110c-lecture21 - Midterm 2 Wed May 19 th Chapters 23.1 –...

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Unformatted text preview: Midterm 2 Wed May 19 th . Chapters 23.1 – 4, 24, 25.1 – 6, and 26.1 – 7&11. Closed book and closed notes. 3” x 5” index card. Review session? Practice midterm2 on website Problem Set #5 Due Wed June 2 nd . Final Exam Tue June 8 th at 1:00-3:00 pm in 1006 Giedt. Chapters 19 – 29? Thermodynamic vs Apparent Equilibrium Constants CH 3 COOH H + + CH 3 COO- COOH CH c c c c K 3 − + = 5 3 2 3 10 74 . 1 − − + ± − + × = = = COOH CH COOH CH a c c c a a a K γ c I ± γ ln ± γ 2 ± γ c K 1 1 1.74E-05 0.001-0.04 0.96 0.93 1.87E-05 0.010-0.11 0.90 0.81 2.15E-05 0.050-0.21 0.81 0.65 2.67E-05 K a is a constant regardless of concentration (I c ) or 2 ± γ . K c depends on both I c and 2 ± γ and is only equal to K a when 2 ± γ = 1. c a K K 2 ± = γ c B r I z z T k q q − + − + ± − = − = 173 . 1 8 ln ε πε κ γ ] [ ] ][ [ HAc Ac H K c − + = Chapter 26: Chemical Equilibria ( ) ( ) ( ) ( ) g Z g Y g B g A Z Y B A ν ν ν ν + = + 1. Change in Free Energy of Reaction ( ∆ r G) B B A A Z Z Y Y rxn P P P P Q ν ν ν ν = EQ B B A A Z Z Y Y P P P P P K ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ν ν ν ν rxn r r Q RT G G ln + ∆ = ∆ P r K RT G ln − = ∆ ∆ r Gº = ν Y ∆ f G(Y) + ν Z ∆ f G(Z) – ν A ∆ f G(A) – ν B ∆ f G(B) (Table 26.1) 2. Free Energy (G) vs. Extent of reaction ( ξ ) ( ) ( ) ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + + ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − − + + − = P RT P RT G G G NO O N ξ ξ ξ ξ ξ ξ ξ ξ ξ 1 2 ln 2 1 1 ln 1 2 1 2 4 2 ( ) , = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∆ P T EQ G G ∂ξ ∂ ξ G tot ξ EQ P P eq K P K + = 4 ξ 3. Equilibrium constant (K P ) vs Temperature (van’t Hoff Eqn) R H T K r P eq 1 ln ∆ − = ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ∂ ∂ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ∆ − = 1 2 1 2 1 1 ln T T R H K K r 2 ln RT H T K r P eq ∆ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ Note, Henry’s Law constant (k H,2 ) and kinetic rate constant (k) obey similar eqn. Chapter 23: Phase Equilibria Clapeyron Eqn l l l g g g dT S dP V dT S dP V µ µ = − = − = V T H V S V V S S dT dP trs trs trs trs l g l g ∆ ∆ = ∆ ∆ = − − = V T T T H P P fus fus ∆ − ∆ + ≈ 1 1 2 1 2 ) ( Sol → liq ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ∆ − = 1 2 1 2 1 1 ln T T R H P P vap Clausius-Clapeyron g l dn dn = − liq → gas Liq → Vap liq gas l l g g gas liq n n G µ µ − = ∆ → ( ) ( ) H H H H H dT dP dT dP vap vap fus vap sub vap sub ∆ ∆ + ∆ = ∆ ∆ = Chapter 24: Liquid-liquid Solutions Initial Final P * 1 P 1 n 1 liq sln Solvent(1) + Solute(2) → Solution Partial Molar Quantities .. , 2 2 .. , 1 1 P T P T n G n G ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠...
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110c-lecture21 - Midterm 2 Wed May 19 th Chapters 23.1 –...

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