thermodynamics (owlbook chapter 5) owlbook deadline 10/29/12

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  • Slide 1
  • Thermodynamics (OWLBook Chapter 5) OWLBook Deadline 10/29/12
  • Slide 2
  • 5.1 Energy and Work
  • Slide 3
  • Slide 4
  • Kinetic Energy, E K The energy associated with an object by virtue of its motion. m = mass (kg) v = velocity (m/s) Potential Energy, E P The energy an object has by virtue of its position in a field of force, such as gravitaitonal, electric or magnetic field. Gravitational potential energy is given by the equation m = mass (kg) g = gravitational constant (9.80 m/s 2 ) h = height (m)
  • Slide 5
  • Units/Conversions/Conservat ion Law of Conservation of Energy Energy may be converted from one form to another, but the total quantity of energy remains constant.
  • Slide 6
  • Thermodynamic System The substance under study in which a change occurs is called the thermodynamic system (or just system). Thermodynamic Surroundings Everything else in the vicinity is called the thermodynamic surroundings (or just the surroundings).
  • Slide 7
  • Signs and Conventions
  • Slide 8
  • Enthalpy ( H ) Enthalpy, H An extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction. Extensive Property: A property that depends on the amount of substance. Mass and volume are extensive properties.
  • Slide 9
  • Endothermic or Exothermic
  • Slide 10
  • Slide 11
  • Lets do quick the OWLBook Check 5.2.1
  • Slide 12
  • Slide 13
  • Applying Stoichiometry to Heats of Reaction
  • Slide 14
  • Enthalpy Change and Stoichiometry Sulfur, S 8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kJ of heat per gram of sulfur (S 8 ) at constant pressure. o Write the balanced chemical equation o Write a thermochemical equation for this reaction, expressing the change in enthalpy in kJ/mole of S 8. o If 48 grams of sulfur is burned in an excess of oxygen, how much heat is produced?
  • Slide 15
  • A bomb calorimeter
  • Slide 16
  • Energy and State Changes
  • Slide 17
  • Lets do this and create the phase change diagram for this problem.
  • Slide 18
  • Enthalpy and Reactions Constant Pressure!
  • Slide 19
  • Remember Energy is Conserved 0 = q system + q surroundings q surroundings = (mass of calorimeter contents) x (specific heat) x T H = q system = -q surroundings
  • Slide 20
  • Specific Heat Capacity, s (or specific heat) o The quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one Kelvin) at constant pressure.
  • Slide 21
  • Slide 22
  • A piece of zinc weighing 35.8 g was heated from 20.00C to 28.00C. How much heat was required? The specific heat of zinc is 0.388 J/(gC).
  • Slide 23
  • Lets do OWLBook Tutorial 5.4.3.
  • Slide 24
  • Hesss Law Enthalpy is a State Function. It is the starting point and ending point that matter, NOT how you get there.
  • Slide 25
  • Hesss Law If you flip a reaction, the sign on H changes. If you multiply a reaction by a number, H gets multiplied by that same number. You can use any number or combination of reactions to get from point A-reactants to point B products. Used when reactions are slow, hard to measure, or even theoretical (never been done before).
  • Slide 26
  • Suppose we want DH for the reaction 2C(graphite) + O 2 (g) 2CO(g) It is difficult to measure directly. However, two other reactions are known: 2CO 2 (g) 2CO(g) + O 2 (g); H = 566.0 kJ C(graphite) + O 2 (g) CO 2 (g); H = -393.5 kJ In order for these to add to give the reaction we want, we must multiply the first reaction by 2. Note that we also multiply H by 2.
  • Slide 27
  • Copyright Cengage Learning. All rights reserved. 6 | 27 Our Goal = 2C(graphite) + O 2 (g) 2CO(g) 2C(graphite) + 2O 2 (g) 2CO 2 (g); H = -787.0 kJ 2CO 2 (g) 2CO(g) + O 2 (g); H = 566.0 kJ 2 C(graphite) + O 2 (g) 2 CO(g); H = 1353.0 kJ
  • Slide 28
  • Slide 29
  • Copyright Cengage Learning. All rights reserved. 6 | 29 What is the enthalpy of reaction, H, for the reaction of calcium metal with water? Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH - (aq) + H 2 (g) This reaction occurs very slowly, so it is impractical to measure H directly. However, the following facts are known: H + (aq) + OH - (aq) H 2 O(l); H = 55.9 kJ Ca(s) + 2H + (aq) Ca 2+ (aq) + H 2 (g); H = 543.0 kJ
  • Slide 30
  • Standard Enthalpy Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: o 1 atm pressure and the specified temperature (usually 25C). o These standard conditions are indicated with a degree sign (). When reactants in their standard states yield products in their standard states, the enthalpy of reaction is called the standard enthalpy of reaction, H. ( H is read delta H zero. )
  • Slide 31
  • The standard enthalpy of formation, H f , is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states. H f for an element in its reference and standard state is zero. For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction H 2 (g) + 1 / 2 O 2 (g) H 2 O(l) H f = 285.8 kJ Other H f values are found in various reference tables (not necessary to be memorized, except for H f for an element in its reference and standard state
  • Slide 32
  • Remember, these are for formation of these substances from their constituent elements in reference states.
  • Slide 33
  • Using Standard Enthalpies of Formation. We want H for the reaction: CH 3 OH(l) CH 3 OH(g) H vap = +38.0 kJ
  • Slide 34
  • Methyl alcohol, CH 3 OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate H o for the following reaction: 2CH 3 OH(aq) + O 2 (g) 2HCHO(aq) + 2H 2 O(l) Standard enthalpies of formation, : CH 3 OH(aq):-245.9 kJ/mol HCHO(aq):-150.2 kJ/mol H 2 O(l):-285.8 kJ/mol