1. Chapter 6 Energy Thermodynamics pp

3. Enthalpy is a STATE FUNCTION . State functions are PATH-INDEPENDENT

6. Potential energy Heat

7. Potential energy Heat

9. System Surroundings Energy  E < 0

10. System Surroundings Energy  E >0

21. Figure 6.5 A Coffee-Cup Calorimeter Made of Two Styrofoam Cups

23. Z5e 237 Table 6.1

34. N 2 2O 2 O 2 2NO 68 kJ 2NO 2 180 kJ -112 kJ H (kJ)

43. Example Given Calculate  Hº for this reaction. Ans. . .  Hº= +77.9kJ  Hº= +495 kJ  Hº= +435.9kJ Cut all reactions in 1/2 Reverse #s 2 & 3 Add up to = -426 kJ = exo thermic (not -426.5 d/t SF)

48. Figure 6.9. A Schematic Diagram of the Energy Changes for the Reaction CH 4 (g) + 2O 2 (g )  CO 2 (g ) + 2H 2 O (l)

53. Figure 6.10 A Pathway for the Combustion of Ammonia

56. Calculating  H o reaction using standard enthalpies of formation - Values for  H o f are looked up in tables such as Appendix 4 in your book

57. Calculate  H o rxn for CaCO 3(s)  CaO (s) + CO 2(g) compound  H o f (kJ/mol) CaCO 3(s) -1206.9 CaO (s) -635.1 CO 2(g) -393.5 [ -635.1 + -393.5 ] - [ -1206.9 ] = 178.3 = endothermic

61. Nitroglycerine is a powerful explosive, giving four different gases when detonated C 3 H 5 (NO 3 ) 3(l)  3 N 2(g) + O 2(g) + 6 CO 2(g) + 5 H 2 O (g) Given that  H o f for nitroglycerine is -364 kJ/mol, and consulting a table for the enthalpies of formation of the other compounds, calculate the enthalpy change when 10.0 g of nitroglycerine is detonated.  H o rxn and Stoichiometry Find enthalpy for 1 mole of NTG (use previous method) Then convert 10.0 g NTG to moles and use mole ratio between that and 1 mole. 10g ÷ 227 g/mol = 0.044 mol

62. C 3 H 5 (NO 3 ) 3(l)  3 N 2(g) + O 2(g) + 6 CO 2(g) + 5 H 2 O (g) calculate ∆H rxn when 10.0 g of NTG is detonated.  H o rxn and Stoichiometry [6•-394 + 5•-242] - [-364] = -3.21 x 10 3 kJ/mol This is for 1 mol. Have 10.0 g, convert to moles 10g ÷ 227 g/mol = 0.044 mol Then 0.044 mol • -3.21 x 10 3 kJ = -141 kJ exothermic