chapter15-01_p - Chapter 5 Chemical Equilibrium Rates of...

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Chapter 5 Chapter 15: Chemical Equilibrium Chemical Equilibrium • Rates of reactions depend on concentrations • Consider products and reactants of a chemical reaction • As reaction progresses, fewer reactants are present slower reaction rate • However, backwards reaction is enhanced: – forward reaction: reactants yield products, – reverse reaction: products original reactants (g) 2NH ) g ( H 3 ) g ( N 3 2 2 Developing chemical Equilibrium: Example • The forward rate decreases as the reactant concentration decreases • The backward rate increases as product concentrations increase • Equilibrium: forward and backward reaction rates are identical Developing chemical Equilibrium: Example Chemical equilibrium : rates of the forward and reverse reactions have become equal • Dynamic equilibrium- products and reactants are constantly exchanging Chemical Equilibrium – Similar to water in a U-shaped tube: constant mixing back and forth through the lower portion of the tube – The forward and reverse reactions occur at the same rate – The system appears to be static (stationary) when, in reality, – It is dynamic – See example: … “reactants” “products” Example: Haber-Bosch Reaction • Haber-Bosch process for producing ammonia from N 2 and H 2 – An equilibrium state is formed with all three species (g) 2NH ) g ( H 3 ) g ( N 3 2 2
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Chapter 5 Haber-Bosch Reaction • Stoichiometry of an equilibrium mixture: – What is the composition of the equilibrium mixture if it contains 0.080 mol NH 3 ? – Consider: 450 o C and 10.0 atmospheres of pressure – 1.000 mol N 2 and – 3.000 mol H 2 in a reaction vessel The reaction occurs: (g) 2NH ) g ( H 3 ) g ( N 3 2 2 Haber-Bosch Reaction – The equilibrium amount of NH 3 was given as 0.080 mol. Therefore, 2x = 0.080 mol NH 3 (x = 0.040 mol) • Using the information given, set up the following table: (g) 2NH ) g ( H 3 ) g ( N 3 2 2 Starting 1.000 3.000 0 Change -x -3x +2x Equilibrium 1.000 - x 3.000 - 3x 2x = 0.080 mol Haber-Bosch Reaction • Using the information given, set up the following table: Equilibrium amount of N 2 = 1.000 - 0.040 = 0.960 mol N 2 Equilibrium amount of H 2 = 3.000 - (3 x 0.040) = 2.880 mol H 2 Equilibrium amount of NH 3 = 2x = 0.080 mol NH 3 (g) 2NH ) g ( H 3 ) g ( N 3 2 2 Starting 1.000 3.000 0 Change -x -3x +2x Equilibrium 1.000 - x 3.000 - 3x 2x = 0.080 mol The Equilibrium Constant • “Position of the equilibrium” of a reaction under any given set of conditions – The ratio of products to remaining reactants is constant at constant temperature – The numerical value of this ratio is called the equilibrium constant for the given reaction
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This note was uploaded on 04/07/2010 for the course CHEMISTRY chem 130 taught by Professor Mr during the Spring '10 term at University of Michigan.

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chapter15-01_p - Chapter 5 Chemical Equilibrium Rates of...

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