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Chem 162-2011 Lecture 19

Chem 162-2011 Lecture 19 - CHEMISTRY 162-2011 LECTURE 19 Ed...

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CHEMISTRY 162-2011 LECTURE 19 Ed Tavss, PhD ANNOUNCEMENTS E-MAIL ATTENDANCE Sign in EXAMS QUIZ We already had two lecture quizzes.  Three to go. MISCELLANEOUS Chem 161-2011 Lecture 19 1
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PLAN FOR TODAY CHAPTER 17 - SPONTANEITY, ENTROPY AND FREE ENERGY 17.7 The Dependence of ∆G o and K eq on Temperature CHAPTER 4 – CHEMICAL REACTIONS IN AQUEOUS SOLUTIONS 4.4 Reactions involving oxidation and reduction CHAPTER 18 - ELECTROCHEMISTRY 18.1 Half-Reactions 18.2 The Half-Reaction Method of Balancing Redox Equations Chem 161-2011 Lecture 19 2
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CHAPT ER 16 - SPONTANEITY, ENTROPY AND FREE ENERGY First law of thermodynamics: Energy can neither be created nor destroyed. Second law of thermodynamics: The entropy of the universe always increases in a spontaneous process. Third law of thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. Entropy describes the number of arrangements (positions and/or energy levels) available to a system. Entropy is (officially called chaos, disorder, randomness) positional freedom, or positional availability, or disorder, or being unconfined positionally or escape from positional confinement or freedom from positional confinement ΔG o = Σn p ΔG o f(products) - Σn r ΔG o f(reactants) ΔH o = Σn p ΔH o f(products) - Σn r ΔH o f(reactants) ΔS o system = Σn p S o products - Σn r S o reactants ΔS system = Σn p S products - Σn r S reactants (used for ΔS system anytime ) ΔS system = ΔH system /T (used for ΔS system only when at equilibrium) ΔS surr = ΔH surr /T = -ΔH system /T (used for ΔS surr anytime) ΔS univ = ΔS sys + ΔS surr Four ways to find ΔG o : (1) ΔG o = ΔH o - TΔS o (or ΔG = ΔH - TΔS) At equilibrium ΔG = 0 0 = (ΔH - TΔS) ΔS system = ΔH system /T Trouton’s rule: At normal B.P., ΔS vaprization ≈ 87J mol -1 K -1 (2) ΔG o = Σn p ΔG o f(products) - Σn r ΔG o f(reactants) (3) ΔG T o = ΔG 1 o + ΔG 2 o (i.e., adding equations [Hess’s law]) (4) ΔG o = -RT ln(K) ΔG = ΔG o + RT ln(Q) (Q can be any pressure or concentration) w max = ΔG lnK eq = ((-∆H o /R) x 1/T) + ∆S o /R Plot straight line: Y = mX + b ln(K 2 /K 1 ) = -(∆H o /R)((1/T 2 ) – (1/T 1 )): van’t Hoff equation ln(P 2 /P 1 ) = -(∆H vapn o /R)((1/T 2 ) – (1/T 1 )): Clausius-Clapeyron equation Chem 161-2011 Lecture 19 3
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RELATIONSHIP BETWEEN K eq  AND TEMPERATURE (related to LeChatelier’s Principle) ∆G o formulas can be combined. ∆G o = -RTlnK eq ∆G o = ∆H o - T∆S o -RTlnK eq = ∆H o - T∆S o lnK eq = -∆H o /RT + ∆S o /R lnK eq = ((-∆H o /R) x 1/T) + ∆S o /R Plot straight line: Y = mX + b More commonly we know the K at one temperature, and want to find the K at a second temperature. This can also be solved non-graphically. lnK 2 = ((-∆H o /R) x (1/T 2 )) + ∆S o /R lnK 1 = ((-∆H o /R) x (1/T 1 )) + ∆S o /R Subtracting the second equation from the first: lnK 2 – lnK 1 = (-∆H o /R) x (1/T 2 – 1/T 1 ) ln(K 2 /K 1 ) = -(∆H o /R)((1/T 2 ) – (1/T 1 )) Kimmel 2007 Recitation Notes modified N 2 (g) + 3H 2 (g) 2NH 3 (g) K = 6.0 x 10 5 at 298.15 K. ∆H o f NH 3 = -46.11 kJ/mol What is value of K at 600 K?
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