1. Subjective Question # 1 Kinetics

2. A.Increase the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l) Increase temperature Increase [HCl] Add a Catalyst Increase Surface Area of CaCO 3

3. B.Measure the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l)

4. B.Measure the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l) mass

5. B.Measure the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l) mass[HCl]

6. B.Measure the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l) mass[HCl]volume

7. B.Measure the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l) mass[HCl]volume[CaCl 2 ]

8. B.Measure the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l) mass[HCl]volume[CaCl 2 ]can’t

9. B.Measure the rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) +H 2 O (l) mass[HCl]volume[CaCl 2 ]can’t over time Measure the decrease in mass of an open container Measure the increase in pressure of an closed container

10. C.Calculate the Rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) + H 2 O (l) Mass (g) 82.0781.8481.7181.6681.6481.63 Time (s) 01530456075 1.Calculate the rate in grams HCl/min

11. C.Calculate the Rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) + H 2 O (l) Mass (g) 82.0781.8481.7181.6681.6481.63 Time (s) 01530456075 1.Calculate the rate in grams HCl/min

12. C.Calculate the Rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) + H 2 O (l) Mass (g) 82.0781.8481.7181.6681.6481.63 Time (s) 01530456075 1.Calculate the rate in grams HCl/min (82.07 - 81.63) g CO 2 75 s

13. C.Calculate the Rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) + H 2 O (l) Mass (g) 82.0781.8481.7181.6681.6481.63 Time (s) 01530456075 1.Calculate the rate in grams HCl/min (82.07 - 81.63) g CO 2 x 1 mole 75 s 44.0 g

14. C.Calculate the Rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) + H 2 O (l) Mass (g) 82.0781.8481.7181.6681.6481.63 Time (s) 01530456075 1.Calculate the rate in grams HCl/min (82.07 - 81.63) g CO 2 x 1 mole x 2 mole HCl 75 s 44.0 g 1 mole CO 2

15. C.Calculate the Rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) + H 2 O (l) Mass (g) 82.0781.8481.7181.6681.6481.63 Time (s) 01530456075 1.Calculate the rate in grams HCl/min (82.07 - 81.63) g CO 2 x 1 mole x 2 mole HCl x 36.5 g 75 s 44.0 g 1 mole CO 2 1 mole

16. C.Calculate the Rate CaCO 3(s) + 2HCl (aq) → CO 2(g) + CaCl 2(aq) + H 2 O (l) Mass (g) 82.0781.8481.7181.6681.6481.63 Time (s) 01530456075 1.Calculate the rate in grams HCl/min (82.07 - 81.63) g CO 2 x 1 mole x 2 mole HCl x 36.5 g x 60 s = 0.56 g/min 75 s 44.0 g 1 mole CO 2 1 mole 1 min

17. D.Collision Theory More Collisions Harder Collisions Lower Ea

18. Subjective Question # 2 Equilibrium

19. When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g)

20. When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g) I C E

21. When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C E0.300 M

22. When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.300 M

23. When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.100 M0.250 M0.300 M

24. When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.100 M0.250 M0.300 M Equilibrium concentrations go in the equilibrium equation! Keq= [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

25. When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.100 M0.250 M0.300 M Equilibrium concentrations go in the equilibrium equation! Keq= [SO 3 ] 2 (0.3) 2 = [SO 2 ] 2 [O 2 ](0.1) 2 (0.25)

26. If 4.00 moles of CO, 4.00 moles H 2 O, 6.00 moles CO 2, and 6.00 moles H 2 are placed in a 2.00 L container at 670 o C, Keq = 1.0 CO (g) + H 2 O (g) ⇄ CO 2(g) + H 2(g) Is the system at equilibrium? If not, how will it shift in order to get there? Calculate all equilibrium concentrations. Get Molarities 2.00 M2.00 M3.00 M3.00 M Calculate a Kt Kt=(3)(3)=2.25 (2)(2) Not in equilibriumShifts left!

27. Do an ICE chart CO (g) + H 2 O (g) ⇄ CO 2(g) + H 2(g) I 2.00 M2.00 M3.00 M3.00 M C+x+x-x-x E2.00 + x2.00 + x3.00 - x3.00 - x Keq=(3 - x) 2 =1.0 (2 + x) 2 Square root 3 - x =1.0 2 + x 3 - x = 2 + x 1=2x x = 0.50 M [CO 2 ]=[H 2 ]=3.00 - 0.50 = 2.50 M [CO]=[H 2 O]=2.00 + 0.50 = 2.50 M

28. Subjective Question # 3 Solubility

29. 200.0 mL 0.10 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl 2(s) ⇌ Pb 2+ +2Cl - 2000.10 M3000.20 M 500500 0.040 M0.12 M TIP=[Pb 2+ ][Cl - ] 2 TIP=[0.040][0.12] 2 =5.8 x 10 -4 Ksp=1.2 x 10 -5 TIP > Ksp ppt forms

30. Calculate the maximum number of grams BaCl 2 that will dissolve in 0.50 L of 0.20 M AgNO 3 solution. AgCl (s) ⇄ Ag + + Cl - 0.20 M Ksp=[Ag + ][Cl - ] 1.8 x 10 -10 = [0.20][Cl - ] [Cl - ] =9.0 x 10 -10 M BaCl 2(s) ⇄ Ba 2+ + 2Cl - 4.5 x 10 -10 M9.0 x 10 -10 M 0.50 L x 4.5 x 10 -10 molex 208.3 g = 4.7 x 10 -8 g 1 L mole

31. PbCl 2(s) ⇌ Pb 2+ +2Cl - Ksp = 4s 3

32. Subjective Question # 4 to 6 Acids

33. HClStrong Acid HCl  H + + Cl -0.10 M pH = -Log[H + ] =1.0 No ICE HFWeak Acid HF ⇌ H + + F - I0.10 M00 CxxxCxxx E0.10 - xxx small Ka x 2 =3.5 x 10 -4 x =0.005916 M 0.10 pH = -Log[0.005916] =2.23

34. NaOH Strong Base Ba(OH) 2  Ba 2+ + 2OH - 0.20 M 0.40 M pOH = -Log[OH - ] =0.40 No ICE NH 3 Weak Base NH 3 + H 2 O ⇌ NH 4 + + OH - I0.20 M 0 0 Cx x x E0.20 - x x x small Kb x 2 =Kb = Kw= 1.0 x 10 -14 = 1.786 x 10 -5 0.20 Ka 5.6 x 10 -10 x =0.001890 M pOH = -Log[0.001890] =2.73 pH=11.27

35. Subjective Question 7 & 8 Redox

36. Review of Cells ElectrochemicalElectrolytic Is a power supplyRequires power supply Spontaneous (+ ve)Nonspontaneous(-ve) Makes electricityMakes chemicals Reduction is highest on ChartReduction is the –ve

37. For all cells: Cations migrate to the cathode, which is the site of reduction. Anions migrate to the anode, which is the site of oxidation. Electrons travel through the wire from anode to cathode.

38. Complete the Chart Electrochemical Cell: Zn, Zn(NO 3 ) 2 II Cu, CuSO 4 Anode: Reaction: Cathode: Reaction: E 0 =

39. Complete the Chart Electrochemical Cell: Zn, Zn(NO 3 ) 2 II Cu, CuSO 4 Anode:Zn Reaction: Cathode:Cu Reaction: E 0 = Higher on reduction Chart

40. Complete the Chart Electrochemical Cell: Zn, Zn(NO 3 ) 2 II Cu, CuSO 4 Anode:Zn Reaction: Zn → Zn 2+ + 2e - 0.76 v Cathode:Cu Reaction: E 0 = Higher on reduction Chart

41. Complete the Chart Electrochemical Cell: Zn, Zn(NO 3 ) 2 II Cu, CuSO 4 Anode:Zn Reaction: Zn (s) → Zn 2+ + 2e - 0.76 v Cathode:Cu Reaction: Cu 2+ + 2e - → Cu (s) 0.34 v E 0 = Higher on reduction Chart

42. Complete the Chart Electrochemical Cell: Zn, Zn(NO 3 ) 2 II Cu, CuSO 4 Anode:Zn Reaction: Zn (s) → Zn 2+ + 2e - 0.76 v Cathode:Cu Reaction: Cu 2+ + 2e - → Cu (s) 0.34 v E 0 =1.10 v Higher on reduction Chart

43. Electrolytic Cell: Molten AlCl 3 Anode:Reaction: Cathode:Reaction:

44. Electrolytic Cell: Molten AlCl 3 Al 3+ Cl - Anode:CReaction: Cathode:CReaction:

45. Electrolytic Cell: Molten AlCl 3 Al 3+ Cl - Anode:CReaction: Cathode:CReaction: Put the vowels together: Anode Anion Oxidation

46. Electrolytic Cell: Molten AlCl 3 Al 3+ Cl - Anode:CReaction: 2Cl - → Cl 2 + 2e - -1.36 v Cathode:CReaction: Put the vowels together: Anode Anion Oxidation Oxidation of Anion

47. Electrolytic Cell: Molten AlCl 3 Anode:CReaction: 2Cl - → Cl 2 + 2e - -1.36 v Cathode:CReaction: Put the consonants together: Cathode Cation Reduction

48. Electrolytic Cell: Molten AlCl 3 Anode:CReaction: 2Cl - → Cl 2 + 2e - -1.36 v Cathode:CReaction: Al 3+ + 3e - → Al -1.66 v Put the consonants together: Cathode Cation Reduction Reduction of Cation

49. Electrolytic Cell: Molten AlCl 3 Anode:CReaction: 2Cl - → Cl 2 + 2e - -1.36 v Cathode:CReaction: Al 3+ + 3e - → Al -1.66 v E 0 = -3.02 v

50. Electrolytic Cell: 1M AlCl 3 Anode:Reaction: Cathode:Reaction: E 0 = MTV=

51. Electrolytic Cell: 1M AlCl 3 Anode:CReaction:H 2 O → 1 / 2 O 2 + 2H + + 2e - Cathode:CReaction: E 0 = MTV=

52. Electrolytic Cell: 1M AlCl 3 Anode:CReaction:H 2 O → 1 / 2 O 2 + 2H + + 2e - Cathode:CReaction:2H 2 O + 2e - → H 2 + 2OH - E 0 = MTV=

53. Electrolytic Cell: 1M AlCl 3 Anode:CReaction:H 2 O → 1 / 2 O 2 + 2H + + 2e - Cathode:CReaction:2H 2 O + 2e - → H 2 + 2OH - E 0 = -3.02 v MTV=

54. Electrolytic Cell: 1M AlCl 3 Anode:CReaction:H 2 O → 1 / 2 O 2 + 2H + + 2e - Cathode:CReaction:2H 2 O + 2e - → H 2 + 2OH - E 0 = -3.02 v MTV=+3.02 v