Notebook

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SOS Exam-AID
CHM 1311
Brought to you by Jeremy
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Topics in this Exam-AID 1

• The Fundamentals
• Balancing Chemical Equations
• Stoichiometry
• Composition by mass Determining Chemical
  Equations
• Trivial things profs might ask to trip you up
  (molality, mole fraction, w/w)
• Molecular Geometry

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Topics in this Exam-AID 2

• Thermochemistry
• Enthalpy
• Heats of Reaction and Heats of Formation
• Calorimetry
• Entropy
• Free Energy, ?G
• Equilibrium and Equilibrium Constants for
• Gases
• Solutions
• Acids/Bases
• Solubility

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Topics in this Exam-AID 3

• Electrochemistry
• Redox Reactions
• Cell potentials
• Kinetics
• Rate equations
• Rate-Determining Steps
• Quantum Numbers

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Balancing Chemical Equations 1

• Unbalanced
• Ca(OH)2 H3PO4 Ca3(PO4)2 H2O
• Balancing the atom which appears the fewest times
  on each side
• Then balance the other atoms
• It can help to make a table

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Balancing Chemical Equations 2

• Balanced
• 3 Ca(OH)2 2 H3PO4 ? Ca3(PO4)2 6 H2O
• It helps to know what the products of your
  reactions are!
• Unbalanced
• C6H5COOH O2 ? ??

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Balancing Chemical Equations 3

• C6H5COOH O2 ? H2O CO2
• Balance the atom that shows up the fewest times
  on each side

C6H5COOH O2 ? 3 H2O 7 CO2
LS RS C 7
7 H 6
6 O 4
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C6H5COOH 15/2 O2 ? 3 H2O 7 CO2
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Stoichiometry

• Ratios and Recipes
• 2 cups Baking Mix 1 Cup Chocolate chips ? 2
  (terrible) Chocolate Chip Cupcakes
• 2 moles H2 1 mole O2 ? 2 moles H2O
• 212 ratio

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Stoichiometry

• 0.005 mols H2 Y moles O2 ? Z moles H2O
• 2 moles H2 1 mole O2 ? 2 moles
  H2O
• Because the ratios are the same, you can divide
  the equations in order to figure out the number
  of moles you need

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Composition by Mass 1

• Given a chemical formula
• C6H12O6
• Pretend you have one mole of the molecule, and
  multiple moles of the component atoms
• 1 mole C6H12O6
• 6 moles C, 12 moles H, 6 moles O

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Composition by Mass 2

• How much do 6 moles of C weigh?
• 6 moles 12.011g/mol 72.066g
• How much do 12 moles of H weigh?
• How much do 6 moles of O weigh?

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Composition by mass 3

• How much would one mole of the substance weigh?
• The molar mass of C6H12O6 is 180.16g/mol
• How much do each of the components weigh?
• 6 moles of C weigh (6 12.011), or 72.066g/mol
• Take percentages
• 72.066/180.16 40.0

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Determining Chemical Formulas

• Steps
• Assume one mole of molecule
• Use molar ratios to determine how many moles of
  atoms you have
• Convert from moles to mass
• To determine formula, do the opposite
• Assume 100g, find mass of each element
• Convert from mass to moles
• Use the number of moles to create molar ratios to
  make your formula

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Determining Chemical Formulas

• You have a substance made of
• 40 Ca, 12 C, and 48 O, molar mass of 100.0869
  g/mol
• Step One Assume 100g of substance, find mass of
  each element
• 40 Ca 100g 40g
• 12 C 100g 12g
• 48 O 100g 48g

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Determining Chemical Formulas 2

• Step Two Convert from mass to moles
• Remember moles mass / molar mass
• 40g Ca / 40.078g/mol 0.998 mole Ca
• 12g C / 12.011g/mol 0.999 mole C
• 48g O ??? moles O

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Determining Chemical Formulas 3

• Step Three Use the number of moles to create
  molar ratios
• Take the smallest number of moles you have and
  divide into all the other numbers
• 0.998 mole Ca, 0.999 mole C, 3.00 moles O
  0.998 0.998 0.998
• 1 mole Ca, 1 mole C, 3 moles O
• If you don't use the smallest number...
• 0.998 mole Ca, 0.999 mole C, 3.00 moles O
  3 3 3
• 0.33 mole Ca? 0.33 mole C? 1 mole O?

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Determining Chemical Formulas 4

• Step 4 Use the simplest formula to find the true
  formula
• Find the molar mass of the simplest formula
• The molecular weight of the true compound will be
  provided
• Make it match by multiplying the number of atoms
  by an integer
• Why multiply?

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Determining Chemical Formulas 5

• You have an empirical formula CH2O, mm
  30.03g/mol
• The formula of the true compound is 60.06g/mol
• What is the true formula???
• 30.03 2 60.06
• CH2O 2 C2H4O2

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Trivial Things

• Molality
• Moles of Solute / Mass of Solvent (not
  solution)
• Mole Fraction A
• Moles of A / Moles of A Moles of B
• m/w (or w/w)
• (mass of solute / mass of solution) 100

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Wrap-Up!

• We discussed
• Balancing chemical equations
• Stoichiometry
• composition by mass
• Determining chemical formula
• Any questions?

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Molecular Geometry

• Sadly, molecular geometry is mostly memorization
• Especially the names of the configurations
• Key Tricks to drawing molecules with proper
  geometry
• Figure out the electron arrangement first
• Electrons repel, and a tetrahedron provides the
  most space for four pairs of electrons
• Memorize all the names though /

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Molecular Geometry
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Why do Reactions Happen?

• Thermochemistry and Kinetics!
• Thermochemistry can be broken into
• Enthalpy (?H) and Entropy (S)
• The interaction between the two form Free Energy
  (?G)
• Free Energy, Enthalpy and Entropy also affect
  equilibrium (K)

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Thermochemistry - Enthalpy

• Change in Enthalpy (?H)
• Change in Internal Energy (?U)
• Work done/by on the system(W)
• Most times, work 0 , ?H ?U, except for gases
• ?H ?U W
• If so, change in internal energy is
• ?H ?U q
• Where q is heat energy transferred

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All about q

• q is all about heat transfer
• If q is negative, heat is given off
• If q is positive, heat is absorbed
• Reactions that give off heat usually happen
• Reactions that take up heat usually don't
• How do you figure out what q is?

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Calorimetry

• Experimentally q mc ?T
• c tells you how much heat the material holds per
  gram
• Check the units of c, usually in J/gK but
  sometimes in J/molK
• If so, q nC ?T
• If you are given a calorimetry question, c will
  be given to you
• Except for water. It's 4.18 J/gK

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Hess's Law

• Finding ?Ho is a pain.
• Luckily, Hess's Law states you can find ?H using
  any combination of reactions as long as they add
  up to the reaction you want
• C(s) O2 ? CO2(g) ?H ??
• C(s) ½O2(g) ? CO(g) ?H -110.5
  kJ
• CO(g) ½O2(g) ? CO2(g) ?H -283.0 kJ
• Think of it like a building

Shamelessly ripped from UOttawa, Pell/Mayer 2009
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Hess's Law

• Taken one step further, you can also take the ?H
  of formation of the molecules in the equation
• ?H of formation is ?Hf
• ?Hf is the energy it takes to form the molecule
  from the constituent elements
• ?Hf of all products - ?Hf of all reactants ?H
  for reaction

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Hess's Law

• CH4(g) ?Hf -74.87 kJO2 (g)
  ?Hf 0 kJH2O(g)
  ?Hf -241.83 kJCO2(g)
  ?Hf-393.509 kJ
• CH4(g) 2O2(g) ? 2H2O(g) CO2(g)
• ?H?Hf (H2O(g) CO2(g)) - ?Hf (CH4(g) O2(g))

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?H Endothermic or Exothermic?

• When ?H is negative, the reaction is exothermic
• Heat is given off to the atmosphere
• Good for the reaction!
• When ?H is positive, the reaction is endothermic
• Heat is absorbed from the atmosphere
• Not so good.

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When does Work Matter?

• Remember, W ?PV
• When you have a gas that expands, it cools off
  and does work
• If you just measured the temperature change, you
  would get a false reading
• Therefore, ?H ?U q w
• q P?V

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Entropy

• Entropy is a measure of disorder in the universe
• Having more things will make them more disordered
• Having a molecule that can twist and turn makes
  it more disordered
• Gases are more disordered than solids

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Enthalpy, Entropy and Free Energy

• ?Go is free energy
• ?Go ?Ho - T?So
• ?So is entropy (J/molK)
• ?S and ?H don't change with T
• ?Go is for T 298K, standard conditions, 1 atm
• Just like ?H, negative is good!

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Finding ?Go, ?Ho, ?So

• If ?Go isn't given, you can calculate it from ?H
  and ?S
• You can also calculate it from ?Gof from a table
  of ?Go by using Hess's law

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Finding ?Go, ?Ho, ?So
?Hfo (kJ/mol) ?Gfo (kJ/mol) ?So (J/molK)
CO(g) -110.5 -137.2 197.7
Cl2(g) 0 0 223.1
COCl2(g) -218.8 -204.6 283.5
?Go ?Gfo (products reactants)
-204.6 137.2 -67.4 kJ

?Go ?Hfo (products reactants) - T ?So
(products - reactants) -218.8kJ
110.5kJ (298) (283.5-223.1-197.7)J/molK
-108.3 kJ 40915.4J -108.3 kJ
40.91 kJ -67.4 kJ
Pirated from Waterloo Chemistry Pages
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Finding ?Go, ?Ho, ?So

• H2(g) N2(g) ? 2NH3(g)
• ?Hf for NH3(g) is -46.1kJ/mol?S for H2(g) is
  130.7J/molK, for N2(g) is 191.56J/molK, and for
  NH3(g) is 192.77J/molK
• What is ?Go , assuming standard conditions?

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?Go and ?G?

• ?Go is great! And useless by itself
• It is only for T 298K
• ?G (no o) is the value for the actual condition
• ?Go lets you get ?G!
• ?Go ?Ho - T ?So
• ?G ?H - T ?S

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?G and Reaction Favourability

• If ?G negative, reaction likely to happen If
  ?G positive, reaction will not happen it might
  happen in the reverse direction
• By plugging values into ?G ?H - T ?S, you can
  find ?G 0
• At ?G 0, the reaction is in equilibrium

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Decomposition in Equilibrium

• 2AlCl3 (s) ? 2Al(s) 3 Cl2 (g) At what
  temperature will the decomposition of AlCl3(s) be
  in equilibrium?

Enthalpy (kJ/mol) Free Energy (kJ/mol) Entropy(J/molK)
Cl2 (g) 0 223.08 0
Al (s) 0 28.3 0
AlCl3 -706.63 109.29 -630.0
Cl2 (g) Chlorine gas 0 223.08 0
Al(s) Aluminum solid 0 28.3 0
AlCl3 (s) Aluminum Chloride -705.63 109.29
-630.0
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Decomposition in Equilibrium

• ?H ?Hf (products) ?Hf (reactants)
• 2 (-706.63) 0
• -1413.26 kJ/mol?S ?S (products) ?S
  (reactants) 2(-630.0) 0 -1260
  J/molK ?G ?H - T ?S 0 -1413.26 (T)
  1.26 T 1121 K

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Equilibrium Constant K

• K is the equilibrium constant it is temperature
  dependent
• Given balanced formula
• aA bB ? cC dD
• K Cc Dd
  
  Aa
  Bb
• In the case of gases Kp pCc pDd
  
  p Aa pBb
• K gt 1 is product favoured
• Klt 1 is reactant favoured

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K

• Keq /Kc K for equation (c for concentration)
• Ksp K for dissolved substances (sp for
  solubility product)
• Ka /Kb K for an acid / K for a base
• ALL Ks are calculated the same way! Do not
  include solids/pure liquids in the calculation.
• K Cc Dd
  
  Aa
  Bb
• Except for gases, which can be calculated
  with pressure

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Kp and Gases

• For gases, Kp pCc pDd
  
  
  p Aa pBb
• Pressures are just easier to use, especially
  since
• Total Pressure Sum of the individuals pressures
  of the gases
• Kp Kc (RT)n
• P RTn/V

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Examples of K

• H2SO4(aq) H2O(l) ? HSO4-(aq) H3O(aq)
• Ka HSO4- H3O / H2SO4
• 2SO2(g) O2(g) ? 2SO3(g)
• Kp pSO32 / pSO22 O2
• CaCO3(s) ? Ca2(aq) CO32-(aq)
• Ksp Ca CO3

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Put example here
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K, ?G, and Equilibrium

• When ?G 0, the reaction was at equilibrium
• How can this be related to K?
• ?G ?Go RT ln K
• ?Go -RT ln K
• ?Go -600 kJ/mol
• T 200K T
  1200K RlnK 3
  RlnK 0.5

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Q and Lechatalier's Principle

• Lechatalier's principle is like a seesaw
• Reactants ?
  Products
• Where ?G is experimental, while ?Go is for
  standard conditions
• Q is experimental, while K is for theoretical
  equilibrium

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Q

• If the experimental conditions thereoretical
  equilibrium, Q K
• I2(g) H2(g) ? 2HI(g) at some temperature/
  pressure
• For this T and P, Kc 1
• Therefore 1 HI2 / H2 I2
• You have 10 moles of HI, two moles of H, and two
  moles of I in the vessel. Which way will the
  reaction go?

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Q

• Q HI2 / H2 I2
• 102 / 2 2
• Q 100/4 25
• if Q K, the reaction would be at equilibrium
• Q gt K, the reaction will go towards reactants
• Q lt K, the reaction will go towards products
• ?G ?Go RT ln Q

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Q in a Ksp Problem

• PbCl2 is highly insoluble in water. The Ksp of
  PbCl2 is 1.6 10-5. PbCl2 is added to 0.1L of
  water at 298K, 1atm pressure. a) How much PbCl2
  dissolves?b) What is ?Go?
• 0.058g of NaCl are added to the solution.
• c) Find Q.
• d) What is ?G right when the NaCl is added?
• e) How much PbCl2 is left dissolved?

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Q in a Ksp Problem

• a) How much PbCl2 dissolves?Ksp 1.6 10-5

Keep in mind that ICE tables are for
concentration! Therefore the concentration of the
ions is x and 2x, not actual amount.
PbCl2(s) ? Pb2(aq) 2Cl-(aq)I /
0 0C / x
2xE / x
2x
Ksp (x) (2x)2 1.6 10-5 4x3x
0.0159 Therefore 0.00159 moles of PbCl2 dissolve
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Q in a Ksp Problem

• b) What is ?Go??G ?Go RT ln Q?G 0At
  equilibrium right now, Q Ksp 1.6 10-5T
  298 KR 8.314Therefore, ?Go - RT ln 1.6 10
  -5
• ?Go 27359 kJ/mol

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Q in a Ksp Problem

• 0.0058g of NaCl are added to the solution.
• c) Find Q.mmNaCl 58.44g/molTherefore 0.001
  mols are added to the solution

PbCl2(s) ? Pb2(aq)
2Cl-(aq)I / 0.00159
(0.003180.001) 0.1
0.1
Q 0.0159 (0.0418)2 2.8 10-5
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Q in a Ksp Problem

• d) What is ?G right when the NaCl is added?

?G ?Go RT ln QQ 2.8 10-5?Go 27359
kJ/mol?G 27359 8.314 298 ln 2.8 10-5?G
1385 kJ/mol
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Q in a Ksp Problem

• e) How much PbCl2 is left dissolved?
• Ksp 1.6 10-5

• PbCl2(s) ? Pb2(aq)
  2Cl-(aq)I / 0.0159
  0.0418C / -x
  -2xE / 0.0159- x
  0.0418 -2xKsp (0.0159-x)
  (0.0418-2x)2

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. . .

• Ksp (0.0159-x) (0.0418-2x)2
• Is not exactly solvable with the math we know
• Using a simplification
• Ax3 Bx2 Cx D Ax3 D
• X 0.0057
• 0.0159- x 0.0418 -2x 0.0102 M
  Pb2 0.0304M Cl-
• 0.00102 Pb

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Van't Hoff Clausius-Clapeyron

• Van't Hoff
• Clausius-Clapeyron

Note H is kJ/mol, while S is J/molK
This equation allows you to find one K if given
another K
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Using Van't Hoff and Clausius-Clapeyron

• You have the reaction
• CaCO3 ? CaO CO2
• At 400o C, K 3.610-6
• At 500o C, K 2.2 10-4
• What is ?Ho, ?So, ?Go, and K at 500o C?

Taken from UOttawa, Prof St-Amant,, past
midterms
60
Acid-Base Equilibrium

• How do Ka, Kb, Kw, pKa, pKb, pH, and pOH all
  relate?
• H2O(l) H2O(l) ? H3O(aq) OH-(aq)
• The K of this reaction is 1.010-14
• Hence Kw 1.010-14

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Acid-Base Equilibrium

• Whenever you add acid/base to a solution, it
  reacts with the water to form H3O or OH-
• In neutral water, H3O is 10-7 OH- is 10-7
• pH - log H3O 7
• pOH - log OH- 7
• pH pOH 14
• Adding acid increases H3O , so pH decreases
• Adding base decreases H3O, so pH increases

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Acid-Base Equilibrium

• In the same vein, pKa pKb 14
• HA H2O ? H3O A- Ka Some
• A- H2O ? HA OH- Kb Some
• H2O H2O ? H3O OH- Kw 10-14
• Ka Kb Kw

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Henderson-Hasselbalch Equation

• Instead of using an ICE table, you can simplify
  with
• This assumes that acid gt 100Ka

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A Buffer Example

• Calculate the pH of 1.00L of a buffer system
  composed of 0.90M acetic acid, CH3COOH and 0.60M
  sodium acetate, NaCH3COO
• a) initially
• b) after the addition of 0.1L of 0.5M NaOH
• The Kb of acetate is 5.610-10

Taken from UOttawa, Prof St-Amant,, past
midterms
66
A Buffer Example

• First, find Ka Ka Kb 10-14
• Ka 10-14/5.610-10 1.8 10-5
• We can calculate the pH of the solution before
  the addition of NaOH, using the
  Henderson-Hasselbalch equation, since acid gt Ka

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A Buffer Example

• pH pKa log A- / HA - log 1.8
  10-5 log ( 0.6 / 0.9)
• 4.57

CH3COOH H2O ? CH3COO H3O
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A Buffer Example

• b) After the addition of 0.1L of 5M NaOH? (How
  many moles of NaOH is that?)

CH3COOH H2O ? CH3COO H3O
I (0.9 - 0.5)/1.1 / (0.60.5)/1.1
0 C -x /
x x E 0.36 -x
/ 1x x

Step 1 Neutralize acid/baseStep 2 Add
conjugate base/acid to other sideStep 3
Recalculate concentrationsStep 4 Finish your
ace table
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A Buffer Example
CH3COOH H2O ? CH3COO H3O

• E 0.36 -x / 1x
  x Ka 1.8 10-5 (1x) (x) /
  (0.36-x)0 6.5 10-6 -1.000018 x -x2
• X 0.0000065 molspH H3O - log 0.0000065
• 5.19

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Electrochem Redox Reactions

• Redox reactions are composed of
• Reduction and Oxidation
• LEO (goes) GERLoss of Electrons is
  OxidationGain of Electrons is Reduction
• Reduction must be paired with Oxidation
• When do you know if something is
  oxidized/reduced?

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Oxidation States

• 1. Check the Oxidation states!
• Unbalanced
• C6H5NO2 Sn ? C6H5NH2 Sn2

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Redox Reactions

• 2. Begin to write out half-reactions
• Sn ? Sn2 2e-
• C6H5NO2 6e- ? C6H5NH2
• 3. Balance atoms besides O and H
• 4. Balance charge with H-
• Sn ? Sn2 2e-
• C6H5NO2 6e- 6H ? C6H5NH2

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Redox Reactions

• 5. Balance LS and RS with H2O
• Sn ? Sn2 2e-
• C6H5NO2 6e- 6H ? C6H5NH2 2H2O
• 6. Add your reactions together so e- cancels out
• 3Sn ? 3Sn2 6e-
• C6H5NO2 6e- 6H ? C6H5NH2 2H2O

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Redox Example

• Balance the following equation
• S2O62-(aq) HClO2(aq) ? SO42-(aq) Cl2(g)

Taken from UOttawa, Prof St-Amant,, past
midterms
75
Cell Potentials

• Oxidation potential Reduction Potential
• Li(aq) e- ? Li(s)
  Eo - 3.04 VCu2(aq) 2e- ? Cu(s)
  Eo 0.34 V
• 2 Li (s) ? 2Li(aq) 2e-
  Eo 3.04 V Cu2(aq) 2e- ?
  Cu(s)
  Eo 0.34 V
  3.34 V
• ?Go - nFEocell ?G
  - nFEcell

They do NOT change when you multiply the molar
coefficients!
76
Kinetics

• Thermodynamics is not the only thing that makes
  reactions go! Kinetics play a large role too
• Kinetics tell you how fast reactions go
• Kinetically, some reactions proceed so slowly,
  they don't proceed at all
• Given A B ? C
• Rate law k Am Bn
• Rate laws are determined experimentally!
• The order of the reaction is the sum of the
  coefficients

77
Reaction Rates

• All reaction rates must be determined
  experimentally
• You did this in one of your labs
• You can measure the initial rate of change of the
  concentration of the reactants/problems
• Experiment 1 2 3 4
• Ao 0.100 0.200 0.200 0.100
• Bo 0.100 0.100 0.300 0.100
• Co 0.100 0.100 0.100 0.400
• Rate 0.100 0.800 7.200 0.400

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Integrated Rate Law

• The whole point of the integrated rate law is to
  turn the rate law into something that can be
  modeled with a linear equation
• From the linear equation, you can easily
  determine K, and figure out other points on the
  line
• If 0th order A -kt AoIf 1st order ln
  A -kt ln AoIf 2nd order 1/A kt
  1/Ao

79
Integrated Rate Laws
Attempt to graph H2O2 vs Time
Time H2O2 ln H2O2 1/H2O2
0 1 0 1
120 0.91 -0.09 1.1
300 0.78 -0.25 1.28
600 0.59 -0.53 1.69
1200 0.37 -0.99 2.7
1800 0.22 -1.51 4.55
2400 0.13 -2.04 7.69
3000 0.082 -2.5 12.2
6000 0.05 -3 20
80
Integrated Rate Laws
Attempt to graph 1/H2O2 vs Time
Time H2O2 ln H2O2 1/H2O2
0 1 0 1
120 0.91 -0.09 1.1
300 0.78 -0.25 1.28
600 0.59 -0.53 1.69
1200 0.37 -0.99 2.7
1800 0.22 -1.51 4.55
2400 0.13 -2.04 7.69
3000 0.082 -2.5 12.2
6000 0.05 -3 20
81
Integrated Rate Laws
Attempt to graph ln H2O2 vs Time
Time H2O2 ln H2O2 1/H2O2
0 1 0 1
120 0.91 -0.09 1.1
300 0.78 -0.25 1.28
600 0.59 -0.53 1.69
1200 0.37 -0.99 2.7
1800 0.22 -1.51 4.55
2400 0.13 -2.04 7.69
3000 0.082 -2.5 12.2
6000 0.05 -3 20
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RDS

• Chemical reactions are actually a series of
  complex reaction reactions involving many
  intermediates
• The speed of the overall reaction is limited by
  the formation of the slowest intermediate
• Think about an assembly line
• The rate law reflects the interactions of this
  slowest step
• If rate k A, molecule A must be in the
  slowest step, and no other molecules
• If rate k A2, two molecules of A are involved
  in the slowest step, and no others

83
RDS and Catalysts

• There are many reasons the RDS might be slow
• Usually, because it is not as thermodynamically
  favourable
• This is reflected in the activation energy of the
  step
• How much energy to stay in a reactive state
• Catalysts change the reactive state itself, or
  make it easier to get into the reactive state, so
  the reaction can proceed

84
Quick Quantum Number Overview

• Your Quantum numbers are n, l, mn, and ml
• n Principle, what shell you are in
• S P D F
• l Angular, n-1
• mn Magnetic, - L to L
• ms Spin, 1/2 or -1/2

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Quick Quantum Number Overview
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Quick Quantum Number Overview