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Le Châtelier’s principle

Le Châtelier’s principle. Kc =. The 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.

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Le Châtelier’s principle

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  1. Le Châtelier’s principle

  2. Kc = The 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 If Kc is about 1, then reactants and products are about equal but not exactly since they may be raise to different exponents Reactants  Products [Products] [Reactants]

  3. Stresses to equilibria • Changes in reactant or product concentrations is one type of “stress” on an equilibrium • Other stresses are temperature, and pressure.

  4. 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 (

  5. Summary of Le Chatelier’s principle • Amounts of products and reactants: equilibrium shifts to compensate • N2  H2 E.g. N2 + 3H2 2 NH3 + 92 kJ N2 + 3H2 2NH3 + 92 kJ shift right N2 + 3H2 2NH3 + 92 kJ N2 + 3H2 2NH3 + 92 kJ N2 + 3H2 2NH3 + 92 kJ shift left N2 + 3H2 2NH3 + 92 kJ N2 + 3H2 2NH3 + 92 kJ Temperature: equilibrium shifts to compensate:  Heat N2 + 3H2 2NH3 + 92 kJ shift left N2 + 3H2 2NH3 + 92 kJ N2 + 3H2 2NH3 + 92 kJ Pressure (due to decreased volume): increase in pressure favors side with fewer molecules

  6. Part II. Equilibria involving sparingly soluble salts • Ag+ + CO3-2 Ag2CO3-2 • 2H+ + CO3-2 H2O + CO2

  7. Part II. Equilibria involving sparingly soluble salts • Ag+ + Cl- AgCl • Ag+ + 2NH3 Ag(NH3)2 + heat NH3+ H+ (NH4)

  8. Questions • Omit part 1 • Pg 256 omit part 3

  9. Lab 20- Acids & Bases Titration

  10. A. Neutralization • Chemical reaction between an acid and a base. • Products are a salt (ionic compound) and water.

  11. Neutralization ACID + BASE  SALT + WATER HCl + NaOH  NaCl + H2O strong strong neutral HC2H3O2 + NaOH  NaC2H3O2 + H2O weak strong basic • Salts can be neutral, acidic, or basic. • Neutralization does not mean pH = 7.

  12. standard solution unknown solution Titration • Titration • Analytical method in which a standard solution is used to determine the concentration of an unknown solution.

  13. B. Titration • Equivalence point (endpoint) • Point at which equal amounts of H3O+ and OH- have been added. • Determined by… • indicator color change • dramatic change in pH

  14. B. Titration moles H3O+ = moles OH- MVn = MVn M: Molarity V: volume n: # of H+ ions in the acid or OH- ions in the base

  15. B. Titration • 42.5 mL of 1.3MNaOHare required to neutralize 50.0 mL of KHP. Find the molarity of KHP. H3O+ M = ? V = 50.0 mL n = 1 OH- M = 1.3M V = 42.5 mL n = 1 MV = MV M(50.0mL) =(1.3M)(42.5mL) M = 1.11 M H2SO4

  16. Procedures • Data you need for part B • Molarity of NaOH • Trial 1 [ .115 M] • Trial 2 [.116 M] • Trial 3 [.117 M]

  17. Procedures Formulas M = mol/liters moles H3O+ = moles OH- MV (H3O+) = MV (OH-) % KHP = mass of KHP/ Mass of mixture

  18. 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

  19. A. Fill a Burett with standarizedNaOH • 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

  20. calculations • Determine moles of base used • Determine moles of KHP • Determine mass of KHp • Determine % of KHP in unknown

  21. Questions • 223 and 224 • Questions 1 -3

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