Ch 6_Part 3_110811_updated

Ch 6_Part 3_110811_updated - 11/9/2011 Solving Equilibrium...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.) 22 H ( ) F ( ) 2HF( ) g g g  S E P (cont.) First : Make sure the reaction is balanced! Second : Construct the equilibrium expression: H ( ) F ( ) 2HF( ) g g g       2 2 HF 1 M 1.15 10 HF C K      
Background image of page 2
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: 22 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 HF 1 M 1.15 10 HF 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!
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
11/9/2011 4 S E P (cont.) Continuing with the example: We need to solve this last equation for x .
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/18/2012 for the course CHEM 142B taught by Professor John during the Fall '11 term at University of Washington.

Page1 / 12

Ch 6_Part 3_110811_updated - 11/9/2011 Solving Equilibrium...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online