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Chapter 5

Chapter 5. Thermochemistry. Definitions. Thermochemistry - the study of how energy in the form of heat is consumed and produced by chemical reactions. Thermochemical Reaction : Example H 2 ( g ) + O 2 ( g ) 2 H 2 O ( l ) + heat

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Chapter 5

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  1. Chapter 5 Thermochemistry

  2. Definitions • Thermochemistry - the study of how energy in the form of heat is consumed and produced by chemical reactions. • ThermochemicalReaction: Example H2(g) + O2(g) 2 H2O (l) + heat • Energy is anything having the capacity to do work or to transfer heat. • Work is Force X Distance • Thermodynamics The study of energy and its transformation from one form to another.

  3. Energy Examples Definition-Energy is anything with the capacity to do work, or create heat. • Food • Gasoline • Electricity • An apple on a tree • A baseball moving

  4. Energy Units Units of Energy • Joules = kg(m/s)2 • Calories, an older unit; the energy to increase one gram of water one deg. C • Calories, unit of food energy = kcal

  5. DefinitionsContinued • Heat is energy transferred between objects because of a difference in their temperatures. • Thermodynamics is the study of relationship between chemical reactions and changes in heat energy. • Heat transfer is the process of heat energy flowing from one object into another

  6. Two Types of Energy Potential:due to position or composition - can be converted to work PE = mgh (m = mass, kg, g = force of gravity(9.8 m/s2), and h = vertical distance in meters) • These units multiplied together = joule • (chemical energy is a form of potential energy) • Kinetic:due to motion of the object KE = 1/2 mv 2 (m = mass(kg), v = velocity(m/s)) • These units multiplied together also equal a joule also. • Joule = kg(m/s)2

  7. Kinetic vs. Potential Energy

  8. Potential Energy: A State Function • A State Function depends only on the present state of the system - not how it arrived there. • It is independent of pathway.

  9. Other State Functions? • Heat (a form of energy)? • Work? Altitude? • Altitude?

  10. Other State Functions? • Heat (a form of energy)? Yes, not path dependent • Work? • Altitude?

  11. Other State Functions? • Heat (a form of energy)? Yes, not path dependent • Work? No, work depends on path • Altitude?

  12. Other State Functions? • Heat (a form of energy)? Yes, not path dependent • Work? No, work depends on path • Altitude? Yes, does not depend on path • Enthalpy ΔH,?

  13. Other State Functions? • Heat (a form of energy)? Yes, not path dependent • Work? No, work depends on path • Altitude? Yes, does not depend on path • Enthalpy ΔH,? Yes, does not depend on path

  14. The Nature of Energy • Energy is conserved: the First Law of thermodynamics is called the Law of Conservation of Energy which states that energy cannot be created nor destroyed, but can be converted from one form to another.

  15. Energy at the Molecular Level Kinetic energy at the molecular level depends on the mass and velocity of the particle. Since velocity depends on temperature, then kinetic energy does too, because matter cannot be created nor destroyed. Another way to say this is that absolute temperature and kinetic energy are directly proportional to each other. Another important form of energy is potential energy at the atomic-molecular level arises from electrostatic interactions. PotentialEnergycan be converted into kinetic, for example an apple falling from a tree.

  16. Electrostatic Potential Energy Coulombicattraction, not gravitationalforce, determines the potential energy of matter at the atomic level. The energy is determined by multiplying the charge in coulombs of the two particles and dividing by the distance squared in meters. 2 • Eel is the electrostatic potential energy • Q is the charge in coulombs • d is the distance between particles in meters

  17. Electrostatic Potential Energy Bond length Energy of an ionic bond Bonds contain potential energy Energy required to break bonds Energy released when bonds are created

  18. Crystal Lattice of NaCl Ionic compounds do notexist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as latticesites. In an ionic compound the ions organize in such a way as to minimize repulsive and maximize attractive forces. This is an example of Coulomb’s Law in action.

  19. The Covalent Bond - + + - The covalent bond in an example of a Columbic attraction. The arrangement of charged particles in a covalent bond organized in such a way as to minimize repulsiveand maximizeattractive forces to give the lowest potential energy.

  20. Terms Describing Energy Transfer • System: the part of the universe that is the focus of a thermodynamic study. • Example a beaker or test tube in the lab • Surroundings: everything in the universe that is not part of the system. • Universe = System + Surroundings • An isolated system exchanges neither energy nor matter with the surroundings.

  21. Examples

  22. Heat Flow • In an exothermic process, heat flows from a system into its surroundings. • In an endothermic process, heat flows from the surroundings into the system

  23. Phase Changes

  24. Internal Energy • The internal energy of a system is the sum of all the KE and PE of all of the components of the system.

  25. First Law of Thermodynamics • The first law of thermodynamics states that the energy gained or lost by a system must equal the energy lost or gained by surroundings. ΔE = q + w (mathematical statement) • The calorie (cal) is the amount of heat necessary to raise the temperature of 1 g of water 1oC. • The joule (J) is the SI unit of energy; 4.184 J = 1 cal.

  26. Energy Flow Diagrams

  27. Change in Internal Energy E = q + w • E = change in system’s internal energy • q = heat • Endothermic +q • Exothermic -q • Expansion –w (since system is losing energy to do work) • Compression +w

  28. First Law Problem Find the change in energy of a system when 12 j of energy flows into the system while the system is doing 8 j of work on the surroundings.

  29. First Law Problem Find the change in energy of a system when 12 j of energy flows into the system while the system is doing 8 j of work on the surroundings. ΔE = q + w ↔ ΔE = + 12 – 8 = 4 j

  30. PV Work Expansion Facts Atm = 14 lb/in2 W = F x d ↓ A P = F/A F = PA h W = PA x d A ΔV = A x d W = P ΔV ΔV = Vfinal – Vinitial

  31. PV Work Facts Expansion ΔV = Vfinal – Vinitial ΔV = A x d h A W = F x d P = F/A F = PA Since expansion is defined as negative work and ΔV is positive for expansion, then we change the sign of PΔV to–PΔV, Substituting W = PxAxd W = PΔV ΔE = q - P ΔV

  32. PV Work Facts Expansion ΔV = Vfinal – Vinitial ΔV = A x d h A W = F x d P = F/A F = PA Since expansion is defined as negative work and ΔV is positive for expansion, then we change the sign of PΔV to–PΔV, Substituting W = PxAxd W = PΔV ΔE = q - P ΔV (L-atm = 101.3 j)

  33. Calculation of Work Calculate the work in L•atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm (Note that as the gas expands, it does work on its surroundings. Energy flows out of the gas, so it is a negative quantity)

  34. Calculation of Work Calculate the work in L•atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm (Note that as the gas expands, it does work on its surroundings. Energy flows out of the gas, so it is a negative quantity) W = -pΔV W = -18atm(72-54)L = -320L-atm j -320L-atm = -3.2 j 101.3 L-atm

  35. Calculation of Work Calculate the work in L•atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm (Note that as the gas expands, it does work on its surroundings. Energy flows out of the gas, so it is a negative quantity) Note: The result is negative, which we would predict relative to expansion. This is the reason that work is -pΔV, to give the correct sign for expansion. W = -pΔV W = -18atm(72-54)L = -320L-atm j -320L-atm = -3.2 j 101.3 L-atm

  36. Enthalpy and Change in Enthalpy • Enthalpy(H) = E + PV ( mathematical definition) • Change in Enthalpy(H) = E + PV • At constant P, qP = E + PV, therefore qP = H • At Constant V, qv = ΔE, since ΔV = 0 • H = change in enthalpy: an energy flow as heat (at constant pressure) • H > 0, Endothermic; H < 0, Exothermic

  37. Heating Curves

  38. Heat Capacities • Molar heat capacity (cp) is the heat required to raise the temperature of 1 mole of a substance by 1oC at constant pressure. • q = ncpT • Specific heat (cs) is the heat required to raise the temperature of 1 gram of a substance by 1oC at constant pressure. • Heat capacity (Cp) is the quantity of heat needed to raise the temperature of some specific object by 1oC at constant pressure.

  39. Phase Change and Energy • Molar heat of fusion (Hfus) - the heat required to convert 1 mole of a solid substance at its melting point to 1 mole of liquid. • q = nHfus • Molar heat of vaporization (Hvap) - the heat required to convert 1 mole of a substance at its boiling point to 1 mole of vapor. • q = nHvap

  40. WATER THERMO VALUES • Ice H2O (s) 2.06 j/g-°C • Water H2O (l) 4.184 j/g-°C • Steam H2O (g) 1.86 j/g-°C • Heat of Fusion (melting) 334.0 j/g • Heat of vaporization (evaporation) 2257 j/g

  41. Practice During a strenuous workout, a student generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat?

  42. Practice During a strenuous workout, a student generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? g 2257 j

  43. Practice During a strenuous workout, a student generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? g 10 3 j 2257 j kj

  44. Practice During a strenuous workout, a student generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? g 10 3 j 2000 kj 2257 j kj

  45. Practice During a strenuous workout, a student generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? g 10 3 j 2000 kj = 886 g water 2257 j kj

  46. Practice5.53 From Text Exactly 10 mL of water at 25oC was added to a hot iron skillet. All of the water was converted into steam at 100oC. If the mass of the pan was 1.20 kg and the molar heat capacity of iron is 25.19 J/mol•oC, what was the temperature change of the skillet?

  47. Sample Problem Solution mole-°C 25.19 j

  48. Sample Problem Solution mole-°C 55.85 g 25.19 j mole

  49. Sample Problem Solution mole-°C 55.85 g kg 25.19 j mole 103 g

  50. Sample Problem Solution mole-°C 55.85 g kg 25.19 j mole 103 g 1.20 kg

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