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Chapter 14 Lecture 1

# Chapter 14 Lecture 1 - I 2 2I Slow Step 2I H 2 2HI Based on...

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Chapter 14 Lecture 1 Scene 1: H 2 I 2 H 2 I 2 Watch and wait, maybe measure the kinetics Scene 2: Three possible out comes: H 2 I 2 H 2 I 2 or Mixture, HI … H 2 I 2 or HI HI HI …. All of these possibilities are given by the chemical reaction: H 2 + I 2 2HI Equilibrium Constant, K, and Equilibrium constant expression are: ] ][ [ ] [ 2 2 2 I H HI K = General form: aA + bB cC + dD b a d c B A D C K ] [ ] [ ] [ ] [ =

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Example: Initial Concentration Equilibrium Concentration [HI] 0.0 1.573 [H 2 ] 1.0 0.213 [I 2 ] 1.0 0.213 K = Points: 1) You would get the same K no matter what the initial concentrations were. 2) Rate of forward and backward reaction are the same. Back to Chapter 13 Equilibrium in Reaction Mechanisms Overall reaction H 2 + I 2 2HI By what mechanism does it proceed to equilibrium? Out of the clear blue propose: Fast equilibrium

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Unformatted text preview: I 2 2I Slow Step 2I + H 2 2HI Based on slow step: Reaction Rate is: For equilibrium Rate forward = Rate backwards Find the overall rate equation: Problem 13.65-66 The following reaction: 2NO + Cl 2 2NOCl Has an experimentally determined rate law as: R = k[NO] 2 [Cl 2 ] Propose a mechanism (who are they kidding?) 2NO N 2 O 2 Fast N 2 O 2 + Cl 2 2NOCl Slow Use the mechanism to prove the rate law. What to do with Gases: N 2 (g) + 3H 2 (g) 2NH 3 (g) 3 2 2 2 3 H N NH P P P P K = If V and T are constant amount of gas can be expressed as moles/V or a gas concentration 3 2 2 2 3 ] ][ [ ] [ H N NH K c = To convert Ks Use Kp = Kc(RT) n...
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Chapter 14 Lecture 1 - I 2 2I Slow Step 2I H 2 2HI Based on...

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