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notes_Chapter_6 - Chapter 6 Chapter Chemical Equilibrium...

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Chapter 6 Chapter 6 Chemical Equilibrium Chemical Equilibrium
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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 LeChatelier’s Principle 6.9 Equilibria Involving Real Gases
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Nitrogen dioxide shown immediately after expanding
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Figure 6.1: Reaction of 2NO 2 ( g ) and N 2 O 4 ( g ) over time in a closed vessel
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Reddish brown nitrogen dioxide, NO 2 (g)
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Reaching Equilibrium on the Macroscopic and Molecular Level N 2 O 4 (g) 2 NO 2 (g) Colorless Brown
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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
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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
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Reaction Direction and the Relative Sizes of Q and K
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Initial and Equilibrium Concentrations for the N 2 O 4 -NO 2 System at 100°C 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
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Figure 6.2: Changes in concentration with time for the reaction H 2 O (g) + CO (g) H 2 (g) + CO 2 (g)
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Molecular model: When equilibrium is reached, how many molecules of H 2 O, CO, H 2 , and CO 2 are present?
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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)
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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)
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Figure 6.5: Concentration profile for the reaction
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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
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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 2 O] (0.613 mol/L) (1.839 mol/L)3 (0.387 mol/L) (0.387 mol/L) This is the reciprocal of K: 1 K = = ____________ mol2/L2 1 0.0393 L2/mol2 a) Calculate the value of the equilibrium constant at 927oC for: 1/3 CO (g) + H 2 (g) = 1/3 CH 4 (g) + 1/3 H 2 O (g) K = = [CO]1/3[H 2 ] [H 2 O]1/3 [CH 4 ]1/3 (0.387mol/L)1/3 (0.387 mol/L)1/3 (0.613 mol/L)1/3 (1.839 mol/L) K = = 0.340 L2/3/mol2/3 = (0.0393L2/mol2)1/3 (0.729) (0.729) (0.850)(1.839)
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Summary: Some Characteristics of the Equilibrium Expression The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original reaction.
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