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Prentice-Hall © 2008General Chemistry: Chapter 7Slide 1 of 50 Chapter 7: Thermochemistry Philip Dutton University of Windsor, Canada Prentice-Hall © 2008.

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Presentation on theme: "Prentice-Hall © 2008General Chemistry: Chapter 7Slide 1 of 50 Chapter 7: Thermochemistry Philip Dutton University of Windsor, Canada Prentice-Hall © 2008."— Presentation transcript:

1 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 1 of 50 Chapter 7: Thermochemistry Philip Dutton University of Windsor, Canada Prentice-Hall © 2008 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition

2 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 2 of 50 Contents 7-1Getting Started: Some Terminology 7-2Heat 7-3Heats of Reaction and Calorimetry 7-4Work 7-5The First Law of Thermodynamics 7-6Heats of Reaction:  U and  H 7-7The Indirect Determination of  H: Hess’s Law

3 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 3 of 50 Contents 7-8Standard Enthalpies of Formation 7-9Fuels as Sources of Energy Focus on Fats, Carbohydrates, and Energy Storage

4 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 4 of Getting Started: Some Terminology System Surroundings

5 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 5 of 50 Terminology Energy, U –The capacity to do work. Work –Force acting through a distance. Kinetic Energy –The energy of motion.

6 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 6 of 50 Energy Kinetic Energy ek =ek = 1 2 m·v2m·v2 [e k ] = kg m 2 s2s2 = J w = F·dF·d [w ] = kg m s2s2 = J m Work

7 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 7 of 50 Energy Potential Energy –Energy due to condition, position, or composition. –Associated with forces of attraction or repulsion between objects. Energy can change from potential to kinetic.

8 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 8 of 50 Energy and Temperature Thermal Energy –Kinetic energy associated with random molecular motion. –In general proportional to temperature. –An intensive property. Heat and Work –q and w. –Energy changes.

9 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 9 of 50 Heat Energy transferred between a system and its surroundings as a result of a temperature difference. Heat flows from hotter to colder. –Temperature may change. –Phase may change (an isothermal process).

10 Heat transfer ThemochemistryDia 10 /55

11 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 11 of 50 Units of Heat Calorie (cal) –The quantity of heat required to change the temperature of one gram of water by one degree Celsius. Joule (J) –SI unit for heat 1 cal = J

12 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 12 of 50 Heat Capacity The quantity of heat required to change the temperature of a system by one degree. –Molar heat capacity. System is one mole of substance. –Specific heat capacity, c. System is one gram of substance –Heat capacity Mass · specific heat. q = mc  T q = CTCT

13 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 13 of 50 Conservation of Energy In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed. q system + q surroundings = 0 q system = -q surroundings

14 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 14 of 50 Determination of Specific Heat

15 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 15 of 50 Example 7-2 Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. q lead = -q water q water = mc  T = (50.0 g)(4.184 J/g °C)( )°C q water = 1423 J q lead = J = mc  T = (150.0 g)(c)( )°C c lead = Jg -1 °C -1

16 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 16 of Heats of Reaction and Calorimetry Chemical energy. –Contributes to the internal energy of a system. Heat of reaction, q rxn. –The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature.

17 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 17 of 50 Heats of Reaction Exothermic reactions. –Produces heat, q rxn < 0. Endothermic reactions. –Consumes heat, q rxn > 0. Calorimeter –A device for measuring quantities of heat.

18 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 18 of 50 Bomb Calorimeter q rxn = -q cal q cal = q bomb + q water + q wires +… Define the heat capacity of the calorimeter: q cal =  m i c i  T = C  T all i heat

19 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 19 of 50 Using Bomb Calorimetry Data to Determine a Heat of Reaction. The combustion of g sucrose, in a bomb calorimeter, causes the temperature to rise from to 28.33°C. The heat capacity of the calorimeter assembly is 4.90 kJ/°C. (a)What is the heat of combustion of sucrose, expressed in kJ/mol C 12 H 22 O 11 (b)Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories. Example 7-3

20 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 20 of 50 Example 7-3 Calculate q calorimeter : q cal = C  T = (4.90 kJ/°C)( )°C = (4.90)(3.41) kJ = 16.7 kJ Calculate q rxn : q rxn = -q cal = kJ per g Example 7-3

21 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 21 of 50 Example 7-3 Calculate q rxn in the required units: q rxn = -q cal = kJ g = kJ/g g 1.00 mol = kJ/g = · 10 3 kJ/mol q rxn (a) Calculate q rxn for one teaspoon: 4.8 g 1 tsp = (-16.5 kJ/g)(q rxn (b) )( )= -19 cal/tsp 1.00 cal J Example 7-3

22 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 22 of 50 Coffee Cup Calorimeter A simple calorimeter. –Well insulated and therefore isolated. –Measure temperature change. q rxn = -q cal See example 7-4 for a sample calculation.

23 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 23 of Work In addition to heat effects chemical reactions may also do work. Gas formed pushes against the atmosphere. Volume changes. Pressure-volume work.

24 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 24 of 50 Pressure Volume Work w = F · d = (P · A) · h = P  V w = -P ext  V

25 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 25 of 50 Example 7-3 Assume an ideal gas and calculate the volume change: V i = nRT/P i = 1.02 L V f = nRT/P f = 1.88 L Example 7-5 Calculating Pressure-Volume Work. Suppose the gas in the previous figure is mol He at 298 K. How much work, in Joules, is associated with its expansion at constant pressure. (P i = kPa, P f = kPa)  V = L = 0.86 L R = J/mol K

26 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 26 of 50 Example 7-3 Calculate the work done by the system: w = -P f ·  V = -( kPa)(0.86 L) = J Example 7-5 Hint: If you use pressure in kPa and volume in liter (L) you get Joules directly.

27 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 27 of The First Law of Thermodynamics Internal Energy, U. –Total energy (potential and kinetic) in a system. Translational kinetic energy. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons.

28 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 28 of 50 Maxwell distributions of a gas with M=50 g/mol

29 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 29 of 50 First Law of Thermodynamics A system contains only internal energy. –A system does not contain heat or work. –These only occur during a change in the system. Law of Conservation of Energy –The energy of an isolated system is constant  U = q + w

30 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 30 of 50 First Law of Thermodynamics

31 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 31 of 50 State Functions Any property that has a unique value for a specified state of a system is said to be a State Function. Water at K and 1.00 atm is in a specified state. d = g/mL This density is a unique function of the state. It does not matter how the state was established.

32 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 32 of 50 Functions of State U is a function of state. –Not easily measured.  U has a unique value between two states. –Is easily measured.

33 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 33 of 50 Path Dependent Functions Changes in heat and work are not functions of state. –Remember example 7-5, w =  J in a one step expansion of gas: –Consider 2.40 atm to 1.80 atm and finally to 1.30 atm. – kPa, kPa, and kPa w = – ·(1.36  1.02) – ·(1.88  1.36) [kPa · L] =  J

34 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 34 of 50 p*V work of 0.1 mol of He gas: 2 steps J J

35 ThermochemistryDia 35 /55 p*V work of 0.1 mol of He gas: 3 steps Pointp(kPa) V(l)V(l) StepsW (J)

36 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 36 of Heats of Reaction:  U and  H Reactants → Products U i U f  U = U f - U i  U = q rxn + w In a system at constant volume:  U = q rxn + 0 = q rxn = q v But we live in a constant pressure world! How does q p relate to q v ?

37 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 37 of 50 Heats of Reaction

38 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 38 of 50 Heats of Reaction q V = q P + w We know that w = - P ·  V and  U = q v, therefore:  U = q P - P ·  V q P =  U + P ·  V These are all state functions, so define a new function. Let H = U + P · V Then  H = H f – H i =  U +  (P · V) If we work at constant pressure and temperature:  H =  U + P ·  V = q P

39 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 39 of 50 Comparing Heats of Reaction q P = -566 kJ/mol =  H P ·  V = P ·(V f – V i ) = RT· (n f – n i ) = -2.5 kJ  U =  H - P ·  V = kJ/mol = q V

40 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 40 of 50 Changes of State of Matter H 2 O (l) → H 2 O(g)  H = 44.0 kJ at 298 K Molar enthalpy of vaporization: Molar enthalpy of fusion: H 2 O (s) → H 2 O(l)  H = 6.01 kJ at K

41 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 41 of 50 Example 7-3 Example 7-8 Break the problem into two steps: Raise the temperature of the liquid first then completely vaporize it. The total enthalpy change is the sum of the changes in each step. Enthalpy Changes Accompanying Changes in States of Matter. Calculate  H for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 25.0°C. = (50.0 g)(4.184 J/g °C)( )°C g 18.0 g/mol 44.0 kJ/mol Set up the equation and calculate: q P = mc H 2 O  T + n·  H vap = 3.14 kJ kJ = 125 kJ

42 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 42 of 50 Standard States and Standard Enthalpy Changes Define a particular state as a standard state. Standard enthalpy of reaction,  H ° –The enthalpy change of a reaction in which all reactants and products are in their standard states. Standard State –The pure element or compound in its reference state (g, l, s) at standard pressure and temperature (101.3 kPa, 0 ° C) (STP).

43 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 43 of 50 Enthalpy Diagrams 2H 2 + O 2 2H 2 O

44 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 44 of Indirect Determination of  H: Hess’s Law  H is an extensive property. –Enthalpy change is directly proportional to the amount of substance in a system. N 2 (g) + O 2 (g) → 2 NO(g)  H = kJ ½N 2 (g) + ½O 2 (g) → NO(g)  H = kJ  H changes sign when a process is reversed NO(g) → ½N 2 (g) + ½O 2 (g)  H = kJ

45 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 45 of 50 Hess’s Law Hess’s law of constant heat summation –If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N 2 (g) + O 2 (g) → NO 2 (g)  H = kJ ½N 2 (g) + ½O 2 (g) → NO(g)  H = kJ NO(g) + ½O 2 (g) → NO 2 (g)  H = kJ

46 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 46 of 50 Hess’s Law Schematically

47 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 47 of 50 The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. The standard enthalpy of formation of a pure element in its reference state is 0. Hf°Hf° 7-8 Standard Enthalpies of Formation

48 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 48 of 50 Standard Enthalpies of Formation

49 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 49 of 50 Standard Enthalpies of Formation

50 TermokémiaSlide 50 of 50 2C(grafit) + 3H 2 (g) + ½O 2 (g)  C 2 H 5 OH(l)

51 H 2 O (g) and (l) 298 K 1 atm TermokémiaSlide 51 of 50

52 C(graphite) + O 2 (g) = CO 2 (g) 298 K 1 atm TermokémiaSlide 52 of 50

53 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 53 of 50 Standard Enthalpies of Reaction  H overall = -2  H f ° NaHCO 3 +  H f ° Na 2 CO 3 +  H f ° CO 2 +  H f ° H 2 O

54 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 54 of 50 Enthalpy of Reaction  H rxn = ∑  H f ° products - ∑  H f ° reactants

55 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 55 of 50 Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions

56 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 56 of Fuels as Sources of Energy Fossil fuels. –Combustion is exothermic. –Non-renewable resource. –Environmental impact.

57 Prentice-Hall © 2008General Chemistry: Chapter 7Slide 57 of 50 Chapter 7 Questions 1, 2, 3, 11, 14, 16, 22, 24, 29, 37, 49, 52, 63, 67, 73, 81


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