notes_Chapter_6 - Chapter 6 Chapter 6 Chemical Equilibrium...

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Unformatted text preview: Chapter 6 Chapter 6 Chemical Equilibrium Chemical Equilibrium Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications of the Equilibrium Constant 6.7 Solving Equilibrium Problems 6.8 LeChateliers Principle 6.9 Equilibria Involving Real Gases Nitrogen dioxide shown immediately after expanding Figure 6.1: Reaction of 2NO 2 ( g ) and N 2 O 4 ( g ) over time in a closed vessel Reddish brown nitrogen dioxide, NO 2 (g) Reaching Equilibrium on the Macroscopic and Molecular Level N 2 O 4 (g) 2 NO 2 (g) Colorless Brown The State of Equilibrium At equilibrium: rate fwd = rate rev rate fwd = k fwd [N 2 O 4 ] rate rev = k rev [NO 2 ] 2 For the Nitrogen dioxide - dinitrogen tetroxide equilibrium: N 2 O 4 (g, colorless) = 2 NO 2 (g, brown) k fwd [N 2 O 4 ] = k rev [NO 2 ] 2 k fwd [NO 2 ] 2 k rev [N 2 O 4 ] = = K eq 1) Small k N 2 (g) + O 2 (g) 2 NO (g) K = 1 x 10-30 2) Large k 2 CO (g) + O 2 (g) 2 CO 2 (g) K = 2.2 x 10 22 3) Intermediate k 2 BrCl (g) Br 2 (g) + Cl 2 (g) K = 5 Writing the Reaction Quotient or Mass-Action Expression Q = mass-action expression or reaction quotient Q = Product of the Reactant Concentrations Product of the Product Concentrations For the general reaction: a A + b B c C + d D Q = [C] c [D] d [A] a [B] b Example: The Haber process for ammonia production: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Q = [NH 3 ] 2 [N 2 ][H 2 ] 3 Reaction Direction and the Relative Sizes of Q and K Initial and Equilibrium Concentrations for the N 2 O 4-NO 2 System at 100C Initial Equilibrium Ratio [N 2 O 4 ] [N 2 O 4 ] [N 2 O 4 ] [NO 2 ] [NO 2 ] [NO 2 ] 2 0.1000 0.0000 0.0491 0.1018 0.211 0.0000 0.1000 0.0185 0.0627 0.212 0.0500 0.0500 0.0332 0.0837 0.211 0.0750 0.0250 0.0411 0.0930 0.210 Figure 6.2: Changes in concentration with time for the reaction H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Molecular model: When equilibrium is reached, how many molecules of H 2 O, CO, H 2 , and CO 2 are present? Figure 6.3: H 2 O and CO are mixed in equal numbers H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Figure 6.4: Changes with time in the rates of forward and reverse reactions H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Figure 6.5: Concentration profile for the reaction Like Example 6.1 (P 201) - I e following equilibrium concentrations were observed for the action between CO and H 2 to form CH 4 and H 2 O at 927oC. CO (g) + 3 H 2 (g) = CH 4 (g) + H 2 O (g) CO] = 0.613 mol/L [CH 4 ] = 0.387 mol/L H 2 ] = 1.839 mol/L [H 2 O] = 0.387 mol/L Calculate the value of K at 927 o C for this reaction. Calculate the value of the equilibrium constant at 927 C for: K = = = _______ L2/mol2 [CO] [H 2 ]3 [CH 4 ][H 2 O] (0.387 mol/L) (0.387 mol/L) (0.613 mol/L)(1.839 mol/L)3 Like Example 6.1 (P 201) - II b) Calculate the value of the equilibrium constant at 927oC for: H 2 O (g) + CH 4 (g) = CO (g) + 3 H 2 (g) K = = = _____ mol2/L2 [CO] [H 2 ]3 [CH 4 ] [H...
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This note was uploaded on 03/22/2011 for the course CHEM 142 taught by Professor Zoller,williamh during the Spring '07 term at University of Washington.

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notes_Chapter_6 - Chapter 6 Chapter 6 Chemical Equilibrium...

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