Post on 02-Jan-2016
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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