1. Copyright © 2011 Pearson Education, Inc. Chapter 6 Thermochemistry Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Copyright © 2011 Pearson Education, Inc. Chemical Hand Warmers Most hand warmers work by using the heat released from the slow oxidation of iron 4 Fe(s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) The amount your hand temperature rises depends on several factors the size of the hand warmer the size of your glove, etc. mainly, the amount of heat released by the reaction 2 Tro: Chemistry: A Molecular Approach, 2/e

3. Copyright © 2011 Pearson Education, Inc. Nature of Energy Even though chemistry is the study of matter, energy affects matter Energy is anything that has the capacity to do work Work is a force acting over a distance Energy = Work = Force x Distance Heat is the flow of energy caused by a difference in temperature Energy can be exchanged between objects through contact collisions 3 Tro: Chemistry: A Molecular Approach, 2/e

4. Copyright © 2011 Pearson Education, Inc. Energy, Heat, and Work You can think of energy as a quantity an object can possess or collection of objects You can think of heat and work as the two different ways that an object can exchange energy with other objects either out of it, or into it 4 Tro: Chemistry: A Molecular Approach, 2/e

5. Copyright © 2011 Pearson Education, Inc. Classification of Energy Kinetic energy is energy of motion or energy that is being transferred Thermal energy is the energy associated with temperature thermal energy is a form of kinetic energy 5 Tro: Chemistry: A Molecular Approach, 2/e

6. Copyright © 2011 Pearson Education, Inc. Classification of Energy Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object energy stored in the structure of a compound is potential 6 Tro: Chemistry: A Molecular Approach, 2/e

7. Copyright © 2011 Pearson Education, Inc. Manifestations of Energy 7 Tro: Chemistry: A Molecular Approach, 2/e

8. Copyright © 2011 Pearson Education, Inc. Some Forms of Energy Electrical kinetic energy associated with the flow of electrical charge Heat or thermal energy kinetic energy associated with molecular motion Light or radiant energy kinetic energy associated with energy transitions in an atom Nuclear potential energy in the nucleus of atoms Chemical potential energy due to the structure of the atoms, the attachment between atoms, the atoms’ positions relative to each other in the molecule, or the molecules, relative positions in the structure 8 Tro: Chemistry: A Molecular Approach, 2/e

9. Copyright © 2011 Pearson Education, Inc. Conservation of Energy The Law of Conservation of Energy states that energy cannot be created nor destroyed When energy is transferred between objects, or converted from one form to another, the total amount of energy present at the beginning must be present at the end 9 Tro: Chemistry: A Molecular Approach, 2/e

10. Copyright © 2011 Pearson Education, Inc. System and Surroundings We define the system as the material or process within which we are studying the energy changes within We define the surroundings as everything else with which the system can exchange energy with What we study is the exchange of energy between the system and the surroundings Surroundings System Surroundings System 10 Tro: Chemistry: A Molecular Approach, 2/e

11. Copyright © 2011 Pearson Education, Inc. Comparing the Amount of Energy in the System and Surroundings During Transfer Conservation of energy means that the amount of energy gained or lost by the system has to be equal to the amount of energy lost or gained by the surroundings 11 Tro: Chemistry: A Molecular Approach, 2/e

12. Copyright © 2011 Pearson Education, Inc. The amount of kinetic energy an object has is directly proportional to its mass and velocity KE = ½mv 2 Units of Energy 12 Tro: Chemistry: A Molecular Approach, 2/e

13. Copyright © 2011 Pearson Education, Inc. Units of Energy joule (J) is the amount of energy needed to move a 1-kg mass a distance of 1 meter 1 J = 1 N∙m = 1 kg∙m 2 /s 2 calorie (cal) is the amount of energy needed to raise the temperature of one gram of water 1°C kcal = energy needed to raise 1000 g of water 1°C food Calories = kcals Energy Conversion Factors 1 calorie (cal)=4.184 joules (J) (exact) 1 Calorie (Cal)=1000 cal = 1 kcal = 4184 J 1 kilowatt-hour (kWh)=3.60 x 10 6 J 13 Tro: Chemistry: A Molecular Approach, 2/e

14. Copyright © 2011 Pearson Education, Inc. Energy Use 14 Tro: Chemistry: A Molecular Approach, 2/e

15. Copyright © 2011 Pearson Education, Inc. The First Law of Thermodynamics Law of Conservation of Energy Thermodynamics is the study of energy and its interconversions The First Law of Thermodynamics is the Law of Conservation of Energy This means that the total amount of energy in the universe is constant You can therefore never design a system that will continue to produce energy without some source of energy 15 Tro: Chemistry: A Molecular Approach, 2/e

16. Copyright © 2011 Pearson Education, Inc. Energy Flow and Conservation of Energy Conservation of energy requires that the sum of the energy changes in the system and the surroundings must be zero  Energy universe = 0 =  Energy system +  Energy surroundings  Is the symbol that is used to mean change  final amount – initial amount 16 Tro: Chemistry: A Molecular Approach, 2/e

17. Copyright © 2011 Pearson Education, Inc. Internal Energy The internal energy is the sum of the kinetic and potential energies of all of the particles that compose the system The change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end a state function is a mathematical function whose result only depends on the initial and final conditions, not on the process used  E = E final – E initial  E reaction = E products − E reactants 17 Tro: Chemistry: A Molecular Approach, 2/e

18. Copyright © 2011 Pearson Education, Inc. State Function You can travel either of two trails to reach the top of the mountain. One is long and winding, the other is shorter but steep. Regardless of which trail you take, when you reach the top you will be 10,000 ft above the base. The distance from the base to the peak of the mountain is a state function. It depends only on the difference in elevation between the base and the peak, not on how you arrive there! 18 Tro: Chemistry: A Molecular Approach, 2/e

19. Copyright © 2011 Pearson Education, Inc. Energy Diagrams 19 Energy diagrams are a “graphical” way of showing the direction of energy flow during a process Internal Energy initial final energy added  E = + Internal Energy initial final energy removed  E = ─ If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be + If the final condition has a smaller amount of internal energy than the initial condition, the change in the internal energy will be ─ Tro: Chemistry: A Molecular Approach, 2/e

20. Copyright © 2011 Pearson Education, Inc. Energy Flow When energy flows out of a system, it must all flow into the surroundings When energy flows out of a system,  E system is ─ When energy flows into the surroundings,  E surroundings is + Therefore: ─  E system =  E surroundings Surroundings  E + System  E ─ 20 Tro: Chemistry: A Molecular Approach, 2/e

21. Copyright © 2011 Pearson Education, Inc. Energy Flow When energy flows into a system, it must all come from the surroundings When energy flows into a system,  E system is + When energy flows out of the surroundings,  E surroundings is ─ Therefore:  E system = ─  E surroundings Surroundings  E ─ System  E + 21 Tro: Chemistry: A Molecular Approach, 2/e

22. Copyright © 2011 Pearson Education, Inc. energy released  E rxn = ─ energy absorbed  E rxn = + Energy Flow in a Chemical Reaction The total amount of internal energy in 1mol of C(s) and 1 mole of O 2 (g) is greater than the internal energy in 1 mole of CO 2 (g) at the same temperature and pressure In the reaction C(s) + O 2 (g) → CO 2 (g), there will be a net release of energy into the surroundings −  E reaction =  E surroundings In the reaction CO 2 (g) → C(s) + O 2 (g), there will be an absorption of energy from the surroundings into the reaction  E reaction = −  E surroundings Internal Energy CO 2 (g) C(s), O 2 (g) Surroundings System C + O 2 → CO 2 System CO 2 →C + O 2 22 Tro: Chemistry: A Molecular Approach, 2/e

23. Copyright © 2011 Pearson Education, Inc. Energy Exchange Energy is exchanged between the system and surroundings through heat and work q = heat (thermal) energy w = work energy q and w are NOT state functions, their value depends on the process  E = q + w 23 Tro: Chemistry: A Molecular Approach, 2/e

24. Copyright © 2011 Pearson Education, Inc. Energy Exchange Energy is exchanged between the system and surroundings through either heat exchange or work being done 24 Tro: Chemistry: A Molecular Approach, 2/e

25. Copyright © 2011 Pearson Education, Inc. The white ball has an initial amount of 5.0 J of kinetic energy As it rolls on the table, some of the energy is converted to heat by friction The rest of the kinetic energy is transferred to the purple ball by collision Heat & Work 25 Tro: Chemistry: A Molecular Approach, 2/e

26. Copyright © 2011 Pearson Education, Inc. energy change for the white ball,  E = KE final − KE initial = 0 J − 5.0 J = −5.0 J kinetic energy transferred to purple ball, w = −4.5 J kinetic energy lost as heat, q = −0.5 J q + w = (−0.5 J) + (−4.5 J) = −5.0 J =  E On a smooth table, most of the kinetic energy is transferred from the white ball to the purple – with a small amount lost through friction energy change for the white ball,  E = KE final − KE initial = 0 J − 5.0 J = −5.0 J kinetic energy transferred to purple ball, w = −3.0 J kinetic energy lost as heat, q = −2.0 J q + w = (−2.0 J) + (−3.0 J) = −5.0 J =  E On a rough table, most of the kinetic energy of the white ball is lost through friction – less than half is transferred to the purple ball 26 Heat & Work Tro: Chemistry: A Molecular Approach, 2/e

27. Copyright © 2011 Pearson Education, Inc. Heat, Work, and Internal Energy In the previous billiard ball example, the  E of the white ball is the same for both cases, but q and w are not On the rougher table, the heat loss, q, is greater q is a more negative number But on the rougher table, less kinetic energy is transferred to the purple ball, so the work done by the white ball, w, is less w is a less negative number The  E is a state function and depends only on the velocity of the white ball before and after the collision in both cases it started with 5.0 kJ of kinetic energy and ended with 0 kJ because it stopped q + w is the same for both tables, even though the values of q and w are different 27 Tro: Chemistry: A Molecular Approach, 2/e

28. Copyright © 2011 Pearson Education, Inc. Example 6.1: If the burning of the fuel in a potato cannon performs 855 J of work on the potato and produces 1422 J of heat, what is  E for the burning of the fuel? the unit is correct, the sign make sense as the fuel should lose energy during the reaction q potato = 855 J, w potato = 1422 J  E fuel, J Check: Solution: Concept Plan: Relationships: Given: Find: q, w EE q system = −q surroundings, w system = −w surroundings,  E = q + w q, w potato q, w fuel 28 Tro: Chemistry: A Molecular Approach, 2/e

29. Copyright © 2011 Pearson Education, Inc. Practice – Reacting 50 mL of H 2 (g) with 50 mL of C 2 H 4 (g) produces 50 mL of C 2 H 6 (g) at 1.5 atm. If the reaction produces 3.1 x 10 2 J of heat and the decrease in volume requires the surroundings do 7.6 J of work on the gases, what is the change in internal energy of the gases? 29 Tro: Chemistry: A Molecular Approach, 2/e

30. Copyright © 2011 Pearson Education, Inc. If the reaction produces 3.1 x 10 2 J of heat and the decrease in volume requires the surroundings do 7.6 J of work on the gases, what is the change in internal energy of the gases? the units are correct, the sign is reasonable as the amount of heat lost in the reaction is much larger than the amount of work energy gained q reaction = −310 J, w surroundings = −7.6 J  E gases, J Check: Solution: Concept Plan: Relationships: Given: Find: q, w EE q system = −q surroundings, w system = −w surroundings,  E = q + w w surrounding w gases 30 Tro: Chemistry: A Molecular Approach, 2/e

31. Copyright © 2011 Pearson Education, Inc. Heat Exchange Heat is the exchange of thermal energy between the system and surroundings Heat exchange occurs when system and surroundings have a difference in temperature Temperature is the measure of the amount of thermal energy within a sample of matter Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature thermal equilibrium 31 Tro: Chemistry: A Molecular Approach, 2/e

32. Copyright © 2011 Pearson Education, Inc. Quantity of Heat Energy Absorbed: Heat Capacity When a system absorbs heat, its temperature increases The increase in temperature is directly proportional to the amount of heat absorbed The proportionality constant is called the heat capacity, C units of C are J/°C or J/K q = C x  T The larger the heat capacity of the object being studied, the smaller the temperature rise will be for a given amount of heat 32 Tro: Chemistry: A Molecular Approach, 2/e

33. Copyright © 2011 Pearson Education, Inc. Factors Affecting Heat Capacity The heat capacity of an object depends on its amount of matter usually measured by its mass 200 g of water requires twice as much heat to raise its temperature by 1°C as does 100 g of water The heat capacity of an object depends on the type of material 1000 J of heat energy will raise the temperature of 100 g of sand 12 °C, but only raise the temperature of 100 g of water by 2.4 °C 33 Tro: Chemistry: A Molecular Approach, 2/e

34. Copyright © 2011 Pearson Education, Inc. Specific Heat Capacity Measure of a substance’s intrinsic ability to absorb heat The specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C C s units are J/(g∙°C) The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C 34 Tro: Chemistry: A Molecular Approach, 2/e

35. Copyright © 2011 Pearson Education, Inc. Specific Heat of Water The rather high specific heat of water allows water to absorb a lot of heat energy without a large increase in its temperature The large amount of water absorbing heat from the air keeps beaches cool in the summer without water, the Earth’s temperature would be about the same as the moon’s temperature on the side that is facing the sun (average 107 °C or 225 °F) Water is commonly used as a coolant because it can absorb a lot of heat and remove it from the important mechanical parts to keep them from overheating or even melting it can also be used to transfer the heat to something else because it is a fluid 35 Tro: Chemistry: A Molecular Approach, 2/e

36. Copyright © 2011 Pearson Education, Inc. Quantifying Heat Energy The heat capacity of an object is proportional to its mass and the specific heat of the material So we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object Heat = (mass) x (specific heat) x (temp. change) q = (m) x (C s ) x (  T) 36 Tro: Chemistry: A Molecular Approach, 2/e

37. Copyright © 2011 Pearson Education, Inc. Example 6.2: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from −8.0 °C to 37.0 °C? q = m ∙ C s ∙  T C s = 0.385 J/gºC (Table 6.4) the unit is correct, the sign is reasonable as the penny must absorb heat to make its temperature rise T 1 = −8.0 °C, T 2 = 37.0 °C, m = 3.10 g q, Jq, J Check: Check Solution: Follow the conceptual plan to solve the problem Conceptual Plan: Relationships: Strategize Given: Find: Sort Information C s m,  T q 37 Tro: Chemistry: A Molecular Approach, 2/e

38. Copyright © 2011 Pearson Education, Inc. Practice – Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C (water’s specific heat is 4.18 J/g  ºC) 38 Tro: Chemistry: A Molecular Approach, 2/e

39. Copyright © 2011 Pearson Education, Inc. Practice – Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C q = m ∙ C s ∙  T C s = 4.18 J/g  C (Table 6.4) T 1 = 49 °C, T 2 = 29 °C, m = 7.40 g q, Jq, J Check: Check Solution: Follow the concept plan to solve the problem Conceptual Plan: Relationships: Strategize Given: Find: Sort Information C s m,  T q 39 the unit is correct, the sign is reasonable as the water must release heat to make its temperature fall Tro: Chemistry: A Molecular Approach, 2/e

40. Copyright © 2011 Pearson Education, Inc. Heat Transfer & Final Temperature When two objects at different temperatures are placed in contact, heat flows from the material at the higher temperature to the material at the lower temperature Heat flows until both materials reach the same final temperature The amount of heat energy lost by the hot material equals the amount of heat gained by the cold material q hot = −q cold m hot  C s,hot   T hot = −(m cold  C s,cold   T cold ) 40 Tro: Chemistry: A Molecular Approach, 2/e

41. Copyright © 2011 Pearson Education, Inc. 4. Heat lost by the metal = heat gained by water 1. Heat lost by the metal > heat gained by water 2. Heat gained by water > heat lost by the metal 3. Heat lost by the metal > heat lost by the water 4. Heat lost by the metal = heat gained by water 5. More information is required A piece of metal at 85 °C is added to water at 25 °C, the final temperature of the both metal and water is 30 °C. Which of the following is true? 41 Tro: Chemistry: A Molecular Approach, 2/e

42. Copyright © 2011 Pearson Education, Inc. q = m ∙ C s ∙  T C s, Al = 0.903 J/gºC, C s, H2O = 4.18 J/gºC(Table 6.4) Example 6.3: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures m Al = 32.5 g, T al = 45.8 °C, m H20 = 105.3 g, T H2O = 15.4 °C T final, °C Check: Solution: Conceptual Plan: Relationships: Given: Find: C s, Al m Al, C s, H2O m H2O  T Al = k   T H2O 42 q Al = −q H2O T final Tro: Chemistry: A Molecular Approach, 2/e

43. Copyright © 2011 Pearson Education, Inc. Practice – A hot piece of metal weighing 350.0 g is heated to 100.0 °C. It is then placed into a coffee cup calorimeter containing 160.0 g of water at 22.4 °C. The water warms and the copper cools until the final temperature is 35.2 °C. Calculate the specific heat of the metal and identify the metal. 43 Tro: Chemistry: A Molecular Approach, 2/e

44. Copyright © 2011 Pearson Education, Inc. 44 Practice – Calculate the specific heat and identify the metal from the data q = m x C s x  T; q metal = −q H2O the units are correct, the number indicates the metal is copper metal: 350.0 g, T 1 = 100.0 °C, T 2 = 35.2 °C H 2 O: 160.0 g, T 1 = 22.4 °C, T 2 = 35.2°C, C s = 4.18 J/g  °C C s, metal, J/g  ºC Check: Solution: Concept Plan: Relationships: Given: Find: m, C s,  T q Tro: Chemistry: A Molecular Approach, 2/e

45. Copyright © 2011 Pearson Education, Inc. Pressure –Volume Work PV work is work caused by a volume change against an external pressure When gases expand,  V is +, but the system is doing work on the surroundings, so w gas is ─ As long as the external pressure is kept constant ─ Work gas = External Pressure x Change in Volume gas w = ─P  V to convert the units to joules use 101.3 J = 1 atm∙L 45 Tro: Chemistry: A Molecular Approach, 2/e

46. Copyright © 2011 Pearson Education, Inc. Example 6.4: If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done? the unit is correct; the sign is reasonable because when a gas expands it does work on the surroundings and loses energy V 1 = 0.100 L, V 2 = 1.85 L, P = 1.00 atm w, Jw, J Check: Solution: Conceptual Plan: Relationships: Given: Find: P,  V w 101.3 J = 1 atm∙L 46 Tro: Chemistry: A Molecular Approach, 2/e

47. Copyright © 2011 Pearson Education, Inc. Practice – A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is released by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? (1 cal = 4.18 J, 101.3 J = 1.00 atmL) 47 Tro: Chemistry: A Molecular Approach, 2/e

48. Copyright © 2011 Pearson Education, Inc. Practice – A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is released by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? so both  E and q are −, and  E > q, w must be −; and when w is − the system is expanding, so V 2 should be greater than V 1 ; and it is  E = −68.3 kJ, q = −15.8 kcal, V 1 = 10.0 L, P = 1.00 atm V 2, L Check: Solution: Conceptual Plan: Relationships: Given: Find:  E = q + w, w = −P  V, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L q,  E w V2V2 P, V 1  E = −6.83 x 10 4 J, q = −6.604 x 10 4 J, V 1 = 10.0L, P = 1.00 atm V 2, L 48 Tro: Chemistry: A Molecular Approach, 2/e

49. Copyright © 2011 Pearson Education, Inc. Exchanging Energy Between System and Surroundings Exchange of heat energy q = mass x specific heat x  Temperature Exchange of work w = −Pressure x  Volume 49 Tro: Chemistry: A Molecular Approach, 2/e

50. Copyright © 2011 Pearson Education, Inc. Measuring  E, Calorimetry at Constant Volume Because  E = q + w, we can determine  E by measuring q and w In practice, it is easiest to do a process in such a way that there is no change in volume, so w = 0 at constant volume,  E system = q system In practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction – so instead, we measure the temperature change in the surroundings use insulated, controlled surroundings q system = −q surroundings The surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water q surroundings = q calorimeter = ─q system 50 Tro: Chemistry: A Molecular Approach, 2/e

51. Copyright © 2011 Pearson Education, Inc. Bomb Calorimeter Used to measure  E because it is a constant volume system The heat capacity of the calorimeter is the amount of heat absorbed by the calorimeter for each degree rise in temperature and is called the calorimeter constant C cal, kJ/ºC 51 Tro: Chemistry: A Molecular Approach, 2/e

52. Copyright © 2011 Pearson Education, Inc. Example 6.5: When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C. If C cal = 4.90 kJ/°C, find  E for burning 1 mole q cal = C cal x  T = −q rxn MM C 12 H 22 O 11 = 342.3 g/mol the units are correct, the sign is as expected for combustion 1.010 g C 12 H 22 O 11, T 1 = 24.92 °C, T 2 = 28.33 °C, C cal = 4.90 kJ/°C  E rxn, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: q rxn, mol EE C cal,  T q cal q rxn 2.9506 x 10 −3 mol C 12 H 22 O 11,  T = 3.41°C, C cal = 4.90 kJ/°C  E rxn, kJ/mol 52 Tro: Chemistry: A Molecular Approach, 2/e

53. Copyright © 2011 Pearson Education, Inc. Practice – When 1.22 g of HC 7 H 5 O 2 (MM 122.12) is burned in a bomb calorimeter, the temperature rises from 20.27 °C to 22.67 °C. If  E for the reaction is −3.23 x 10 3 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C? 53 Tro: Chemistry: A Molecular Approach, 2/e

54. Copyright © 2011 Pearson Education, Inc. Practice – When 1.22 g of HC 7 H 5 O 2 is burned in a bomb calorimeter, the temperature rises from 20.27 °C to 22.67 °C. If  E for the reaction is −3.23 x 10 3 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C? q cal = C cal x  T = −q rxn MM HC 7 H 5 O 2 = 122.12 g/mol the units and sign are correct 1.22 g HC 7 H 5 O 2, T 1 =20.27 °C, T 2 =22.67 °C,  E= −3.23 x 10 3 kJ/mol C cal, kJ/°C Check: Solution: Conceptual Plan: Relationships: Given: Find: q cal,  T C cal mol,  E q rxn q cal 9.990 x 10 −3 mol HC 7 H 5 O 2,  T = 2.40 °C,  E= −3.23 x 10 3 kJ/mol C cal, kJ/°C 54 Tro: Chemistry: A Molecular Approach, 2/e

55. Copyright © 2011 Pearson Education, Inc. Enthalpy The enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume H is a state function H = E + PV The enthalpy change,  H, of a reaction is the heat evolved in a reaction at constant pressure  H reaction = q reaction at constant pressure Usually  H and  E are similar in value, the difference is largest for reactions that produce or use large quantities of gas 55 Tro: Chemistry: A Molecular Approach, 2/e

56. Copyright © 2011 Pearson Education, Inc. Endothermic and Exothermic Reactions When  H is ─, heat is being released by the system Reactions that release heat are called exothermic reactions When  H is +, heat is being absorbed by the system Reactions that absorb heat are called endothermic reactions Chemical heat packs contain iron filings that are oxidized in an exothermic reaction ─ your hands get warm because the released heat of the reaction is transferred to your hands Chemical cold packs contain NH 4 NO 3 that dissolves in water in an endothermic process ─ your hands get cold because the pack is absorbing your heat 56 Tro: Chemistry: A Molecular Approach, 2/e

57. Copyright © 2011 Pearson Education, Inc. For an exothermic reaction, the surrounding’s temperature rises due to release of thermal energy by the reaction This extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat During the course of a reaction, old bonds are broken and new bonds are made The products of the reaction have less chemical potential energy than the reactants The difference in energy is released as heat Molecular View of Exothermic Reactions 57 Tro: Chemistry: A Molecular Approach, 2/e

58. Copyright © 2011 Pearson Education, Inc. Molecular View of Endothermic Reactions In an endothermic reaction, the surrounding’s temperature drops due to absorption of some of its thermal energy by the reaction During the course of a reaction, old bonds are broken and new bonds are made The products of the reaction have more chemical potential energy than the reactants To acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products 58 Tro: Chemistry: A Molecular Approach, 2/e

59. Copyright © 2011 Pearson Education, Inc. Enthalpy of Reaction The enthalpy change in a chemical reaction is an extensive property the more reactants you use, the larger the enthalpy change By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(g)  H = −2044 kJ 59 Tro: Chemistry: A Molecular Approach, 2/e

60. Copyright © 2011 Pearson Education, Inc. Example 6.7: How much heat is evolved in the complete combustion of 13.2 kg of C 3 H 8 (g)? 1 kg = 1000 g, 1 mol C 3 H 8 = −2044 kJ, Molar Mass = 44.09 g/mol the sign is correct and the number is reasonable because the amount of C 3 H 8 is much more than 1 mole 13.2 kg C 3 H 8, q, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: molkJkgg 60 Tro: Chemistry: A Molecular Approach, 2/e

61. Copyright © 2011 Pearson Education, Inc. Practice – How much heat is evolved when a 0.483 g diamond is burned? (  H combustion = −395.4 kJ/mol C) 61 Tro: Chemistry: A Molecular Approach, 2/e

62. Copyright © 2011 Pearson Education, Inc. Practice – How much heat is evolved when a 0.483 g diamond is burned? 1 mol C = −395.4 kJ, Molar Mass = 12.01 g/mol the sign is correct and the number is reasonable because the amount of diamond is less than 1 mole 0.483 g C,  H = −395.4 kJ/mol C q, kJ/mol Check: Solution: Concept Plan: Relationships: Given: Find: molkJg 62 Tro: Chemistry: A Molecular Approach, 2/e

63. Copyright © 2011 Pearson Education, Inc. Measuring  H Calorimetry at Constant Pressure Reactions done in aqueous solution are at constant pressure open to the atmosphere The calorimeter is often nested foam cups containing the solution q reaction = ─ q solution = ─(mass solution x C s, solution x  T)   H reaction = q constant pressure = q reaction to get  H reaction per mol, divide by the number of moles 63 Tro: Chemistry: A Molecular Approach, 2/e

64. Copyright © 2011 Pearson Education, Inc. Example 6.8: What is  H rxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 (g) if 0.158 g Mg reacts in 100.0 mL of solution and changes the temperature from 25.6 °C to 32.8 °C? the sign is correct and the value is reasonable because the reaction is exothermic 0.158 g Mg, 100.0 mL, T 1 = 25.6 °C, T 2 = 32.8 °C, assume d = 1.00 g/mL, C s = 4.18 J/g∙°C  H, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: q rxn, mol HH C s,  T, m q sol’n q rxn q sol’n = m x C s, sol’n x  T = −q rxn, MM Mg= 24.31 g/mol 6.499 x 10 −3 mol Mg, 1.00 x 10 2 g,  T = 7.2 °C,,C s = 4.18 J/g∙°C  H, kJ/mol 64 Tro: Chemistry: A Molecular Approach, 2/e

65. Copyright © 2011 Pearson Education, Inc. Practice – What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added to a 50.00 mL sample of 0.250 NaOH(aq). The initial temperature is 19.50 °C and the  H rxn is −57.2 kJ/mol NaOH. (assume the density of the solution is 1.00 g/mL and the specific heat of the solution is 4.18 J/g∙°C) 65 Tro: Chemistry: A Molecular Approach, 2/e

66. Copyright © 2011 Pearson Education, Inc. q sol’n = m x C s, sol’n x  T = −q rxn, 0.250 mol NaOH = 1 L Practice – What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added to a 50.00 mL sample of 0.250 NaOH(aq). The initial temperature is 19.50 °C. the sign is correct and the value is reasonable, because the reaction is exothermic we expect the final temperature to be higher than the initial 50.00 mL of 0.250 M HCl, 50.00 mL of 0.250 M NaOH T 1 = 19.50 °C, d = 1.00 g/mL, C s = 4.18 J/g∙°C,  H = −57.2 kJ/mol T 2, °C Check: Solution: Conceptual Plan: Relationships: Given: Find: 1.25 x 10 −2 mol NaOH, 1.00 x 10 2 g, T 1 = 19.50 °C,  H =−57.2 kJ/mol, C s = 4.18 J/g∙°C T 2, °C q rxn q sol’n HH TT C s, q sol’n, m T2T2 66 Tro: Chemistry: A Molecular Approach, 2/e

67. Copyright © 2011 Pearson Education, Inc. Relationships Involving  H rxn When reaction is multiplied by a factor,  H rxn is multiplied by that factor because  H rxn is extensive C(s) + O 2 (g) → CO 2 (g)  H = −393.5 kJ 2 C(s) + 2 O 2 (g) → 2 CO 2 (g)  H = 2(−393.5 kJ) = −787.0 kJ If a reaction is reversed, then the sign of  H is changed CO 2 (g) → C(s) + O 2 (g)  H = +393.5 kJ 67 Tro: Chemistry: A Molecular Approach, 2/e

68. Copyright © 2011 Pearson Education, Inc. Relationships Involving  H rxn Hess’s Law If a reaction can be expressed as a series of steps, then the  H rxn for the overall reaction is the sum of the heats of reaction for each step 68 Tro: Chemistry: A Molecular Approach, 2/e

69. Copyright © 2011 Pearson Education, Inc. [3 NO 2 (g)  3 NO(g) + 1.5 O 2 (g)]  H° = (+174 kJ) [1 N 2 (g) + 2.5 O 2 (g) + 1 H 2 O(l)  2 HNO 3 (aq)]  H° = (−128 kJ) [2 NO(g)  N 2 (g) + O 2 (g)]  H° = (−183 kJ) 3 NO 2 (g) + H 2 O(l)  2 HNO 3 (aq) + NO(g)  H° = − 137 kJ Example: Hess’s Law Given the following information: 2 NO(g) + O 2 (g)  2 NO 2 (g)  H° = −116 kJ 2 N 2 (g) + 5 O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq)  H° = −256 kJ N 2 (g) + O 2 (g)  2 NO(g)  H° = +183 kJ Calculate the  H° for the reaction below: 3 NO 2 (g) + H 2 O(l)  2 HNO 3 (aq) + NO(g)  H° = ? [2 NO 2 (g)  2 NO(g) + O 2 (g)] x 1.5  H° = 1.5(+116 kJ) [2 N 2 (g) + 5 O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq)] x 0.5  H° = 0.5(−256 kJ) [2 NO(g)  N 2 (g) + O 2 (g)]  H° = −183 kJ 69 Tro: Chemistry: A Molecular Approach, 2/e

70. Copyright © 2011 Pearson Education, Inc. Practice – Hess’s Law Given the following information: Cu(s) + Cl 2 (g)  CuCl 2 (s)  H° = −206 kJ 2 Cu(s) + Cl 2 (g)  2 CuCl(s)  H° = − 36 kJ Calculate the  H° for the reaction below: Cu(s) + CuCl 2 (s)  2 CuCl(s)  H° = ? kJ 70 Tro: Chemistry: A Molecular Approach, 2/e

71. Copyright © 2011 Pearson Education, Inc. Practice – Hess’s Law Given the following information: Cu(s) + Cl 2 (g)  CuCl 2 (s)  H° = −206 kJ 2 Cu(s) + Cl 2 (g)  2 CuCl(s)  H° = − 36 kJ Calculate the  H° for the reaction below: Cu(s) + CuCl 2 (s)  2 CuCl(s)  H° = ? kJ CuCl 2 (s)  Cu(s) + Cl 2 (g)  H° = +206 kJ 2 Cu(s) + Cl 2 (g)  2 CuCl(s)  H° = − 36 kJ Cu(s) + CuCl 2 (s)  2 CuCl(s)  H° = +170. kJ 71 Tro: Chemistry: A Molecular Approach, 2/e

72. Copyright © 2011 Pearson Education, Inc. Standard Conditions The standard state is the state of a material at a defined set of conditions pure gas at exactly 1 atm pressure pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest  usually 25 °C substance in a solution with concentration 1 M The standard enthalpy change,  H°, is the enthalpy change when all reactants and products are in their standard states The standard enthalpy of formation,  H f °, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements the elements must be in their standard states the  H f ° for a pure element in its standard state = 0 kJ/mol  by definition 72 Tro: Chemistry: A Molecular Approach, 2/e

73. Copyright © 2011 Pearson Education, Inc. Standard Enthalpies of Formation 73 Tro: Chemistry: A Molecular Approach, 2/e

74. Copyright © 2011 Pearson Education, Inc. Formation Reactions Reactions of elements in their standard state to form 1 mole of a pure compound if you are not sure what the standard state of an element is, find the form in Appendix IIB that has a  H f ° = 0 because the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions 74 Tro: Chemistry: A Molecular Approach, 2/e

75. Copyright © 2011 Pearson Education, Inc. Writing Formation Reactions Write the Formation Reaction for CO(g) The formation reaction is the reaction between the elements in the compound, which are C and O C + O → CO(g) The elements must be in their standard state there are several forms of solid C, but the one with  H f ° = 0 is graphite oxygen’s standard state is the diatomic gas C(s, graphite) + O 2 (g) → CO(g) The equation must be balanced, but the coefficient of the product compound must be 1 use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient C(s, graphite) + ½ O 2 (g) → CO(g) 75 Tro: Chemistry: A Molecular Approach, 2/e

76. Copyright © 2011 Pearson Education, Inc. Write the formation reactions for the following: CO 2 (g) Al 2 (SO 4 ) 3 (s) C(s, graphite) + O 2 (g)  CO 2 (g) 2 Al(s) + 3/8 S 8 (s, rhombic) + 6 O 2 (g)  Al 2 (SO 4 ) 3 (s) 76 Tro: Chemistry: A Molecular Approach, 2/e

77. Copyright © 2011 Pearson Education, Inc. Calculating Standard Enthalpy Change for a Reaction Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products The  H° for the reaction is then the sum of the  H f ° for the component reactions  H° reaction =  n  H f °(products) −  n  H f °(reactants)  means sum n is the coefficient of the reaction 77 Tro: Chemistry: A Molecular Approach, 2/e

78. Copyright © 2011 Pearson Education, Inc. CH 4 (g) → C(s, graphite) + 2 H 2 (g)  H° = + 74.6 kJ 2 H 2 (g) + O 2 (g) → 2 H 2 O(g)  H° = −483.6 kJ CH 4 (g)+ 2 O 2 (g)→ CO 2 (g) + H 2 O(g) C(s, graphite) + 2 H 2 (g) → CH 4 (g)  H f °= − 74.6 kJ/mol CH 4 C(s, graphite) + O 2 (g) → CO 2 (g)  H f °= −393.5 kJ/mol CO 2 H 2 (g) + ½ O 2 (g) → H 2 O(g)  H f °= −241.8 kJ/mol H 2 O CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g)  H° = −802.5 kJ  H° = [(  −393.5 kJ)+ 2(−241.8 kJ)− ((−74.6 kJ)+ 2(0 kJ))] = −802.5 kJ 78  H° = [(  H f ° CO 2 (g) + 2∙  H f ° H 2 O(g) )− (  H f ° CH 4 (g) + 2∙  H f ° O 2 (g) )] Tro: Chemistry: A Molecular Approach, 2/e

79. Copyright © 2011 Pearson Education, Inc. Example: Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l) 1. Write formation reactions for each compound and determine the  H f ° for each 2 C(s, gr) + H 2 (g)  C 2 H 2 (g)  H f ° = +227.4 kJ/mol C(s, gr) + O 2 (g)  CO 2 (g)  H f ° = −393.5 kJ/mol H 2 (g) + ½ O 2 (g)  H 2 O(l)  H f ° = −285.8 kJ/mol 79 Tro: Chemistry: A Molecular Approach, 2/e

80. Copyright © 2011 Pearson Education, Inc. 2 C 2 H 2 (g)  4 C(s) + 2 H 2 (g)  H° = 2(−227.4) kJ Example: Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l) 4 C(s) + 4 O 2 (g)  4CO 2 (g)  H° = 4(−393.5) kJ 2 H 2 (g) + O 2 (g)  2 H 2 O(l)  H° = 2(−285.8) kJ 2.Arrange equations so they add up to desired reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l)  H = −2600.4 kJ 80 Tro: Chemistry: A Molecular Approach, 2/e

81. Copyright © 2011 Pearson Education, Inc. Example: Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l)  H° reaction =  n  H f °(products) −  n  H f °(reactants)  H rxn = [(4  H CO2 + 2  H H2O ) – (2  H C2H2 + 5  H O2 )]  H rxn = [(4(−393.5) + 2(−285.8)) – (2(+227.4) + 5(0))]  H rxn = −2600.4 kJ 81 Tro: Chemistry: A Molecular Approach, 2/e

82. Copyright © 2011 Pearson Education, Inc. 82 Example 6.12: How many kg of octane must be combusted to supply 1.0 x 10 11 kJ of energy? MM octane = 114.2 g/mol, 1 kg = 1000 g the units and sign are correct, the large value is expected 1.0 x 10 11 kJ mass octane, kg Check: Solution: Conceptual Plan: Relationships: Given: Find:  H f °’s  H rxn ° Write the balanced equation per mole of octane kJ mol C 8 H 18 g C 8 H 18 kg C 8 H 18 from above C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g) Look up the  H f ° for each material in Appendix IIB  H°rxn = −5074.1 kJ Tro: Chemistry: A Molecular Approach, 2/e

83. Copyright © 2011 Pearson Education, Inc. Practice – Calculate the  H° for decomposing 10.0 g of limestone, CaCO 3, under standard conditions. CaCO 3 (s) → CaO(s) + O 2 (g) 83 Tro: Chemistry: A Molecular Approach, 2/e

84. Copyright © 2011 Pearson Education, Inc. Practice – Calculate the  H° for decomposing 10.0 g of limestone, CaCO 3 (s) → CaO(s) + O 2 (g) MM limestone = 100.09 g/mol, the units and sign are correct 10.0 g CaCO 3  H°, kJ Check: Solution: Conceptual Plan: Relationships: Given: Find:  H f °’s  H° rxn per mol CaCO 3 g CaCO 3 mol CaCO 3 kJ from above CaCO 3 (s) → CaO(s) + O 2 (g)  H° rxn = +572.7 kJ 84 Tro: Chemistry: A Molecular Approach, 2/e

85. Copyright © 2011 Pearson Education, Inc. 85 Energy Use and the Environment In the United States, each person uses over 10 5 kWh of energy per year Most comes from the combustion of fossil fuels combustible materials that originate from ancient life C(s) + O 2 (g) → CO 2 (g)  H° rxn = −393.5 kJ CH 4 (g) +2 O 2 (g) → CO 2 (g) + 2 H 2 O(g)  H° rxn = −802.3 kJ C 8 H 18 (g) +12.5 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g)  H° rxn = −5074.1 kJ Fossil fuels cannot be replenished At current rates of consumption, oil and natural gas supplies will be depleted in 50–100 yrs Tro: Chemistry: A Molecular Approach, 2/e

86. Copyright © 2011 Pearson Education, Inc. Energy Consumption 86 The distribution of energy consumption in the United States The increase in energy consumption in the United States Tro: Chemistry: A Molecular Approach, 2/e

87. Copyright © 2011 Pearson Education, Inc. 87 The Effect of Combustion Products on Our Environment Because of additives and impurities in the fossil fuel, incomplete combustion, and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy Therefore fossil fuel emissions contribute to air pollution, acid rain, and global warming Tro: Chemistry: A Molecular Approach, 2/e

88. Copyright © 2011 Pearson Education, Inc. 88 Global Warming CO 2 is a greenhouse gas it allows light from the sun to reach the earth, but does not allow the heat (infrared light) reflected off the Earth to escape into outer space  it acts like a blanket CO 2 levels in the atmosphere have been steadily increasing Current observations suggest that the average global air temperature has risen 0.6 °C in the past 100 yrs. Atmospheric models suggest that the warming effect could worsen if CO 2 levels are not curbed Some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitats Tro: Chemistry: A Molecular Approach, 2/e

89. Copyright © 2011 Pearson Education, Inc. 89 Tro: Chemistry: A Molecular Approach, 2/e

90. Copyright © 2011 Pearson Education, Inc. Our greatest unlimited supply of energy is the sun New technologies are being developed to capture the energy of sunlight parabolic troughs, solar power towers, and dish engines concentrate the sun’s light to generate electricity solar energy used to decompose water into H 2 (g) and O 2 (g); the H 2 can then be used by fuel cells to generate electricity H 2 (g) + ½ O 2 (g) → H 2 O(l)  H° rxn = −285.8 kJ Hydroelectric power Wind power 90 Renewable Energy Tro: Chemistry: A Molecular Approach, 2/e

91. Copyright © 2011 Pearson Education, Inc. 91 Tro: Chemistry: A Molecular Approach, 2/e