Download 3 - College of Arts and Sciences

11/14/2011
Brief Review for Exam 3
Chapter4
Molarity
& Solution Stoichiometry
Chapter 5
Thermo chemistry
Chapter 6
Electrons
Chapter 4
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What is a Solution?
A HOMOGENEOUS Mixture
of TWO or MORE Substances
•
•
In which the CONCENTRATION varies
Concentration is expressed in terms of
Solvent + Solute = Solution
MOLARITY
ELECTROLYTES
• ELECTROLYTE – A substance that
dissolves in water to produce IONS
Example: HCl(aq), NaOH(aq), NaCl(aq)
• NON ELECTROLYTE A substance that
DOES NOT produce IONS, remain as
molecules, when dissolved in water.
1. STRONG ELECTROLYTES
A substance that completely ionizes
Strong Acids [7]
Strong Bases [7]
Soluble Salts
2. WEAK ELECTROLYTES
A substance that does NOT completely ionize
Example: sugar
Molarity (M)
FORMULA WEIGHT
is the Conversion Factor between
GRAMS & MOLES
Memorize the definition
Molarity =
moles of solute
Liters of solution
or
MOLARITY
is the Conversion Factor Between
Moles Solute = Molarity x Liters of solution
LITERS & MOLES
moles = M x V
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NaCl (solid) + H2O (liquid) NaCl (aq)
58.5 grams + water
Molarity =
58.5 grams solid NaCl + water
moles of solute
Liters of solution
molesNaCl = 58.5 grams x
1mole
= 1.00mol
58.5grams
Formula weight of NaCl= 23.0 + 35.5= 58.5 g/mol
moles of solute
1.00
Molarity =
=
= 1.00M
Liters of solution 1.00
For Example
Pour 51.0 mL of 2.00 M NaOH(aq) into
50.0 mL of 2.00 M HCl(aq)
1. Is there a Limiting Reagent ?
If so, what is it?
------------------------------------
2. What is the Theoretical Yield
of water in grams?
-------------------------------
3. If 5.85 grams of NaCl is formed,
what is the % Yield ?
a Solution
What is in solution?
a One Liter Salt Solution
NaCl is completely soluble in water so
1 NaCl (solid) + water 1 Na+(aq) +1 Cl-(aq)
No NaCl (solid) is in solution 
1.00 mol of Na+(aq) & 1.00 mol of Cl -(aq) are in
solution for a total of 2.00 moles of ions
Molarity of Na (aq) + = 1.00 M
Molarity of Cl(aq) − = 1.00M
How are Moles determined from Molarity?
Moles of Solute = Molarity x (Volume in Liters)
------------------------------------Calculate the number of moles of HCl
in 50.0 mL of 2.00 M HCl(aq)
Moles = M x V = (0.0500)x(2.00) = 0.100
-----------------------------------Calculate the number of moles of NaOH
in 51.0 mL of 2.00 M NaOH
Moles = M x V = (0.0510)x(2.00) = 0.102
Write & Balance Reaction
For Chemical Reactions in Aqueous Solutions
CONSIDER
1. Theoretical Yield
2. % Yield
3. Limiting Reagent
USING MOLARITY
moles of solute
Molarity =
Liters of solution
1 HCl (aq) + 1 NaOH (aq) 1 H2O + 1 NaCl (aq)
or
0.1HCl(aq) + 0.102 NaOH(aq) 0.1H2O + 0.1NaCl(aq)
What is the Limiting
Reagent ?
--------------------------
What is the Theoretical
Yield of water?
----------------------
5.85 grams of NaCl
formed, % Yield is
HCl(aq)
-----------------------0.1 mole = 1.8 grams
-----------------------5.85 g NaCl = 0.1 mole
Therefore 100% yield
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Molarity =
moles of solute
Liters of solution
Preparing solutions of known concentrations
1. Weigh solute and calculate moles
2. Add water until desired volume of solution obtained
Preparing solutions By Dilution
(adding water does not change the number of moles of solute)
M1 V1 = M2 V2 = moles of Solute
Chapter 5
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•
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Preparing solutions By Dilution
Start with 50.0 mL of 2.00 M HCl(aq)
Add water until have 100.0 mL of solution
What is the Molarity of the new solution?
(1) M1 V1 = M2 V2 = moles of Solute
(2) (2.00)(50.0) = ( M2)(100.0) = ???
(3) M2 = 1.00
Energy
Thermo chemistry
5.1 Omit
5.2 Omit
5.3 Omit
5.4 Enthalpy of Reactions
5.5 Calorimetry
5.6 Hess’s Law
5.7 Enthalpies of Formation
HEAT LOST = HEAT GAIN
Units of Energy: 1 cal = 4.184 joule
Energy is constant (system + surroundings)
------------------------------------------------------
System = the portion of the universe that we
single out for study
Surroundings = everything outside the
system
Physical Changes
Part 1
Something is gaining Heat
While Something else looses Heat.
If you know one of these
then you know the other
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Physical Changes
Part 2
Energy Change WITHIN a state {No Phase Change}
Specific heat =
quantity of heat transferred
(grams of substance) x (temperatu re change)
How much Energy required to heat 1.0
gram of ice at –10oC to steam at 110oC
Ice Ice Liq Liq Gas Gas
-10oC 0oC
0oC
100oC 100oC
110oC
The heat required to raise the temperature of one gram
of a substance by one degree C
For Water : S.H. = 4.18
What information do you need to work
this problem
JOULES
(Grams) ( ∆T)
How much Energy required to heat 1.0
gram of ice at –10oC to steam at 110oC
How much Energy required to heat 1.0
gram of ice at –10oC to ice at 0oC ?
• Heat of fusion = 6.008 kJ / mole
• Heat of vaporization = 40.67 kJ / mole
• Specific heat: Ice 2.092 J / g - K
•
Liq 4.184 J / g – K
•
Steam 1.841 J / g - K
Heat required to melt 1.0 grams of H2O ?
(1 . 0 g H 2 O ) x
(0.0555 mole) x
1 mole
= 0 . 0555 mole
18.0
g
6.01 KJ
= 0.3338885 KJ
1 mole
Let UNITS solve the problem.
Joules = (Specific Heat)x(grams)x(change in Temp)
= (2.1) x (1.0) x (10) = 21 Joules
How much Energy required to heat 1.0
gram of water at 0oC to water at 100oC
Let UNITS solve the problem.
Joules = (Specific Heat)x(grams)x(change in Temp)
= (4.18) x (1.0) x (100) = 418 Joules
∆H = 0.33 kJ
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Heat required to vaporize 1.0 grams of H2O
(1 . 0 g H 2 O ) x
(0.0555 mole) x
1 mole
= 0 . 0555 mole
18.0
g
40.67 KJ
= 0.592777 KJ
1 mole
How much Energy required to heat 1.0
gram of steam at 100oC to steam at 110oC
Let UNITS solve the problem.
Joules = (Specific Heat)x(grams)x(change in Temp)
= (1.84) x (1.0) x (10) = 18 Joules
∆H = 0.59 kJ
How much Energy required to heat 1.0
gram of ice at –10oC to steam at 110oC
CHEMICAL REACTIONS
Part 1
Reactants Products +/-ENERGY
Determination of Heats of Reaction
Add all the numbers
Using
21 J + 0.33 kJ + 418 J + 0.59 kJ + 18 J
THE DIRECT METHOD
EXPERIMENTAL
Go to Lab and use a Calorimeter
How many significant figures in answer ?
EXPERIMENTAL
Two (2) types of Calorimeters
In Both Types
Heat LOST = Heat GAINED
1. OPEN
{ to the atmosphere
&
2. CLOSED {to the atmosphere
(mainly for gas reactions)
1.435 g of naphthalene (C10H8)was burned in a constant
volume bomb calorimeter. The temperature of the water
rose from 20.17 to 25.84 oC. If the mass of the water
was exactly 2000 g and the heat capacity of the
calorimeter was 1.80kJ/oC find the heat of combustion
Write and balance reaction
1 C10H8 + 12 O2 10 CO2 + 4 H2O + HEAT
Heat Lost by =
Chemical
Reaction
Heat Gain by
1. Water +
2. Calorimeter
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HEAT LOST = HEAT GAIN
1. Heat Gain by Water = S. H. x grams x Temp
1 C10H8 + 12 O2 10 CO2 + 4 H2O
∆H = ??
Heat lost by 1.435 g of naphthalene = 5.76 x 104 J
qwater = (4.184)(2000)(25.84 - 20.17) = 4.74 x 104 J
MW of naphthalene (C10H8) = 128.2 g / mol
2. Heat Gain by Calorimeter = Heat Cap x Temp Change
1.435 g / (128.2 g / mol) = 0.01119 mole
qCalorimeter = (1800)(25.84 - 20.17) = 1.02x 104 J
How much heat for 1 mole of naphthalene ?
5.76 x 104 J / 0.01119 mole = 5.1458895 x 106 J/mol
Total Heat Gained = Water + Calorimeter
= 4.74 x 104 +1.02 x 104 = 5.76 x 104 J
Heat of combustion of naphthalene
∆H = - 5.15 x 103 kJ/mole
CHEMICAL REACTIONS
Part 2
Calculate the heat given off for reaction
N2(gas) + 3 H2(gas) 2 NH3 (gas)
Example 1:
Determination of Heats of Reaction
Using
THE INDIRECT METHOD
MATHEMATICAL
Using HESS’S Law:
ENTHALPY CHANGES ARE ADDITIVE
Example 2: Calculate the heat of vaporization of CS2
Given ∆Hf for CS2 (liq) = 88 kJ / mole
and ∆Hf for CS2 (gas) = 117 kJ / mole
Given: ∆Hf for NH3 (gas)= -46.19 kJ / mole
∆Hf for N2(gas) = ?
∆Hf for H2(gas) = ?
Therefore heat given off = 46.19 kJ x 2 =
92.38kJ
Calculate [using Hess’ Law] the heat
of reaction for CO(g) + ½ O2(g) → CO2(g)
Example 3:
What DATA Do You Need From Table ?
(1) C + 2 S CS2 (liq) = 88 kJ
(2) C + 2 S CS2 (gas) = 117 kJ
(3) CS2 (liq) C + 2 S = - 88 kJ
Add equations (2) and (3)
CS2 (liq) CS2 (gas)
∆Hvao = 117 – 88 = 29kJ
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∆Hf From Table
From Enthalpy of Formation Table
Substance
∆Hf (kJ/mol)
- 393.5
- 110.5
Formula
Carbon dioxide
CO2
Carbon monoxide
CO
WRITE AND BALANCE REACTIONS
Formation of CO (g) is :
1. C(s)+ ½ O2(g) → CO(g) ∆H = - 110.5 kJ
Write & Balance FORMATION Reactions
Formation of CO2(g) is :
1. C(s)+ ½ O2(g) → CO(g) ∆H = - 110.5 kJ
2.
C(s)+ O2(g) → CO2(g) ∆H = - 393.5 kJ
Want CO (g) + ½ O2(g) → CO2(g)
1. C(s)+ ½ O2(g) → CO(g) ∆H = - 110.5 kJ
REWRITE Eq 1
1b. CO (g) → C(s) + ½ O2(g)
∆H = + 110.5 kJ
also
2. C(s)+ O2(g) → CO2(g) ∆H = - 393.5 kJ
2.
C(s)+ O2(g) → CO2(g) ∆H = - 393.5 kJ
CO (g) → C(s) + ½ O2(g)
∆H = + 110.5 kJ
C(s)+ O2(g) → CO2(g) ∆H = - 393.5 kJ
Add Equations To Get :
CO (g) + ½ O2(g) → CO2(g)
Add ∆H ‘s To Get :
∆H = + 110.5 kJ - 393.5 kJ = -283 kJ
Chapter 6
ELECTRONS
1. HOW MANY ARE THERE ?
2. ELECTRON CONFIGURATION
3. ORBITAL DIAGRAM
• HUND’S Rule
4. QUANTUM NUMBERS (FOUR)
• PAULI Principle
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ORBITAL DIAGRAMS
Review
Number of electrons
H
He
Li
Be
B
C
Electron Configuration
1S1
1S2
1S2
1S2
1S2
1S2
(1 )
(2 )
(3 )
(4 )
(5 )
(6 )
2S1
2S2
2S2 2P1
2S2 2P2
P 6.62 How many unpaired electrons in
each of the following atoms ?
(a) C
(b) Cl
(c) Ti
(d) Ga
(e) Rh
(f) Po
1S
H (↑ )
He ( ↑↓)
Li ( ↑↓)
Be ( ↑↓)
B ( ↑↓)
C ( ↑↓)
2S
(↑ )
( ↑↓)
( ↑↓)
( ↑↓)
2P
3S
(↑ )( ) ( )
(↑ )(↑ )( ) NOTE!
QUANTUM NUMBERS
Each electron is assigned FOUR
1. The Principal Quantum Number,
n
n = 1, 2, 3, 4, 5, 6, or 7
2. The Angular Momentum Quantum Number l
l = n – 1, n – 2, ….
(a)2 unpaired
(b) 1unpaired
(c) 2 unpaired
(d) 1 unpaired
(e) 3 unpaired
(f) 2 unpaired
ml
3. The Magnetic Quantum Number,
m = - l to + l
4. The Spin Quantum Number
S
S = + ½ or - ½
Quantum Numbers
N (7) 1s2
(↑↓ )
2s2
( ↑↓)
2p3
( ↑)
2
2
(a) n = 2 m = ½
(a)
1
1
(b) n = 5 l = 3
(b) 14
+1
(c) n = 4 l = 3 m = -3
(c)
2
½
(d) n = 4 l =1 m = 1
(d)
2
2
2
l
0
0
1
0
0
-1
0
½ -½
½
½
m =0
0
s =½ -½
2
(↑ ) (↑ )
n =1 1
=0 0
P6.56 What is the maximum number of
electrons in an atom that can the following
quantum numbers
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If an atom absorbs energy, an
electron in a lower energy level will
jump to a higher energy level.
RELATIONSHIP BETWEEN
C λ and ν
Relationship Between Εnergy
wavelength (λ
λ) and frequency
(ν
ν)
C = λ x ν
UNITS Solve Problems !!
Ε ∝ ν (directly)
Ε ∝ 1 / λ (inversely)
1
meters
= meters ×
sec
sec
Which wave has the higher frequency?
If λ = 1000 meters (AM Radio)
what is the frequency (ν
ν ) of
radiation ?
UNITS Solve Problems !!
C = λ x ν
The lower wave has a longer wavelength
(greater distance between peaks). The
lower wave has the lower frequency, and
the upper one has the higher frequency
3.00 x108
ν=
C
λ
=
meters
1
= 1000 meters × ?
sec
sec
3.00x108 m / s
= 3.00 x 105 sec - 1
1000
m
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