# 27 Oct. 2010 - PowerPoint PPT Presentation

27 Oct. 2010

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27 Oct. 2010

## 27 Oct. 2010

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1. 27 Oct. 2010 • Take out Homework • Objective: SWBAT calculate work done on a system and energy change for a heat transfer. • Do now: Describe how you would make 1.00 L of a 0.8 M LiCl solution from solid LiCl.

2. Agenda • Do now • Homework answers • Thermochemistry notes and examples Part 1 • Homework: p. 263 #13, 15, 17, 19 (ELS) • Read p. 239-245

3. Thermochemistry

4. Thermodynamics: study of interconversion of heat and other kinds of energy • First Law of Thermodynamics: energy can be converted from one form to another, but can not be created or destroyed • We can not accurately measure total energy of a system • Instead, we measure changes in energy ΔE

5. ΔEsys + ΔEsurr = 0 • ΔEsys=- ΔEsurr • energy gained in one place must be lost somewhere else • In chem, we are usually interested in just the ΔE to the system we are studying: • ΔE=q+w • q=heat exchange between systems • w=work done on or by the system

6. Sign Conventions: heat and work

7. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process (-q)is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process (+q)is any process in which heat has to be supplied to the system from the surroundings.

8. Work, Pressure, and Volume Compression Expansion +V (increase) -V (decrease) -w results +w results Esystemdecreases Esystemincreases Work has been done on the system by the surroundings Work has been done by the system on the surroundings

9. Ex 1 A gas expands in volume from 2.0 L to 6.0 L at a constant temperature. Calculate the work done by the gas if it expands a) against a vacuum b) against a constant pressure of 1.2 atm 1 L·atm = 101.3 J Joule = unit of heat

10. Ex 2 • A gas expands from 264 mL to 971 mL at constant temperature. Calculate the work done (in J) by the gas if it expands • against a vacuum • against a constant pressure of 4.00 atm

11. Ex 3 The work done when gas is compressed in a cylinder is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Determine the signs for q and w. Calculate the energy change for this process. ΔE=q+w

12. Homework • p. 263 #13, 15, 17, 19 (ELS) • Read p. 239-245

13. 28 Oct. 2010 Objective: SWBAT calculate heat change to a system given a thermochemical equation. Do now: The work done when gas is compressed in a cylinder is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Determine the signs for q and w. Calculate the energy change for this process. ΔE=q+w

14. Agenda • Do now • Homework check • Thermochemistry part 2 Homework: p. 263 #24, 25, 27 (JMS) Read p. 245-252, read lab handout and do pre-lab questions

15. A gas expands and does P-V work on the surroundings equal to 279 J. At the same time, it absorbs 216 J of heat from the surroundings. What is the change in energy of the system?

16. Enthalpy of Chemical Reactions

17. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = DE + PDV DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.4

18. H2O (s) H2O (l) DH = 6.01 kJ Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

19. DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

20. 2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) DH = -6.01 kJ DH = 6.01 kJ/mol ΔH = 6.01 kJ DH = 2 mol x 6.01 kJ/mol= 12.0 kJ Thermochemical Equations • The stoichiometric coefficients always refer to the number of moles of a substance • If you reverse a reaction, the sign of DH changes • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. The physical states of all reactants and products must be specified in thermochemical equations.

21. Example 1 • Given the thermochemical equation 2SO2(g) + O2(g)  2SO3(g) DH = -198.2kJ/mol calculate the heat evolved when 87.9 g of SO2 is converted to SO3.

22. Example 2 • Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air: P4(s) + 5O2(g)  P4O10(s) DH= -3013kJ/mol

23. Homework • p. 263 #24, 25, 26 • Read LAB, DO prelab, read in book….

24. 1 Nov. 2010 • Take out pre-lab questions • Objective: SWBAT calculate the amount of heat supplied to the surroundings (q) by a reusable chemical hand warmer. • Do now: For the chemical hand warmer, predict the sign convention for heat (q) and work (w).

25. Lab • With your partner, carefully follow the directions. Hand warmer pouch and metal disk: 3.42 grams • Collect all data, results and answers to questions in your lab notebook. • Lab notebooks collected tomorrow: 2 lab grades.

26. 2 Nov. 2010 • Objective: SWBAT relate ΔE to ΔH and complete calorimetric calculations. • Do now: Type your second trial results into the spreadsheet. Homework: Lab notebooks due tomorrow – 2 lab grades p. 263 #27, 28, 31, 32, 34, 36 (JR)

27. Comparing ΔH and ΔE • ΔE = ΔH – RTΔn • R = 8.314J/K·mol • T = temperature in Kelvin • n = number of moles • For gas reactions, the change in volume can be very large because hot gases expand. • For reactions not involving gas, ΔH and ΔE are often very similar.

28. Ex 1 • Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC 2CO(g) + O2(g)  2CO2(g) ΔH = -566.0 kJ/mol

29. Ex 2 • What is ΔE for the formation of 1 mole of CO at 1 atm and 25oC? C(graphite) + ½O2(g)  CO(g) ΔH = -110.5 kJ/mol

30. Calorimetry • the measurement of heat changes • q=CΔt • C = heat capacity of a substance: the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree Celsius. (=m*s) • q=msΔt Ex. 1: A 466 g sample of water is heated from 8.50oC to 74.60oC. Calculate the amount of heat absorbed (in kJ) by the water.

31. Ex. 2 • 1.435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter. The temperature of the water rose from 20.28oC to 25.95oC. If the heat capacity of the bomb plus water was 10.17 kJ/oC, calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion).

32. Ex. 3 • A lead pellet having a mass of 26.47 g at 89.98oC was placed in a constant pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50oC to 23.17oC. What is the specific heat of the lead pellet?

33. Homework Lab notebooks due tomorrow – 2 lab grades p. 263 #27, 28, 31, 32, 34, 36 (JR)

34. 3 Nov. 2010 • Take out Homework and Lab notebook • Objective: SWBAT calculate the standard enthalpy of formation and reaction for compounds. • Do now: An iron bar of mass 869 g cools from 94oC to 5oC. Calculate the heat released (in kJ) by the metal. specific heat of iron=0.444 J/g·oC

35. Agenda • Do now • Homework check • Standard Enthalpy of Formation/Reaction notes/problems Homework: p. 264 #42, 43, 46, 51, 53, 58, 63 (SR) Read p. 258-261: Heats of solution/dilution Tomorrow: Practice problems/review Bring your book to class! Test Monday (ch. 6)

36. Announcement • Tutor needed for 10th grade chem student. • You will earn community service hours. • Interested? See Ms. Kern.

37. Standard Enthalpy of Formation and Reaction

38. DH0 (O2) = 0 DH0 (O3) = 142 kJ/mol DH0 (C, graphite) = 0 DH0 (C, diamond) = 1.90 kJ/mol f f f f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.

39. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = DH0 DH0 rxn rxn dDH0 (D) aDH0 (A) bDH0 (B) cDH0 (C) DH0 (products) f f f f f - DH0 (reactants) S S = f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

40. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

41. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3267 kJ/mol C6H6 6DH0 (H2O) -6535 kJ f 2 mol DH0 (reactants) DH0 (products) f f Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

42. Practice Problem • Calculate the heat of combustion for the following reaction: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) The standard enthalpy of formation of ethylene (C2H4) is 52.3 kJ/mol. Then, calculate per mole of ethylene combusted.

43. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn What if the reaction is not a single step? Calculate the standard enthalpy of formation of CS2 (l) given that:

44. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn.

45. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn.

46. Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements: 2C(graphite) + H2(g)  C2H2(g) Here are the equations for each step: C(graphite) + O2(g)  CO2(g) DH0=-393.5 kJ/mol H2(g) + 1/2O2(g)  H2O(l) DH0=-285.8 kJ/mol 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l) DH0= -2598.8 kJ/mol

47. Homework • p. 264 #42, 43, 46, 51, 53, 58, 63 • Bring your book to class!

48. 4 Nov. 2010 • Take out homework: p. 264 #42, 43, 46, 51, 53, 58, 63 • Objective: SWBAT review thermochemistry for a test on Monday! • Do now: Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements: 2C(graphite) + H2(g)  C2H2(g) Here are the equations for each step: C(graphite) + O2(g)  CO2(g) DH0=-393.5 kJ/mol H2(g) + 1/2O2(g)  H2O(l) DH0=-285.8 kJ/mol 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l) DH0= -2598.8 kJ/mol

49. Agenda • Do now • Homework • Chapter 6 Problem Set Work Time Homework: Finish problem set: Mon. Test: Mon. Bonus: Tues.