Subjective Question # 1 Kinetics

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

D.Collision Theory More Collisions Harder Collisions Lower Ea

Subjective Question # 2 Equilibrium

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

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

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

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

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

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

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

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!

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 E x x x x Keq=(3 - x) 2 =1.0 (2 + x) 2 Square root 3 - x = x 3 - x = 2 + x 1=2x x = 0.50 M [CO 2 ]=[H 2 ]= = 2.50 M [CO]=[H 2 O]= = 2.50 M

Subjective Question # 3 Solubility

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

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 M Ksp=[Ag + ][Cl - ] 1.8 x = [0.20][Cl - ] [Cl - ] =9.0 x M BaCl 2(s) ⇄ Ba Cl x M9.0 x M 0.50 L x 4.5 x molex g = 4.7 x g 1 L mole

PbCl 2(s) ⇌ Pb 2+ +2Cl - Ksp = 4s 3

Subjective Question # 4 to 6 Acids

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

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

Subjective Question 7 & 8 Redox

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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=

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 = v MTV=

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 = v MTV=+3.02 v