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