AP Chemistry Super Saturday Review I tried to include as much review material as possible in this session.
Download ReportTranscript AP Chemistry Super Saturday Review I tried to include as much review material as possible in this session.
Slide 1
AP Chemistry
Super Saturday
Review
I tried to include as much review material
as possible in this session. Work on practice
tests and review the Bozeman videos for
other material.
Slide 2
AP Chemistry
Super Saturday
Review
5 Essentials
1. Know the basics – writing formulas,
writing and balancing equations,
dimensional analysis, atomic theory,
acid-base theories (Arrhenius and Bronsted
Lowry), VSEPR, kinetic molecular theory,
and collision theory.
Slide 3
5 Essentials
2. Atomic and Molecular Structures
Atomic structures (like electron configurations)
will help explain relationships on the periodic
table which explains many physical and chemical
properties. Molecular structures involves Lewis
structures and VSEPR to determine shapes
which describes polarities thus describing
intermolecular forces which describes many
physical properties.
Slide 4
5 Essentials
3. Stoichiometric Calculations
Basic stoichiometry, limiting reactants,
titration calculations, and empirical formulas.
4. Principles of chemical kinetics, equilibrium,
And thermodynamics
Kinetics- describes the speed in which
substances react
Equilibrium – used to determine the extent of
a reaction or the composition at equilibrium
Thermodynamics – explains why chemical reactions
happen in terms of kinetic and potential energies
Slide 5
5 Essentials
5. Representation and Interpretation
Be able to draw what is happening at the
molecular level and read and interpret graphs
and data tables
Slide 6
Net Ionic Equations
Graphic: Wikimedia Commons User Tubife
Slide 7
Solubility Rules – AP Chemistry
All sodium, potassium, ammonium, and
nitrate salts are soluble in water.
Memorization of other “solubility rules”
is beyond the scope of this course and
the AP Exam.
What dissociates (“breaks apart”) –
Aqueous solutions of the following:
• All strong acids (HCl, HBr, HI, HNO3,
H2SO4, and HClO4)
• Strong bases (group I and II hydroxides)
• Soluble salts
Slide 8
Write the net ionic for the following:
1. Solutions of lead nitrate and potassium
chloride are mixed
2. Solutions of sulfuric acid and
potassium hydroxide are mixed.
3. Solid sodium hydroxide is mixed with
acetic acid
Slide 9
Big Idea #6:Chemical Equilibrium
2NO2(g) 2NO(g) + O2(g)
Sketch a graph of change in concentration
vs. Time for the reaction above
Slide 10
2NO2(g) 2NO(g) + O2(g)
Be able to explain the variance in slope
Slide 11
Law of Mass Action
For the reaction:
jA + kB lC + mD
l
K
[C ] [ D ]
m
j
k
[ A] [B ]
Where K is the equilibrium constant,
and is unitless
Slide 12
Product Favored Equilibrium
Large values for K signify the reaction is
“product favored”
When equilibrium is achieved, most
reactant has been converted to product
Slide 13
Reactant Favored Equilibrium
Small values for K signify the reaction is
“reactant favored”
When equilibrium is achieved, very little
reactant has been converted to product
Slide 14
Writing an Equilibrium Expression
Write the equilibrium expression for the
reaction:
2NO2(g) 2NO(g) + O2(g)
K = ???
2
K
[ NO ] [ O 2 ]
[ NO 2 ]
2
Slide 15
Conclusions about Equilibrium Expressions
The equilibrium expression for a reaction
is the reciprocal for a reaction written in
reverse
2NO2(g) 2NO(g) + O2(g)
2
K
[ NO ] [ O 2 ]
[ NO 2 ]
2
2NO(g) + O2(g) 2NO2(g)
K
'
1
K
[ NO 2 ]
2
2
[ NO ] [ O 2 ]
Slide 16
Conclusions about Equilibrium Expressions
When the balanced equation for a reaction
is multiplied by a factor n, the equilibrium
expression for the new reaction is the
original expression, raised to the nth power.
2NO2(g) 2NO(g) + O2(g)
2
K
[ NO ] [ O 2 ]
[ NO 2 ]
2
NO2(g) NO(g) + ½O2(g)
1
1
K K
1
2
[ NO ][ O 2 ] 2
[ NO 2 ]
Slide 17
If the equilibrium constant for A + B
2C
2A + 2B is
A) 0.584
B) 4.81
C) 0.416
D) 23.1
E) 0.208
Answer: D
C is 0.208 then the equilibrium constant for
Slide 18
Equilibrium Expressions Involving Pressure
For the gas phase reaction:
3H2(g) + N2(g) 2NH3(g)
KP
P
NH
2
3
3
( PN 2 )( PH 2 )
PNH 3 , PN 2 , PH 2 are equilibriu m partial
K p K ( RT )
n
pressures
Slide 19
Heterogeneous Equilibria
The position of a heterogeneous equilibrium
does not depend on the amounts of pure solids
or liquids present
Write the equilibrium expression for
the reaction:
PCl5(s) PCl3(l) + Cl2(g)
Pure
solid
Pure
liquid
K [ Cl 2 ]
K p PCl 2
Slide 20
The Reaction Quotient
For some time, t, when the system is not at
equilibrium, the reaction quotient, Q takes the
place of K, the equilibrium constant, in the
law of mass action.
jA + kB lC + mD
l
Q
[C ] [ D ]
m
j
k
[ A] [ B ]
Slide 21
Significance of the Reaction Quotient
If Q = K, the system is at equilibrium
If Q > K, the system shifts to the left,
consuming products and forming reactants
until equilibrium is achieved
If Q < K, the system shifts to the right,
consuming reactants and forming products
until equilibrium is achieved
Slide 22
If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a
one-liter container, which direction would the reaction
initially proceed?
A)
To the left.
B)
To the right.
C)
The above mixture is the equilibrium mixture.
D)
Cannot tell from the information given.
Slide 23
LeChatelier’s Principle
When a system at
equilibrium is placed under
stress, the system will
undergo a change in such
a way as to relieve that
stress and restore a
state of equilibrium.
Henry Le Chatelier
Slide 24
Le Chatelier Translated:
When you take something away from a
system at equilibrium, the system shifts
in such a way as to replace some of what
you’ve taken away.
When you add something to a system at
equilibrium, the system shifts in such a
way as to use up some of what you’ve
added.
Slide 25
•Do FRQ #1
Slide 26
Acid
Equilibrium
and pH
Søren Sørensen
Slide 27
Acid/Base Definitions
Arrhenius Model
Acids produce hydrogen ions in aqueous
solutions
Bases produce hydroxide ions in
aqueous solutions
Bronsted-Lowry Model
Acids are proton donors
Bases are proton acceptors
Slide 28
Acid Dissociation
HA
Acid
H+ +
AProton
Conjugate
base
Ka
[ H ][ A ]
[ HA ]
Alternately, H+ may be written in its
hydrated form, H3O+ (hydronium ion)
Slide 29
Dissociation Constants: Strong Acids
Acid
Formula
Perchloric
Hydriodic
Hydrobromic
Hydrochloric
Nitric
Sulfuric
HClO4
HI
HBr
HCl
HNO3
H2SO4
Hydronium ion
H3O+
Conjugate
Base
ClO4-
Ka
NO3HSO4-
Very large
Very large
Very large
Very large
Very large
Very large
H2O
1.0
IBrCl-
Slide 30
Dissociation Constants: Weak Acids
Acid
Iodic
Oxalic
Sulfurous
Phosphoric
Citric
Nitrous
Formula
Conjugate
Base
Ka
HIO3
IO3-
1.7 x 10-1
H2SO3
HSO3-
1.5 x 10-2
H2C6H5O7-
7.1 x 10-4
H2C2O4
H3PO4
H3C6H5O7
HNO2
HC2O4-
5.9 x 10-2
H2PO4-
7.5 x 10-3
NO2-
4.6 x 10-4
F-
3.5 x 10-4
Hydrofluoric
HF
Formic
HCOOH
HCOO-
1.8 x 10-4
Benzoic
C6H5COOH
C6H5COO-
6.5 x 10-5
H2CO3
HCO3-
4.3 x 10-7
Acetic
Carbonic
Hypochlorous
Hydrocyanic
CH3COOH
HClO
HCN
CH3COO-
1.8 x 10-5
ClO-
3.0 x 10-8
CN-
4.9 x 10-10
Slide 31
Reaction of Weak Bases with Water
The base reacts with water, producing its
conjugate acid and hydroxide ion:
CH3NH2 + H2O CH3NH3+ + OH-
Kb = 4.38 x 10-4
K b 4.38 x 10
4
[ C H 3 N H 3 ][ O H ]
[C H 3 N H 2 ]
Slide 32
Kb for Some Common Weak Bases
Many students struggle with identifying weak
bases and their conjugate acids.What patterns
do you see that may help you?
Formula
Conjugate
Acid
Kb
NH3
NH4+
1.8 x 10-5
Methylamine
CH3NH2
CH3NH3+
4.38 x 10-4
Ethylamine
C2H5NH2
C2H5NH3+
5.6 x 10-4
Diethylamine
(C2H5)2NH
(C2H5)2NH2+
1.3 x 10-3
Triethylamine
(C2H5)3N
(C2H5)3NH+
4.0 x 10-4
Hydroxylamine
HONH2
HONH3+
1.1 x 10-8
Base
Ammonia
Hydrazine
H2NNH2
H2NNH3+
3.0 x 10-6
Aniline
C6H5NH2
C6H5NH3+
3.8 x 10-10
Pyridine
C5H5N
C5H5NH+
1.7 x 10-9
Slide 33
Reaction of Weak Bases with Water
The generic reaction for a base reacting
with water, producing its conjugate acid and
hydroxide ion:
B + H2O BH+ + OH
Kb
[ B H ][ O H ]
[B]
(Yes, all weak bases do this – DO NOT
try to make this complicated!)
Ex. Write the reaction of ammonia
with water
Slide 34
Self-Ionization of Water
H2O + H2O
H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7
Kw is a constant at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14
Slide 35
Calculating pH, pOH
pH = -log10(H3O+)
pOH = -log10(OH-)
Relationship between pH and pOH
pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH
Slide 36
Calculate the pH of a
0.1M HCl solution?
(answer =1)
Slide 37
A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M
solution of acetic acid,
-5 ?
HC
H
O
,
K
=
1.8
x
10
2 3 2
a
(answer=4.52)
Slide 38
A Weak Base Equilibrium Problem
What is the pH of a 0.50 M
solution of ammonia, NH3,
Kb = 1.8 x 10-5 ? (answer=9.48)
Slide 39
Acid-Base Properties of Salts
To determine if a salt is acidic or
basic, determine the stronger parent.
Examples:
KCl
NH4Cl
NaC2H3O2
NaCl
KNO3
Slide 40
Acid-Base Properties of Salts
If both parents are weak:
Type of Salt
Examples Comment
Cation is the
conjugate acid of a
weak base, anion is
conjugate base of a
weak acid
NH4C2H3O2 Cation is acidic,
NH4CN
Anion is basic
pH of
solution
See below
IF Ka for the acidic ion is greater than Kb for
the basic ion, the solution is acidic
IF Kb for the basic ion is greater than Ka for
the acidic ion, the solution is basic
IF Kb for the basic ion is equal to Ka for the
acidic ion, the solution is neutral
Slide 41
Buffered Solutions
A solution that resists a change in
pH when either hydroxide ions or
protons are added.
Buffered solutions contain either:
A weak acid and its salt
A weak base and its salt
Slide 42
Acid/Salt Buffering Pairs
The salt will contain the anion of the acid,
and the cation of a strong base (NaOH, KOH)
Weak Acid
Formula
of the acid
Hydrofluoric
HF
Formic
HCOOH
Benzoic
C6H5COOH
Acetic
Carbonic
Propanoic
Hydrocyanic
CH3COOH
H2CO3
HC3H5O2
HCN
Example of a salt of the
weak acid
KF – Potassium fluoride
KHCOO – Potassium formate
NaC6H5COO – Sodium benzoate
NaH3COO – Sodium acetate
NaHCO3 - Sodium bicarbonate
NaC3H5O2 - Sodium propanoate
KCN - potassium cyanide
Slide 43
Base/Salt Buffering Pairs
The salt will contain the cation of the base,
and the anion of a strong acid (HCl, HNO3)
Formula of
the base
Example of a salt of the weak
acid
NH3
NH4Cl - ammonium chloride
Methylamine
CH3NH2
CH3NH2Cl – methylammonium chloride
Ethylamine
C2H5NH2
C2H5NH3NO3 - ethylammonium nitrate
Aniline
C6H5NH2
C6H5NH3Cl – aniline hydrochloride
Base
Ammonia
Pyridine
C5H5N
C5H5NHCl – pyridine hydrochloride
Slide 44
Titration of an Unbuffered Solution
13
12
11
10
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 45
Titration of a Buffered Solution
13
12
11
A solution that is
0.10 M CH3COOH and
0.10 M NaCH3COO is
titrated with
0.10 M NaOH
10
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 46
Comparing Results
Gra ph
14
12
10
Buffered
pH
8
6
4
Unbuffered
2
0
0
5
10
15
20
25
mL 0.10 M NaOH
30
35
40
45
Slide 47
Henderson-Hasselbalch Equation
[ A ]
pK a log
pH pK a log
[ HA ]
[ base
[ acid
[ BH ]
pK b log
pOH pK b log
[
B
]
[ acid
[ base
]
]
]
]
This is an exceptionally powerful tool that can be
used in your problem solving.
Slide 48
Title:
13
12
11
10
9
Endpoint is above
pH 7
pH
8
7
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 49
Title:
13
12
11
10
9
pH
8
7
Endpoint is at
pH 7
A solution that is
0.10 M HCl is
titrated with
0.10 M NaOH
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 50
Title:
13
12
A solution that is
0.10 M NaOH is
titrated with
0.10 M HCl
11
10
9
pH
8
7
Endpoint is at
pH 7
It is important to
recognize that
titration curves are
not always
increasing from left
to right.
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters HCl (0.10 M)
30.00
35.00
40.00
45.00
Slide 51
Title:
13
12
11
10
9
pH
8
7
6
5
Endpoint is below
pH 7
4
A solution that is
0.10 M HCl is
titrated with
0.10 M NH3
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NH3 (0.10 M)
30.00
35.00
40.00
45.00
Slide 52
Solubility
Equilibria
Lead (II) iodide precipitates when potassium
iodide is mixed with lead (II) nitrate.
Graphic: Wikimedia Commons user PRHaney
Slide 53
Solving Solubility Problems
For the salt AgI at 25C, Ksp = 1.5 x 10-16
Answer = solubility of AgI in mol/L = 1.2 x 10-8 M
Slide 54
Solving Solubility Problems
For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5
Answer = solubility of PbCl2 in mol/L = 1.6 x 10-2 M
Slide 55
Solving Solubility with a Common Ion
For the salt AgI at 25C, Ksp = 1.5 x 10-16
What is its solubility in 0.05 M NaI?
Answer = solubility of AgI in mol/L = 3.0 x 10-15 M
Slide 56
Big Idea #5:
Spontaneity, Entropy
and Free Energy
Slide 57
G = H - TS
Spontaneity, Entropy
and Free Energy
Slide 58
Spontaneous Processes and Entropy
First Law
• “Energy can neither be created nor
destroyed"
• The energy of the universe is constant
Spontaneous Processes
• Processes that occur without outside
intervention
• Spontaneous processes may be fast or
slow
– Many forms of combustion are fast
– Conversion of diamond to graphite is
slow
Slide 59
Which of the following reactions
is spontaneous?
• H2(g) + I2(g) ↔ 2HI
• Br2 + Cl2 ↔ 2BrCl
• HF + H2O ↔ F- + H30+
Kc=49
Kc=6.9
Kc=6.8x10-4
Slide 60
Second Law of
Thermodynamics
"In any spontaneous process there is
always an increase in the entropy of
the universe"
Ssolid < Sliquid << Sgas
Slide 61
Calculating Free Energy
Method #1
For reactions at constant temperature:
0
G
=
0
H
-
0
TS
Slide 62
Calculating Free Energy: Method #2
An adaptation of Hess's Law:
Cdiamond(s) + O2(g) CO2(g)
G0 = -397 kJ
Cgraphite(s) + O2(g) CO2(g)
G0 = -394 kJ
Cdiamond(s) + O2(g) CO2(g)
G0 = -397 kJ
CO2(g) Cgraphite(s) + O2(g)
G0 = +394 kJ
Cdiamond(s) Cgraphite(s)
G0 = -3 kJ
Slide 63
Calculating Free Energy
Method #3
Using standard free energy of formation (Gf0):
G
0
n
p
G
0
f ( p ro d u cts )
n
r
G
0
f ( reactan ts )
Gf0 of an element in its standard state is zero
Slide 64
Free Energy and Equilibrium
Equilibrium point occurs at the
lowest value of free energy available
to the reaction system
At equilibrium, G = 0 and Q = K
G0
G0 = 0
G0 < 0
G0 > 0
K
K = 1
K > 1
K < 1
Slide 65
Bond Energy
• Breaking bonds require energy (+)
• Forming bonds releases energy (-)
• If the reaction A + B → C is exothermic,
which is larger – the energy needed to
break the bonds or the energy released
when forming the bonds?
Slide 66
•TRY FRQ #2
Slide 67
Big Idea #4:
Kinetics, Rates,
and
Rate Laws
Slide 68
Reaction Rate
The change in concentration of a reactant or
product per unit of time
Rate
[ A ] at time t 2 [ A ] at time t1
t 2 t1
Rate
[ A]
t
Slide 69
2NO2(g) 2NO(g) + O2(g)
Reaction Rates:
1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products
3. Are proportional
stoichiometrically
Slide 70
2NO2(g) 2NO(g) + O2(g)
[NO2]
t
Reaction Rates:
4. Are equal to the
slope tangent to
that point
5. Change as the
reaction proceeds,
if the rate is
dependent upon
concentration
[ N O2 ]
t
constant
Slide 71
Rate Laws
Differential rate laws express (reveal) the
relationship between the concentration of
reactants and the rate of the reaction.
The differential rate law is usually just
called “the rate law.”
Integrated rate laws express (reveal) the
relationship between concentration of
reactants and time
Slide 72
Writing a (differential) Rate Law
Problem - Write the rate law, determine the
value of the rate constant, k, and the overall
order for the following reaction:
2 NO(g) + Cl2(g) 2 NOCl(g)
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
Slide 73
Writing a Rate Law
Part 1 – Determine the values for the exponents
in the rate law: R = k[NO]x[Cl2]y
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant
while [NO] doubles. The rate quadruples,
so the reaction is second order with
respect to [NO]
R = k[NO]2[Cl2]y
Slide 74
Writing a Rate Law
Part 2 – Determine the value for k, the rate
constant, by using any set of experimental data:
R = k[NO]2[Cl2]
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
1.43 x 10
6
2
m ol
m ol
k 0.250
0.250
Ls
L
L
m ol
1.43 x 10 6
k
3
0.250
2
m ol L3
L
5
9.15 x 10
3
2
L
s
m
ol
m
ol
s
Slide 75
Writing a Rate Law
Part 3 – Determine the overall order for the
reaction.
R = k[NO]2[Cl2]
2 + 1 = 3
The reaction is 3rd order
Overall order is the sum of the exponents,
or orders, of the reactants
Slide 76
Determining Order with
Concentration vs. Time data
(the Integrated Rate Law)
Zero Order: time vs . concentrat ion is linear
First Order: time vs . ln( concentrat ion ) is linear
Second Order: time vs .
1
concentrat ion
is linear
Slide 77
Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Problem: Find the
integrated rate law
and the value for the
rate constant, k
A graphing calculator
with linear regression
analysis greatly
simplifies this process!!
(Click here to download my Rate Laws program
for theTi-83 and Ti-84)
Slide 78
Time vs. [H2O2]
Regression results:
y = ax + b
a = -2.64 x 10-4
b = 0.841
r2 = 0.8891
r = -0.9429
Time (s)
[H2O2]
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Slide 79
Time vs. ln[H2O2]
Time (s)
ln[H2O2]
0
0
120
-0.0943
300
-0.2485
600
-0.5276
Regression results:
1200
-0.9943
y = ax + b
a = -8.35 x 10-4
b = -.005
r2 = 0.99978
r = -0.9999
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Slide 80
Time vs. 1/[H2O2]
Regression results:
y = ax + b
a = 0.00460
b = -0.847
r2 = 0.8723
r = 0.9340
Time (s)
1/[H2O2]
0
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
Slide 81
And the winner is… Time vs. ln[H2O2]
1. As a result, the reaction is 1st order
2. The (differential) rate law is:
R k [ H 2O 2 ]
3. The integrated rate law is:
ln[ H 2 O 2 ] kt ln[ H 2 0 2 ] 0
4. But…what is the rate constant, k ?
Slide 82
Finding the Rate Constant, k
Method #2: Obtain k from the linear
regresssion analysis.
Regression results:
4
1
slope 8 . 32 x 10 s
y = ax + b
a = -8.35 x 10-4
Now remember:
b = -.005
2 = 0.99978
r
ln[ H 2 O 2 ] kt ln[ H 2 0 2 ] 0
r = -0.9999
k = -slope
k = 8.35 x 10-4s-1
Slide 83
Rate Laws Summary
Rate Law
Integrated
Rate Law
Plot that
produces a
straight line
Relationship of
rate constant
to slope of
straight line
Zero Order
First Order
Second Order
Rate = k
Rate = k[A]
Rate = k[A]2
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
1
[ A]
[A] versus t
ln[A] versus t
1
kt
1
[ A ]0
versus t
[ A]
Slope = -k
Slope = -k
[ A ]0
0.693
Half-Life
t1 / 2
2k
t1 / 2
k
Slope = k
t1 / 2
1
k [ A ]0
Slide 84
Reaction Mechanism
The reaction mechanism is the series of
elementary steps by which a chemical
reaction occurs.
The sum of the elementary steps
must give the overall balanced equation
for the reaction
The mechanism must agree with the
experimentally determined rate law
Slide 85
Rate-Determining Step
In a multi-step reaction, the
slowest step is the rate-determining
step. It therefore determines the
rate of the reaction.
The experimental rate law must agree
with the rate-determining step
Slide 86
Identifying the Rate-Determining Step
For the reaction:
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the
rate-determining (slowest) step?
Step #1
H2(g) + 2NO(g) N2O(g) + H2O(g)
Step #2
N2O(g) + H2(g) N2(g) + H2O(g)
Step #1 agrees with the experimental rate law
Slide 87
Identifying Intermediates
For the reaction:
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
Which species in the reaction mechanism are
intermediates (do not show up in the final,
balanced equation?)
Step #1
H2(g) + 2NO(g) N2O(g) + H2O(g)
Step #2
N2O(g) + H2(g) N2(g) + H2O(g)
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
N2O(g) is an intermediate
Slide 88
Collision Model
Key Idea: Molecules must collide to react.
However, only a small fraction of collisions
produces a reaction. Why?
Slide 89
Collision Model
Collisions must have
sufficient energy to
produce the reaction
(must equal or
exceed the
activation energy).
Colliding particles must be correctly oriented
to one another in order to produce a
reaction.
Slide 90
Factors Affecting Rate
Increasing temperature always increases
the rate of a reaction.
Particles collide more frequently
Particles collide more energetically
Increasing surface area increases the
rate of a reaction
Increasing Concentration USUALLY increases
the rate of a reaction
Presence of Catalysts, which lower the
activation energy by providing alternate
pathways
Slide 91
• TRY FRQ #3
Slide 92
Big Idea #2
Intermolecular Forces
Slide 93
Relative Magnitudes of Forces
The types of bonding forces vary in their
strength as measured by average bond
energy.
Strongest
Covalent bonds (400 kcal/mol)
Hydrogen bonding (12-16 kcal/mol )
Dipole-dipole interactions (2-0.5 kcal/mol)
Weakest
London forces (less than 1 kcal/mol)
Slide 94
London Dispersion Forces
The temporary separations of
charge that lead to the
London force attractions are
what attract one nonpolar
molecule to its neighbors.
London forces increase with
the size of the molecules.
Fritz London
1900-1954
Synonyms: “Induced dipoles”, “dispersion
forces”, and “dispersion-interaction forces”
Slide 95
London Dispersion Forces
Slide 96
Dipole-Dipole
• Forces of attraction between
two polar molecules
• Permanent dipoles
Slide 97
Hydrogen Bonding
Special type of dipole-dipole in which
hydrogen bonds with N, O, F.
Hydrogen bonding between ammonia and water
Slide 98
Hydrogen Bonding in DNA
Thymine hydrogen bonds to Adenine
H 3C
O
H 2N
N
OH
HO
NH
O
HO
N
O
O
O
N
T
A
OH
P
O
O
P
HO
O
N
N
OH
Slide 99
Boiling point as a measure of intermolecular
attractive forces
Slide 100
TRY FRQ #4
Slide 101
Big Idea #1: Periodic Trends
Atomic Radius
Definition: Half of the distance
between nuclei in covalently bonded
diatomic molecule
Radius decreases across a period
Increased effective nuclear charge
due to decreased shielding
Radius increases down a group
Each row on the periodic table adds
a “shell” or energy level to the atom
Slide 102
Table of
Atomic
Radii
Slide 103
Period Trend:
Atomic Radius
Slide 104
Ionization Energy
Definition: the energy required to remove an
electron from an atom
Increases for successive electrons taken from
the same atom
Tends to increase across a period
Electrons in the same quantum level do not
shield as effectively as electrons in inner levels
Irregularities at half filled and filled
sublevels due to extra repulsion of electrons
paired in orbitals, making them easier to
remove
Tends to decrease down a group
Outer electrons are farther from the
nucleus and easier to remove
Slide 105
Ionization Energy: the energy required to
remove an electron from an atom
Increases for successive electrons taken from the
same atom
Tends to increase across a period
Electrons in the same quantum level do not
shield as effectively as electrons in inner
levels
Irregularities at half filled and filled
sublevels due to extra repulsion of electrons
paired in orbitals, making them easier to
remove
Tends to decrease down a group
Outer electrons are farther from the nucleus
Slide 106
Table of 1st Ionization
Energies
Slide 107
Periodic Trend:
Ionization Energy
Slide 108
Electronegativity
Definition: A measure of the ability of an
atom in a chemical compound to attract
electrons
o Electronegativity tends to increase
across a period
o As radius decreases, electrons get
closer to the bonding atom’s nucleus
o Electronegativity tends to decrease
down a group or remain the same
o As radius increases, electrons are
farther from the bonding atom’s
nucleus
Slide 109
Periodic Table of Electronegativities
Slide 110
Periodic Trend:
Electronegativity
Slide 111
Summary of
Periodic Trends
Slide 112
Ionic Radii
Cations
Positively charged ions formed when
an atom of a metal loses one or
more electrons
Smaller than the corresponding
atom
Negatively charged ions formed
when nonmetallic atoms gain one
Anions
or more electrons
Larger than the corresponding
atom
Slide 113
Table
of Ion
Sizes
Slide 114
Determine the element
Answer: Hg
Slide 115
Determine the element
Answer: Na and Mg
Slide 116
•TRY FRQ #5
Slide 117
Electrochemistry
Slide 118
Electrochemistry Terminology #1
Oxidation – A process in which an
element attains a more positive
oxidation state
Na(s) Na+ + eReduction – A process in which an
element attains a more negative
oxidation state
Cl2 + 2e- 2Cl-
Slide 119
Electrochemistry Terminology #2
An old memory device for oxidation
and reduction goes like this…
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction
Slide 120
Electrochemistry Terminology #4
Anode
The electrode where
oxidation occurs
Cathode
The electrode where
reduction occurs
Memory device:
Reduction
at the
Cathode
Slide 121
Galvanic (Electrochemical) Cells
Spontaneous redox
processes have:
A positive
cell potential, E0
A negative free
energy change, (-G)
G=-nFE
Slide 122
Zn - Cu
Galvanic
Cell
From a table
of reduction
potentials:
Zn2+ + 2e- Zn
Cu2+ + 2e- Cu
E = -0.76V
E = +0.34V
Slide 123
Zn - Cu
Galvanic
Cell
The less positive,
or more negative
reduction potential
becomes the
oxidation…
Zn Zn2+ + 2eCu2+ + 2e- Cu
E = +0.76V
E = +0.34V
Zn + Cu2+ Zn2+ + Cu
E0 = + 1.10 V
Slide 124
Line
Notation
An abbreviated
representation of
an electrochemical
cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode
Anode
|
material
solution
||
Cathode
solution
|
Cathode
material
Slide 125
Calculating G0 for a Cell
G0 = -nFE0
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-
Zn + Cu2+ Zn2+ + Cu
G (2 m ol e )(96 485
0
E0 = + 1.10 V
coulom bs
m ol e
)(1.10
G 212267 Joules 212 kJ
0
Joules
C oulom b
)
Slide 126
???
Concentration
Cell
Both sides have
the same
components but
at different
concentrations.
Step 1: Determine which side undergoes
oxidation, and which side undergoes reduction.
Slide 127
???
Anode
Concentration
Cell
Cathode
Both sides have
the same
components but
at different
concentrations.
The 1.0 M Zn2+ must decrease in concentration, and
the 0.10 M Zn2+ must increase in concentration
Zn2+ (1.0M) + 2e- Zn
(reduction)
Zn Zn2+ (0.10M) + 2eZn2+ (1.0M) Zn2+ (0.10M)
(oxidation)
Slide 128
Electrolytic
Processes
Electrolytic
processes are
NOT spontaneous.
They have:
A negative
cell potential, (-E0)
A positive free
energy change, (+G)
Slide 129
Solving an Electroplating Problem
Q: What mass of copper is plated out when a
current of 10 amps is passed for 30 minutes
through a solution of Cu2+? (Amp=C/sec)
Cu2+ + 2e- Cu
10 C
sec
1800sec 1 mol e-
1 mole Cu 63.5 g Cu
1 mole Cu
2
mol
e
96 485 C
= 5.94g Cu
Slide 130
• Good Luck on the
Exam!! Try your best.
You have worked hard
and will do great! I
am very proud of each
one of you.
•
Mrs. L
AP Chemistry
Super Saturday
Review
I tried to include as much review material
as possible in this session. Work on practice
tests and review the Bozeman videos for
other material.
Slide 2
AP Chemistry
Super Saturday
Review
5 Essentials
1. Know the basics – writing formulas,
writing and balancing equations,
dimensional analysis, atomic theory,
acid-base theories (Arrhenius and Bronsted
Lowry), VSEPR, kinetic molecular theory,
and collision theory.
Slide 3
5 Essentials
2. Atomic and Molecular Structures
Atomic structures (like electron configurations)
will help explain relationships on the periodic
table which explains many physical and chemical
properties. Molecular structures involves Lewis
structures and VSEPR to determine shapes
which describes polarities thus describing
intermolecular forces which describes many
physical properties.
Slide 4
5 Essentials
3. Stoichiometric Calculations
Basic stoichiometry, limiting reactants,
titration calculations, and empirical formulas.
4. Principles of chemical kinetics, equilibrium,
And thermodynamics
Kinetics- describes the speed in which
substances react
Equilibrium – used to determine the extent of
a reaction or the composition at equilibrium
Thermodynamics – explains why chemical reactions
happen in terms of kinetic and potential energies
Slide 5
5 Essentials
5. Representation and Interpretation
Be able to draw what is happening at the
molecular level and read and interpret graphs
and data tables
Slide 6
Net Ionic Equations
Graphic: Wikimedia Commons User Tubife
Slide 7
Solubility Rules – AP Chemistry
All sodium, potassium, ammonium, and
nitrate salts are soluble in water.
Memorization of other “solubility rules”
is beyond the scope of this course and
the AP Exam.
What dissociates (“breaks apart”) –
Aqueous solutions of the following:
• All strong acids (HCl, HBr, HI, HNO3,
H2SO4, and HClO4)
• Strong bases (group I and II hydroxides)
• Soluble salts
Slide 8
Write the net ionic for the following:
1. Solutions of lead nitrate and potassium
chloride are mixed
2. Solutions of sulfuric acid and
potassium hydroxide are mixed.
3. Solid sodium hydroxide is mixed with
acetic acid
Slide 9
Big Idea #6:Chemical Equilibrium
2NO2(g) 2NO(g) + O2(g)
Sketch a graph of change in concentration
vs. Time for the reaction above
Slide 10
2NO2(g) 2NO(g) + O2(g)
Be able to explain the variance in slope
Slide 11
Law of Mass Action
For the reaction:
jA + kB lC + mD
l
K
[C ] [ D ]
m
j
k
[ A] [B ]
Where K is the equilibrium constant,
and is unitless
Slide 12
Product Favored Equilibrium
Large values for K signify the reaction is
“product favored”
When equilibrium is achieved, most
reactant has been converted to product
Slide 13
Reactant Favored Equilibrium
Small values for K signify the reaction is
“reactant favored”
When equilibrium is achieved, very little
reactant has been converted to product
Slide 14
Writing an Equilibrium Expression
Write the equilibrium expression for the
reaction:
2NO2(g) 2NO(g) + O2(g)
K = ???
2
K
[ NO ] [ O 2 ]
[ NO 2 ]
2
Slide 15
Conclusions about Equilibrium Expressions
The equilibrium expression for a reaction
is the reciprocal for a reaction written in
reverse
2NO2(g) 2NO(g) + O2(g)
2
K
[ NO ] [ O 2 ]
[ NO 2 ]
2
2NO(g) + O2(g) 2NO2(g)
K
'
1
K
[ NO 2 ]
2
2
[ NO ] [ O 2 ]
Slide 16
Conclusions about Equilibrium Expressions
When the balanced equation for a reaction
is multiplied by a factor n, the equilibrium
expression for the new reaction is the
original expression, raised to the nth power.
2NO2(g) 2NO(g) + O2(g)
2
K
[ NO ] [ O 2 ]
[ NO 2 ]
2
NO2(g) NO(g) + ½O2(g)
1
1
K K
1
2
[ NO ][ O 2 ] 2
[ NO 2 ]
Slide 17
If the equilibrium constant for A + B
2C
2A + 2B is
A) 0.584
B) 4.81
C) 0.416
D) 23.1
E) 0.208
Answer: D
C is 0.208 then the equilibrium constant for
Slide 18
Equilibrium Expressions Involving Pressure
For the gas phase reaction:
3H2(g) + N2(g) 2NH3(g)
KP
P
NH
2
3
3
( PN 2 )( PH 2 )
PNH 3 , PN 2 , PH 2 are equilibriu m partial
K p K ( RT )
n
pressures
Slide 19
Heterogeneous Equilibria
The position of a heterogeneous equilibrium
does not depend on the amounts of pure solids
or liquids present
Write the equilibrium expression for
the reaction:
PCl5(s) PCl3(l) + Cl2(g)
Pure
solid
Pure
liquid
K [ Cl 2 ]
K p PCl 2
Slide 20
The Reaction Quotient
For some time, t, when the system is not at
equilibrium, the reaction quotient, Q takes the
place of K, the equilibrium constant, in the
law of mass action.
jA + kB lC + mD
l
Q
[C ] [ D ]
m
j
k
[ A] [ B ]
Slide 21
Significance of the Reaction Quotient
If Q = K, the system is at equilibrium
If Q > K, the system shifts to the left,
consuming products and forming reactants
until equilibrium is achieved
If Q < K, the system shifts to the right,
consuming reactants and forming products
until equilibrium is achieved
Slide 22
If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a
one-liter container, which direction would the reaction
initially proceed?
A)
To the left.
B)
To the right.
C)
The above mixture is the equilibrium mixture.
D)
Cannot tell from the information given.
Slide 23
LeChatelier’s Principle
When a system at
equilibrium is placed under
stress, the system will
undergo a change in such
a way as to relieve that
stress and restore a
state of equilibrium.
Henry Le Chatelier
Slide 24
Le Chatelier Translated:
When you take something away from a
system at equilibrium, the system shifts
in such a way as to replace some of what
you’ve taken away.
When you add something to a system at
equilibrium, the system shifts in such a
way as to use up some of what you’ve
added.
Slide 25
•Do FRQ #1
Slide 26
Acid
Equilibrium
and pH
Søren Sørensen
Slide 27
Acid/Base Definitions
Arrhenius Model
Acids produce hydrogen ions in aqueous
solutions
Bases produce hydroxide ions in
aqueous solutions
Bronsted-Lowry Model
Acids are proton donors
Bases are proton acceptors
Slide 28
Acid Dissociation
HA
Acid
H+ +
AProton
Conjugate
base
Ka
[ H ][ A ]
[ HA ]
Alternately, H+ may be written in its
hydrated form, H3O+ (hydronium ion)
Slide 29
Dissociation Constants: Strong Acids
Acid
Formula
Perchloric
Hydriodic
Hydrobromic
Hydrochloric
Nitric
Sulfuric
HClO4
HI
HBr
HCl
HNO3
H2SO4
Hydronium ion
H3O+
Conjugate
Base
ClO4-
Ka
NO3HSO4-
Very large
Very large
Very large
Very large
Very large
Very large
H2O
1.0
IBrCl-
Slide 30
Dissociation Constants: Weak Acids
Acid
Iodic
Oxalic
Sulfurous
Phosphoric
Citric
Nitrous
Formula
Conjugate
Base
Ka
HIO3
IO3-
1.7 x 10-1
H2SO3
HSO3-
1.5 x 10-2
H2C6H5O7-
7.1 x 10-4
H2C2O4
H3PO4
H3C6H5O7
HNO2
HC2O4-
5.9 x 10-2
H2PO4-
7.5 x 10-3
NO2-
4.6 x 10-4
F-
3.5 x 10-4
Hydrofluoric
HF
Formic
HCOOH
HCOO-
1.8 x 10-4
Benzoic
C6H5COOH
C6H5COO-
6.5 x 10-5
H2CO3
HCO3-
4.3 x 10-7
Acetic
Carbonic
Hypochlorous
Hydrocyanic
CH3COOH
HClO
HCN
CH3COO-
1.8 x 10-5
ClO-
3.0 x 10-8
CN-
4.9 x 10-10
Slide 31
Reaction of Weak Bases with Water
The base reacts with water, producing its
conjugate acid and hydroxide ion:
CH3NH2 + H2O CH3NH3+ + OH-
Kb = 4.38 x 10-4
K b 4.38 x 10
4
[ C H 3 N H 3 ][ O H ]
[C H 3 N H 2 ]
Slide 32
Kb for Some Common Weak Bases
Many students struggle with identifying weak
bases and their conjugate acids.What patterns
do you see that may help you?
Formula
Conjugate
Acid
Kb
NH3
NH4+
1.8 x 10-5
Methylamine
CH3NH2
CH3NH3+
4.38 x 10-4
Ethylamine
C2H5NH2
C2H5NH3+
5.6 x 10-4
Diethylamine
(C2H5)2NH
(C2H5)2NH2+
1.3 x 10-3
Triethylamine
(C2H5)3N
(C2H5)3NH+
4.0 x 10-4
Hydroxylamine
HONH2
HONH3+
1.1 x 10-8
Base
Ammonia
Hydrazine
H2NNH2
H2NNH3+
3.0 x 10-6
Aniline
C6H5NH2
C6H5NH3+
3.8 x 10-10
Pyridine
C5H5N
C5H5NH+
1.7 x 10-9
Slide 33
Reaction of Weak Bases with Water
The generic reaction for a base reacting
with water, producing its conjugate acid and
hydroxide ion:
B + H2O BH+ + OH
Kb
[ B H ][ O H ]
[B]
(Yes, all weak bases do this – DO NOT
try to make this complicated!)
Ex. Write the reaction of ammonia
with water
Slide 34
Self-Ionization of Water
H2O + H2O
H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7
Kw is a constant at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14
Slide 35
Calculating pH, pOH
pH = -log10(H3O+)
pOH = -log10(OH-)
Relationship between pH and pOH
pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH
Slide 36
Calculate the pH of a
0.1M HCl solution?
(answer =1)
Slide 37
A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M
solution of acetic acid,
-5 ?
HC
H
O
,
K
=
1.8
x
10
2 3 2
a
(answer=4.52)
Slide 38
A Weak Base Equilibrium Problem
What is the pH of a 0.50 M
solution of ammonia, NH3,
Kb = 1.8 x 10-5 ? (answer=9.48)
Slide 39
Acid-Base Properties of Salts
To determine if a salt is acidic or
basic, determine the stronger parent.
Examples:
KCl
NH4Cl
NaC2H3O2
NaCl
KNO3
Slide 40
Acid-Base Properties of Salts
If both parents are weak:
Type of Salt
Examples Comment
Cation is the
conjugate acid of a
weak base, anion is
conjugate base of a
weak acid
NH4C2H3O2 Cation is acidic,
NH4CN
Anion is basic
pH of
solution
See below
IF Ka for the acidic ion is greater than Kb for
the basic ion, the solution is acidic
IF Kb for the basic ion is greater than Ka for
the acidic ion, the solution is basic
IF Kb for the basic ion is equal to Ka for the
acidic ion, the solution is neutral
Slide 41
Buffered Solutions
A solution that resists a change in
pH when either hydroxide ions or
protons are added.
Buffered solutions contain either:
A weak acid and its salt
A weak base and its salt
Slide 42
Acid/Salt Buffering Pairs
The salt will contain the anion of the acid,
and the cation of a strong base (NaOH, KOH)
Weak Acid
Formula
of the acid
Hydrofluoric
HF
Formic
HCOOH
Benzoic
C6H5COOH
Acetic
Carbonic
Propanoic
Hydrocyanic
CH3COOH
H2CO3
HC3H5O2
HCN
Example of a salt of the
weak acid
KF – Potassium fluoride
KHCOO – Potassium formate
NaC6H5COO – Sodium benzoate
NaH3COO – Sodium acetate
NaHCO3 - Sodium bicarbonate
NaC3H5O2 - Sodium propanoate
KCN - potassium cyanide
Slide 43
Base/Salt Buffering Pairs
The salt will contain the cation of the base,
and the anion of a strong acid (HCl, HNO3)
Formula of
the base
Example of a salt of the weak
acid
NH3
NH4Cl - ammonium chloride
Methylamine
CH3NH2
CH3NH2Cl – methylammonium chloride
Ethylamine
C2H5NH2
C2H5NH3NO3 - ethylammonium nitrate
Aniline
C6H5NH2
C6H5NH3Cl – aniline hydrochloride
Base
Ammonia
Pyridine
C5H5N
C5H5NHCl – pyridine hydrochloride
Slide 44
Titration of an Unbuffered Solution
13
12
11
10
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 45
Titration of a Buffered Solution
13
12
11
A solution that is
0.10 M CH3COOH and
0.10 M NaCH3COO is
titrated with
0.10 M NaOH
10
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 46
Comparing Results
Gra ph
14
12
10
Buffered
pH
8
6
4
Unbuffered
2
0
0
5
10
15
20
25
mL 0.10 M NaOH
30
35
40
45
Slide 47
Henderson-Hasselbalch Equation
[ A ]
pK a log
pH pK a log
[ HA ]
[ base
[ acid
[ BH ]
pK b log
pOH pK b log
[
B
]
[ acid
[ base
]
]
]
]
This is an exceptionally powerful tool that can be
used in your problem solving.
Slide 48
Title:
13
12
11
10
9
Endpoint is above
pH 7
pH
8
7
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 49
Title:
13
12
11
10
9
pH
8
7
Endpoint is at
pH 7
A solution that is
0.10 M HCl is
titrated with
0.10 M NaOH
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Slide 50
Title:
13
12
A solution that is
0.10 M NaOH is
titrated with
0.10 M HCl
11
10
9
pH
8
7
Endpoint is at
pH 7
It is important to
recognize that
titration curves are
not always
increasing from left
to right.
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters HCl (0.10 M)
30.00
35.00
40.00
45.00
Slide 51
Title:
13
12
11
10
9
pH
8
7
6
5
Endpoint is below
pH 7
4
A solution that is
0.10 M HCl is
titrated with
0.10 M NH3
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NH3 (0.10 M)
30.00
35.00
40.00
45.00
Slide 52
Solubility
Equilibria
Lead (II) iodide precipitates when potassium
iodide is mixed with lead (II) nitrate.
Graphic: Wikimedia Commons user PRHaney
Slide 53
Solving Solubility Problems
For the salt AgI at 25C, Ksp = 1.5 x 10-16
Answer = solubility of AgI in mol/L = 1.2 x 10-8 M
Slide 54
Solving Solubility Problems
For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5
Answer = solubility of PbCl2 in mol/L = 1.6 x 10-2 M
Slide 55
Solving Solubility with a Common Ion
For the salt AgI at 25C, Ksp = 1.5 x 10-16
What is its solubility in 0.05 M NaI?
Answer = solubility of AgI in mol/L = 3.0 x 10-15 M
Slide 56
Big Idea #5:
Spontaneity, Entropy
and Free Energy
Slide 57
G = H - TS
Spontaneity, Entropy
and Free Energy
Slide 58
Spontaneous Processes and Entropy
First Law
• “Energy can neither be created nor
destroyed"
• The energy of the universe is constant
Spontaneous Processes
• Processes that occur without outside
intervention
• Spontaneous processes may be fast or
slow
– Many forms of combustion are fast
– Conversion of diamond to graphite is
slow
Slide 59
Which of the following reactions
is spontaneous?
• H2(g) + I2(g) ↔ 2HI
• Br2 + Cl2 ↔ 2BrCl
• HF + H2O ↔ F- + H30+
Kc=49
Kc=6.9
Kc=6.8x10-4
Slide 60
Second Law of
Thermodynamics
"In any spontaneous process there is
always an increase in the entropy of
the universe"
Ssolid < Sliquid << Sgas
Slide 61
Calculating Free Energy
Method #1
For reactions at constant temperature:
0
G
=
0
H
-
0
TS
Slide 62
Calculating Free Energy: Method #2
An adaptation of Hess's Law:
Cdiamond(s) + O2(g) CO2(g)
G0 = -397 kJ
Cgraphite(s) + O2(g) CO2(g)
G0 = -394 kJ
Cdiamond(s) + O2(g) CO2(g)
G0 = -397 kJ
CO2(g) Cgraphite(s) + O2(g)
G0 = +394 kJ
Cdiamond(s) Cgraphite(s)
G0 = -3 kJ
Slide 63
Calculating Free Energy
Method #3
Using standard free energy of formation (Gf0):
G
0
n
p
G
0
f ( p ro d u cts )
n
r
G
0
f ( reactan ts )
Gf0 of an element in its standard state is zero
Slide 64
Free Energy and Equilibrium
Equilibrium point occurs at the
lowest value of free energy available
to the reaction system
At equilibrium, G = 0 and Q = K
G0
G0 = 0
G0 < 0
G0 > 0
K
K = 1
K > 1
K < 1
Slide 65
Bond Energy
• Breaking bonds require energy (+)
• Forming bonds releases energy (-)
• If the reaction A + B → C is exothermic,
which is larger – the energy needed to
break the bonds or the energy released
when forming the bonds?
Slide 66
•TRY FRQ #2
Slide 67
Big Idea #4:
Kinetics, Rates,
and
Rate Laws
Slide 68
Reaction Rate
The change in concentration of a reactant or
product per unit of time
Rate
[ A ] at time t 2 [ A ] at time t1
t 2 t1
Rate
[ A]
t
Slide 69
2NO2(g) 2NO(g) + O2(g)
Reaction Rates:
1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products
3. Are proportional
stoichiometrically
Slide 70
2NO2(g) 2NO(g) + O2(g)
[NO2]
t
Reaction Rates:
4. Are equal to the
slope tangent to
that point
5. Change as the
reaction proceeds,
if the rate is
dependent upon
concentration
[ N O2 ]
t
constant
Slide 71
Rate Laws
Differential rate laws express (reveal) the
relationship between the concentration of
reactants and the rate of the reaction.
The differential rate law is usually just
called “the rate law.”
Integrated rate laws express (reveal) the
relationship between concentration of
reactants and time
Slide 72
Writing a (differential) Rate Law
Problem - Write the rate law, determine the
value of the rate constant, k, and the overall
order for the following reaction:
2 NO(g) + Cl2(g) 2 NOCl(g)
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
Slide 73
Writing a Rate Law
Part 1 – Determine the values for the exponents
in the rate law: R = k[NO]x[Cl2]y
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant
while [NO] doubles. The rate quadruples,
so the reaction is second order with
respect to [NO]
R = k[NO]2[Cl2]y
Slide 74
Writing a Rate Law
Part 2 – Determine the value for k, the rate
constant, by using any set of experimental data:
R = k[NO]2[Cl2]
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
1.43 x 10
6
2
m ol
m ol
k 0.250
0.250
Ls
L
L
m ol
1.43 x 10 6
k
3
0.250
2
m ol L3
L
5
9.15 x 10
3
2
L
s
m
ol
m
ol
s
Slide 75
Writing a Rate Law
Part 3 – Determine the overall order for the
reaction.
R = k[NO]2[Cl2]
2 + 1 = 3
The reaction is 3rd order
Overall order is the sum of the exponents,
or orders, of the reactants
Slide 76
Determining Order with
Concentration vs. Time data
(the Integrated Rate Law)
Zero Order: time vs . concentrat ion is linear
First Order: time vs . ln( concentrat ion ) is linear
Second Order: time vs .
1
concentrat ion
is linear
Slide 77
Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Problem: Find the
integrated rate law
and the value for the
rate constant, k
A graphing calculator
with linear regression
analysis greatly
simplifies this process!!
(Click here to download my Rate Laws program
for theTi-83 and Ti-84)
Slide 78
Time vs. [H2O2]
Regression results:
y = ax + b
a = -2.64 x 10-4
b = 0.841
r2 = 0.8891
r = -0.9429
Time (s)
[H2O2]
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Slide 79
Time vs. ln[H2O2]
Time (s)
ln[H2O2]
0
0
120
-0.0943
300
-0.2485
600
-0.5276
Regression results:
1200
-0.9943
y = ax + b
a = -8.35 x 10-4
b = -.005
r2 = 0.99978
r = -0.9999
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Slide 80
Time vs. 1/[H2O2]
Regression results:
y = ax + b
a = 0.00460
b = -0.847
r2 = 0.8723
r = 0.9340
Time (s)
1/[H2O2]
0
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
Slide 81
And the winner is… Time vs. ln[H2O2]
1. As a result, the reaction is 1st order
2. The (differential) rate law is:
R k [ H 2O 2 ]
3. The integrated rate law is:
ln[ H 2 O 2 ] kt ln[ H 2 0 2 ] 0
4. But…what is the rate constant, k ?
Slide 82
Finding the Rate Constant, k
Method #2: Obtain k from the linear
regresssion analysis.
Regression results:
4
1
slope 8 . 32 x 10 s
y = ax + b
a = -8.35 x 10-4
Now remember:
b = -.005
2 = 0.99978
r
ln[ H 2 O 2 ] kt ln[ H 2 0 2 ] 0
r = -0.9999
k = -slope
k = 8.35 x 10-4s-1
Slide 83
Rate Laws Summary
Rate Law
Integrated
Rate Law
Plot that
produces a
straight line
Relationship of
rate constant
to slope of
straight line
Zero Order
First Order
Second Order
Rate = k
Rate = k[A]
Rate = k[A]2
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
1
[ A]
[A] versus t
ln[A] versus t
1
kt
1
[ A ]0
versus t
[ A]
Slope = -k
Slope = -k
[ A ]0
0.693
Half-Life
t1 / 2
2k
t1 / 2
k
Slope = k
t1 / 2
1
k [ A ]0
Slide 84
Reaction Mechanism
The reaction mechanism is the series of
elementary steps by which a chemical
reaction occurs.
The sum of the elementary steps
must give the overall balanced equation
for the reaction
The mechanism must agree with the
experimentally determined rate law
Slide 85
Rate-Determining Step
In a multi-step reaction, the
slowest step is the rate-determining
step. It therefore determines the
rate of the reaction.
The experimental rate law must agree
with the rate-determining step
Slide 86
Identifying the Rate-Determining Step
For the reaction:
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the
rate-determining (slowest) step?
Step #1
H2(g) + 2NO(g) N2O(g) + H2O(g)
Step #2
N2O(g) + H2(g) N2(g) + H2O(g)
Step #1 agrees with the experimental rate law
Slide 87
Identifying Intermediates
For the reaction:
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
Which species in the reaction mechanism are
intermediates (do not show up in the final,
balanced equation?)
Step #1
H2(g) + 2NO(g) N2O(g) + H2O(g)
Step #2
N2O(g) + H2(g) N2(g) + H2O(g)
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
N2O(g) is an intermediate
Slide 88
Collision Model
Key Idea: Molecules must collide to react.
However, only a small fraction of collisions
produces a reaction. Why?
Slide 89
Collision Model
Collisions must have
sufficient energy to
produce the reaction
(must equal or
exceed the
activation energy).
Colliding particles must be correctly oriented
to one another in order to produce a
reaction.
Slide 90
Factors Affecting Rate
Increasing temperature always increases
the rate of a reaction.
Particles collide more frequently
Particles collide more energetically
Increasing surface area increases the
rate of a reaction
Increasing Concentration USUALLY increases
the rate of a reaction
Presence of Catalysts, which lower the
activation energy by providing alternate
pathways
Slide 91
• TRY FRQ #3
Slide 92
Big Idea #2
Intermolecular Forces
Slide 93
Relative Magnitudes of Forces
The types of bonding forces vary in their
strength as measured by average bond
energy.
Strongest
Covalent bonds (400 kcal/mol)
Hydrogen bonding (12-16 kcal/mol )
Dipole-dipole interactions (2-0.5 kcal/mol)
Weakest
London forces (less than 1 kcal/mol)
Slide 94
London Dispersion Forces
The temporary separations of
charge that lead to the
London force attractions are
what attract one nonpolar
molecule to its neighbors.
London forces increase with
the size of the molecules.
Fritz London
1900-1954
Synonyms: “Induced dipoles”, “dispersion
forces”, and “dispersion-interaction forces”
Slide 95
London Dispersion Forces
Slide 96
Dipole-Dipole
• Forces of attraction between
two polar molecules
• Permanent dipoles
Slide 97
Hydrogen Bonding
Special type of dipole-dipole in which
hydrogen bonds with N, O, F.
Hydrogen bonding between ammonia and water
Slide 98
Hydrogen Bonding in DNA
Thymine hydrogen bonds to Adenine
H 3C
O
H 2N
N
OH
HO
NH
O
HO
N
O
O
O
N
T
A
OH
P
O
O
P
HO
O
N
N
OH
Slide 99
Boiling point as a measure of intermolecular
attractive forces
Slide 100
TRY FRQ #4
Slide 101
Big Idea #1: Periodic Trends
Atomic Radius
Definition: Half of the distance
between nuclei in covalently bonded
diatomic molecule
Radius decreases across a period
Increased effective nuclear charge
due to decreased shielding
Radius increases down a group
Each row on the periodic table adds
a “shell” or energy level to the atom
Slide 102
Table of
Atomic
Radii
Slide 103
Period Trend:
Atomic Radius
Slide 104
Ionization Energy
Definition: the energy required to remove an
electron from an atom
Increases for successive electrons taken from
the same atom
Tends to increase across a period
Electrons in the same quantum level do not
shield as effectively as electrons in inner levels
Irregularities at half filled and filled
sublevels due to extra repulsion of electrons
paired in orbitals, making them easier to
remove
Tends to decrease down a group
Outer electrons are farther from the
nucleus and easier to remove
Slide 105
Ionization Energy: the energy required to
remove an electron from an atom
Increases for successive electrons taken from the
same atom
Tends to increase across a period
Electrons in the same quantum level do not
shield as effectively as electrons in inner
levels
Irregularities at half filled and filled
sublevels due to extra repulsion of electrons
paired in orbitals, making them easier to
remove
Tends to decrease down a group
Outer electrons are farther from the nucleus
Slide 106
Table of 1st Ionization
Energies
Slide 107
Periodic Trend:
Ionization Energy
Slide 108
Electronegativity
Definition: A measure of the ability of an
atom in a chemical compound to attract
electrons
o Electronegativity tends to increase
across a period
o As radius decreases, electrons get
closer to the bonding atom’s nucleus
o Electronegativity tends to decrease
down a group or remain the same
o As radius increases, electrons are
farther from the bonding atom’s
nucleus
Slide 109
Periodic Table of Electronegativities
Slide 110
Periodic Trend:
Electronegativity
Slide 111
Summary of
Periodic Trends
Slide 112
Ionic Radii
Cations
Positively charged ions formed when
an atom of a metal loses one or
more electrons
Smaller than the corresponding
atom
Negatively charged ions formed
when nonmetallic atoms gain one
Anions
or more electrons
Larger than the corresponding
atom
Slide 113
Table
of Ion
Sizes
Slide 114
Determine the element
Answer: Hg
Slide 115
Determine the element
Answer: Na and Mg
Slide 116
•TRY FRQ #5
Slide 117
Electrochemistry
Slide 118
Electrochemistry Terminology #1
Oxidation – A process in which an
element attains a more positive
oxidation state
Na(s) Na+ + eReduction – A process in which an
element attains a more negative
oxidation state
Cl2 + 2e- 2Cl-
Slide 119
Electrochemistry Terminology #2
An old memory device for oxidation
and reduction goes like this…
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction
Slide 120
Electrochemistry Terminology #4
Anode
The electrode where
oxidation occurs
Cathode
The electrode where
reduction occurs
Memory device:
Reduction
at the
Cathode
Slide 121
Galvanic (Electrochemical) Cells
Spontaneous redox
processes have:
A positive
cell potential, E0
A negative free
energy change, (-G)
G=-nFE
Slide 122
Zn - Cu
Galvanic
Cell
From a table
of reduction
potentials:
Zn2+ + 2e- Zn
Cu2+ + 2e- Cu
E = -0.76V
E = +0.34V
Slide 123
Zn - Cu
Galvanic
Cell
The less positive,
or more negative
reduction potential
becomes the
oxidation…
Zn Zn2+ + 2eCu2+ + 2e- Cu
E = +0.76V
E = +0.34V
Zn + Cu2+ Zn2+ + Cu
E0 = + 1.10 V
Slide 124
Line
Notation
An abbreviated
representation of
an electrochemical
cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode
Anode
|
material
solution
||
Cathode
solution
|
Cathode
material
Slide 125
Calculating G0 for a Cell
G0 = -nFE0
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-
Zn + Cu2+ Zn2+ + Cu
G (2 m ol e )(96 485
0
E0 = + 1.10 V
coulom bs
m ol e
)(1.10
G 212267 Joules 212 kJ
0
Joules
C oulom b
)
Slide 126
???
Concentration
Cell
Both sides have
the same
components but
at different
concentrations.
Step 1: Determine which side undergoes
oxidation, and which side undergoes reduction.
Slide 127
???
Anode
Concentration
Cell
Cathode
Both sides have
the same
components but
at different
concentrations.
The 1.0 M Zn2+ must decrease in concentration, and
the 0.10 M Zn2+ must increase in concentration
Zn2+ (1.0M) + 2e- Zn
(reduction)
Zn Zn2+ (0.10M) + 2eZn2+ (1.0M) Zn2+ (0.10M)
(oxidation)
Slide 128
Electrolytic
Processes
Electrolytic
processes are
NOT spontaneous.
They have:
A negative
cell potential, (-E0)
A positive free
energy change, (+G)
Slide 129
Solving an Electroplating Problem
Q: What mass of copper is plated out when a
current of 10 amps is passed for 30 minutes
through a solution of Cu2+? (Amp=C/sec)
Cu2+ + 2e- Cu
10 C
sec
1800sec 1 mol e-
1 mole Cu 63.5 g Cu
1 mole Cu
2
mol
e
96 485 C
= 5.94g Cu
Slide 130
• Good Luck on the
Exam!! Try your best.
You have worked hard
and will do great! I
am very proud of each
one of you.
•
Mrs. L