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Chapter 6 Thermochemistry

Chemistry: A Molecular Approach , 2nd Ed. Nivaldo Tro. Chapter 6 Thermochemistry. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. Chemical Hand Warmers. Most hand warmers work by using the heat released from the slow oxidation of iron

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Chapter 6 Thermochemistry

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

  2. Chemical Hand Warmers • Most hand warmers work by using the heat released from the slow oxidation of iron 4 Fe(s) + 3 O2(g) → 2 Fe2O3(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 Tro: Chemistry: A Molecular Approach, 2/e

  3. 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 Tro: Chemistry: A Molecular Approach, 2/e

  4. 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 Tro: Chemistry: A Molecular Approach, 2/e

  5. Classification of Energy • Kinetic energyis 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 Tro: Chemistry: A Molecular Approach, 2/e

  6. 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 Tro: Chemistry: A Molecular Approach, 2/e

  7. Manifestations of Energy Tro: Chemistry: A Molecular Approach, 2/e

  8. 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 Tro: Chemistry: A Molecular Approach, 2/e

  9. 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 Tro: Chemistry: A Molecular Approach, 2/e

  10. Surroundings Surroundings System System 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 Tro: Chemistry: A Molecular Approach, 2/e

  11. 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 Tro: Chemistry: A Molecular Approach, 2/e

  12. Units of Energy • The amount of kinetic energy an object has is directly proportional to its mass and velocity • KE = ½mv2 Tro: Chemistry: A Molecular Approach, 2/e

  13. 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∙m2/s2 • 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 Tro: Chemistry: A Molecular Approach, 2/e

  14. Energy Use Tro: Chemistry: A Molecular Approach, 2/e

  15. The First Law of ThermodynamicsLaw 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 Tro: Chemistry: A Molecular Approach, 2/e

  16. 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 • DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings • D Is the symbol that is used to mean change • final amount – initial amount Tro: Chemistry: A Molecular Approach, 2/e

  17. Internal Energy • The internal energyis 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 functionis a mathematical function whose result only depends on the initial and final conditions, not on the process used • DE = Efinal – Einitial • DEreaction = Eproducts − Ereactants Tro: Chemistry: A Molecular Approach, 2/e

  18. 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! Tro: Chemistry: A Molecular Approach, 2/e

  19. final energy added DE = + Internal Energy initial initial energy removed DE = ─ Internal Energy final Energy Diagrams • 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 Energy diagrams are a “graphical” way of showing the direction of energy flow during a process

  20. Surroundings DE + System DE ─ Energy Flow • When energy flows out of a system, it must all flow into the surroundings • When energy flows out of a system, DEsystem is ─ • When energy flows into the surroundings, DEsurroundings is + • Therefore: ─ DEsystem= DEsurroundings Tro: Chemistry: A Molecular Approach, 2/e

  21. Surroundings DE ─ System DE + Energy Flow • When energy flows into a system, it must all come from the surroundings • When energy flows into a system, DEsystem is + • When energy flows out of the surroundings, DEsurroundings is ─ • Therefore: DEsystem= ─DEsurroundings Tro: Chemistry: A Molecular Approach, 2/e

  22. energy absorbed DErxn = + energy released DErxn = ─ Energy Flow in a Chemical Reaction • The total amount of internal energy in 1mol of C(s) and 1 mole of O2(g) is greater than the internal energy in 1 mole of CO2(g) • at the same temperature and pressure • In the reaction C(s) + O2(g) → CO2(g), there will be a net release of energy into the surroundings • −DEreaction = DEsurroundings • In the reaction CO2(g) → C(s) + O2(g), there will be an absorption of energy from the surroundings into the reaction • DEreaction = − DEsurroundings C(s), O2(g) Internal Energy CO2(g) Surroundings System CO2 →C + O2 System C + O2→ CO2 Tro: Chemistry: A Molecular Approach, 2/e

  23. 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 DE = q + w Tro: Chemistry: A Molecular Approach, 2/e

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

  25. Heat & Work • 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 Tro: Chemistry: A Molecular Approach, 2/e

  26. Heat & Work 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, DE = KEfinal − KEinitial = 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 = DE energy change for the white ball, DE = KEfinal − KEinitial = 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 = DE 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 Tro: Chemistry: A Molecular Approach, 2/e

  27. Heat, Work, and Internal Energy • In the previous billiard ball example, the DE 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 DE 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 Tro: Chemistry: A Molecular Approach, 2/e

  28. q, w q, w potato q, w fuel DE Example 6.1: If the burning of the fuel in a potato cannon performs 855 J of work on the potato andproduces 1422 J of heat, what is DE for the burning of the fuel? Given: Find: qpotato = 855 J, wpotato = 1422 J DEfuel,J Concept Plan: Relationships: qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w Solution: Check: the unit is correct, the sign make sense as the fuel should lose energy during the reaction Tro: Chemistry: A Molecular Approach, 2/e

  29. Practice – Reacting 50 mL of H2(g) with 50 mL of C2H4(g) produces 50 mL of C2H6(g) at 1.5 atm. If the reaction produces 3.1 x 102 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? Tro: Chemistry: A Molecular Approach, 2/e

  30. q, w w surrounding DE w gases If the reaction produces 3.1 x 102 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? Given: Find: qreaction = −310 J, wsurroundings = −7.6 J DEgases,J Concept Plan: Relationships: qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w Solution: Check: 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 Tro: Chemistry: A Molecular Approach, 2/e

  31. Heat Exchange • Heatis 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 Tro: Chemistry: A Molecular Approach, 2/e

  32. 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 DT • The larger the heat capacity of the object being studied, the smaller the temperature rise will be for a given amount of heat Tro: Chemistry: A Molecular Approach, 2/e

  33. 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 Tro: Chemistry: A Molecular Approach, 2/e

  34. 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 • Cs • units are J/(g∙°C) • The molar heat capacityis the amount of heat energy required to raise the temperature of one mole of a substance 1°C Tro: Chemistry: A Molecular Approach, 2/e

  35. 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 Tro: Chemistry: A Molecular Approach, 2/e

  36. 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 (Cs) x (DT) Tro: Chemistry: A Molecular Approach, 2/e

  37. Cs m, DT q 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? • Sort Information Given: Find: T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g q,J q = m ∙ Cs ∙ DT Cs = 0.385 J/g•ºC (Table 6.4) • Strategize Conceptual Plan: Relationships: • Follow the conceptual plan to solve the problem Solution: • Check Check: the unit is correct, the sign is reasonable as the penny must absorb heat to make its temperature rise Tro: Chemistry: A Molecular Approach, 2/e

  38. 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) Tro: Chemistry: A Molecular Approach, 2/e

  39. Cs m, DT q Practice – Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C • Sort Information Given: Find: T1= 49 °C, T2 = 29 °C, m = 7.40 g q,J q = m ∙ Cs ∙ DT Cs = 4.18 J/gC (Table 6.4) • Strategize Conceptual Plan: Relationships: • Follow the concept plan to solve the problem Solution: • Check Check: 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. 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 qhot = −qcold mhotCs,hotThot = −(mcoldCs,coldTcold) Tro: Chemistry: A Molecular Approach, 2/e

  41. Heat lost by the metal > heat gained by water Heat gained by water > heat lost by the metal Heat lost by the metal > heat lost by the water Heat lost by the metal = heat gained by water 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? • Heat lost by the metal > heat gained by water • Heat gained by water > heat lost by the metal • Heat lost by the metal > heat lost by the water • Heat lost by the metal = heat gained by water • More information is required Tro: Chemistry: A Molecular Approach, 2/e 41

  42. Cs, AlmAl, Cs, H2O mH2O DTAl = kDTH2O 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? Given: Find: mAl = 32.5 g, Tal = 45.8 °C, mH20 = 105.3 g, TH2O = 15.4 °C Tfinal,°C q = m ∙ Cs ∙ DT Cs, Al = 0.903 J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4) Conceptual Plan: Relationships: qAl = −qH2O Tfinal Solution: the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures Check: Tro: Chemistry: A Molecular Approach, 2/e

  43. 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. Tro: Chemistry: A Molecular Approach, 2/e

  44. m, Cs, DT q Practice – Calculate the specific heat and identify the metal from the data Given: Find: metal: 350.0 g, T1 = 100.0 °C, T2 = 35.2 °C H2O: 160.0 g, T1 = 22.4 °C, T2 = 35.2°C, Cs = 4.18 J/g °C Cs , metal,J/gºC Concept Plan: Relationships: q = m x Cs x DT; qmetal = −qH2O Solution: Check: the units are correct, the number indicates the metal is copper Tro: Chemistry: A Molecular Approach, 2/e

  45. Pressure –Volume Work • PV work is work caused by a volume change against an external pressure • When gases expand, DV is +, but the system is doing work on the surroundings, so wgas is ─ • As long as the external pressure is kept constant ─ Workgas = External Pressure x Change in Volumegas w = ─PDV • to convert the units to joules use 101.3 J = 1 atm∙L Tro: Chemistry: A Molecular Approach, 2/e

  46. P, DV w 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? Given: Find: V1 = 0.100 L, V2 = 1.85 L, P = 1.00 atm w,J Conceptual Plan: Relationships: 101.3 J = 1 atm∙L Solution: Check: the unit is correct; the sign is reasonable because when a gas expands it does work on the surroundings and loses energy Tro: Chemistry: A Molecular Approach, 2/e

  47. 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 atm•L) Tro: Chemistry: A Molecular Approach, 2/e 47

  48. 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? Given: Find: DE = −68.3 kJ, q = −15.8 kcal, V1 = 10.0 L, P = 1.00 atm V2, L DE = −6.83 x 104 J, q = −6.604 x 104J, V1 = 10.0L, P = 1.00 atm V2, L w Conceptual Plan: Relationships: q, DE V2 P, V1 DE = q + w, w = −PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L Solution: Check: so both DE and q are −, and DE > q,w must be −; and when w is − the system is expanding, so V2 should be greater than V1; and it is Tro: Chemistry: A Molecular Approach, 2/e 48

  49. Exchanging Energy BetweenSystem and Surroundings • Exchange of heat energy q = mass x specific heat x DTemperature • Exchange of work w = −Pressure x DVolume Tro: Chemistry: A Molecular Approach, 2/e

  50. Measuring DE, Calorimetry at Constant Volume • Because DE = q + w, we can determine DE 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, DEsystem = qsystem • 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 • qsystem = −qsurroundings • The surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water qsurroundings = qcalorimeter = ─qsystem Tro: Chemistry: A Molecular Approach, 2/e

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