7/1/2014
Chem 1312 Final has 115 points (similar to others)
WHAT TO STUDY ?
1. 50 multiple choice questions ( 2 pts each)
30 questions from “old” material {Chapters 10 – 14
20 questions from “new” material {Chap 15 – 17 & 20
2. 15 nomenclature questions ( 1 pt each)
At the end of each chapter
•
Strategies in Chemistry
•
Summary & Key Terms
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•
Key to final exam will be posted in glass case
outside room 100 after exam is over
A 90  up
C
70  79
B 80  89
D
60  69
You Will Have Five (5) Grades
Exam 1, 2, 3, 4 + Assignments
I will Average These Five (5) grades
{add grades & divide by five}
One Exam {1,2, or 3} Grade Can be
REPLACED By Exam 4 (Final Exam) If
Necessary
No, I Do Not Drop a Grade!!!!!!!
PRESSURE
(force per unit area)
Measured with
1. Barometers
2. Manometers
Open
Closed
F
P
A
Also
•
Old Exams
•
Web problems & Exams
CHAP 10:
Gases (Ideal & Real)
STANDARD CONDITIONS
for Gases
STP
Standard temperature
0C = 273.15 K
Standard pressure
1 atm
THE IDEAL GAS EQUATION
PV = n R T
and its APPLICATIONS
• GAS DENSITY
Density 
grams
P MW

volume R T
• MOLECULAR WEIGHT
MW 
grams gRT

mole
VP
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Gas Mixtures & Partial Pressures
PV = n R T
Pi V = ni R T
PT V = nT R T
Pi
n
 i
PT nT
or
1. Collecting Gases over Water
Pi   i Ptotal
i 
ni
nTotal
REAL GASES
and the
van der Waals Equation
nRT
n 2a

V  nb
V 2
Corrects for
molecular volume
PTotal = (Pgas + Pwater) = (ngas + nwater ) R T
2. Gas Volumes in Chemical Reactions
1 N2 (gas) + 3 H2 (gas) = 2 NH3 (gas)
where i is the mole fraction
P 
APPLICATIONS OF THE IDEAL GAS
EQUATION
Corrects for
molecular attraction
1 volume reacts with 3 volumes to produce 2 volumes
Chapter 11 Liquids & Solids
INTER & INTRA
MOLECULAR FORCES
• INTRA molecular forces are Chemical
Bonds which are formed between elements
• INTER molecular forces aare the attractive
forces between molecules and ions that
determine bulk properties of matter
types of Intermolecular Forces
1. Ion–dipole
Some Properties of Liquids
2. Dipole–dipole
3. London Forces => Instantaneous
induced–dipole (dispersion forces)
4. Hydrogen “bonds.”
Vapor Pressure
Boiling Point
Viscosity
Surface Tension
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FEATURES OF A PHASE DIAGRAM
• Triple point: temperature and pressure at which all
three phases are in equilibrium.
• Vapor-pressure curve: generally as pressure
increases, temperature increases.
• Critical point: critical temperature and pressure for
the gas.
• Melting point curve: as pressure increases, the
solid phase is favored if the solid is more dense than the
liquid.
• Normal melting point: melting point at 1 atm.
CHAP 12: Properties of Solutions
Supercritical Fluid
• A state of matter beyond the critical point
that is neither liquid nor gas
• SOLUBILITY RULES – Chapter 4
• SOLUBILITY – factors that affect
– Pressure (HENRY’S LAW)
– Temperature
• CONCENTRATION
– Molecular weight
• COLLIGATIVE PROPERTIES
– Molecular solutions
– Ionic solutions
• COLLOIDS
CONCENTRATION
is either based on weight or volume
1. ppm (parts per million) (based on weight)
2. ppb (parts per billion) (based on weight)
3. Mole Fraction ………. (based on weight)
4. Molarity ……………. (based on volume)
5. Molality …………….. (based on weight)
Solutions, Mixtures, & Colloids
(a) A MIXTURE is heterogeneous
(b) A SOLUTION is homogeneous
(c) A colloidal dispersion is between a
solution and a mixture
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COLLIGATIVE PROPERTIES
(determination of molecular weight
and/or van’t Hoff i facror)
COLLIGATIVE PROPERTIES
(determination of molecular weight)
of non Electrolytes
Vapor Pressure LOWERING
∆P = x2 P°1
Boiling Point ELEVATION
∆Tb = Kb m
Freezint Point LOWERING
∆Tf = Kf m
Osmotic Pressure
Chap 14:
π = MRT
of Electrolytes
Vapor Pressure LOWERING
∆P = i x2 P°1
Boiling Point ELEVATION
∆Tb = i Kb m
Freezint Point LOWERING
∆Tf = i Kf m
Osmotic Pressure
Part I:
The Rates of Reaction Depends On
CHEMICAL KINETICS
CHEMICAL KINETICS DEALS WITH
1. How FAST {Speed like miles per hour
and
2. By what MECHANISM
Does a
Reaction Take Place ?
Determine the rate law for
π = i MRT
1.
2.
3.
4.
Nature of Reactants {fixed
Concentration of Reactants {variable
Temperature {variable
Etc. {variable
• Catalysts
• Particle Size
• Photochemical
All radioactive decay is 1st order
ln
[ A ]0
 kt
[ A]
Half-life {t1/2 }
is defined as the time required for onehalf of a reactant to react
[ A ]0
ln
 k t 1/2
[ 12 A] 0
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TEMPERATURE AND RATE
Arrhenius developed a mathematical equation
between k (rate constant) and Ea (energy of activation)
AS TEMPERATURE INCREASES, SO
DOES THE REACTION RATE.
THIS IS BECAUSE k IS TEMPERATURE
DEPENDENT.
Part II
Reaction mechanism.
• Reactions occur through several discrete steps
The following mechanism has been proposed
for the gas phase reaction of H2 with ICl
• Each of these is known as an elementary reaction or
elementary process
• The molecularity of a process tells how many
molecules are involved in the process
------------------------------------------------------------------------------------------------------------------------
The Collision Model
Molecules can only react if they collide
H2 (g) + ICl (g)  HI (g) + HCl
HI (g) + ICl  I2(g) + HCl (g)
(a) H2 (g) + 2 ICl (g)  I2 (g) + 2 HCl
(b) HI
(c) Step 1: Rate = k [H2][ICl]
Step 2 : Rate = k[HI][ICl]
(d) Slowest step determines rate law therefore
reaction is 1st order wrt H2 and ICl and 2nd
order overall
(a)
(b)
(c)
(d)
H2 (g) + ICl (g)  HI (g) + HCl
HI (g) + ICl  I2(g) + HCl (g)
Write the balanced equation for the reaction
Identify intermediates in mechanism
Write rate laws for each elementary step
If the 1st step is the slowest what is the rate
law for the overall reaction ?
EQUILIBRIUM Chapters 15 -17
Molecular ---------------- Chap 15
Ionic (Weak Acid / Base ) - Chap 16
Ionic (“Insoluble” Salts) - Chap 17
•
•
Homogeneous Equilibrium
Heterogeneous Equilibrium
• Equilibrium Calculations
• Le Châtelier’s Principle
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The equilibrium-constant expression depends
only on the stoichiometry not on the mechanism
Productsp
K
Reactantsr
Pure Solids and Liquids do not have a concentration
so they do not appear in the equilibrium expression
Catalysts increase the rate of both the
forward and reverse reactions
• Equilibrium is achieved faster, but the
equilibrium composition remains unaltered
• Therefore Catalysts have No effect on the
equilibrium concentrations
A Change in Temperature Changes
the equilibrium constant
Endothermic processes are favored with increase
Exothermic processes are favored with decrease
Chapter 16 is a continuation of chapter 15
For WEAK Acids & WEAK Bases
• What is an Acid ?
• What is a Base ?
1. ARRHENIUS Acid / Base
2. BRONSTED-LOWRY Acid / Base
3. LEWIS Acid / Base
The Initial Change Equilibrium method
for 2 SO2 (g) + O2 (g) = 2 SO3 (g)
moles SO2
moles O2
Initially
1.00
1.00
Change
- 0.925
- ½ (0.925)
Equilibrium 0.075
0.537
K eq 
moles SO3
0
+0.925
0.925
2

0.925 
[ so3 ]2


 2.8 x10 2
[ so2 ]2 [o2 ]  0.075  2  0.537 

 

Relationship between Kc and Kp
Kp = Kc (RT)n
Where
n = (moles of gaseous product)
− (moles of gaseous reactant)
Kp = Kc
For H2(gas) + I2(gas) = 2 HI(gas)
WHY ?
pH and the ion product constant of water
pH = –log [H+]
pOH = –log [OH-]
pK = - log Kw
Kw = [H+] [OH–] = 1.0 x10–14
pH + pOH = 14
When [H+] = [OH–] have a “neutral” solution
and pH = pOH = 7
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pH of some common fluids
•
•
•
•
•
•
•
•
•
•
•
•
Gastric juice in stomach
Lemon juice
Vinegar
Grapefruit juice
Orange juice
Urine
Saliva
Milk
Blood
Tears
Milk of Magnesia
Household Ammonia
1.0 – 2.0
2.4
3.0
3.2
3.5
4.8 – 7.5
6.4 – 6.9
6.5
7.35 – 7.45
7.4
10.6
11.5
Acid–Base Properties of Salts
SALTS IN WATER EITHER FORM
neutral solutions
basic solutions
or acidic solutions
Will the reaction
Ba2+ (aq) + SO42– (aq)  BaSO4(s)
occur?
If
ACID
Ka
7.1 x 10 –4
HF
4.5 x 10 –4
HNO2
1.7 x 10 –4
HCO2H (formic)
–5
CH3CO2H (acetic) 1.8 x 10
4.9 x 10 –10
HCN
CONJ. BASE
F–
–
NO2
HCO2 –
CH3CO2 –
CN –
Kb
1.4 x 10 –11
2.2 x 10 –11
5.9 x 10 –11
5.6 x 10 –10
2.0 x 10 –5
Relationship between Ka and Kb
Ka x Kb = Kw
Buffer Solution: is a solution of
• a Weak acid
or
• a Weak base
and
• its salt
BOTH ACID (OR BASE) & SALT
MUST BE PRESENT
Ba2+ (aq) + SO42– (aq)  BaSO4(s)
Ksp = [Ba+2] [SO–]
[Ba+2] [SO–] > Ksp it will
Moles Ba2+ = 0.0010M x 0.050L = 5 x 10-5
[Ba2+ ][SO42- ] = (5 x 10-4)(5 x 10-5) = 2.5 x 10-8
Moles SO42- = 0.00010M x 0.050L = 5 x 10-6
From Table: Ksp = 1.1 x10-10
[Ba2+ ] = (5 x 10-4)
[SO42- ] =( 5 x 10-5)
{100mL of solution
“
Since the ion product (2.5 x 10-8) is greater than the
Ksp, (1.1 x10-10)
a precipate of barium sulfate is expected
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The concentration of calcium ion in the blood plasma is
0.0025 M. If the concentration of oxalate ion is 1.0 x 107 do you expect calcium oxalate to precipitate?
Henderson–Hasselbalch equation.
Consider the equilibrium constant expression for
the dissociation of a generic weak acid, HA:
Ksp for calcium oxalate is 2.3 x 10-9
Since the ion product (2.5x10-10) is smaller than
Ksp, (2.3 x 10-9)
H3O+ + A
HA + H2O
[Ca2+][C2O42-] = (0.0025)(1.0 x 10-7) = 2.5x10-10
Ka =

[H3O+] [A] = [H O+] [A ]
3
[HA]
[HA]
Taking the negative log of both sides
log Ka = log [H3O+] + log
do not expect precipitation to occur
pKa
pH
base
[A]
[HA]
.
acid
Chap 20 ELECTROCHEMISTRY
Two Parts To
ELECTROCHEMISTRY
“Ordinary” Chemical Reactions produce Heat
Part 1 SPONTANEOUS Process
Zn(s) + HCl(aq) → H2(g) + ZnCl2(aq) + HEAT
A Chemical Reaction Produces Voltage
BATTERYS
----------------------------------------------------------------------------------------------
Electrochemical Reactions produce Voltage
Zn(s) + HCl(aq) → H2(g) + ZnCl2(aq) + VOLTS
Two “types” of batteries
Part 2
NONSpontaneous Process
Voltage causes a Chemical Reaction
ELECTROLYSIS
Two (2) electrolytes with a Salt Bridge
1. Two (2) electrolytes  Uses a salt bridge
Negative electrode on
Zn/ Zn2+ // H+ / H2
left by convention
Zn  Zn2+ + 2 e---------------------------------------------------------------------------------------------------
2. One (1) electrolyte  Does Not need a salt bridge
Pb / H2SO4 (aq) / PbO2
Pb  PbSO4
PbO2  PbSO4
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Zn/ Zn2+ // H+ / H2
ANODE: Zn
(the negative electrode)
Zn (s)  Zn 2+ (aq) + 2eCATHODE: H2 {positive electrode
2 H+(aq) + 2e-  H2(gas)
The overall electrochemical reaction
Zn(s) + 2 H+(aq)  Zn2+ (aq) + H2(gas)
ANODE: Pb
(the negative electrode)
Pb (s) + SO42- (aq)  PbSO4 (s) + 2e-
One (1) Electrolyte: Sulfuric Acid
No need for a salt bridge
ELECTROLYSIS
External force causes chemical reactions
CATHODE: PbO2 (lead IV oxide) {positive electrode
PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- 
PbSO4(s) + 2H2O(l)
The overall electrochemical reaction
PbO2(s) + Pb (s) + 2 SO42-(aq) + 4 H+(aq) 
2 PbSO4(s) + 2 H2O(l)
How Much Ni would be electroplated if 1 Faraday
was passed through cell?
1 Ni+2 + 2e  1 Ni
Two (2) Faradays would produce 1 mole (58.71grams)
so one (1) Faraday would produce ½ mole (29.36 gms)
--------------------------------
How long would it take for 29.36 grams to be deposited?
1 Faraday = 96,500 coulombs
Coulombs = Amp- seconds
Q = I (amps) x t (sec)
How Much Ni
would be
electroplated
on the Steel
How long would it take for 29.36 grams of Ni to be
electroplated if 2 amps were passed through the cell?
1 Faraday = 96,500 coulombs
Coulombs = Amp- seconds
Q = I (amps) x t (sec)
96,500 coulombs = 2 (amps) x ? (sec)
Time = ½ (96,500) = 48,250 sec
= 804 minutes
= 13.4 hours
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Will Fe 2+(aq) oxidize Al(s) ?
Or……will the following reaction occur?
3 Fe 2+ (aq) + 2 Al (s)  3 Fe (s) + 2 Al+3 (aq)
In order to answer that question, calculate E°cell
2e- + Fe 2+ (aq)  Fe (s) Ered = -0.44
Eoxid = + 1.66
Al (s)  Al+3 (aq) + 3eEcell = Eoxid + Ered = +
Since Ecell is positive, YES the reaction will occur
---------------------------------------------------------
Balancing Oxidation-Reduction Reactions
In Acidic Solution
1. Write ion half reactions
2. Balance atoms
3. Add water (if needed)
4. Add electrons
In Basic Solution {one more step is added
5. OH- is added to equation
End Review
Example of Balancing OxidationReduction Reaction:
NO2- (aq) + Al (s)  NH3 (g) + Al(OH)4- (aq)
in Acidic Solution
The two half reactions are
NO2-(aq)  NH3(g)
Al(s)  Al(OH)4- (aq)
7 H+(aq) + NO2-(aq)  NH3(g) + 2 H2O
4 H2O +Al(s)  Al(OH)4- (aq) + 4 H+(aq)
NO2-(aq)  NH3(g)
Al(s)  Al(OH)4- (aq)
Adding water
NO2
 NH3 (gas) + 2 H2O
4 H2O +Al (s)  Al(OH)4- (aq)
Adding H+
7 H+ (aq) + NO2- (aq)  NH3 (gas) + 2 H2O
4 H2O +Al (s)  Al(OH)4- (aq) + 4 H+ (aq)
- (aq)
6e + 7H+(aq) + NO2-(aq)  NH3(aq) + 2H2O
8H2O +2Al (s)  2Al(OH)4- (aq) + 8H+ (aq) + 6e
Adding electrons
6e + 7
+ NO2- (aq) NH3 (g) + 2 H2O
4 H2O +Al(s)  Al(OH)4- (aq) + 4 H+(aq) + 3e
To balance electrons multiply 2nd Eq by 2
H+ (aq)
8H2O +2Al (s)  2Al(OH)4- (aq) + 8H+ (aq) + 6e
Adding two half reactions gives:
6 H2O+ NO2-(aq) +2Al(s) NH3(g)+ 2Al(OH)4-(aq)+ H+(aq)
Which is Balanced in acidic solution
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Balancing Oxidation-Reduction
Reactions in Basic Solution
OH-(aq) + 5H2O + NO2-(aq) +2Al(s) 
NH3(gas) + 2Al(OH)4- (aq)
6H2O + NO2- + 2Al  NH3 + 2Al(OH)4- + H+
The equation is now balanced in basic
solution
Are you sure ?
Count elements on both sides of reaction
Add OH-(aq) to BOTH sides
OH-(aq)  OH-(aq)
OH- + 5 H2O + NO2- + 2Al  NH3 + 2 Al(OH)4-
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