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Overall Objectives – GC II. Thermodynamics: Directionality of Chemical Reactions (Chapter 18) Liquids, Solids, and Materials (Chapter 11) Solutes and Solutions (Chapter 15) Chemical Kinetics: Rates of Reactions (Chapter 13) Chemical Equilibrium (Chapter 14) Acids and Bases (Chapter 16)
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Overall Objectives – GC II • Thermodynamics: Directionality of Chemical Reactions (Chapter 18) • Liquids, Solids, and Materials (Chapter 11) • Solutes and Solutions (Chapter 15) • Chemical Kinetics: Rates of Reactions (Chapter 13) • Chemical Equilibrium (Chapter 14) • Acids and Bases (Chapter 16) • Additional Aqueous Equilibria (Chapter 17) • Electrochemistry (Chapter 19) All chapters share themes of thermodynamics and equilibria.
Main ‘Thread’– GC II • Thermodynamics: Directionality of Chemical Reactions (Chapter 18) • Chemical Kinetics: Rates of Reactions (Chapter 13) • Chemical Equilibrium (Chapter 14) Will a reaction proceed? How fast will it go? How far will it go?
Chemistry: The Molecular Science Moore, Stanitski, and Jurs Chapter 18: Thermodynamics: Directionality of Chemical Reactions Chemistry: The Science in Context Gilbert, Kriss, Davies Chapter 13: Entropy and Free Energy Chemistry: Principles and Reactions Masterton & Hurley Chapter 17: Spontaneous Processes
Thermodynamics: Directionality of Chemical Reactions (Chapter 18) Will a reaction proceed? • Introduction (18.1, 18.2) • What is a spontaneous reaction? • What is entropy (ΔS)? • Qualitative Guidelines for Entropy, ΔS (18.3) • Entropy and Criteria for spontaneity (18.5) • Second Law of thermodynamics • Gibbs Free Energy, ΔG (18.6) • The ultimate criteria for spontaneity • Free Energy ΔG and Equilibria Constants (18.7)
Spontaneous? • What is a spontaneous process? A spontaneous reaction will occur “by itself” , at a particular temperature and pressure, without exertion of any outside forces. • Which of these reactions are spontaneous (at 1atm and 25°C)? • 1. Reaction of sodium metal in water • Na (s) + H2O (l) Na+(aq) + OH- (aq) + H2(g) • 2. Iron rusting • 4Fe (s) + 3O2(g) + 6H2O (l) 4Fe(OH)3 (s) • 3. Mixing of two ideal gases (helium and argon) • 4. Melting of ice in your soda on a hot day • H2O (s) H2O (l) SLOW!
The Energy Factor: Enthalpy, ΔHMany spontaneous processes proceed with a decrease of energy: exothermic reaction, net bond-making, more stable products, “down-hill”. 1. Reaction of sodium metal in water Na (s) + H2O (l) Na+(aq) + OH-(aq) + H2 (g) 2. Rusting 4Fe (s) + 3O2(g) + 6H2O (l) 4Fe(OH)3 (s) 3. Mixing of two ideal gases (He + Ar) 4. Melting of ice in your soda on a hot day H2O (s) H2O (l) (ΔH< O, exothermic) (ΔH< O, exothermic) (ΔH= O) (ΔH> O, endothermic)
The Randomness Factor: Entropy (∆S) Nature tends to move spontaneously from a state of lower probability to one of higher probability, or ordered to more random. Second Law of thermodynamics A process occurs spontaneously if it results in an INCREASE in the ENTROPY of the UNIVERSE!!! Entropy (S)is a measure of nanoscale disorder. • dispersal of energy • dispersal of matter probability driven
Entropy (∆S): Dispersal of Energy Exothermic reactions: • E is transferred to the surroundings. • Bond (potential) E spread over many particles rather than concentrated over a few. • E is distributed more randomly.
Entropy (∆S): Dispersal of Energy Consider 2 atoms (A and B) and 2 packets of energy (* *). Possible distributions of energy: A** or A*,B* or B** Now consider contact with atoms C and D: How are the 2 packets of energy dispersed? A** B C D A* B* C D A* B C* D A* B C D* A B** C D A B* C* D A B* C D* A B C** D A B C* D* A B C D** Only 3/10 (30%) have the energy in atoms A and B like the original state. There is 70% chance that E will spread out over more atoms. E concentration is much less likely.
Entropy (∆S): Dispersal of Matter Matter tends to disperse (like E ): • Matter is dispersed: • Pure systems mix. • Particles occupy a larger space. Gases expand (if no barrier). 50% chance that a molecule will be in a given flask. ForN molecules the chance that ALL will be in the top flask = (½)N (½)2 = 0.25 (½)1000 = 8 x 10-31
Entropy (∆S): Dispersal of Matter • Matter is dispersed: • Pure systems mix. • Particles occupy a larger space. dice marbles CLOSED OPEN ordered random SPONTANEOUS
Qualitative Prediction Of ΔSsystem Entropy usually increases in the following situations: • gas >> liquid > solid (of the same substance) • temperature increases (more motion) • number of moles increases (ReactantsProducts) • total volume increases • solid or liquid substance dissolving
Qualitative Prediction Of ΔSsystem Sgas >> Sliquid > Ssolid more freedom = more disorder = higher S gas liquid solid S°(I2) 261 - 116 S°(Br2) 245 152 - S°(Cl2) 223 - - room temperature Weaker bond = larger S (Less tightly bound = easier motion)
Qualitative Guidelines for Entropy Solid or liquid dissolving Larger matter dispersal. S usually* increases: Substance S° (pure) S° (aq) CH3COOH(l) 160 179 NH4NO3(s) 151 260
What Happens When Gases Dissolve? Gas-molecule motion becomes restricted. ΔS < 0.
Entropy and Temperature A perfect crystal at 0K has an entropy of 0. 0K
Predicting Entropy Changes Will entropy increase or decrease for the following processes? Increase entropy if…. number of moles increases (ReactantsProducts) 2 CO(g) + O2(g) → 2 CO2(g) Decrease 3 non-identical gas molecules →2 identical gas molecules. Less disorder. NaCl(s) → Na+(aq) + Cl-(aq) Each crystal “unit” → 2 ions dispersed in water. Much more disorder. Increase H2O(l) → H2O(s) Decrease Disorganized water molecules → more organized ice molecules. Less disorder.
Sign of ΔSsystem Predict (qualitatively) the sign of ΔSsystem for the following: CaCO3(s) CaO (s) + CO2 (g) CS2(l) CS2(g) 2 Hg(l) + O2(g) 2 HgO (s) 2 Na2O2(s) + 2 H2O (l) 4 NaOH (aq) + O2(g) + + - +
(ΔSsystem) alone good enough? 1. Melting of ice in your soda on a hot day H2O (s) H2O (l) Spontaneous and (ΔSsystem > 0) 2. Freezing of water on a cold day (-15oC) H2O (l) H2O (s) Spontaneous and (ΔSsystem < 0)
Gibbs Free Energy Neither entropy (S), nor enthalpy (H ), alone can predict whether a reaction is spontaneous. Spontaneous reactions can: - be exothermic or endothermic. -increase or decrease the Ssystem. The Gibbs free energy (G), combines H and S. ΔGcan predict if a reaction is product favored or not.
Gibbs Free Energy For a constant T, ΔG is equal to: ΔG = ΔH – TΔS(T in K). If G: Decreases(ΔG < 0) a reaction is spontaneous. Increases(ΔG > 0) a reaction is nonspontaneous. (at constant P and T).
Free Energy of System (∆Gsystem) ∆Gsystem = (∆Hsystem - T ∆Ssystem ) < 0 Ultimate Criteria for Spontaneity If ∆Gsystem < 0 (negative) process is spontaneous or exergonic If ∆Gsystem > 0 (positive) process is not spontaneous or endergonic If ∆Gsystem = 0 process is in equilibrium
Spontaneity - Driving Forces • Thermodynamic Criterea for Reaction Spontaneity • ΔH, the change in enthalpy ΔH < 0 ΔH < 0, tends to make a reaction spontaneous • ΔS, the change in entropy ΔS > 0, tends to make a reaction spontaneous • ΔG, the change in free energy ΔG < 0, will be spontaneous .
The Effect of T on Reaction Direction Since ΔG = ΔH – TΔS Sign of ΔH Sign of ΔS Product Favored? Negative (exothermic) Positive Yes Negative (exothermic) Negative Yes at low T; No at high T Positive (endothermic) Positive No at low T;Yes at high T Positive (endothermic) Negative No ΔG = ΔH – TΔS
Putting it all together (∆G = ∆H- T ∆S) qualitative prediction Classify reaction into 4 types : (always spontaneous, never spontaneous, spontaneous high T, spontaneous at low T) a) 2 O3 (g) 3 O2(g) ∆H< 0, ∆S = ∆G = b) 2 C(graphite)+ 2 H2 (g) C2H4 (g) ∆H> 0, ∆S = ∆G = c) N2(g) + 3H2(g) 2 NH3(g) ∆H< 0, ∆S = ∆G = d) 2 H2O (g) 2 H2 (g) + O2 (g) ∆H> 0, ∆S = ∆G = + - - + - + - low T - high T DEMO: rubber band
The Effect of T on Reaction Direction ΔG = ΔH − TΔS Reactions go from non-spontaneous (+ΔG) to spontaneous (-ΔG) when, TΔS > ΔH ∆H +, ∆S+ Need to maximize Entropy term. T> ΔH ΔS
The Effect of T on Reaction Direction At what T will the following reaction be spontaneous? CH4(g) + H2O(g) CO(g) + 3 H2(g) ΔH° = 206.10 kJ ΔS° = 214.63 J K-1 ΔG = ΔH − TΔS Spontaneous at higher T. Switch occurs at ΔG = 0, when two opposing terms are equal 0= ΔH − TΔS T= ΔH/ΔS T = 206.10 kJ/mol = 960 K 0.21463 kJ/Kmol Spontaneous > 960K
The Effect of T on Reaction Direction • Do Now: • Calculate the temperature at which the following reaction will occur spontaneously: Fe2O3(s) +3H2(g) 2Fe(s) +3H2O(g) • At what T will the following reaction be spontaneous? H2O (l) H2O (g) ΔH° = 98.8 kJ ΔS° = 141.5 J K-1 698K 373K, 100°C ΔH° = 40.7 kJ ΔS° = 109 J K-1
Reactions that Reach Equilibrium ? ΔG° = −RT ln K° 2 NO2(g) N2O4(g) • Keq° = 6.75 [products]m[N2O4] [reactants]n[NO2]2 Keq= = ΔG° = -8.314 JK-1 ∙ 298K ∙ ln(6.75) - + ΔG° = -4730 J or 4.73kJ
Reactions that Reach Equilibrium ? ΔG° = −RT ln K° PbCl2(s) Pb2+(aq) + 2Cl-(aq) [products] [Pb+2][Cl-]2 [reactants] [PbCl2] Keq= = • Keq° = 1.7 x 10-5 What is the ∆G (positive or negative?) of the above dissolution equation, at 25°C? ∆G = +27.3 kJ