Ch 6_Part 3_110811_updated

# Ch 6_Part 3_110811_updated - Solving Equilibrium...

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11/9/2011 1 Solving Equilibrium Problems (6.7) Determining Equilib. Concentrations Comparing Q and K In the previous lecture we encountered some simple problems where we could use basic algebra to determine the value of the equilibrium constant. In this lecture we want to explore more complex equilibrium chemistry. One can view this entire lecture as introducing you to two skills: Given K, determine the concentration of species at equilibrium. Comparing Q and K to determine the direction in which a reaction is proceeding. Solving Equilibrium Problems

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11/9/2011 2 Let’s consider the following reaction (K C = 1.15 x 10 2 ): Initial concentrations for each species are (with subscript “0” indicating initial concentration): [H 2 ] 0 = 1.00 M [F 2 ] 0 = 2.00 M [HF] 0 = 0 M What are the concentrations of each species at equilibrium? S E P (cont.) 2 2 H ( ) F ( ) 2HF( ) g g g  S E P (cont.) First : Make sure the reaction is balanced! Second : Construct the equilibrium expression: 2 2 H ( ) F ( ) 2HF( ) g g g  2 2 2 2 HF 1 M 1.15 10 H F 1 M 1 M C K   
11/9/2011 3 S E P (cont.) Third : Using the stoichiometry of the reaction, derive expressions for concentrations at equilibrium. You can think through this process by considering the initial , change in , and equilibrium concentrations. Easy to do using an “ I C E ” table: 2 2 H ( ) F ( ) 2HF( ) g g g  Species Initial Conc .(M) Change in Conc. (M) Equilib. Conc. (M) H 2 1.00 -x 1.00 - x F 2 2.00 -x 2.00 - x HF 0 2x 2x S E P (cont.) Fourth : Substitute equilibrium concentrations into the equilibrium expression and solve. 2 2 2 2 HF 1 M 1.15 10 H F 1 M 1 M C K    Species Equilib. Conc. (M) H 2 1.00 - x F 2 2.00 - x HF 2x By “solve” we mean solve for x, which we can use to determine equilib. concentrations!

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11/9/2011 4 S E P (cont.) Continuing with the example: We need to solve this last equation for x .
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