Le Châtelier’s principle
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Transcript of Le Châtelier’s principle
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Le Châtelier’s principle
Le Châtelier’s principle
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The significance of Kc valuesThe significance of Kc values
If Kc is small (0.001 or lower), [products] must be small, thus forward reaction is weak
If Kc is large (1000 or more), [products] must be large, thus forward reaction is strong
[Products]Kc =
[Reactants]
Reactants Products
If Kc is about 1, then reactants and products are about equal
but not exactly since they may be raise to different exponents
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Stresses to equilibriaStresses to equilibriaChanges in reactant or product
concentrations is one type of “stress” on an equilibrium
Other stresses are temperature, and pressure.
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The response of equilibria to these stresses is explained by Le Chatelier’s principle:If an equilibrium in a system is upset, the system will tend to react in a direction that will reestablish equilibrium
Thus we have: 1) Equilibrium, 2) Disturbance of equilibrium, 3) Shift to restore equilibrium
Le Chatelier’s principle predicts how an equilibrium will shift (
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N2 + 3H2 2NH3 + 92 kJ
N2 + 3H2 2NH3 + 92 kJN2 + 3H2 2NH3 + 92 kJN2 + 3H2 2NH3 + 92 kJ
N2 + 3H2 2NH3 + 92 kJN2 + 3H2 2NH3 + 92 kJ
N2 + 3H2 2NH3 + 92 kJN2 + 3H2 2NH3 + 92 kJN2 + 3H2 2NH3 + 92 kJ
Summary of Le Chatelier’s principleSummary of Le Chatelier’s principleE.g. N2 + 3H2 2 NH3 + 92 kJ
Pressure (due to decreased volume): increase in pressure favors side with fewer molecules
Amounts of products and reactants: equilibrium shifts to compensate
N2
H2
Temperature: equilibrium shifts to compensate:
Heat
shift right
shift left
shift left
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Part II. Equilibria involving sparingly soluble salts
Part II. Equilibria involving sparingly soluble salts
Ag+ + CO3-2 Ag2CO3
-2
• 2H+ + CO3-2 H2O + CO2
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Part II. Equilibria involving sparingly soluble salts
Part II. Equilibria involving sparingly soluble salts
Ag+ + Cl- AgCl
• Ag+ + 2NH3 Ag(NH3)2 + heat
NH3 + H+ (NH4)
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QuestionsQuestions
Omit part 1 Pg 256 omit part 3
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Titration Titration
Lab 20- Acids & BasesLab 20- Acids & Bases
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A. NeutralizationA. Neutralization
Chemical reaction between an acid and a base.
Products are a salt (ionic compound) and water.
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NeutralizationNeutralization
ACID + BASE SALT + WATER
HCl + NaOH NaCl + H2O
HC2H3O2 + NaOH NaC2H3O2 + H2O
• Salts can be neutral, acidic, or basic.• Neutralization does not mean pH = 7.
weak
strong strong
strong
neutral
basic
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Titration Titration
Titration• Analytical method
in which a standard solution is used to determine the concentration of an unknown solution.
standard solution
unknown solution
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Equivalence point (endpoint)• Point at which equal amounts
of H3O+ and OH- have been added.
• Determined by…• indicator color change
B. TitrationB. Titration
• dramatic change in pH
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B. TitrationB. Titration
moles H3O+ = moles OH-
MV n = MV nM: MolarityV: volumen: # of H+ ions in the acid
or OH- ions in the base
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B. TitrationB. Titration
42.5 mL of 1.3M NaOH are required to neutralize 50.0 mL of KHP. Find the molarity of KHP.
H3O+
M = ?V = 50.0 mLn = 1
OH-
M = 1.3MV = 42.5 mLn = 1
MV = MVM(50.0mL)
=(1.3M)(42.5mL)
M = 1.11 M H2SO4
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ProceduresProcedures
Data you need for part B• Molarity of NaOH
• Trial 1 [ .115 M] • Trial 2 [.116 M] • Trial 3 [.117 M]
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ProceduresProcedures
FormulasM = mol/liters
moles H3O+ = moles OH-
MV (H3O+) = MV (OH-)
% KHP = mass of KHP/ Mass of mixture
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Part B. Standardization of unknown acid
Part B. Standardization of unknown acid
• Add 2.5 grams of unknown acid into 3 Erlenmeyer flasks• Use analytical balance to 4 sig figs
• Add 100 ml of D.I water into each flask• Add 2 drops of Phenolphthalein soln
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A. Fill a Burett with standarized NaOH• Slowly add NaOH into 1st flask, while
swirling flask• See fig 20-2 and read procedures for proper titration
Record amount of NaOH used to titrate
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calculationscalculations
Determine moles of base usedDetermine moles of KHPDetermine mass of KHpDetermine % of KHP in unknown
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QuestionsQuestions
223 and 224 Questions 1 -3