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Bell Ringer 2NO(g) + O 2 (g) 2NO 2 (g)  H = -27 kcal Reaction Progress Energy AB CD Source: 2004 VA Chemistry EOC Exam Which graph represents the reaction.

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Presentation on theme: "Bell Ringer 2NO(g) + O 2 (g) 2NO 2 (g)  H = -27 kcal Reaction Progress Energy AB CD Source: 2004 VA Chemistry EOC Exam Which graph represents the reaction."— Presentation transcript:

1 Bell Ringer 2NO(g) + O 2 (g) 2NO 2 (g)  H = -27 kcal Reaction Progress Energy AB CD Source: 2004 VA Chemistry EOC Exam Which graph represents the reaction shown above?

2 Homework Answers 1.A, B 2.B 3.C 4.B 5.A 6.C 7.Inter 8.Intra 9.Dispersion forces 10.Boiling Point 11.Amorphous 12.Higher 13.Dipole-dipole 14.Freezing 15.Hydrogen bonding 16.30ºC 17.1.0 atm 18.40ºC 19.47ºC 20.0.4 atm

3 Practice with Enthalpy Calculate  H for the following reactions: 2 KBr (s) + I 2 (g)2 KI (s) + Br 2 (l) 2 HCl (g) + 2 Ag2 AgCl (s) + H 2 (g)  H total = [2mol (-327.9 kJ/mol) + 1mol (0 kJ/mol)] - [2mol (-393.8 kJ/mol) + 1mol (0 kJ/mol)]  H total = 131.8 kJ  H total = [2mol (-127.01 kJ/mol) + 1mol (0 kJ/mol)] - [2mol (- 92.31kJ/mol) + 2mol (0 kJ/mol)]  H total = -69.40 kJ ENDOTHERMIC EXOTHERMIC

4 Entropy and Gibb’s Free Energy They work the same way!! “PRODUCTS – REACTANTS”  S total =  S products -  S reactants  G total =  G products -  G reactants  S > 0, entropy increases (more disorder)  S < 0, entropy decreases (less disorder)  G > 0, not spontaneous  G < 0, spontaneous  G = 0, equilibrium

5 How do H, S, and G Relate?  G =  H - T  S Potassium bromide reacts with iodine at 345K. Will this reaction occur spontaneously? 2 KBr (s) + I 2 (s)2 KI (s) + Br 2 (l)  H total = [2mol (-327.9 kJ/mol) + 1mol (0 kJ/mol)] - [2mol (-393.8 kJ/mol) + 1mol (0 kJ/mol)]  H total = 131.8 kJ  S total = [2mol (106.3 J/mol K) + 1mol (152.21 J/mol K)] - [2mol(95.9 J/mol K) + 1mol (116.14 J/mol K)]  S total = 56.87 J/ K

6 112,000 J = 112 kJ How do H, S, and G Relate?  G =  H - T  S Potassium bromide reacts with iodine at 345K. Will this reaction occur spontaneously? 2 KBr (s) + I 2 (s)2 KI (s) + Br 2 (l)  H total = 131.8 kJ  S total = 56.87 J/ K  G = 131,800 J – (345K)(56.87 J/K) DG = 131800 J – 19620.15 J  G = 112179.85 J NOT SPONTANEOUS! x 1 kJ 1000 J =131,800 J

7 Properties of Water Ms. Besal 3/24-27/2006

8 Characteristics of Water Bent Shape Bent Shape Hydrogen Bonding Hydrogen Bonding Liquid at Room Temperature Liquid at Room Temperature Requires great amount of energy to change (raise or lower) temperature Requires great amount of energy to change (raise or lower) temperature Expands as a solid Expands as a solid High boiling point High boiling point High surface tension High surface tension High heat of vaporization High heat of vaporization “universal” solvent “universal” solvent O H H

9 How does the flow of heat change water? 0ºC 100ºC A SOLID B C D E LIQUID GAS MELT FREEZE VAPORIZATION CONDENSATION TIME TEMP.  H fus = 6.02 kJ/mol  H vap = 40.7 kJ/mol

10 Phase Diagram of Water B temperature pressure A C D solid liquid gas. FREEZE MELT CONDENSATION VAPORIZATION SUBLIMATION DEPOSITION TcTc E

11 B temperature pressure A C D solid liquid gas. A: Triple point – the temperature and pressure at which gas, liquid, and solid form of a substance all exist in equilibrium. B: Melting curve C: Vaporization curve D: Sublimation curve E: Critical point – the temperature and pressure above which gas cannot be liquefied no matter how much pressure is applied. E


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