1. Equilibrium Chemistry

2. Equilibrium A + B  AB We may think that all reactions change all reactants to products, or the reaction has gone to completion But in reality, products may start to change back into reactants; the reaction is reversible. A + B ↔ AB A double sided arrow means the reaction is reversible

3. CO 2 + H 2 O  C 6 H 12 O 6 + O 2 O 2 + C 6 H 12 O 6  CO 2 + H 2 O A reversible reaction! Photosynthesis Respiration Reactants are forming products and products have enough energy to form reactants

4. Equilibrium Equilibrium is the state where the forward and reverse reactions balance because they are occurring at equal rates. NO OVERALL CHANGE! Old guy steps up as fast as escalator moves down --- they happen at the same rate

5. The Law of Chemical Equilibrium –the point in the reaction where a ratio of reactant and product concentration has a constant value, K eq Keq < 1reactants favored Keq = 1neither is favored Keq > 1products favored Concentration of reactants is greater than that of products Concentration of products is greater than that of reactants

6. If Keq = 2.4, which is more favored? A. products B. reactants C. neither is favored D. not enough information

7. Homogeneous equilibrium –States of all compounds are the same H 2 (g) + I 2 (g) ↔ 2HI (g) Heterogeneous equilibrium –States of compounds are different CaCO 3 (g) ↔ CaO (s) + CO 2 (g) Remember we only pay attention to the GASES and AQUEOUS SOLUTIONS when we calculate K eq ! In the solid state, isn’t dissolved, doesn’t drive the reaction rate

8. Writing Equilibrium Constants We only use gases and aqueous solutions when we write the equilibrium constant!! The coefficients become superscripts! Coefficients represent number of moles

9. What is the correct K eq ? N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

10. What is the correct the K eq ? 2 NbCl 4 (g) ↔ NbCl 3 (g) + NbCl 5 (g)

11. What is the correct Keq? H 2 O (l) ↔ H 2 O (g)

12. Calculating Equilibrium Constants The equilibrium concentrations of the reactants and products may be used to calculate K eq Example: 2NO (g) + Br 2(g) ↔ 2NOBr (g) [NOBr] = 0.0474 mol/L [NO] = 0.312 mol/L [Br 2 ] = 0.259 mol/L Step 1: Write the equilibrium constant expression (use gases and aqueous solutions only!) Step 2: Substitute the known values into the equation and solve for K eq

13. Chemical Equilibrium Problems I Write the equilibrium constant expression (K eq ) for these homogeneous equations. 1. N 2 O 4 (g) ↔ 2 NO 2 (g)

14. 2. CO (g) + 3H 2(g) ↔ CH 4 (g) + H 2 O (g) 3. 2 H 2 S (g) ↔ 2 H 2 (g) + S 2 (g)

15. Write the equilibrium constant expression (K eq ) for these heterogeneous equations. 4. C 10 H 8 (s) ↔ C 10 H 8 (g) 5. CaCO 3 (s) ↔ CaO (s) + CO 2 (g)

16. 6. C (s) + H 2 O (g) ↔ CO (g) + H 2 (g) 7. FeO (s) + CO (g) ↔ Fe (s) + CO 2(g)

17. 1. Calculate the Keq for the following equation using the data: [N 2 O 4 ] = 0.0613 mol/L [NO 2 ] = 0.0627 mol/L N 2 O 4 (g) ↔ 2NO 2 (g) A. 0.98 B. 15.59 C. 0.064 D. 1.02 E. 2.05

18. 2. [CO] = 0.0613 mol/L [H 2 ] = 0.1839 mol/L [CH 4 ]=0.0387 mol/L [H 2 O]=0.0387 mol/L CO (g) + H 2 (g) ↔ CH 4 (g) + H 2 O (g) A. 0.25 B. 0.72 C. 0.04 D. 3.93 E. 0.02

19. 3. [H 2 ]=1.5 mol/L [N 2 ]=2.0 mol/L [NH 3 ]=1.8 mol/L 3H 2 (g) + N 2(g) ↔ 2NH 3 (g) A. 0.48 B. 0.24 C. 2.5 D. 0.40

20. 4. If the Keq = 0.48, we know that the equilibrium favors… a. reactants b. products c. neither

21. 5. [Mg]=2 mol/L[HCl]=3 mol/L [MgCl]=6 mol/L [H 2 ]=3 mol/L 2 Mg (s) + 2 HCl (g) ↔ 2 MgCl (g) + H 2 (g) A. 3 B. 12

22. 6. Since Keq = 3, we know that the equilibrium favors a. reactants b. products c. neither

23. 7. [H 2 ]=0.52 mol/L [I 2 ]=0.23 mol/L [HI]=1.7 mol/L H 2 (g) + I 2 (g) ↔ 2HI (g) A. 202 B. 0.04 C. 24.16

24. 8. Since Keq= 24.16, we know that the equilibrium favors a. reactants b. products c. neither

25. LeChatelier’s Principle

26. Concentration Change A + B ↔ C + D Add more A A and B are more likely to collide, B goes down (consumed), C and D are increased (more produced) Add more A and B C and D are increased (more produced)  Moves in the direction to relieve the stress Add more C, Consume extra D, increase the amount of A and B produced

27. Temperature Change Stress the reaction, the reaction will move in the way to relieve the stress A + B + heat ↔ C + D + E Take away heat, the reaction will shift right Reaction rate of A+ B slows down when heat is taken away, C+D+E speeds up

28. Pressure Change N 2(g) + 3H 2(g) ↔ 2NH 3(g) Increase pressure, forcing molecules closer together, relieve stress by shifting to the side with fewer molecules Lower the number of molecules, we will relieve the pressure ↔

29. Significance of the Reaction Quotient  If Q = K, the system is at equilibrium  If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved  If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

30. LeChatelier’s Principle states that if a stress is applied to a chemical reaction in equilibrium, the system will move in the direction (forward or reverse) that relieves the stress. There are 3 stresses that can be applied to a reversible chemical reaction in equilibrium: 1. Changes in Concentration 2. Changes in Volume/Pressure 3. Changes in Temperature Translated: The system undergoes a temporary shift in order to restore equilibrium. LeChatelier’s Principle

31. EX. A closed container of N 2 O 4 and NO 2 is at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium. left Concentration If concentration of a compound is increased (compound is added) to a reaction at equilibrium, the reaction will shift in the opposite direction of the added compound Add = Away

32. Concentration Continued A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. right If concentration of a compound is decreased (compound is removed) from a reaction at equilibrium, the reaction will shift in the same direction of the removed compound Remove = Towards

33. Changes in Concentration If the concentration of a reactant or product is increased in a reaction in equilibrium, the equilibrium will shift in the opposite direction of the added substance. CO + 3H 2 CH 4 + H 2 O In this reaction, if more CO is added, the equilibrium point of reaction will shift to the right, and there will be more CH 4 and H 2 O produced. If the reactant or product of a chemical reaction is removed, the equilibrium will shift in the direction of the substance that was removed. CO + 3H 2 CH 4 + H 2 O In this reaction, if the concentration of the CO is decreased, the equilibrium point of the reaction will shift to the left to replace the CO that was removed and increase the amount of H 2.

34. Changes in Volume A closed container of N 2 O 4 and NO 2 is at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. left Decreasing volume causes and increase in pressure which causes more effective collisions  volume =  pressure  volume =  pressure If volume is decreased, equilibrium shifts to the side with fewest moles Decrease = Least

35. Changes in Volume Continued If volume is increased, equilibrium shifts to the side with most moles Increase = Most 2H 2 + O 2  2H 2 O If number of moles are the same, no shift

36. Changes in Volume If the volume of a container is decreased, the molecules are pushed closer together. This also increases the pressure, and increases the effective collisions of the molecules. Since pressure is a function of the number of gas particles present, the increase in effective collisions produces fewer gas particles then the stress on the reversible reaction is relieved. High pressure, low volume Low pressure, high volume

37. Changes in Volume Continued… Since a mole of a substance represents a molecule, and the number of moles of reactant is compared to the number of moles of product. The number of molecules on each side of the equation can quickly be determined. In this equation: CO + 3H 2 CH 4 + H 2 O 1 mole + 3 moles 1 mole + 1 mole 4 molecules 2 molecules If the volume is decreased on this equation, the equilibrium point of the reaction will move to the right which is the direction of the fewest moles (least molecules).

38. Changes in Temperature All chemical reactions are either exothermic or endothermic. In a reversible reaction, if the forward reaction is exothermic, then the reverse reaction will be endothermic. Exothermic and endothermic reactions are identified by the Δ H value at the end of the reaction. Exothermic reactions are always negative and endothermic reactions are always positive because: ΔH = H reactants – H products If the reactants release heat during a reaction, the heat ends up on the product side and becomes a negative number.

39. Changes in Temperature EX: A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right If heat is added, reaction shifts away If heat is removed, reaction shifts toward Example 2: ∆H = -206.5 kJ CO (g) + 3H 2 (g)  CH 4 (g) + H 2 O (g) If heat is added, reaction will shift left If heat is removed, reaction will shift right

40. LeChatelier’s Principle Practice H 2 + I 2 ↔ HI Add more I 2 Shift: left or right CO + H 2 ↔ CH 4 + H 2 O Take away H 2 O Shift: left or right H 2 + I 2 ↔ HI Add more HI Shift: left or right CO + H 2 ↔ CH 4 + H 2 O Add more CH 4 Shift: left or right