Lecture+Ch+10+c

# Lecture+Ch+10+c - Ch 10 Le Chteliers principle 60:40 Text...

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Ch 10 – Le Châtelier’s principle

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60:40 ? Assigned homework will be basis for how I write the tests. I will use same format for >80 % of text problems (mas o menos!) Text : electronic
Electronic Homework c02 Molarity - Unit Quiz 02 5 Apr 2009 at 11:00 PM c02 Solution Stoichiometry - Unit Quiz 02 5 Apr 2009 at 11:00 PM c10 Chemical Equilibrium - Equilibrium Constant 10 Mastery 5 Apr 2009 at 11:00 PM c10 Chemical Equilibrium - Equilibrium Concentrations 10 Mastery 12 Apr 2009 at 11:00 PM c10 Chemical Equilibrium - Stoichiometric Relationships 10 Mastery 12 Apr 2009 at 11:00 PM c10 Solubility Products - Solubility 10 Mastery 12 Apr 2009 at 11:00 PM c10 Solubility Products - Common Ion Effect 10 Mastery 19 Apr 2009 at 11:00 PM

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Ch 10 – Recap Last time we talked about : The REACTION QUOTIENT , Q , based on any non-equilibrium conditions Q = [C] [D] [A] [B] A + B C + D non-equilibrium [ ]’s used, in the book as ( ) Q is the “crystal ball” since it can predict a reaction’s future. Eg, if Q < K eq Q α [products] [reactants] must get bigger must get smaller Q −Δ PCl 5 + Δ Cl 2 PCl 3 0.100 M 0.000 M 0.000 M + Δ PCl 3 + Δ Cl 2 0.100 Δ PCl 5 And how to set up an ICE table, based on the balance equation PCl 3 (g) + Cl 2 (g) PCl 5 (g) C hange E C I I nitial E nding
= Δ * Δ (0.100 - Δ) Solving for the change!! Ch 10 – The reaction quotient K eq = [PCl 3 ][Cl 2 ] [PCl 5 ] = 0.030 Δ PCl 5 = Δ Cl 2 Δ PCl 3 = Δ = PCl 3 (g) + Cl 2 (g) PCl 5 (g) Δ = b + b 2 4ac 2a Obtaining the final concentrations PCl 5 = 0.100 – 0.042 = 0.058 M PCl 3 = 0.042 M Cl 2 = 0.042 M −Δ PCl 5 + Δ Cl 2 PCl 3 0.100 M 0.000 M 0.000 M + Δ PCl 3 + Δ Cl 2 0.100 Δ PCl 5 E C I And checking our results [.042][.042] [.058] = 0.0304 We use this info to solve for the change in concentrations at equilibrium!! By describing the equilibrium constant in terms of the ending values Remember you are allowed to round a term in the equilibrium expression as long as the neglected value is less than 5% of the term.

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Ch 10 – Le Châtelier’s principle Henri Louis Le Châtelier, ca. 1888 We have seen what a reaction does to establish equilibrium based on a set of initial conditions. We will now discuss what happens when we “perturb” (or mess with) a system already at equilibrium. If a change is imposed on a system at equilibrium which disrupts the equilibrium position, the system will react in a manner that tends to reduce that change. Le Châtelier’s principle There are three changes which we will see affect an equilibrium position: 1 – change in amount of species 2 – change in total pressure 3 – change in temperature
Change in amount of species present: Let’s examine the “water/gas shift” equilibrium reaction from before: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) K C = 3.6 × 10 2 We see a certain set of concentrations, or pressures, developing as the reaction establishes equilibrium. As long as T is constant, K C is constant . What happens if we add CO

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Lecture+Ch+10+c - Ch 10 Le Chteliers principle 60:40 Text...

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