2. CHEM 100 Fall 2012. Page- 1 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: Chemistry 100(02) Fall 2012 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

3. 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.http://moodle.latech.edu/ 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. Tro 2nd Text Book & Resources

4. CHEM 100 Fall 2012. Page- 3 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 Chapter 5. Gases

5. CHEM 100 Fall 2012. Page- 4 Chapter 6. KEY CONCEPTS: Thermochemistry Kinetic energy and potential energy Energy units and unit conversion Conservation of energy: heat and work Thermodynamic terms: system, surroundings, thermal equilibrium, exothermic, endothermic, and state function Heat capacity and calorimetry Internal energy and enthalpy Thermochemical equations Thermostoichiometric factors Standard enthalpy change for a reaction,  H o Enthalpy change from bond enthalpies Calorimetry and thermal energy transferred during a reaction Hess's law and enthalpy Standard molar enthalpies of formation  H o of a reaction. Chemical fuels and heating Main components of food Caloric intake

6. CHEM 100 Fall 2012. Page- 5 Hess’s law energy A + B C C D + B F E + The value of H for the reaction is the same whether it occurs directly or in a series of steps.  Hoverall =  H1 +  H2 +  H3 + · · ·

7. CHEM 100 Fall 2012. Page- 6 What is Hess's Law of Summation of Heat? To get  H for new reactions. Two methods? 1st method (indirect): new  H is calculated by adding  Hs of several other reactions. 2nd method (direct): Where Hf Hf (  H of formation) of reactants and products are used to calculate  H of a reaction.

8. CHEM 100 Fall 2012. Page- 7 EXAMPLE C H 4(g ) + 2 O 2(g)  CO 2(g) + 2 H 2 O (l) CH 4(g)  C (s) + 2 H 2(g)  H 1 2 O 2(g)  2 O 2(g)  H 2 C (s) + O 2(g)  CO 2(g)  H 3 2 H 2(g) + O 2(g)  2 H 2 O (l)  H 4 CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (l)  H overall =  H 1 +  H 2 +  H 3 +  H 4

9. CHEM 100 Fall 2012. Page- 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!

10. CHEM 100 Fall 2012. Page- 9 Calculating enthalpies: 1st method Thermochemical equations can be combined to calculate  H rxn. Example.Example. 2C (graphite) + O 2 (g) 2CO (g) This cannot be directly determined because CO 2 is always formed. However, we can measure the following: C (graphite) + O 2 (g) CO 2 (g)  H rxn = -393.51 kJ 2CO (g) + O 2 (g) 2CO 2 (g)  H rxn = -565.98 kJ

11. CHEM 100 Fall 2012. Page- 10 Calculating enthalpies By combining the two equations, we can determine the  H rxn we want. 2 [ C (graphite) + O 2 (g) CO 2 (g) ]  H rxn = -787.02 kJ 2CO 2 (g) 2CO (g) + O 2 (g)  H rxn = +565.98 kJ Note. Because we need 2 moles of CO 2 to be produced in the top reaction, the equation and its  H rxn were doubled.

12. CHEM 100 Fall 2012. Page- 11 Calculating enthalpies Now all we need to do is to add the two equations together. 2C (graphite) + 2O 2 (g) 2CO 2 (g)  H rxn = -787.02 kJ 2CO 2 (g) 2 CO (g) + O 2 (g)  H rxn = +565.98 kJ 2C (graphite) + O 2 (g) 2 CO (g)  H rxn = -221.04 kJ Note. The 2CO 2 cancel out, as does one of the O2 O2 on the right-hand side.

13. CHEM 100 Fall 2012. Page- 12 Method 1: Calculate  H for the reaction: so 2 (g) + 1/2 O 2 (g) + H 2 O(g) -----> H 2 SO 4 (l)  H = ? Other reactions: SO 2 (g) ------> S(s) + O 2 (g)  H = 297kJ H 2 SO 4 (l)------> H 2 (g) + S(s) + 2O 2 (g)  H = 814 kJ H 2 (g) +1/2O 2 (g) -----> H 2 O(g)  H = -242 kJ

14. CHEM 100 Fall 2012. Page- 13 SO 2 (g) ------> S(s) + O 2 (g);  H 1 = 297 kJ - 1 H 2 (g) + S(s) + 2O 2 (g) ------> H 2 SO 4 (l) H2 H2 = -814 kJ - 2 H 2 O(g) ----->H 2 (g) + 1/2 O 2 (g) H3 H3 = +242 kJ - 3 ______________________________________ SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) -----> H 2 SO 4 (l)  H =  H 1 +  H 2 + H3H3  H = +297 - 814 + 242  H = -275 kJ

15. CHEM 100 Fall 2012. Page- 14 Problem: Determine the  H rxn for the following reaction: C(s) + ½ O 2 (g)  CO(g) Useful information: Eq. #1C(s) + O 2 (g)  CO 2 (g)  H = - 394 kJ Eq. #2CO(g) + ½ O 2 (g)  CO 2 (g)  H = - 283 kJ Hess’s Law: Indirect Method Problem

16. CHEM 100 Fall 2012. Page- 15 Problem: Determine the  H rxn for the following reaction: C(s) + ½ O 2 (g)  CO(g) Solution Strategy: 1. Arrange Equations #1 and #2 so, when summed, they equal C(s) + ½ O 2 (g)  CO(g) 2. If Equations #1 and/or #2 are reversed, then the sign of  H is changed. 3. If Equations #1 and/or #2 are multiplied or divided to obtain correct stoichiometric quantities, then  H is also multiplied or divided by the same factor. Hess’s Law: Indirect Method Problem

17. CHEM 100 Fall 2012. Page- 16 Problem: Determine the  H rxn for the following reaction: C(s) + ½ O 2 (g)  CO(g) Useful information: Eq. #1C(s) + O 2 (g)  CO 2 (g)  H = - 394 kJ Eq. #2CO(g) + 1/2 O 2 (g)  CO 2 (g)  H = - 283 kJ Solution: C(s) + O 2 (g)  CO 2 (g)  H = - 394 kJ ReverseCO 2 (g)  CO(g) + 1/2 O 2 (g)  H = + 283 kJ C(s) + ½ O 2 (g)  CO(g) C(s) + ½ O 2 (g)  CO(g)  H = - 111 kJ

18. CHEM 100 Fall 2012. Page- 17 Problem: Determine the  H rxn for the following reaction: 2 B(s) + 3 H 2 (g)  B 2 H 6 (g) Useful Information:  H (kJ) 2 B(s) + 3/2 O 2 (g)  B 2 O 3 (g)–1273 B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (g) + 3 H 2 O(g) - 2035 H 2 (g) + ½ O 2 (g)  H 2 O(l)- 286.0 H 2 O(l)  H 2 O(g) + 44.0

19. CHEM 100 Fall 2012. Page- 18 Problem: Determine the  H rxn for the following reaction: 2 B(s) + 3 H 2 (g)  B 2 H 6 (g) Answer:  H (kJ) 2 B(s) + 3/2 O 2 (g)  B 2 O 3 (g)–1273 B 2 O 3 (g) + 3 H 2 O(g)  B 2 H 6 (g) + 3 O 2 (g) + 2035 (reversed) 3 (H 2 (g) + ½ O 2 (g)  H 2 O(l)) 3 (- 286.0) 3 (H 2 O(l)  H 2 O(g)) 3 (+ 44.0) 2 B(s) + 3 H 2 (g)  B 2 H 6 (g) +36.0 kJ

20. CHEM 100 Fall 2012. Page- 19 1)For the formation reaction, H 2(g) + 1/2O 2(g) ---> H 2 O (g) H f = -286 kJ What is the H for the reverse reaction? H 2 O (g) ---> H 2(g) + 1/2O 2(g) H rex = ? b) What is the H for the reaction 2H 2 O (g) ---> 2H 2(g) + O 2(g) H rex = ?

21. CHEM 100 Fall 2012. Page- 20 2) 2NOCl (g) = 2NO (g) +Cl 2 (g) ;  H rxn1 0 = 75.56 kJ/mol 2NO (g) + O 2(g) = 2NO 2 (g) ;  H rxn2 0 = -113.05 kJ/mol 2NO 2(g) = N 2 O 4 (g) ;  H rxn3 0 = -58.03 kJ/mol 2NO (g) +Cl 2 (g) = 2NOCl (g)  H rxn1’ 0 = N 2 O 4 (g) = 2NO 2(g)  H rxn2’ 0 = 2NO 2 (g) = 2NO (g) + O 2(g)  H rxn3’ 0 = Sum of the a), b) and c):  H rxn 0 = What is the  H rxn 0 for N 2 O 4 (g) + Cl 2 (g) = 2NOCl (g) + O 2(g) ;  H rxn 0 = ???

22. CHEM 100 Fall 2012. Page- 21 3) The  H f 0 standard enthalpies of formation of SO 2 and SO 3 are -297 and -396 respectively. What is  H f 0 of O 2(g) ? SO 3 (g) -> SO 2 (g) + 1 / 2 O 2 (g) Write the expression:  H rex = [  n (  H° f ) Products] - [  n (  H° f ) reactants]: SO 3 (g) -> SO 2 (g) + 1 / 2 O 2 (g);  H rex = ?

23. CHEM 100 Fall 2012. Page- 22 4) The  H f 0 standard enthalpies of formation of H 2 O, H 2 S, and SO 2 are -285.8, -20.6 and -296.8 kJ/mol respectively. What is  H f 0 of S(s)? 2H 2 S (g)+ SO 2 (g) -> 3 S(s) + 2 H 2 O(l) Write the expression:  H rex = [  n (  H° f ) Products] - [  n (  H° f ) reactants]: 2H 2 S (g)+ SO 2 (g) -> 3 S(s) + 2 H 2 O(l);  H rex = ?

24. CHEM 100 Fall 2012. Page- 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. FormationFormation reactions Reactions in which compounds are formed from elements. 2 H2 H2 (g) + O2 O2 2 H2O H2O (l)  H rxn = -571.66 kJ

25. CHEM 100 Fall 2012. Page- 24 Calculation of  H o  H o =  c   H f o products –  c   H f o reactants

26. CHEM 100 Fall 2012. Page- 25 Example What is the value of  H rx for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) from Appendix J Text C 6 H 6(l)  H f o = + 49.0 kJ/mol O 2(g)  H f o = 0 CO 2(g)  H f o = - 393.5 H 2 O (g)  H f o = - 241.8  H rx  c   H f o  product –  c  H f o  reactants

27. CHEM 100 Fall 2012. Page- 26 Example What is the value of  H rx for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) from Appendix J Text C 6 H 6(l)  H f o = + 49.0 kJ/mol; O 2(g)  H f o = 0 CO 2(g)  H f o = - 393.5; H 2 O (g)  H f o = - 241.8  H rx  c  H f o  product -  c  H f o  reactants  H rx  - 393.5) + 6(- 241.8)  product -  2(+ 49.0 ) + 15(0)  reactants kJ/mol = - 6.2708  10 3 kJ

28. CHEM 100 Fall 2012. Page- 27 Standard enthalpy of formation HfoHfoHfoHfo Enthalpy change that results from one mole of a substance being formed from its elements. All elements are at their standard states. The Hfo Hfo of an element in its standard state has a value of zero.

29. CHEM 100 Fall 2012. Page- 28 Problem: Using Hess’s law (direct method) calculate the ∆H˚(enthalpy change) forthe following reaction: Problem: Using Hess’s law (direct method) calculate the ∆H˚(enthalpy change) for the following reaction: H 2 O(g) + C(graphite)  H 2 (g) + CO(g) Using Standard Enthalpy Values to Determine ΔH o for a Chemical Reaction

30. CHEM 100 Fall 2012. Page- 29 Answer: 1.Have a balanced reaction: H 2 O(g) + C(graphite)(s)  H 2 (g) + CO(g) 2. From the thermodynamic tables, you find that ∆H˚ of H 2 O vapor = - 242 kJ/mol H 2 (g) + 1/2 O 2 (g)  H 2 O(g) ∆H˚ of CO(g) = - 111 kJ/mol C(s) + 1/2 O 2 (g)  CO(g)

31. CHEM 100 Fall 2012. Page- 30 Answer continued:  H rxn o = ? for H 2 O(g) + C(graphite)  H 2 (g) + CO(g) 3. Use Hess’s law (direct): ∆H rxn o =  ∆H o f (products) -  ∆H o f(reactants)  H rxn o = {[0 + (-111 kJ/mol x 1 mol)] - [(-242 kJ/mol x 1 mol) + 0]} Answer:  H rxn o = + 131 kJ –This is an endothermic process, for  H o is positive. –The surroundings temperature DECREASED, for the overall energy of the system INCREASED.

32. CHEM 100 Fall 2012. Page- 31 PROBLEM: Calculate the heat of combustion of methanol, ∆H o, for CH 3 OH(g) + 3/2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) ∆H rxn o =  ∆H o f (products) -  ∆H o f (reactants) ∆H o = {[∆H o (CO 2 ) + 2 ∆H o (H 2 O)] – [(3/2 ∆H o (O 2 ) + ∆H o (CH 3 OH)]} = {[(-393.5 kJ/mol x 1 mol) + 2 mol x (-241.8 kJ/mol)] – [(0 + 1 mol x (-201.5 kJ/mol)]} [(0 + 1 mol x (-201.5 kJ/mol)]} ∆H rxn o = - 675.6 kJ (exothermic)

33. CHEM 100 Fall 2012. Page- 32 Problem: Determine the  H rxn for the following reaction: C(s) + ½ O 2 (g)  CO(g) Useful information: Eq. #1C(s) + O 2 (g)  CO 2 (g)  H = - 394 kJ Eq. #2CO(g) + ½ O 2 (g)  CO 2 (g)  H = - 283 kJ Hess’s Law: Indirect Method Problem

34. CHEM 100 Fall 2012. Page- 33 Problem: Determine the  H rxn for the following reaction: C(s) + ½ O 2 (g)  CO(g) Solution Strategy: 1. Arrange Equations #1 and #2 so, when summed, they equal C(s) + ½ O 2 (g)  CO(g) 2. If Equations #1 and/or #2 are reversed, then the sign of  H is changed. 3. If Equations #1 and/or #2 are multiplied or divided to obtain correct stoichiometric quantities, then  H is also multiplied or divided by the same factor. Hess’s Law: Indirect Method Problem

35. CHEM 100 Fall 2012. Page- 34 Problem: Determine the  H rxn for the following reaction: C(s) + ½ O 2 (g)  CO(g) Useful information: Eq. #1C(s) + O 2 (g)  CO 2 (g)  H = - 394 kJ Eq. #2CO(g) + 1/2 O 2 (g)  CO 2 (g)  H = - 283 kJ Solution: C(s) + O 2 (g)  CO 2 (g)  H = - 394 kJ ReverseCO 2 (g)  CO(g) + 1/2 O 2 (g)  H = + 283 kJ C(s) + ½ O 2 (g)  CO(g) C(s) + ½ O 2 (g)  CO(g)  H = - 111 kJ

36. CHEM 100 Fall 2012. Page- 35 Problem: Determine the  H rxn for the following reaction: 2 B(s) + 3 H 2 (g)  B 2 H 6 (g) Useful Information:  H (kJ) 2 B(s) + 3/2 O 2 (g)  B 2 O 3 (g)–1273 B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (g) + 3 H 2 O(g) - 2035 H 2 (g) + ½ O 2 (g)  H 2 O(l)- 286.0 H 2 O(l)  H 2 O(g) + 44.0

37. CHEM 100 Fall 2012. Page- 36 Problem: Determine the  H rxn for the following reaction: 2 B(s) + 3 H 2 (g)  B 2 H 6 (g) Answer:  H (kJ) 2 B(s) + 3/2 O 2 (g)  B 2 O 3 (g)–1273 B 2 O 3 (g) + 3 H 2 O(g)  B 2 H 6 (g) + 3 O 2 (g) + 2035 (reversed) 3 (H 2 (g) + ½ O 2 (g)  H 2 O(l)) 3 (- 286.0) 3 (H 2 O(l)  H 2 O(g)) 3 (+ 44.0) 2 B(s) + 3 H 2 (g)  B 2 H 6 (g) +36.0 kJ