CHAPTER 5 Principles of Reactivity: Energy and Chemical Reactions

1. CHAPTER 5Principles of Reactivity: Energy and Chemical Reactions

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3. THERMOCHEMISTRY

4. Geothermal power —Wairakei North Island, New Zealand

5. Burning peanuts supply sufficient energy to boil a cup of water. Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) Energy & Chemistry PLAY MOVIE PLAY MOVIE

6. These reactions are PRODUCT FAVORED They proceed almost completely from reactants to products, perhaps with some outside assistance. Energy & Chemistry

7. Energy & Chemistry 2 H2(g) + O2(g) f 2 H2O(g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2f 4 H+ + 4 e- Reduction: 4 e- + O2 + 2 H2O f 4 OH- H2/O2 Fuel Cell Energy, page 262

8. Energy & Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy — • light • electrical • kinetic and potential

9. Potential & Kinetic Energy Potential energy— energy a motionless body has by virtue of its position. PLAY MOVIE

10. Potential Energyon the Atomic Scale • Positive and negative particles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy NaCl — composed of Na+ and Cl- ions.

11. Potential Energyon the Atomic Scale • Positive and negative particles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy PLAY MOVIE

12. Potential & Kinetic Energy Kinetic energy— energy of motion • Translation PLAY MOVIE PLAY MOVIE

13. Potential & Kinetic Energy Kinetic energy— energy of motion. PLAY MOVIE PLAY MOVIE

14. Internal Energy (U) • PE + KE = Internal energy (U) • Internal energy of a chemical system depends on • number of particles • type of particles • temperature

15. Internal Energy (U) PE + KE = Internal energy (U) PLAY MOVIE

16. Internal Energy (U) • The higher the T the higher the internal energy • So, use changes in T (∆T) to monitor changes in energy (∆U).

17. Thermodynamics • Thermodynamics is the science of energy transfer as heat. PLAY MOVIE PLAY MOVIE Heat energy is associated with molecular motions. Energy transfers as heat until thermal equilibrium is established. ∆T measures energy transferred.

18. System and Surroundings • SYSTEM • The object under study • SURROUNDINGS • Everything outside the system

19. T(system) goes down T(surr) goes up Directionality of Energy Transfer • Energy transfer as heat is always from a hotter object to a cooler one. • EXOthermic: energy transfers from SYSTEM to SURROUNDINGS.

20. T(system) goes up T (surr) goes down Directionality of Energy Transfer • Energy transfer at heat is always from a hotter object to a cooler one. • ENDOthermic: heat transfers from SURROUNDINGSto theSYSTEM.

21. Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. • The total energy is unchanged in a chemical reaction. • If PE of products is less than reactants, the difference must be released as KE.

22. Energy Change in Chemical Processes PLAY MOVIE PE of system dropped. KE increased. Therefore, you often feel a T increase.

23. James Joule 1818-1889 UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) But we use the unit called the JOULE 1 cal = exactly 4.184 joules

24. Which has the larger heat capacity? HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C.

25. Specific Heat Capacity How much energy is transferred due to T difference? The heat (q) “lost” or “gained” is related to a) sample mass b) change in T and c) specific heat capacity

26. Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.184 Ethylene glycol 2.39 Al 0.897 glass 0.84 Aluminum

27. Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, what amount of energy (J) is lost by the Al?

28. heat gain/lose = q = (sp. ht.)(mass)(∆T) Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, what amount of energy (J) has been transferred by the Al? where ∆T = Tfinal - Tinitial q = (0.897 J/g•K)(25.0 g)(37 - 310)K q = - 6120 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

29. Heat/Energy TransferNo Change in State q transferred = (sp. ht.)(mass)(∆T)

30. Energy Transfer • Use energy transfer as a way to find specific heat capacity, Cp • 55.0 g Fe at 99.8 ˚C • Drop into 225 g water at 21.0 ˚C • Water and metal come to 23.1 ˚C • What is the specific heat capacity of the metal?

31. Energy Transfer Because of conservation of energy, q(Fe) = –q(H2O) (energy out of Fe = energy into H2O) or q(Fe) + q(H2O) = 0 q(Fe) = (55.0 g)(Cp)(23.1 ˚C – 99.8 ˚C) q(Fe) = –4219 • Cp q(H2O) = (225 g)(4.184 J/K•g)(23.1 ˚C – 21.0 ˚C) q(H2O) = 1977 J q(Fe) + q(H2O) = –4219 Cp + 1977 = 0 Cp = 0.469 J/K•g

32. Energy Transfer with Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) f Liquid water q = (heat of fusion)(mass)

33. Energy Transfer and Changes of State Requires energy (heat). This is the reason a) you cool down after swimming b) you use water to put out a fire. Liquid f Vapor + energy

34. Heating/Cooling Curve for Water Note that T is constant as ice melts

35. +333 J/g +2260 J/g Heat & Changes of State What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC? Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/g

36. Heat & Changes of State What quantity of energy as heat is required to melt 500. g of ice and heat the water to steam at 100 oC? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0 oC to 100 oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J 3. To evaporate water at 100 oC q = (500. g)(2260 J/g) = 1.13 x 106 J 4. Total energy = 1.51 x 106 J = 1510 kJ

37. Abba’s Refrigerator CCR, pages 222

38. Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery

39. Chemical Reactivity But energy transfer also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are product-favored. So, let us consider energy transfer in chemical processes.

40. Transferring Energy as Heat in a Physical Process CO2 (s, -78 oC) f CO2 (g, -78 oC) Energy transfers from surroundings to system in endothermic process. PLAY MOVIE

41. Transferring Energy as Heat in a Physical Process • CO2 (s, -78 oC) f CO2 (g, -78 oC) • A regular array of molecules in a solid f gas phase molecules. • Gas molecules have higher kinetic energy. PLAY MOVIE

42. CO2 gas ∆U = U(final) - U(initial) = U(gas) - U(solid) CO2 solid Energy Level Diagram for Energy Transfer

43. Transferring Energy as Heat in Physical Change • Gas molecules have higher kinetic energy. • Also, WORKis done by the system in pushing aside the atmosphere. CO2 (s, -78 oC) fCO2 (g, -78 oC) Two things have happened!

44. heat energy transferred work done by the system energy change FIRST LAW OF THERMODYNAMICS ∆U = q + w Energy is conserved!

45. energy transfer in (endothermic), +q energy transfer out (exothermic), -q w transfer in (+w) w transfer out (-w) SYSTEM ∆U = q + w

46. ENTHALPY Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ∆Hwhere H = enthalpy and so ∆U = ∆H + w (and w is usually small) ∆H = energy transferred as heat at constant P ≈ ∆U ∆H = change in heat contentof the system ∆H = Hfinal - Hinitial

47. ENTHALPY ∆H = Hfinal - Hinitial If Hfinal > Hinitial then ∆H is positive Process is ENDOTHERMIC If Hfinal < Hinitial then ∆H is negative Process is EXOTHERMIC

48. USING ENTHALPY Consider the formation of water H2(g) + 1/2 O2(g) œ H2O(g) + 241.8 kJ Exothermic reaction — energy is a “product” and∆H = – 241.8 kJ

49. USING ENTHALPY Making liquid H2O from H2 + O2 involves twoexothermic steps. H2 + O2 gas H2O vapor Liquid H2O

50. USING ENTHALPY Making H2O from H2 involves two steps. H2(g) + 1/2 O2(g) f H2O(g) + 242 kJ H2O(g) f H2O(liq) + 44 kJ ----------------------------------------------------------------------- H2(g) + 1/2 O2(g) f H2O(liq) + 286 kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.