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Chemistry 100(02) Fall 2012

Chemistry 100(02) Fall 2012. Instructor: Dr. Upali Siriwardane e-mail : upali@coes.latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M,W, 8:00-9:00 & 11:00-12:00 a.m Tu,Th,F 9:00 - 10:00 a.m.   Test Dates :. October 1 , 2012 (Test 1): Chapter 1 & 2

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Chemistry 100(02) Fall 2012

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  1. Chemistry 100(02) Fall 2012 Instructor: Dr. UpaliSiriwardane e-mail: upali@coes.latech.edu Office: CTH 311 Phone257-4941 Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m Tu,Th,F9:00 - 10:00 a.m.   Test Dates: October 1, 2012 (Test 1): Chapter 1 & 2 October 22, 2012 (Test 2): Chapter 3 & 4 November 14, 2012 (Test 3) Chapter 5 & 6 November 15, 2012 (Make-up test) comprehensive: Chapters 1-6 9:30-10:45:15 AM, CTH 328

  2. REQUIRED: Textbook:Principles of Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro - Pearson Prentice Hall and also purchase the Mastering Chemistry Group Homework, Slides and Exam review guides and sample exam questions are available online: http://moodle.latech.edu/ and follow the course information links. OPTIONAL: Study Guide: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro 2nd Edition Student Solutions Manual: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro2nd Text Book & Resources

  3. Chapter 5. Gases 6.1 Chemical Hand Warmers…………………………………………………………………..231 6.2 The Nature of Energy: Key Definitions…………………………………………………...232 6.3 The First Law of Thermodynamics: There Is No Free Lunch…………………………….234 6.4 Quantifying Heat and Work……………………………………………………………….240 6.5 Measuring for Chemical Reactions: Constant-Volume Calorimetry……………………...246 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure ……………..249 6.7 Constant-Pressure Calorimetry: Measuring……………………………………………….253 6.8 Relationships Involving……………………………………………………………………255 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation……………257 6.1 0 Energy Use and the Environment……………………………………………………….263

  4. Chapter 6. KEY CONCEPTS: Thermochemistry

  5. Hess’s law A + B A + B F energy E + D + B C C The value of H for the reaction is the same whether it occurs directly or in a series of steps. DHoverall = DH1 + DH2 + DH3 + · · ·

  6. What is Hess's Law of Summation of Heat? To get DH for new reactions. Two methods? 1st method (indirect): new DH is calculated by adding DHs of several other reactions. 2nd method (direct): Where DHf ( DH of formation) of reactants and products are used to calculate DH of a reaction.

  7. EXAMPLECH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) CH4(g) C(s) + 2 H2(g)DH1 2 O2(g) 2 O2(g)DH2 C(s) + O2(g) CO2(g)DH3 2 H2(g) + O2(g) 2 H2O(l)DH4 CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DHoverall = DH1 + DH2 + DH3 + DH4

  8. Hess’s law • The thermal energy given off or absorbed in a given change is the same whether it takes place in a single step or several steps. • This is just another way of stating the law of conservation of energy. • If the net change in energy were to differ based on the steps taken, then it would be possible to create energy -- this cannot happen!

  9. Calculating enthalpies: 1st method • Thermochemical equations can be combined to calculate DHrxn. • Example.2C(graphite) + O2 (g) 2CO (g) • This cannot be directly determined because CO2 is always formed. • However, we can measure the following: • C(graphite) + O2 (g) CO2(g) DHrxn= -393.51 kJ • 2CO (g) + O2 (g) 2CO2 (g)DHrxn= -565.98 kJ

  10. Calculating enthalpies By combining the two equations, we can determine the DHrxn we want. 2 [ C(graphite) + O2 (g) CO2(g) ] DHrxn= -787.02 kJ 2CO2 (g) 2CO (g) + O2 (g)DHrxn= +565.98 kJ Note. Because we need 2 moles of CO2 to be produced in the top reaction, the equation and its DHrxn were doubled.

  11. Calculating enthalpies Now all we need to do is to add the two equations together. 2C(graphite) + 2O2 (g) 2CO2(g)DHrxn= -787.02 kJ 2CO2 (g) 2 CO (g) + O2 (g)DHrxn= +565.98 kJ 2 C(graphite) + O2 (g) 2 CO (g)DHrxn= -221.04 kJ Note. The 2CO2 cancel out, as does one of the O2 on the right-hand side.

  12. Method 1: Calculate DH for the reaction: so2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) DH = ? Other reactions: SO2(g) ------> S(s) + O2(g) DH = 297kJ H2SO4(l)------> H2(g) + S(s) + 2O2(g) DH = 814 kJ H2(g) +1/2O2(g) -----> H2O(g) DH = -242 kJ

  13. SO2(g) ------> S(s) + O2(g);DH1 = 297 kJ - 1 H2(g) + S(s) + 2O2(g) ------> H2SO4(l) DH2 = -814 kJ - 2 H2O(g) ----->H2(g) + 1/2 O2(g) DH3= +242 kJ - 3 ______________________________________ SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) DH = DH1 + DH2 + DH3 DH = +297 - 814 + 242 DH = -275 kJ

  14. Hess’s Law: Indirect Method Problem Problem: Determine the DHrxn for the following reaction: C(s) + ½ O2(g)  CO(g) Useful information: Eq. #1 C(s) + O2(g)  CO2(g) DH = - 394 kJ Eq. #2 CO(g) + ½ O2(g)  CO2(g) DH = - 283 kJ

  15. Hess’s Law: Indirect Method Problem Problem: Determine the DHrxn for the following reaction: C(s) + ½ O2(g)  CO(g) Solution Strategy: 1. Arrange Equations #1 and #2 so, when summed, they equal C(s) + ½ O2(g)  CO(g) 2. If Equations #1 and/or #2 are reversed,then the sign of DHis changed. 3. If Equations #1 and/or #2 are multiplied or divided to obtain correct stoichiometric quantities, then DH is also multiplied or divided by the same factor.

  16. Problem: Determine the DHrxn for the following reaction: C(s) + ½ O2(g)  CO(g) Useful information: Eq. #1 C(s) + O2(g)  CO2(g) DH = - 394 kJ Eq. #2 CO(g) + 1/2 O2(g)  CO2(g) DH = - 283 kJ Solution: C(s) + O2(g)  CO2(g) DH = - 394 kJ Reverse CO2(g)  CO(g) + 1/2 O2(g) DH = + 283 kJ C(s) + ½ O2(g)  CO(g) DH = - 111 kJ

  17. Problem: Determine the DHrxn for the following reaction: 2 B(s) + 3 H2(g)  B2H6(g) Useful Information: DH (kJ) 2 B(s) + 3/2 O2(g)  B2O3(g) –1273 B2H6(g) + 3 O2(g)  B2O3(g) + 3 H2O(g) - 2035 H2(g) + ½ O2(g)  H2O(l) - 286.0 H2O(l)  H2O(g) + 44.0

  18. Problem: Determine the DHrxn for the following reaction: 2 B(s) + 3 H2(g)  B2H6(g) Answer: DH (kJ) 2 B(s) + 3/2 O2(g)  B2O3(g) –1273 B2O3(g) + 3 H2O(g)  B2H6(g) + 3 O2(g) + 2035 (reversed) 3 (H2(g) + ½ O2(g)  H2O(l)) 3(- 286.0) 3(H2O(l)  H2O(g)) 3 (+ 44.0) 2 B(s) + 3 H2(g)  B2H6(g) +36.0 kJ

  19. For the formation reaction, H2(g) + 1/2O2(g) ---> H2O(g)Hf= -286 kJ What is the H for the reverse reaction? H2O(g) ---> H2(g) + 1/2O2(g)Hrex = ? b) What is the H for the reaction 2H2O(g) ---> 2H2(g) + O2(g)Hrex = ?

  20. 2) 2NOCl(g) = 2NO(g) +Cl2 (g) ;         DHrxn10 = 75.56 kJ/mol2NO(g) + O2(g) = 2NO2 (g)    ;        DHrxn20 = -113.05 kJ/mol2NO2(g) = N2O4 (g)   ;                     DHrxn30 = -58.03 kJ/mol 2NO(g) +Cl2 (g)  = 2NOCl(g)DHrxn1’0 = N2O4 (g) =  2NO2(g)DHrxn2’0 = 2NO2 (g) =    2NO(g) + O2(g)DHrxn3’0 = Sum of the a), b) and c): DHrxn0 = What is the DHrxn0 for N2O4 (g) + Cl2 (g) =   2NOCl(g) +  O2(g) ;  DHrxn0 = ???

  21. 3) The DHf0 standard enthalpies of formation of SO2 and SO3 are -297 and -396 respectively. What is DHf0of O2(g) ? SO3(g) -> SO2(g) + 1/2O2(g) Write the expression: DHrex= [Sn(DH°f) Products] - [Sn (DH°f) reactants]: SO3(g) -> SO2(g) + 1/2 O2(g); DHrex= ?

  22. 4) The DHf0 standard enthalpies of formation of H2O, H2S, and SO2 are -285.8, -20.6 and -296.8 kJ/mol respectively. What is DHf0of S(s)? 2H2S (g)+ SO2(g) -> 3 S(s) + 2 H2O(l) Write the expression: DHrex= [Sn(DH°f) Products] - [Sn (DH°f) reactants]: 2H2S (g)+ SO2(g) -> 3 S(s) + 2 H2O(l); DHrex= ?

  23. Calculating enthalpies 2nd method • The real problem with using Hess’s law is figuring out what equations to combine. • The most often used equations are those for formation reactions. • Formation reactions • Reactions in which compounds are formed from elements. 2 H2 (g) + O2 (g) 2 H2O (l)DHrxn = -571.66 kJ

  24. Calculation of DHo DHo = Sc DHfoproducts – Sc DHforeactants

  25. ExampleWhat is the value of DHrx for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g) from Appendix J Text C6H6(l)DHfo = + 49.0 kJ/mol O2(g) DHfo = 0 CO2(g) DHfo = - 393.5 H2O(g)DHfo = - 241.8 DHrx = [S c D Hfo]product–[S c  D Hfo]reactants

  26. ExampleWhat is the value of DHrx for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)from Appendix J TextC6H6(l)DHfo = + 49.0 kJ/mol; O2(g) DHfo = 0CO2(g) DHfo = - 393.5; H2O(g)DHfo = - 241.8DHrx = [S c D Hfo]product - [S c  D Hfo]reactants D Hrx = [12(- 393.5) + 6(- 241.8)]product - [2(+ 49.0 ) + 15(0)]reactants kJ/mol = - 6.2708  103 kJ

  27. Standard enthalpy of formation DHfo Enthalpy change that results from one mole of a substance being formed from its elements. All elements are at their standard states. The DHfo of an element in its standard state has a value of zero.

  28. Using Standard Enthalpy Values to Determine ΔHo for a Chemical Reaction Problem: Using Hess’s law (direct method) calculate the ∆H˚(enthalpy change) forthe following reaction: H2O(g) + C(graphite)  H2(g) + CO(g)

  29. Answer: • Have a balanced reaction: H2O(g) + C(graphite)(s) H2(g) + CO(g) 2. From the thermodynamic tables, you find that ∆H˚ of H2O vapor = - 242 kJ/mol H2(g) + 1/2 O2(g) H2O(g) ∆H˚ of CO(g) = - 111 kJ/mol C(s) + 1/2 O2(g) CO(g)

  30. Answer continued: DHrxno = ? for H2O(g) + C(graphite)  H2(g) + CO(g) 3. Use Hess’s law (direct): ∆Hrxno =  ∆Hof (products) -  ∆Hof(reactants) DHrxno= {[0 + (-111 kJ/mol x 1 mol)] - [(-242 kJ/mol x 1 mol) + 0]} Answer: DHrxno = + 131 kJ • This is an endothermic process, for DHo is positive. • The surroundings temperature DECREASED, for the overall energy of the system INCREASED.

  31. PROBLEM: Calculate the heat of combustion of methanol, ∆Ho, for CH3OH(g) + 3/2 O2(g)  CO2(g) + 2 H2O(g) ∆Hrxno =  ∆Hof (products) -  ∆Hof (reactants) ∆Ho = {[∆Ho(CO2) + 2 ∆Ho(H2O)] – [(3/2 ∆Ho(O2) + ∆Ho(CH3OH)]} = {[(-393.5 kJ/mol x 1 mol) + 2 mol x (-241.8 kJ/mol)] – [(0 + 1 mol x (-201.5 kJ/mol)]} ∆Hrxno = - 675.6 kJ (exothermic)

  32. Hess’s Law: Indirect Method Problem Problem: Determine the DHrxn for the following reaction: C(s) + ½ O2(g)  CO(g) Useful information: Eq. #1 C(s) + O2(g)  CO2(g) DH = - 394 kJ Eq. #2 CO(g) + ½ O2(g)  CO2(g) DH = - 283 kJ

  33. Hess’s Law: Indirect Method Problem Problem: Determine the DHrxn for the following reaction: C(s) + ½ O2(g)  CO(g) Solution Strategy: 1. Arrange Equations #1 and #2 so, when summed, they equal C(s) + ½ O2(g)  CO(g) 2. If Equations #1 and/or #2 are reversed,then the sign of DHis changed. 3. If Equations #1 and/or #2 are multiplied or divided to obtain correct stoichiometric quantities, then DH is also multiplied or divided by the same factor.

  34. Problem: Determine the DHrxn for the following reaction: C(s) + ½ O2(g)  CO(g) Useful information: Eq. #1 C(s) + O2(g)  CO2(g) DH = - 394 kJ Eq. #2 CO(g) + 1/2 O2(g)  CO2(g) DH = - 283 kJ Solution: C(s) + O2(g)  CO2(g) DH = - 394 kJ Reverse CO2(g)  CO(g) + 1/2 O2(g) DH = + 283 kJ C(s) + ½ O2(g)  CO(g) DH = - 111 kJ

  35. Problem: Determine the DHrxn for the following reaction: 2 B(s) + 3 H2(g)  B2H6(g) Useful Information: DH (kJ) 2 B(s) + 3/2 O2(g)  B2O3(g) –1273 B2H6(g) + 3 O2(g)  B2O3(g) + 3 H2O(g) - 2035 H2(g) + ½ O2(g)  H2O(l) - 286.0 H2O(l)  H2O(g) + 44.0

  36. Problem: Determine the DHrxn for the following reaction: 2 B(s) + 3 H2(g)  B2H6(g) Answer: DH (kJ) 2 B(s) + 3/2 O2(g)  B2O3(g) –1273 B2O3(g) + 3 H2O(g)  B2H6(g) + 3 O2(g) + 2035 (reversed) 3 (H2(g) + ½ O2(g)  H2O(l)) 3(- 286.0) 3(H2O(l)  H2O(g)) 3 (+ 44.0) 2 B(s) + 3 H2(g)  B2H6(g) +36.0 kJ

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