Post on 14-Jan-2016
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Le Châtelier’s principle
Le Châtelier’s principle
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
Stresses to equilibriaStresses to equilibriaChanges in reactant or product
concentrations is one type of “stress” on an equilibrium
Other stresses are temperature, and pressure.
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 (
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
Part II. Equilibria involving sparingly soluble salts
Part II. Equilibria involving sparingly soluble salts
Ag+ + CO3-2 Ag2CO3
-2
• 2H+ + CO3-2 H2O + CO2
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)
QuestionsQuestions
Omit part 1 Pg 256 omit part 3
Titration Titration
Lab 20- Acids & BasesLab 20- Acids & Bases
A. NeutralizationA. Neutralization
Chemical reaction between an acid and a base.
Products are a salt (ionic compound) and water.
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
Titration Titration
Titration• Analytical method
in which a standard solution is used to determine the concentration of an unknown solution.
standard solution
unknown solution
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
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
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
ProceduresProcedures
Data you need for part B• Molarity of NaOH
• Trial 1 [ .115 M] • Trial 2 [.116 M] • Trial 3 [.117 M]
ProceduresProcedures
FormulasM = mol/liters
moles H3O+ = moles OH-
MV (H3O+) = MV (OH-)
% KHP = mass of KHP/ Mass of mixture
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
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
calculationscalculations
Determine moles of base usedDetermine moles of KHPDetermine mass of KHpDetermine % of KHP in unknown
QuestionsQuestions
223 and 224 Questions 1 -3