1. Dynamic Equilibrium Notes p. 3-4

2. 2 NO 2 (g)  N 2 O 4 (g) Initial concentration N 2 O 4 : 0.40 M Final concentration N 2 O 4 : 0.035 M Initial concentration NO 2 : 0.00 M Final concentration NO 2 : 0.01 M Initial concentration N 2 O 4 : 0.00 M Final concentration N 2 O 4 : 0.035 M Initial concentration NO 2 : 0.08 M Final concentration NO 2 : 0.01 M

3. Equilibrium is reached when concentration stops changing.

4. Equilibrium is reached when reaction rate stops changing.

5. Red rover

9. Chemical Equilibrium Reaction: NH 3 + HCl  NH 4 Cl Watch this demonstration (Click on the link when the PPT is in “play” mode)this demonstration Describe what is happening.

10. Reversible reaction A chemical reaction in which the products can react to re- form the reactants Symbol used to represent reversible reactions: ⇌ or 

11. Example Example: NH 3 + HCl ⇌ NH 4 Cl Forward reaction: NH 3 + HCl  NH 4 Cl Backward reaction: NH 4 Cl  NH 3 + HCl

12. equilibrium Chemical equilibrium exists when: 1. The rate of the forward reaction (  ) is equal to the rate of the reverse reaction (  ) 2. The concentration of the products and reactants remains unchanged

13. LeChatelier’s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress. Henry Le Chatelier

14. In other words, When you add something to a system at equilibrium, the system shifts in such a way as to use up what you’ve added. Add to the reactants, shifts towards the products. Add to the products, shifts towards the reactants.

16. In other words, When you take something away from a system at equilibrium, the system shifts in such a way as to replace what you’ve taken away. Remove from the reactants, shifts towards the reactants. Remove from the products, shifts towards the products.

17. Easy way TO REMEMBER If you ADD, then move AWAY If you TAKE, then move TOWARDS

19. LeChatelier Example #1 A closed container of ice and water at equilibrium. The temperature is raised. Ice + Energy  Water The equilibrium of the system shifts to the _______ to use up the added energy. right

20. LeChatelier Example #2 A closed container of N 2 O 4 and NO 2 at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy  2 NO 2 (g) The equilibrium of the system shifts to the _______ to use up the added NO 2. left

21. LeChatelier Example #3 A closed container of water and its vapor at equilibrium. Vapor is removed from the system. water + Energy  vapor The equilibrium of the system shifts to the _______ to replace the vapor. N 2 O 4 (g) + Energy  2 NO 2 (g) * Mistake on the notes – should be THIS reaction, not N 2 O 4 (g) + Energy  2 NO 2 (g) right

22. Equilibrium Click to watch an episode of Crash Course about equilibrium It is very clear and very helpful.

23. Solutions and Concentrations Notes Page 5

24. SOLUTIONS A homogenous mixture of two or more substances

25. 2 PARTS OF A SOLUTION SOLUTE A solute is the dissolved substance in a solution. Sugar in soda Carbon dioxide in soda Salt in salt water SOLVENT A solvent is the dissolving medium in a solution. Water in salt water Water in soda

28. Molarity = concentration Amount of substance (mol) Volume of mixture (L) Unit = mol/L or M 1.0 M = 1 molar 12.0 M = 12 molar

29. solUTE over solVENT

30. Example 1 1.5 mol of NaCl are dissolved in 1 L of water. What is the molarity of the solution? 1.5 mol NaCl = 1.5 mol/L or 1.5 M 1 L H 2 O

31. Example 2 100 g of NaCl are dissolved in 1 L of water. What is the molarity of the solution? 100 g NaCl |1 mol NaCl = 1.71 mol 58.44 g NaCl 1.71 mol NaCl = 1.71 M 1 L

32. Example 3 100 g of NaCl are dissolved in 250 mL of water. What is the molarity of the solution? 100 g NaCl |1 mol NaCl = 1.71mol 58.44 g NaCl 250 mL / 10 3 mL = 0.25 L 1.71 mol NaCl = 6.84 M 0.25 L

33. Example 4 How many moles of NaCl are in 500 mL of a 0.25 M solution? Basic formula: M = mol L Convert: 500 mL = 0.5 L 0.25 M = mol = 0.5 L x 0.25 M = mol 0.5 L = 0.125 mol NaCl

34. Example 5 How many grams of NaCl are in 500 mL of a 2.25 M solution? Basic formula: M = mol L Convert: 500 mL = 0.5 L 2.25 M = mol = 0.5 L x 2.25 M = mol 0.5 L = 1.125 mol NaCl 1.125 mol NaCl | 58.44 g NaCl = 65.75 g NaCl 1 mol