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Thermochemistry

Thermochemistry. Dr. Ron Rusay. Energy Joules (J) / calorie (cal) : (4.184 J = 1 cal). Can be defined as the capacity to do work. Chemical energy is defined as heat. Name five other types of Energy. Two Types of Energy.

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Thermochemistry

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  1. Thermochemistry Dr. Ron Rusay

  2. Energy Joules (J) / calorie (cal) : (4.184 J = 1 cal) • Can be defined as the capacity to do work. • Chemical energy is defined as heat. • Name five other types of Energy.

  3. Two Types of Energy • Potential: due to an object’s position or material’s composition - which can be converted to work • Kinetic: due to motion of an object • KE = 1/2 mv2 • (m = mass, v = velocity)

  4. Law of Conservation of Energy • Different forms of energy can be inter-converted but can neither be created nor destroyed. • (Euniverse is constant) • Describe three inter-conversions of energy.

  5. Temperature v. Energy • Temperature reflects random motions of particles; i.e. the kinetic energy of a system. • Heat involves a transfer of energy between 2 objects due to different energies and temperature differences. Always: HOT cold

  6. Heat (Energy) Loss

  7. Energy: A State Function • Depends only on the state of the system - not the path of how it arrived at that state. • It is independent of pathway.

  8. System and Surroundings • System: That on which we focus attention • Surroundings: Everything else in the universe • Universe = System + Surroundings

  9. Defining Energy ChangeExo- and Endo- thermic(Exergonic and Endergonic) • Two types of energy change : • Exothermic: Heat flows outof the system (to the surroundings).…negative sign • Endothermic: Heat flows intothe system (from the surroundings).…positive sign

  10. First of Three Laws of Thermodynamics • First Law of Thermodynamics: The energy of the universe is constant or “energy is conserved”.

  11. Heat Capacity(Specific Heat) P O http://chemconnections.org/general/chem120/Flash/specific_heat_s.swf

  12. Terminology • Specific heat capacity heat capacity per gram = J/°C g or J/K g • Molar heat capacity heat capacity per mole = J/°C mol or J/K mol

  13. Calorimeters http://chemconnections.org/general/chem120/Flash/calorimetry_s.html

  14. QUESTION

  15. 0.13 J/g C A common mistake is attempting to solve for the heat capacity instead of specific heat capacity. Using and keeping track of the units of measure takes care of this type of problem. ANSWER ° B) 0.13 J/g C A common mistake is attempting to solve for the heat capacity instead of specific heat capacity. Using and keeping track of the units of measure

  16. Heat Capacities

  17. QUESTION

  18. E) still a liquid. = . = 5.0 kJ, = 4.184 J/g C m D and = 15.5 g. Solving for shows that the m water was not raised to the 100 C necessary for boiling. ANSWER ´ ´ D · ° q s T q s T °

  19. Why can you burn the top of your mouth with hot pizza and not the bottom? (The top & bottom are at the same temperature!) http://www.dailymotion.com/video/x3hfwx_the-science-of-pizza_people

  20. Why can you burn the top of your mouth with hot pizza and not the bottom? (The top & bottom are at the same temperature.) (Cp) on body fat. In obese mice (fat content 52.76% body wt) the heat capacity was 2.65 kJ kg-1 K-1 and in lean mice (fat content 7.55% body wt) the heat capacity was 3.66 kJ kg-1 K-1.

  21. Specific Heat Interactive

  22. Energy diagrams w = - 1500 kJ q = +2000 kJ E  = q + w = +2000kJ + (-1500kJ) = +500kJ Can the system do 3000 kJ of work on the surroundings? Process Path If a system has 2,000 kJ put into it and the system does work of 1500 kJ on the surroundings, what is ?

  23. Energy diagrams w = - 3000 kJ q = +2000 kJ E  = q + w = +2000kJ + (-3000kJ) = -1000kJ Process Path Can the system do 3000 kJ of work on the surroundings? YES, the are many possibilities for the system-surroundings energy transfer

  24. QUESTION

  25. D) 15.2 J. When = 0, then the internal energy equals the work. ANSWER – q

  26. QUESTION

  27. D C) = 35 kJ. Internal energy = + . The heat and work both have a positive sign indicating energy flowed from the system to the surroundings. ANSWER E q w

  28. EnergyThe Gas Combustion Engine http://chemconnections.org/general/movies/html-swf/workversusenergyflow.htm

  29. Enthalpy diagramsqp = H H2O (l) @ 25 oC H H = ? H2O (s) @- 25 oC Process Path If 50.0 g of ice @ -25 oC warms to 25 oC what is H of the process?

  30. http://chemconnections.org/general/movies/HeatingCurves.swf

  31. Hdeposition= (-) H = J or kJ cal or kcal Hcondensation= (-) Hsolidification= (-) f.p. oC cooling cooling heating heating b.p. oC m.p. oC Hvaporization= (+) Hfusion= (+) Hsublimation= (+)

  32. Enthalpies H?s  l  g H2O (l) @ 25 oC H = Cp liq x mass x T H H = 5.22 kJ H2 O (l) @ 0 oC Hfusion = 6.009 kJ/mol H2 O (s) @ 0 oC Hfusion = 16.69 kJ H2 O (s) @- 25 oC H = Cp ice x mass x T H = 2.54 kJ Process Path H = Hice +Hfusion +Hliq If 50.0 g of ice @ -25 oC warms to 25 oC what is H of the process? H = ? H = 24.45 kJ

  33. QUESTION

  34. C) 461 J/g The heat of fusion of the solid is found indirectly from the heat loss of the water. ANSWER

  35. “Heat of Reaction” Change in Enthalpy http://chemconnections.org/general/movies/hesslaw.mov • The heat of any reaction can be calculated from enthalpies of formation of reactants and products.(“Hess’s Law”) • Hrxn° = npHf(products) nrHf(reactants)

  36. QUESTION • 4675 kJ B) -1545 kJ C) -290 kJ • D) -1720 kJ

  37. 290 kJ Remember to multiply the heat of formation of each compound by its coefficient. ANSWER C

  38. Ex. Combustion of MethaneCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

  39. Hrxn° = [1Hf(c) + 2Hf(d)] -[1Hf(a) + 2Hf(b)] - Hrxn°= [-394kJ+(-572kJ)]-[-75kJ+0kJ]= -891kJ

  40. QUESTION

  41. B) 413 kJ Don’t forget to convert grams to moles in this problem. Enthalpy is always in units of kJ/mol. ANSWER –

  42. Exo- and Endo- thermic(Exergonic and Endergonic) • Heat exchange accompanies chemical reactions. • Exothermic: Heat flows outof the system (to the surroundings).…negative sign • Endothermic: Heat flows intothe system (from the surroundings).…positive sign

  43. QUESTION

  44. E) The chemical reaction is absorbing energy (C) and the energy can be calculated in (D). The temperature drops because the reactants are absorbing energy from the solution faster than the surroundings can replace it. ANSWER

  45. ANSWER(s)

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