Lec-11-Chap14-3-Complete

Lec-11-Chap14-3-Complete - Lec-11-Chap14-3-Complete...

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Lec-11-Chap14-3-Complete 9/28/2009 Chemical Equilibrium Concentrations at equilibrium. Le Chatelier’s Principle 1 Lec-11: Chemical Equilibrium Chapter 14 1 Chemical Equilibrium So far we have learned: 1. How to determine the equilibrium constant K c from initial and equilibrium concentrations. 2. How to determine, given initial concentrations of the reactants and the products, the direction in which the reaction will proceed. C+ D+ A+ B+ ab cd         D AB C c K the equilibrium constant expression is For a reaction We now consider how to find the composition of a reaction system at equilibrium, given that K c and the initial concentrations are given. 2
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Lec-11-Chap14-3-Complete 9/28/2009 Chemical Equilibrium Concentrations at equilibrium. Le Chatelier’s Principle 2 Calculating concentrations at equilibrium is 3.76 × 10 –5 at T =1000 K. The experiment is started by injecting 1 mol of I 2 (g) into a 2L container at 1000 K. What are the molar concentrations of I 2 (g) and of I(g) at equilibrium at 1000 K?       2 2 0 1 mol initial number of moles of I mol I 0.5 volume of the container 2 L L In this problem we start with an initial concentration of I 2 (g):     2 Ig 2I g The equilibrium constant K c for the reaction 3 Calculating concentrations at equilibrium           concentrations at equilibrium 2 2 0 2 2 I2 I I c x x K  The equilibrium equation gives To find the concentrations of reactants and products at equilibrium we analyze the changes in the concentrations brought by the reaction:           2 2 0 2 0 initial 0 cha nges equilib riu I g m I I 2 2 xx Notice that x is the number of moles of I 2 per unit volume that are consumed in the reaction. 4
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Lec-11-Chap14-3-Complete 9/28/2009 Chemical Equilibrium Concentrations at equilibrium. Le Chatelier’s Principle 3 Calculating concentrations at equilibrium We can solve this equation for x . We find that x satisfies the equation         22 0 2 0 4 4 16 2 I 48 I c c c c c c K K K x K K K      that has the solution   0 2 40 I cc xx KK         5 5 5 31 3.76 10 3.76 10 3.76 1 16 8 2. 2 17 1 .16 0 0 .5 10 0 mol L L mol x x    With the values given in the problem A concentration cannot be negative!!!! 5 The concentrations of reactant and product at equilibrium are then Calculating concentrations at equilibrium       3 1 0 1 I I 0.498mol I 4.32 2 10 mol L L x x   6
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Lec-11-Chap14-3-Complete 9/28/2009 Chemical Equilibrium Concentrations at equilibrium.
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This note was uploaded on 01/02/2010 for the course CHE 132 taught by Professor Hanson during the Fall '08 term at SUNY Stony Brook.

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Lec-11-Chap14-3-Complete - Lec-11-Chap14-3-Complete...

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