Subjective

Question # 1

Kinetics

A. Increase the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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Increase temperature

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Increase [HCl]

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Add a Catalyst

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Increase Surface Area of CaCO₃

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B. Measure the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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B. Measure the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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mass

B. Measure the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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mass [HCl]

B. Measure the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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mass [HCl] volume

B. Measure the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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mass [HCl] volume [CaCl₂]

B. Measure the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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mass [HCl] volume [CaCl₂] can’t

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B. Measure the rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

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mass [HCl] volume [CaCl₂] can’t

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over time

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Measure the decrease in mass of an open container

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Measure the increase in pressure of an closed container

C. Calculate the Rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

1. Calculate the rate in grams HCl/min

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C. Calculate the Rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

1. Calculate the rate in grams HCl/min

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C. Calculate the Rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

1. Calculate the rate in grams HCl/min

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(82.07 - 81.63) g CO₂

75 s

C. Calculate the Rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

1. Calculate the rate in grams HCl/min

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(82.07 - 81.63) g CO₂ x 1 mole

75 s 44.0 g

C. Calculate the Rate

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

1. Calculate the rate in grams HCl/min

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(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

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

1. Calculate the rate in grams HCl/min

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(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

CaCO_3(s) + 2HCl_(aq) → CO_2(g) + CaCl_2(aq) + H₂O_(l)

1. Calculate the rate in grams HCl/min

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(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

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More Collisions

Harder Collisions

Lower Ea

Subjective

Question # 2

Equilibrium

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.

2SO_{2 (g)} + 1O_{2 (g)} ⇋ 2SO_{3 (g)}

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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.

2SO_{2 (g)} + 1O_{2 (g)} ⇋ 2SO_{3 (g)}

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I

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C

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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.

2SO_{2 (g)} + 1O_{2 (g)} ⇋ 2SO_{3 (g)}

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I 0.400 M 0.400M 0

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C

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E 0.300 M

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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.

2SO_{2 (g)} + 1O_{2 (g)} ⇋ 2SO_{3 (g)}

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I 0.400 M 0.400M 0

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C -0.300 M -0.150 M +0.300 M

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E 0.300 M

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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.

2SO_{2 (g)} + 1O_{2 (g)} ⇋ 2SO_{3 (g)}

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I 0.400 M 0.400M 0

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C -0.300 M -0.150 M +0.300 M

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E 0.100 M 0.250 M 0.300 M

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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.

2SO_{2 (g)} + 1O_{2 (g)} ⇋ 2SO_{3 (g)}

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I 0.400 M 0.400M 0

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C -0.300 M -0.150 M +0.300 M

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E 0.100 M 0.250 M 0.300 M

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Equilibrium concentrations go in the equilibrium equation!

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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.

2SO_{2 (g)} + 1O_{2 (g)} ⇋ 2SO_{3 (g)}

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I 0.400 M 0.400M 0

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C -0.300 M -0.150 M +0.300 M

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E 0.100 M 0.250 M 0.300 M

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Equilibrium concentrations go in the equilibrium equation!

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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 = 1.0

CO_(g) + H₂O_(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.

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Get Molarities

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2.00 M 2.00 M 3.00 M 3.00 M

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Calculate a Kt

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Kt = (3)(3) = 2.25

(2)(2)

Not in equilibrium Shifts left!

Do an ICE chart

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CO_(g) + H₂O_(g) ⇄ CO_2(g) + H_2(g)

I 2.00 M 2.00 M 3.00 M 3.00 M

C +x +x -x -x

E 2.00 + x 2.00 + x 3.00 - x 3.00 - 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

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[CO₂] = [H₂] = 3.00 - 0.50 = 2.50 M

[CO] = [H₂O] = 2.00 + 0.50 = 2.50 M

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Subjective

Question # 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?

PbCl_2(s) ⇌ Pb²⁺ + 2Cl^-

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200 0.10 M 300 0.20 M 500 500

0.040 M 0.12 M

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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

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

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AgCl_(s) ⇄ Ag⁺ + Cl^-

0.20 M

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Ksp = [Ag⁺][Cl^-]

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1.8 x 10^-10 = [0.20][Cl^-]

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[Cl^-] = 9.0 x 10^-10 M

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BaCl_2(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 g

1 L mole

PbCl_2(s) ⇌ Pb²⁺ + 2Cl^-

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Ksp = 4s³

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Subjective

Question # 4 to 6

Acids

HCl Strong Acid HCl → H⁺ + Cl^-

0.10 M 0.10 M

pH = -Log[H⁺] = 1.0 No ICE

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HF Weak Acid HF ⇌ H⁺ + F^-

I 0.10 M 0 0

C x x x

E 0.10 - x x x

small Ka

x² = 3.5 x 10^-4 x = 0.005916 M

0.10

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pH = -Log[0.005916] = 2.23

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

0.20 M 0.40 M

pOH = -Log[OH^-] = 0.40 No ICE

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NH₃ Weak Base NH₃ + H₂O ⇌ NH₄⁺ + OH^-

I 0.20 M 0 0

C x x x

E 0.20 - x x x

small Kb

x² = 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

Subjective

Question 7 & 8

Redox

Review of Cells

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Electrochemical Electrolytic

Is a power supply Requires power supply

Spontaneous (+ ve) Nonspontaneous(-ve)

Makes electricity Makes chemicals

Reduction is highest on Chart Reduction is the –ve

For all cells:

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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.

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Complete the Chart

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

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Anode: Reaction:

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Cathode: Reaction:

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E⁰ =

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Complete the Chart

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

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Anode: Zn Reaction:

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Cathode: Cu Reaction:

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E⁰ =

Higher on reduction Chart

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Complete the Chart

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

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

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Cathode: Cu Reaction:

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E⁰ =

Higher on reduction Chart

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Complete the Chart

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

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

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

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E⁰ =

Higher on reduction Chart

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Complete the Chart

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

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

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

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E⁰ = 1.10 v

Higher on reduction Chart

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�

Electrolytic Cell: Molten AlCl₃

Anode: Reaction:

Cathode: Reaction:

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�

Electrolytic Cell: Molten AlCl₃ Al³⁺ Cl^-

Anode: C Reaction:

Cathode: C Reaction:

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�

Electrolytic Cell: Molten AlCl₃ Al³⁺ Cl^-

Anode: C Reaction:

Cathode: C Reaction:

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Put the vowels together:

Anode

Anion

Oxidation

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�

Electrolytic Cell: Molten AlCl₃ Al³⁺ Cl^-

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

Cathode: C Reaction:

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Put the vowels together:

Anode

Anion

Oxidation

Oxidation of Anion

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�

Electrolytic Cell: Molten AlCl₃

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

Cathode: C Reaction:

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Put the consonants together:

Cathode

Cation

Reduction

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�

Electrolytic Cell: Molten AlCl₃

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

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

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Put the consonants together:

Cathode

Cation

Reduction

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Reduction of Cation

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�

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

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�

Electrolytic Cell: 1M AlCl₃

Anode: Reaction:

Cathode: Reaction:

E⁰ =

MTV =

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�

Electrolytic Cell: 1M AlCl₃

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

Cathode: C Reaction:

E⁰ =

MTV =

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�

Electrolytic Cell: 1M AlCl₃

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

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

E⁰ =

MTV =

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�

Electrolytic Cell: 1M AlCl₃

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

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

E⁰ = -3.02 v

MTV =

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�

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

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