements with all substances in their standard states. Copyright © Cengage Learning. All rights reserved 42

43. Section 6.4 Standard Enthalpies of Formation Conventional Definitions of Standard States  For a Compound  For a gas, pressure is exactly 1 atm.  For a solution, concentration is exactly 1 M.  Pure substance (liquid or solid)  For an Element  The form [N 2 (g), K(s)] in which it exists at 1 atm and 25°C. Copyright © Cengage Learning. All rights reserved 43

44. Section 6.4 Standard Enthalpies of Formation Problem-Solving Strategy: Enthalpy Calculations The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: Δ H° rxn =  n p Δ H f  (products) -  n r Δ H f  (reactants) Elements in their standard states are not included in the ΔH reaction calculations because ΔH f ° for an element in its standard state is zero. Copyright © Cengage Learning. All rights reserved 44

45. Section 6.4 Standard Enthalpies of Formation Calculate Δ H° for the following reaction: 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) Given the following information: Δ H f ° (kJ/mol) Na(s)0 H 2 O(l) –286 NaOH(aq) –470 H 2 (g)0 Δ H° = –368 kJ Copyright © Cengage Learning. All rights reserved 45 EXERCISE!