SubjectiveQuestion # 1

Kinetics

A. Increase the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Increase temperature

Increase [HCl]

Add a Catalyst

Increase Surface Area of CaCO3

B. Measure the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

B. Measure the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

mass

B. Measure the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

mass [HCl]

B. Measure the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

mass [HCl] volume

B. Measure the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

mass [HCl] volume [CaCl2]

B. Measure the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

mass [HCl] volume [CaCl2] can’t

B. Measure the rate CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

mass [HCl] volume [CaCl2] 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

 CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Mass (g) 82.07 81.84 81.71 81.66 81.64 81.63

Time (s) 0 15 30 45 60 75

1. Calculate the rate in grams HCl/min

C. Calculate the Rate

 CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Mass (g) 82.07 81.84 81.71 81.66 81.64 81.63

Time (s) 0 15 30 45 60 75

1. Calculate the rate in grams HCl/min

C. Calculate the Rate  CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Mass (g) 82.07 81.84 81.71 81.66 81.64 81.63

Time (s) 0 15 30 45 60 75

1. Calculate the rate in grams HCl/min

(82.07 - 81.63) g CO2

75 s

C. Calculate the Rate  CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Mass (g) 82.07 81.84 81.71 81.66 81.64 81.63

Time (s) 0 15 30 45 60 75

1. Calculate the rate in grams HCl/min

(82.07 - 81.63) g CO2 x 1 mole 75 s 44.0 g

C. Calculate the Rate  CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Mass (g) 82.07 81.84 81.71 81.66 81.64 81.63

Time (s) 0 15 30 45 60 75

1. Calculate the rate in grams HCl/min

(82.07 - 81.63) g CO2 x 1 mole x 2 mole HCl 75 s 44.0 g 1 mole CO2

C. Calculate the Rate  CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Mass (g) 82.07 81.84 81.71 81.66 81.64 81.63

Time (s) 0 15 30 45 60 75

1. Calculate the rate in grams HCl/min

(82.07 - 81.63) g CO2 x 1 mole x 2 mole HCl x 36.5 g 75 s 44.0 g 1 mole CO2 1 mole

C. Calculate the Rate  CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Mass (g) 82.07 81.84 81.71 81.66 81.64 81.63

Time (s) 0 15 30 45 60 75

1. Calculate the rate in grams HCl/min

(82.07 - 81.63) g CO2 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 CO2 1 mole 1 min

D. Collision Theory

More CollisionsHarder CollisionsLower Ea

SubjectiveQuestion # 2Equilibrium

When 0.800 moles of SO2 and 0.800 moles of O2 are placed

into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq

value.

2SO2 (g) + 1O2 (g) ⇋ 2SO3 (g)

When 0.800 moles of SO2 and 0.800 moles of O2 are placed

into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq

value.

2SO2 (g) + 1O2 (g) ⇋ 2SO3 (g)

I

C

E

When 0.800 moles of SO2 and 0.800 moles of O2 are placed

into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq

value.

2SO2 (g) + 1O2 (g) ⇋ 2SO3 (g)

I 0.400 M 0.400M 0

C

E 0.300 M

When 0.800 moles of SO2 and 0.800 moles of O2 are placed

into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq

value.

2SO2 (g) + 1O2 (g) ⇋ 2SO3 (g)

I 0.400 M 0.400M 0

C -0.300 M -0.150 M +0.300 M

E 0.300 M

When 0.800 moles of SO2 and 0.800 moles of O2 are placed

into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq

value.

2SO2 (g) + 1O2 (g) ⇋ 2SO3 (g)

I 0.400 M 0.400M 0

C -0.300 M -0.150 M +0.300 M

E 0.100 M 0.250 M 0.300 M

When 0.800 moles of SO2 and 0.800 moles of O2 are placed

into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq

value.

2SO2 (g) + 1O2 (g) ⇋ 2SO3 (g)

I 0.400 M 0.400M 0

C -0.300 M -0.150 M +0.300 M

E 0.100 M 0.250 M 0.300 M

Equilibrium concentrations go in the equilibrium equation!

Keq = [SO3]2

[SO2]2[O2]

When 0.800 moles of SO2 and 0.800 moles of O2 are placed

into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq

value.

2SO2 (g) + 1O2 (g) ⇋ 2SO3 (g)

I 0.400 M 0.400M 0

C -0.300 M -0.150 M +0.300 M

E 0.100 M 0.250 M 0.300 M

Equilibrium concentrations go in the equilibrium equation!

Keq = [SO3]2 (0.3)2

=[SO2]

2[O2] (0.1)2(0.25)

If 4.00 moles of CO, 4.00 moles H2O, 6.00 moles CO2, and

6.00 moles H2 are placed in a 2.00 L container at 670 oC,

Keq = 1.0CO(g) + H2O(g) ⇄ CO2(g) + H2(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 M 2.00 M 3.00 M 3.00 M

Calculate a Kt

Kt = (3)(3) = 2.25(2)(2)

Not in equilibrium Shifts left!

Do an ICE chart

CO(g) + H2O(g) ⇄ CO2(g) + H2(g)

I 2.00 M 2.00 M 3.00 M 3.00 MC +x +x -x -xE 2.00 + x 2.00 + x 3.00 - x 3.00 - x

Keq = (3 - x)2 = 1.0(2 + x)2

Square root3 - x = 1.02 + x 3 - x = 2 + x1 = 2xx = 0.50 M

[CO2] = [H2] = 3.00 - 0.50 = 2.50 M

[CO] = [H2O] = 2.00 + 0.50 = 2.50 M

SubjectiveQuestion # 3

Solubility

200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M

NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl-

200 0.10 M 300 0.20 M500 500

0.040 M 0.12 M

TIP = [Pb2+][Cl-]2

TIP = [0.040][0.12] 2

= 5.8 x 10-4

Ksp = 1.2 x 10-5 TIP > Ksp ppt forms

Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 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

BaCl2(s) ⇄ Ba2+ + 2Cl-

4.5 x 10-10 M 9.0 x 10-10 M

0.50 L x 4.5 x 10-10 mole x 208.3 g = 4.7 x 10-8 g1 L mole

PbCl2(s) ⇌ Pb2+ + 2Cl-

Ksp = 4s3

SubjectiveQuestion # 4 to 6

Acids

HCl Strong Acid HCl H+ + Cl-

0.10 M 0.10 MpH = -Log[H+] = 1.0 No ICE

HF Weak Acid HF ⇌ H+ + F-

I 0.10 M 0 0C x x xE 0.10 - x x x

small Kax2 = 3.5 x 10-4 x = 0.005916 M0.10

pH = -Log[0.005916] = 2.23

NaOH Strong Base Ba(OH)2 Ba2+ + 2OH-

0.20 M 0.40 MpOH = -Log[OH-] = 0.40 No ICE

NH3 Weak Base NH3 + H2O ⇌ NH4+ + OH-

I 0.20 M 0 0C x x xE 0.20 - x x x

small Kbx2 = Kb = Kw = 1.0 x 10-14 = 1.786 x 10-5

0.20 Ka 5.6 x 10-10

x = 0.001890 MpOH = -Log[0.001890] = 2.73 pH = 11.27

SubjectiveQuestion 7 & 8

Redox

Review of Cells

Electrochemical Electrolytic Is a power supply Requires power supplySpontaneous (+ ve) Nonspontaneous(-ve)Makes electricity Makes chemicals Reduction is highest on Chart Reduction 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(NO3)2 II Cu, CuSO4

 

Anode: Reaction: 

Cathode: Reaction:

E0 =

 

Complete the Chart Electrochemical Cell: Zn, Zn(NO3)2 II Cu, CuSO4

 

Anode: Zn Reaction: 

Cathode: Cu Reaction:

E0 =

Higher on reduction Chart

 

Complete the Chart Electrochemical Cell: Zn, Zn(NO3)2 II Cu, CuSO4

 

Anode: Zn Reaction: Zn → Zn2+ + 2e- 0.76 v 

Cathode: Cu Reaction:

E0 =

Higher on reduction Chart

 

Complete the Chart Electrochemical Cell: Zn, Zn(NO3)2 II Cu, CuSO4

 

Anode: Zn Reaction: Zn(s) → Zn2+ + 2e- 0.76 v

 

Cathode: Cu Reaction: Cu2+ + 2e- → Cu(s) 0.34 v

E0 =  Higher on reduction Chart

 

Complete the Chart Electrochemical Cell: Zn, Zn(NO3)2 II Cu, CuSO4

 

Anode: Zn Reaction: Zn(s) → Zn2+ + 2e- 0.76 v 

Cathode: Cu Reaction: Cu2+ + 2e- → Cu(s) 0.34 v

E0 = 1.10 v   Higher on reduction Chart

 

   

Electrolytic Cell: Molten AlCl3

 Anode: Reaction: Cathode: Reaction: 

 

 

   

Electrolytic Cell: Molten AlCl3 Al3+ Cl-

 Anode: C Reaction: Cathode: C Reaction: 

 

 

   

Electrolytic Cell: Molten AlCl3 Al3+ Cl-

 Anode: C Reaction: Cathode: C Reaction: 

 Put the vowels together: AnodeAnionOxidation

 

   

Electrolytic Cell: Molten AlCl3 Al3+ Cl-

 Anode: C Reaction: 2Cl- → Cl2 + 2e- -1.36 v

 Cathode: C Reaction: 

 Put the vowels together: AnodeAnionOxidation

Oxidation of Anion

 

   

Electrolytic Cell: Molten AlCl3

 Anode: C Reaction: 2Cl- → Cl2 + 2e- -1.36 v

 Cathode: C Reaction:  

 Put the consonants together: CathodeCationReduction

 

   

Electrolytic Cell: Molten AlCl3

 Anode: C Reaction: 2Cl- → Cl2 + 2e- -1.36 v

 Cathode: C Reaction: Al3+ + 3e- → Al -1.66 v 

 Put the consonants together: CathodeCationReduction

Reduction of Cation

 

   

Electrolytic Cell: Molten AlCl3

 Anode: C Reaction: 2Cl- → Cl2 + 2e- -1.36 v

 Cathode: C Reaction: Al3+ + 3e- → Al -1.66 v E0 = -3.02 v 

 

   

Electrolytic Cell: 1M AlCl3

 Anode: Reaction: Cathode: Reaction: E0 = MTV =

 

   

Electrolytic Cell: 1M AlCl3

 Anode: C Reaction: H2O → 1/2O2 + 2H+ + 2e-

 Cathode: C Reaction: E0 = MTV =

 

   

Electrolytic Cell: 1M AlCl3

 Anode: C Reaction: H2O → 1/2O2 + 2H+ + 2e-

 Cathode: C Reaction: 2H2O + 2e- → H2 + 2OH-

 E0 = MTV =

 

   

Electrolytic Cell: 1M AlCl3

 Anode: C Reaction: H2O → 1/2O2 + 2H+ + 2e-

 Cathode: C Reaction: 2H2O + 2e- → H2 + 2OH-

 E0 = -3.02 v MTV =

 

   

Electrolytic Cell: 1M AlCl3

 Anode: C Reaction: H2O → 1/2O2 + 2H+ + 2e-

 Cathode: C Reaction: 2H2O + 2e- → H2 + 2OH-

 E0 = -3.02 v MTV = +3.02 v