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Chapter 5: Thermochemistry

Chapter 5: Thermochemistry

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Chapter 5: Thermochemistry

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  1. Chapter 5: Thermochemistry AP Chemistry

  2. Energy work The capacity to do WORK or produce HEAT No work

  3. Forms of Energy Potential Energy Stored energy due to position or composition

  4. Gravitational Potential Energy (Eg) Stored energy due to an object’s height above the surface of the earth

  5. Chemical Potential Energy (Echem) Methane CH4 Landfill Gas Flare Energy Stored in the bonds of molecules

  6. Forms of Energy Kinetic Energy Energy due to motion

  7. Mechanical Kinetic Energy (Ek) Ek = ½ mv2 Energy due to an object’s motion

  8. Thermal Kinetic Energy • Energy due to molecular motion

  9. First Law of Thermodynamics AKA The Law of Conservation of Energy Energy cannot be created or destroyed; it can just change forms

  10. James Prescott Joule Wealthy English Beer Baron (1818 - 1889)

  11. Joule’s Chamonix Falls Experiment Potential Energy (mechanical) P.E. = m g h Kinetic Energy (mechanical and thermal) K.E. = ½ m v2 Q = m c DT Law of Conservation of Energy Energy cannot be created or destroyed in only changes forms

  12. How we look at reactions • system: The Reactants and Products • surroundings: The room, container, etc. http://www.weldonryan.com/new_page_1.htm

  13. DE = Efinal- Einitial DE E R P R P Internal Energy, E • The TOTAL energy of the system (potential + kinetic energy) DE E Energy GAINED by system Energy LOST by system

  14. Temperature vs. Heat Temperature— Measure of the average kinetic energy of molecules Cl2, N2, He 100, 300, 700K

  15. Heat (q) Transfer of thermal energy between the system and surroundings from high to low temps

  16. Work (w) A force exerted over a distance w = F•Dd Examples: syringe, hatchback

  17. Internal Energy surroundings • The energy of a system can be changed by a flow of heat (q), work (w) or both System DE = q + w Note: q and w must have a magnitude (#) and a sign (+ or -)

  18. Energy transfer of heat only (no work) EXOTHERMIC ENDOTHERMIC heat flows INTO the system heat EXITS the system the temperature _________ the temperature ________ surroundings system system

  19. ENDOTHERMIC +DE E R P R P Energy transfer of heat only EXOTHERMIC -DE E Energy GAINED by system (+q) Energy LOST by system (-q)

  20. P =1 atm cnst Dd When work is done by a gas F P = w = FDd DV =ADd surroundings A F = PA System wout = PADd wout= PDV Sign of w? Why? -w= PDV w= -PDV

  21. Sign Conventions for q & w— Take the point of view of the system! • +q: • - q: • +w: • -w: heat flows into the system heat flows out of the system work is done ON the system work is done BY the system

  22. Sample Problem: • Calculate DE for an exothermic reaction in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. DE = q + w DE = -15.6 kJ + 1.4 kJ = -14.2 kJ

  23. Units of Energy • calorie • Joule • 1 cal = 4.18 J

  24. State Function—UPPER CASE depends on the initial and final positions; not the path depends on the current state; not how it got there 6 pts either way Panthers Visitors Explain why Energy is a state function

  25. start finish Path Function—lower case depends on the pathbecause they have to do with a transition. Explain why work is a path function Explain why heat is a path function

  26. STATE path • Temperature • Pressure • Volume • Heat Capacity • Density • Energy • Displacement • Altitude • Heat • Work • Distance

  27. H2O2 Ep H2O + O2 Demo: State or Path Function? control add KI add MnO4

  28. Heat absorbed q J C = = DT DT oC How can we calculate heat (q)? • Heat Capacity (C): quantity of heat needed to change it’s temperature by 1 oC an extensive property— depends on amount of the substance

  29. Heat q J c = = m • DT g • oC m • DT How can we calculate heat (q)? 2) specific heat capacity (c) intensive property does not depend on amount can be used to identify a substance The amount of heat needed to raise the temperature of 1 g of substance 1oC cH2O(l) = 4.18 J/g •oC

  30. Heat transfer • In an isolated system the sum of the energy (heat) changes within the system must be zero • q1 + q2 + q3 + ……. = 0

  31. Example • Hot metal block in water: • q metal + qwater = 0 • (CmmmTfinal-Tintial)+(CwmwTfinal-Tinitial) = 0

  32. Pressure • volume = work Energy = q + w Enthalpy (H) Enthalpy a property defined as internal energy + product of pressure and volume H = E + PDV DH = DE + PDV

  33. Comparing DE and DH • Reactions that do not involve gases 2KOH (aq) + H2SO4(aq)  K2SO4 (aq) + 2H2O(l) DH = DE + PDV What is DV? With no gases, DV = 0 sooooooo DH = DE

  34. Comparing DE and DH • Reactions in which the moles of gas does NOT change N2 (g) + O2 (g)  2NO(g) DH = DE + PDV What is DV? same moles of gas, DV = 0 sooooooo DH = DE

  35. Comparing DE and DH—redo this; add derivation • Reactions in which the moles of gas DOES change 2H2 (g) + O2(g)  2H2O (g) DH = DE + PDV What is DV? waaaayyyy smaller than DH so that it’s insignificant and DH ≈ DE

  36. Enthalpy changes for chemical reactions • Given as: ΔrHo = standard reaction enthalpy; unit kJ/mol-rxn • H2(g) + 1/2 O2(g)  H2O(g) • ΔrHo = -241.8 kJ/mol-rxn

  37. Types of Reaction systems OPENCLOSEDINSULATED Enthalpy; Endo/Exo Problems Calorimetry HARD! Engineering/Thermo Classes

  38. Calorimetry: A calorimeter is a device used to experimentally determine the amount of heat associated with a chemical reaction (a device that measures calories)

  39. Consider a system at a constant pressure (an insulated system)—check/redo this derivation Recall: E = q + w and E = H -PDV E = q - PDV q - PDV = H - PDV q - PDV = H - PDV H = q Conclusion: Enthalpy = heat AT A CONSTANT PRESSURE DH = qp

  40. Bomb Calorimetry: used to determine heat content in food/fuels volume constant __________ E = q + w E = q - PDV E = qv = ΔU w = 0 since constant volume

  41. Calculation of energy transfer: • qr (ΔU) + qbomb + qwater = 0, where qr is energy released as heat by reaction.

  42. Hess’s Law: In going from a set of reactants to a set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps

  43. Hess’s Law Problems Use the thermochemical equations to determine the enthalpy for the reaction: C2H4O(l) + 5/2O2(g) 2CO2(g) + 2H2O(g) 2C2H4O(l) + 2H2O(l)2C2H6O(l) + O2(g) DH=610.5KJ 2CO2(g) + 3H2O(l)C2H6O(l) + 3O2(g) DH= 2056.5KJ

  44. NO2(g) + 2H2(g) 1/2N2(g) + 2H20(l) 2NH3(g)N2(g) + 3H2(g) DH= 23KJ 2H2O(l)+NH3(g)NO2(g) + 7/2H2(g) DH=54KJ

  45. Things to Remember 1) If you reverse a reaction, you must change the sign of DH 2) If you multiply the coefficients by an integer, do the same to DH

  46. Standard State • Symbol = o • gases are at 1 atm ex: NO(g) • solutions are at 1.0 M ex: NaCl(aq), Ba2+ • Pure liquids and solids are in their condensed states ex: H2O (l)

  47. Standard Heat of Reaction (DHorxn) The enthalpy taken in / given off in a reaction

  48. Standard Enthalpies of Formation (DHof) • WhenDH values cannot be obtained from constant pressure calorimetry experiments The change in enthalpy to form one mole of a compound from its elements with all substances at their standard states.

  49. Standard Enthalpies of Formation (DHof) • DHof for all elements at standard state = 0 (1 atm, the state they are at 25oC) • DHof of most compounds is negative—find on Thermochemical Ref Chart = 0 O2 (g) , Hg (l) , Fe (s)