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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
I. Spontaneous Processes and Entropy.
A. A spontaneous process occurs without outside intervention; it
may either be fast or slow.
1. Examples:
a) A ball rolls down a hill but never spontaneously rolls back
up the hill.
b) If exposed to air and moisture, steel rusts
spontaneously but the reverse process does not naturally
occur.
c) In nature, processes spontaneously proceed toward the
states that have the highest existing probabilities of
existing.
d)
Ssolid < Sliquid << Sgas
B. Positional Entropy.
For each of the following pairs, choose the substance with the
higher positional entropy (per mole) at a given temperature.
1. Solid CO2 and gaseous CO2
c) A gas fills a container uniformly. It never spontaneously
collects at one end of the container.
d) Heat always flows from a hot object to a cold one. The
reverse process never occurs spontaneously.
e) Wood burns exothermically to form carbon dioxide and
water, but wood is not formed when the two compounds
are heated together.
f) At temperatures below 0 C, water freezes. At
temperatures above 0 C, water melts.
2. N2 gas at 1 atm and N2 at 1.0 x 10
-2
atm
C. Predicting Entropy Changes.
Predict the sign of the entropy change for each of the following
processes.
1. Solid sugar is added to water to form a solution.
2. Originally, chemists thought spontaneity was reflective of
exothermic reactions only.
3. Scientists have since concluded that the common
characteristic of all spontaneous reactions is an increase in
entropy.
a) Entropy (S) is said to be a measure of randomness or
disorder of a system.
* No accurate, simplistic definition exists.
b) Entropy describes the number of possible arrangements
that are available to a system existing in a given state.
2. Iodine vapor condenses on a cold surface to form crystals.
II. Entropy and the Second Law of Thermodynamics.
A. Reviewing the First Law of Thermodynamics.
1. The energy of the universe is constant; energy cannot be
created or destroyed but changed from one form to another.
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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
2. The first law provides a means for accounting for energy, it
gives no hint as to why a particular process occurs in a given
direction.
B. The Second Law of Thermodynamics.
1. This law states that in any spontaneous process there is
always an increase in the entropy of the universe.
2. Energy in the universe is conserved, but entropy is not. The
entropy of the universe is increasing.
III. The Effect of Temperature on Spontaneity.
A. Consider the following:
H2O(l) ------> H2O(g)
Water is the system, everything else is the surroundings.
1. For liquid water to vaporize, heat must be added
(endothermic). Explain in terms of entropy.
Suniv = Ssys + Ssurr
3. To predict spontaneity, we must know the sign of Suniv
 
a. If Sunivthenthe entropy of the universe
increases and the process is spontaneous in the direction
written.

 
b. If Suniv < 1, then the entropy of the universe
2. Explain condensation in terms of entropy.
decreases and the process is spontaneous in the opposite
direction.
c. If Suniv = 0, then the process has no tendency to
occur and the system is at equilibrium.
d. In a living cell, large molecules are assembled from
smaller ones. Is this process consistent with the second
law?
B. Conclusions.
1. Entropy changes in the surroundings are determined by heat
flow and determines the sign.
a. Endothermic:
b. Exothermic:
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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
2. The magnitude of Ssurr depends on the temperature.
a. Endothermic:
b. Exothermic:
C. Summary.
1. For an exothermic process, Ssurrpositive.
2. For an endothermic process, Ssurrnegative.
3.
3. The impact of the transfer of a given quantity of energy as
heat to and from the surroundings will be greater at lower
temperatures.
* Example:
IV. Free Energy.
A. Free energy (G) is a function that combines the system’s
enthalpy and entropy.
G = H - TS
4. Ssurr = _ Hsys
T
5. Determining Ssurr.
* In the metallurgy of antimony, the pure metal is
recovered via different reactions, depending on the
composition of the ore. For example, iron is used to
reduce antimony in the sulfide ores:
Sb2S3(s) + 3Fe(s) ----> 2Sb(s) + 3FeS(s) H = -125 kJ
Carbon is used as the reducing agent for oxide ores:
B. The free energy change (G) is a measure of the spontaneity of
a process and of the useful energy available from it.
G = H - TS
1. All of these quantities reflect the system (anytime in this
chapter that a subscript is not noted, it is understood to be
from the system’s perspective).
2. How does the equation relate to spontaneity?
- G = - H + S
and
Ssurr = - H
T
T
Sb4O6(s) + 6C(s) ----> 4Sb(s) + 6CO(g) H = 778 kJ
Calculate Ssurr for each of these at 25 C and 1 atm.
T
therefore
- G = - H + S = Ssurr + S = Suniv
T
T
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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
3. Conclusion: A process (at constant T and P) is spontaneous
in the direction in which the free energy decreases (-G
means + Suniv).
4. Two functions predict spontaneity.
a. The entropy of the universe (applies to all processes).
b. Free energy (at constant T and P).
* More useful to chemists.
5.
V. Entropy Changes in Chemical Reactions.
A. Standard Molar Entropies, So
1. In thermodynamics it is the change in a certain function that
is usually important.
* Absolute values for H and G cannot be determined.
2. The third law of thermodynamics states that at 0 K, the
entropy of a pure crystal is equal to 0.
* Because this provides a starting point to compare all
other entropies, an absolute entropy scale has meaning.
3. Standard molar entropy, So, is the entropy of one mole of a
substance in its standard state, which occurs when a
substance is heated from 0 K to 298 K at 1 atm of pressure.
B. Predicting Relative So Values of the System.
1. Temperature changes.
* For a given substance, So increases as the temperature
rises.
6. Free Energy and Spontaneity.
* At what temperature is the following process
spontaneous at 1 atm?
Br2(l) -----> Br2(g)
Ho = 31.0 kJ/mol and So = 93.0 J/K*mol
What is the normal boiling point of liquid Br2?
2. Physical states and phase changes.
* For a given substance, So increases as the substance
changes from a solid to a liquid to a gas (the change from
a liquid to a gas is greater than from a solid to a liquid).
3. Dissolution of a solid or a liquid.
a. The entropy of a dissolved solid or liquid solute is
greater than the entropy of the pure solute.
b. The type of solute and solvent and the nature of the
solution process affects the overall entropy change.
4. Dissolution of a gas.
* A gas always becomes more ordered when dissolved in a
liquid and gas.
5. Complexity of the element or compound.
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a. In general, difference in entropy values for substances
in the same phase are based on atomic size and molecular
complexity.
d) 1 mol KBr(s) or 1 mol KBr(aq)
b. For elements within a periodic group, those with higher
molar masses have higher entropy.
f) 1 mol HF(g) or 1 mol HI(g)
c. For compounds, the chemical complexity increases as
the number of atoms (ions) in a compound increases, and
so does the entropy.
d. For larger molecule, entropy increases with chain length.
6. Predicting the Sign of So.
* Predict the sign of So for each of the following
reactions.
a) The thermal decomposition of solid calcium carbonate:
CaCO3(s) -----> CaO(s) + CO2(g)
e) Sea water in midwinter at 2 C or in mid summer at 23 C
C. Calculating the Change in Entropy of a Reaction.
1. Because entropy is a state function, the entropy change for
a given reaction can be calculated by taking the difference
between the standard entropy values of products and those
of the reactants.
Soreaction =npSoproductsnrSoreactants
2. Calculating So.
a. Calculate So at 25 C for the reaction
2NiS(s) + 3O2(g) -----> 2SO2(g) + 2NiO(s)
Use the standard entropy values found in the appendix.
b) The oxidation of SO2 in air:
2SO2(g) + O2(g) -----> 2SO3(g)
7. Predicting Relative Entropy Values.
* Choose the member with the higher entropy in each of
the following pairs, and justify your choice.
a) 1 mol SO2(g) or 1 mol SO3(g)
b) 1 mol CO2(s) or 1 mol CO2(g)
c) 3 mol O2(g) or 2 mol O3(g)
b. Calculate So for the reaction of aluminum oxide by
hydrogen gas:
Al2O3(s) + 3H2(g) -----> 2Al(s) + 3H2O(g)
Use the standard entropy values found in the appendix.
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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
VI. Free Energy and Chemical Reactions.
A. Standard Free Energy.
1. Standard free energy (Go) is the change in the free energy
that will occur if the reactants in their standard states are
converted to the products in their standard states.
2. The value of Go tells nothing about the rate of a reaction,
only its eventual equilibrium position.
2. Solving Go Using Hess’s Law.
* Using the following data (at 25 C)
Cdiamond(s) + O2(g) -----> CO2(g)
Go = -397 kJ
Cgraphite(s) + O2(g) -----> CO2(g) Go = -394 kJ
Calculate Go for the reaction
Cdiamond(s) -----> Cgraphite(s)
B. Calculating Go as a State Function.
1. GoHoTSo
* Consider the reaction
2SO2(g) + O2(g) -----> 2SO3(g)
carried out at 25 C and 1 atm. Calculate Ho and So,
then calculate Go. Use the appendix in this book for
standard values.
3. Standard Free Energy of Formation (Gfo).
Go =npGfoproductsnrGforeactants
* Methanol is a high-octane fuel used in high-performance
racing engines. Calculate Go for the reaction
2CH3OH(g) + 3O2(g) -----> 2CO2(g) + 4H2O(g)
given the free energies of formation in this book.
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3rd 9-weeks Notes (The end is near…in more ways than one.)
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4. Free Energy and Spontaneity.
* A chemical engineer wants to determine the feasibility
of making ethanol (C2H5OH) by reacting water with the
ethylene (C2H4) according to the equation
C2H4(g) + H2O(l) -----> C2H5OH(l)
Is the reaction spontaneous under standard conditions?
3. This leads to the equation
G = Go + RTln(Q)
where Q is the reaction quotient, T is the Kelvin
temperature, R is the universal gas constant 8.314 J/mol K,
Go is the free energy at 1 atm, and G is the free energy at
a specified temperature.
4. Calculating Go.
* One method for synthesizing methanol (CH3OH) involves
reacting carbon monoxide and hydrogen gases:
CO(g) + 2H2(g) -----> CH3OH(l)
Calculate G at 25 C for this reaction where carbon
monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are
converted to liquid methanol.
VII. The Dependence of Free Energy on Pressure.
A. The equilibrium position represents the lowest free energy
value available to a particular reaction.
* Free energy changes throughout the course of a reaction
because it is pressure and concentration dependent.
B. The Effects of Pressure.
1. Consider 1 mole of a gas at a given T.
Slarge volume > Ssmall
volume
or Slow
pressure
> Shigh pressure
2. A more detailed argument (beyond this course) states that
G = Go + RTln(P)
o
where G is the free energy of a gas at 1 atm, G is the free
energy at pressure P, R is the universal gas constant, and
T is the Kelvin temperature.
C. The Meaning of G for a Chemical Reaction.
1. A system achieves the lowest possible free energy by going
to equilibrium, not by going to completion.
* Spontaneous reactions do not go to completion.
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2. Illustration.
4. Conclusions.
a. For nonstandard conditions NOT at equilibrium,
G = Go + RTln(Q)
b. For standard conditions at equilibrium
Go = -RTln(K)
VIII. Free Energy and Equilibrium.
* The Equilibrium Point.
1. The equilibrium point occurs at the lowest value of free
energy available to the reaction.
2. Recall the following:
a. If Q < K, then the reaction as written proceeds to the
right (Q/K <1).
b. If Q > K, then the reaction as written proceeds to the
left (Q/K > 1).
5. Free Energy and Equilibrium I.
* Consider the ammonia synthesis reaction
N2(g) + 3H2(g) <====> 2NH3(g)
where Go = -33 kJ per mole of N2 consumed at 25 C. For
each of the following mixtures of reactants and products
at 25 C, predict the direction in which the system will
shift to reach equilibrium.
a) PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00 x 10
-2
atm
b) PNH3 = 1.00 atm, PN2 = 1.00 atm, PH2 = 1.00 atm
c. If Q = K, then the reaction as written is at equilibrium,
and there is no net reaction in either direction (Q/K = 1).
3. The sign of G and the magnitude of Q/K are related
mathematically.
a. If Q/K < 1, then ln(Q/K) < 0; the reaction proceeds to
the right (G < 0).
b. If Q/K > 1, then ln(Q/K) > 0; the reaction proceeds to
the left (G > 0).
c. If Q/K = 1, then ln(Q/K) = 0; the reaction is at
equilibrium (G = 0).
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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
6. Free Energy and Equilibrium II.
* The overall reaction for the corrosion (rusting) of iron
by oxygen is
4Fe(s) + 3O2(g) <====> 2Fe2O3(s)
Using Hfo and So values in the appendix, calculate the
equilibrium constant for this reaction at 25 C.
IX. Free Energy and Work.
A. For a spontaneous reaction, G is the maximum work obtainable
from the system.
wmax = G
B. For a nonspontaneous process, G is the minimum work that
must be done to the system to make a change happen.
C. For Gsys = Hsys - TSsys, Gsys is the portion of the total
energy change that does the work.
* TSsys is given off as heat and is not usable.
D. In the real world, some free energy is always changed to heat
and is thereby unharnessed. Therefore, no process is 100%
efficient.
1. A spontaneous reaction will occur and can do work on the
surroundings.
2. A nonspontaneous reaction will not occur unless the
surroundings do work on it.
3. A reaction at equilibrium can no longer do work.
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3rd 9-weeks Notes (The end is near…in more ways than one.)
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I. Galvanic Cells.
A. The Galvanic Cell.
1. A galvanic cell is a device in which chemical energy is changed to
electrical energy.
2. A galvanic cell uses a spontaneous redox reaction to produce a
current that can be used to do work. The system does work on the
surroundings.
c)
+
2-
2-
H (aq) + 2CrO4 (aq) ----> Cr2O7 (aq) + H2O(l)
e. Half-Reaction Method for Balancing Redox Reactions.
* Acidic Solution.
-
3+
Cr(s) + NO3 (aq) ----> Cr (aq) + NO(g) [acidic]
3. A Review of Oxidation and Reduction and the Galvanic Cell.
a. A redox reaction involves the transfer of electrons from the
reducing agent to the oxidating agent.
b. Oxidation.
* Involves a loss of electrons.
* Increase in oxidation number.
* “To get more positive.”
* Basic Solution.
* Occurs at the anode of a galvanic cell.
c. Reduction.
* Involves a gain of electrons.
-
-
2-
-
MnO4 (aq) + I (aq) ----> MnO4 (aq) + IO3 (aq) [basic]
* Decrease in oxidation number.
* “To get more negative.”
* Occurs at the cathode of a galvanic cell.
d. Identification of Redox Components.
* Specify which of the following equations represents
oxidation-reduction reactions, and indicate the
oxidizing agent, the reducing agent, the species being
oxidized, and the species being reduced.
a)
CH4(g) + H2O(g) ----> CO(g) + 3H2(g)
b)
2AgNO3(aq) + Cu(s) ----> Cu(NO3)2(aq) + 2Ag(s)
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3rd 9-weeks Notes (The end is near…in more ways than one.)
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f. Major Redox Points to Remember.
* Any redox reaction can be treated as the sum of the
reduction and oxidation half-reactions.
* Mass (atoms) and charge are conserved in each halfreaction.
* Electrons lost in one half-reaction are gained in
the other.
* Even though the half-reactions are treated separately,
electron loss and electron gain occur simultaneously.
4. The Salt Bridge or the Porous Disk.
* These devices allow ion flow to occur (circuit completion)
without mixing the solutions. They are typically made of sodium
sulfate or potassium nitrate.
B. Cell Potential (Ecell)
1. Cell potential (electromotive force, emf) is the driving force in a
galvanic cell that pulls electrons from the reducing agent in one
compartment to the oxidizing agent in the other.
2. The volt (V) is the unit of electrical potential.
3. Electrical charge is measured in coulombs (C).
4. Therefore, a volt is 1 joule of work per coulomb of charge
transferred: 1 V = 1 J/C.
5. A voltmeter is a device which measures cell potential.
II. Standard Reduction Potentials.
A. Standard Cell Potentials.
1. The measured potential of a voltaic cell is affected by changed in
concentration of the reactants as the reaction proceeds and by
energy losses due to heating of the cell and external circuit.
2. In order to compare the output of different cells, the standard
cell potential (Eocell) is obtained at 298 K, 1 atm for gases, 1 M for
solutions, and the pure solid for electrodes.
B. The Standard Hydrogen Electrode.
1. This is considered the reference half-cell electrode, with a potential
equal to 0.00 V.
+
2. It is obtained when platinum is immersed in 1 M H (aq), through
which H2(g) is bubbled.
C. The Standard Electrode (Half-Cell) Potential (Ehalf-cell).
1. A standard electrode potential always refers to the half-reaction
written as a reduction.
-
Oxidized form + ne -----> reduced form
Eohalf-cell
2. Reversing a reaction changes the sign of the potential.
-
Reduced form + ne -----> oxidized form
-Eohalf-cell
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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
3.
o
E
o
cell
= E
b. Calculating an Unknown Eohalf-cell from Eocell.
o
reduction
+ E
oxidation
4. As the potential increases in value (more positive), the tendency to
occur increases (spontaneity occurs). Therefore, Eocell must be
positive if the cell is galvanic (voltaic).
5. Spontaneous reactions occur between a substance and those below it.
* A voltaic cell based on the reaction between aqueous Br2 and
vanadium (III) ions has Eocell = 1.39 V:
3+
2+
+
-
Br2(aq) + 2V (aq) + 2H2O(l) ----> 2VO (aq) + 4H (aq) + 2Br (aq)
What is the standard electrode potential for the reduction of
2+
VO
3+
to V ?
6. Although some half-reactions must be manipulated with coefficients,
NEVER MULTIPLY THE Eocell BY THE COEFFICIENT!!!
7. Refer to the Reduction Table in the book.
8. Problems.
a. Galvanic Cells.
* Consider a galvanic cell based on the reaction
3+
2+
Al (aq) + Mg(s) ----> Al(s) + Mg (aq)
Give the balanced cell reaction and calculate E cell.
o
D. Line Notations.
1. This is a short-hand abbreviation for a galvanic cell.
2. Key Parts:
a. The components of the anode compartment are written to the
left of the cathode compartment.
b. Double vertical lines separate the half-cells and represents the
wire and salt bridge.
c. Within each half-cell, a single vertical line represents a phase
boundary. A comma separates half-cell components in the same
phase.
* Consider a galvanic cell based on the reaction
-
+
-
-
2+
MnO4 (aq) + H (aq) + ClO3 (aq) ----> ClO4 (aq) + Mn (aq) + H2O(l)
d. Half-cell components appear in the same order as in the halfreaction, while electrodes appear at the extreme left and right
of the notation.
Give the balanced cell reaction and calculate Eocell.
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3. Examples:
3. Diagramming Voltaic Cells.
* In one compartment of a voltaic cell, a graphite rod dips into an
acidic solution of K2Cr2O7 and Cr(NO3)3; in the other, a tin bar
dips into a Sn(NO3)2 solution. A KNO3 salt bridge joins the halfcells. The tin electrode is negative relative to the graphite.
Diagram the cell, show balanced equations, and write the cell
notation.
E. The Complete Description of a Galvanic Cell.
1. A cell will always run spontaneously in the direction that produces a
positive cell potential.
2. A complete description of a galvanic cell always includes four items:
a. The cell potential (always positive for a galvanic cell) and the
balanced cell reaction.
b. The direction of electron flow, obtained by inspecting the halfreactions and using the direction that gives a positive Eocell.
c. Designation of the anode and cathode.
4. Description of a Galvanic Cell.
* Describe completely the galvanic cell based on the following halfreactions under standard conditions:
+
-
Ag + e -----> Ag
3+
Fe
-
+ e -----> Fe
2+
Eocell = 0.80 V (1)
Eocell = 0.77 V (2)
In addition, draw the cell and write the line notation.
d. The nature of each electrode and the ions present in each
compartment. A chemically inert conductor is required if none
of the substances participating in the half-reaction is a
conducting solid.
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III. Cell Potential, Electrical Work, and Free Energy.
A. Cell Potential.
1. The work that can be accomplished when electrons are transferred
(emf) is defined in terms of a potential difference (in volts) between
two circuits.
emf = potential difference (V) = work (J)
or 1 V = 1 J/C
charge (C)
c. If Ecell = 0, then G= 0 and the process is at equilibrium.
o
7. Calculating G for a Cell Reaction.
o
* Using the Reduction Table in your book, calculate G for the
reaction
2+
2+
Cu (aq) + Fe(s) ----> Cu(s) + Fe (aq)
Is the reaction spontaneous?
2. Work is viewed from the point of view of the system.
Therefore, E = -w/q
or
-w = qE
3. The maximum work is defined as
wmax = -qEmax
* Maximum work is an impossibility. In any real, spontaneous
process some energy is always wasted-the actual work realized
is always less than the calculated maximum.
4. The Faraday (F).
a. The faraday is defined as the charge of 1 mole of electrons.
b. F = 96,485 C/mol e
8. Predicting Spontaneity.
* Using the Reduction Table in your book, predict whether 1 M
-
5. The purpose of a voltaic cell is to convert the free energy change of
-
3+
HNO3 will dissolve gold metal to form a 1 M Au solution.
a spontaneous reaction into the KE of e moving through an external
circuit.
wmax = G
6. For a galvanic cell,
G = -nFE
max
or
o
o
G = -nFE
max
a. If Ecell > 0, then G< 0 and the process is spontaneous.
b. If Ecell < 0, then G> 0 and the process is nonspontaneous.
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HW # Handouts
B. The Relationship to the Equilibrium Constant.
1. Two relationships:
o
o
G = -RTlnK
G = -nFE
IV. Dependence of Cell Potential on Concentration.
A. The Effects of Concentration on E.
1. Think Le Chatelier’s Principle!
o
cell
2. For the cell reaction
2. Through substitution,
E
o
cell
2+
3+
2Al(s) + 3Mn (aq) ----> 2Al (aq) + 3Mn(s)
= RT/nF lnK
predict whether Ecell is larger or smaller than E
o
3. Calculating K and G from E
o
cases.
.
cell
2+
2+
* When cadmium metal reduces Cu in solution, Cd forms in
o
addition to copper metal. If G = -143 kJ, calculate K at 25 C.
What would be E
o
cell
in a voltaic cell that used this reaction?
3+
2+
3+
2+
E
o
cell
o
cell
= 0.48 V
for the following
a) [Al ] = 2.0 M, [Mn ] = 1.0 M
b) [Al ] = 1.0 M, [Mn ] = 3.0 M
B. The Nernst Equation.
1. The Nernst equation gives the relationship between the cell potential
and the concentrations of cell components.
o
E = E - RT/nF lnQ
at 25 C
o
E = E - 0.0591/n logQ
2. The Effects of Q.
a. When Q < 1 and [reactant] > [product], Ecell > E
o
cell
b. When Q = 1 and [reactant] = [product], Ecell = E
c. When Q > 1 and [reactant] < [product], Ecell < E
.
o
.
cell
o
cell
.
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3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
3. A cell in which the concentrations are not in their standard states
will continue to discharge until equilibrium is reached.
b. Describe the cell based on the following half-reactions:
+
* At this point, Q = K and Ecell = 0.
2+
Zn
4. At equilibrium, the components in the two cell compartments have
the same free energy, and G = 0. The cell can no longer do work!
5. Problems.
a. Consider a cell based on the reaction
2+
2+
Fe(s) + Cu (aq) ----> Fe (aq) + Cu (s)
2+
2+
If [Cu ] = 0.30 M, what [Fe ] is needed to increase Ecell by
0.25 V above E
o
cell
+
-
2+
VO2 + 2H + e -----> VO
where
-
+ 2e -----> Zn
+ H2O
o
E = 1.00 V
o
E = -0.76 V
T = 25 C
+
[VO2 ] = 2.0 M
+
[H ] = 0.50 M
2+
[VO ] = 0.01 M
2+
[Zn ] = 0.1 M
at 25 C?
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3rd 9-weeks Notes (The end is near…in more ways than one.)
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C. Concentration Cells.
1. Concentration cells are constructed with the exact same
half-reactions, with the exception of a difference in concentrations.
2. Voltages are typically small as electrons are transferred from the
cell of higher concentration to the cell of lower concentration.
3. E
o
cell
= 0.00 because conditions are not standard (not 1 M).
4. Calculating the Potential of a Concentration Cell.
-4
B. The Stoichiometry of Electrolysis.
1. Faraday’s Law of Electrolysis: the amount of a substance produced
at each electrode is directly proportional to the amount of electric
charge flowing through the cell.
2. The SI Unit of current is the ampere (A).
3+
* A concentration cell is built using two Au/Au half-cells. In
3+
3. Everything is the same as a galvanic cell except the signs of the
anode and cathode.
3+
half-cell A, [Au ] = 7.0 x 10 M, and in half-cell B, [Au ] =
-2
2.5 x10 M. What is Ecell, and which electrode is negative?
1 ampere = 1 coulomb/second or 1 A = 1 C/s
3. Applying the Relationship Among Current, Time, and Amount of a
Substance.
* Using a current of 4.75 A, how many minutes does it take to plate
1.50 g Cu onto a sculpture from a CuSO4 solution?
V. Electrolysis.
A. The Electrolytic Cell.
1. An electrolytic cell uses electrical energy to drive a nonspontaneous
process.
2. The process is called electrolysis, which involves forcing a current
through a cell to produce a chemical reaction for which the cell
potential is negative.
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HW # Handouts
I. Pressure.
A. The Barometer.
1. A barometer is a device to measure atmospheric pressure.
2. The barometer was invented in 1643 by a student of Galileo, Italian
scientist Evangelista Torricelli.
3. This barometer is constructed by filling a glass tube with liquid
mercury and inverting it in a dish of mercury.
* Diagram:
2. Mercury is used to measure pressure because of its high density. By
way of comparison, the column of water required to measure a given
pressure would be 13.5 times as high as a mercury column used for
the same purpose.
3. Common Units of Pressure.
* 1 atmosphere (atm)
=
760 mm Hg
760 torr
101,324 Pascals (Pa)
101.325 kPa
29.92 in Hg
14.7 lb/in
2
4. Pressure Conversions.
* The pressure of a gas is measured as 49 torr. Represent this
pressure in both atmospheres and pascals.
4. Atmospheric pressure results from the mass of the air being pulled
toward the center of the earth by gravity.
a. The weather affects the pressure reading.
b. Altitude also affects the pressure.
* At sea level, the height of the column of mercury is 760 mm.
* At the peak of Mt. Everest, the column measures 270 mm Hg.
B. Units of Pressure.
1. The open-ended manometer (barometer).
* Diagrams:
II. The Gas Laws of Boyle, Charles, and Avogadro.
A. Boyle’s Law.
1. Named after Irish chemist Robert Boyle (1627-1691).
2. Using a closed-ended J-tube, Boyle found that gas volume is inversely
proportional to its pressure.
3. With technology, Boyle’s law holds precisely only at very low
pressures.
4. If a gas obey Boyle’s law exactly, it is said to be “ideal.”
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5.
Boyle’s Law: PV = k
or
* He was the first person to fill a hot air balloon with hydrogen gas
on the first solo balloon flight.
P1V1 = P2V2.
6. Boyle’s Law I.
* Sulfur dioxide (SO2), a gas that plays a central role in the
formation of acid rain, is found in the exhaust of automobiles and
power plants. Consider a 1.53-L sample of gaseous SO2 at a
3
4
2. He found that the volume of a gas at constant pressure increases
linearly with the temperature of the gas.
a. The temperature was originally plotted in degrees C.
pressure of 5.6 x 10 Pa. If the pressure is changed to 1.5 x 10
Pa at a constant temperature, what will be the new volume of the
gas?
b. A plot of the gases showed that all volumes extrapolated to the
same temperature, -273.2 C.
* This temperature is also known as 0 Kelvin or absolute zero.
* The Kelvin scale was devised by English physicist William
Thomson, also known as Lord Kelvin, 50 years after Charles’s
conclusions.
3.
7. Boyle’s Law II.
* In a study to see how closely gaseous ammonia obey Boyle’s law,
several volume measurements were made at various pressures,
using 1.0 mol NH3 gas at a temperature of 0 C. Using the results
listed below, calculate the Boyle’s law constant for NH3 at the
Charles’s Law:
V/T = k or V1/T1 = V2/T2
* Temperature is in Kelvins.
4. Charles’s Law.
* A sample of gas at 15 C and 1 atm has a volume of 2.58 L. What
volume will the gas occupy at 38 C and 1 atm?
various pressures.
E
x
p
e
r
im
e
n
t
1
2
3
4
5
6
P
r
e
ssu
r
e(a
tm
)
0
.1
3
0
0
0
.2
5
0
0
0
.3
0
0
0
0
.5
0
0
0
0
.7
5
0
0
1
.0
0
0
0
V
o
lu
m
e(L
)
1
7
2
.1
8
9
.2
8
7
4
.3
5
4
4
.4
9
2
9
.5
5
2
2
.0
8
C. Avogadro’s Law.
1. Named in honor of Italian chemist Amadeo Avogadro (1776-1856).
B. Charles’s Law.
1. Named in honor of Jacques Charles (1746-1823).
2. He stated that at the same temperature and pressure, equal volumes
of different gases contain the same number of particles.
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HW # Handouts
3.
Avogadro’s Law: V/n = k
or V1/n1 = V2/n2
4. At 1 atm of pressure and 273 K, 1 mol of any gas occupies 22.4 L.
5. Avogadro’s Law.
* Suppose we have a 12.2-L sample containing 0.50 mol oxygen gas
(O2) at a pressure of 1 atm and a temperature of 25 C. If all
this O2 is converted to ozone (O3) at the same temperature and
pressure, what would be the volume of the ozone?
III. The Ideal Gas Law.
A. Summary.
*
Boyle’s Law:
V = k/P
(at constant T and n)
Charles’s Law:
V = kT
(at constant P and n)
Avogadro’s Law:
V = kn
(at constant T and P).
2. R is the universal gas constant.
* Values.
R = 0.08206 L atm/mol K = 8.314 L kPa/mol K =
62.4 L mm Hg/mol K
3. Real gases behave as ideal gases as they approach low pressures
(below 1 atm) and high temperatures.
4. Ideal Gas Law I.
* A sample of hydrogen gas (H2) has a volume of 8.56 L at a
temperature of 0 C and a pressure of 1.5 atm. Calculate the
moles of H2 molecules present in this sample of gas.
5. Ideal Gas Law II.
* Suppose we have a sample of ammonia gas with a volume of 3.5 L
at a pressure of 1.68 atm. The gas is compressed to a volume of
1.35 L at a constant temperature. Use the ideal gas law to
calculate the final pressure.
B. The Ideal Gas Law.
1. Combining the above laws we obtain the ideal gas law:
6. Ideal Gas Law III.
* A sample of methane gas that has a volume of 3.8 L at 5 C is
heated to 86 C at constant pressure. Calculate its new volume.
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7. Ideal Gas Law IV.
* A sample of diborane gas (B2H6), a substance that bursts into
flames when exposed to air, has a pressure of 345 torr at a
temperature of -15 C and a volume of 3.48 L. If conditions are
changed so that the temperature is 36 C and the pressure is 468
torr, what will be the volume of the sample?
2. Molar volume under these condition for an ideal gas is 22.42 L.
3. Gas Stoichiometry I.
* A sample of nitrogen gas has a volume of 1.75 L at STP. How
many moles of N2 are present?
4. Gas Stoichiometry II.
* Quicklime (CaO) is produced by the thermal decomposition of
calcium carbonate (CaCO3). Calculate the volume of CO2 at STP
8. Ideal Gas Law V.
* A sample containing 0.35 mol argon gas at a temperature of 13 C
and a pressure of 568 torr is heated to 56 C and a pressure of
897 torr. Calculate the change in volume that occurs.
produced from the decomposition of 152 g CaCO3 by the reaction
CaCO3(s) -----> CaO(s) + CO2(g)
IV. Gas Stoichiometry.
A. STP.
1. STP stands for standard temperature and pressure.
a. The temperature = 0 C = 273 K.
b. The pressure is 1 atm or an equivalent.
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5. Gas Stoichiometry III.
* A sample of methane gas having a volume of 2.80 L at 25 C and
1.65 atm was mixed with a sample of oxygen gas having a volume
of 35.0 L at 31 C and 1.25 atm. Them mixture was then ignited
to form carbon dioxide and water. Calculate the volume of CO2
formed at a pressure of 2.50 atm and a temperature of 125 C.
V. Dalton’s Law of Partial Pressure.
A. Dalton’s Law.
1. Named in honor of English scientist John Dalton.
a. He stated that gases mix homogeneously.
b. He also said that each gas in a mixture behaves as if it were the
only gas present.
c. Dalton’ Law: For a mixture of gases in a container, the total
pressure exerted is the sum of the pressures that each gas
would exert if it were alone.
d.
B. Molar Mass of a Gas and Gas Density.
1. Deriving the equations for molar mass of a gas and gas density:
P total = P1 + P2 + P3 +
......
e. The fact that the pressure exerted by an ideal gas is not
affected by the identity (composition) of the gas particle reveal
two things.
* The volume of the individual gas particles must not be
important.
* The forces among the particles must not be important.
2. Dalton’s Law I.
* Mixtures of helium and oxygen are used in scuba diving tanks to
help prevent “the bends.” For a particular dive, 46 L He at 25 C
and 1.0 atm and 12 L O2 at 25 C and 1.0 atm were pumped into a
tank with a volume of 5.0 L. Calculate the partial pressure of
each gas and the total pressure in the tank at 25 C.
2. Gas Density/Molar Mass.
* The density of a gas was measured at 1.50 atm and 27 C and
found to be 1.95 g/L. Calculate the molar mass of the gas.
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B. Mole Fraction.
1. A mole fraction () is the ratio of the number of moles of a given
component in the total number of moles in the mixture.
2. Arriving at the mole fraction equation:
4. Dalton’s Law III.
* The mole fraction of nitrogen in the air is 0.7808. Calculate the
partial pressure of N2 in air when the atmospheric pressure is
760. torr.
5. Gas Collection over Water.
* A sample of solid potassium chlorate (KClO3) was heated in a test
tube and decomposed by the following reaction:
2KClO3(s) -----> 2KCl(s) + 3O2(g)
The oxygen produced was collected by water displacement at 22
C at a total pressure of 754 torr. The volume of the gas
collected was 0.650 L, and the vapor pressure of water at 22 C is
21 torr. Calculate the partial pressure of O2 in the gas collected
3. Dalton’s Law II.
* The partial pressure of oxygen was observed to be 1.56 torr in
air with a total atmospheric pressure of 743 torr. Calculate
the mole fraction of O2 present.
and the mass of KClO3 in the sample that was decomposed.
23
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VI. The Kinetic Molecular Theory of Gases.
A. KMT.
1. The kinetic molecular theory (KMT) is a simple model that attempts
to explain the properties of an ideal gas.
2. From HONORS CHEM: Kinetic Theory Revisited.
* Recall that gases consist of hard, spherical particles, usually atoms or
molecules.
a. Because the particles are so small and the distances between
them are so great, their individual volumes are insignificant.
* The volume of a gas is mostly empty space.
b. Property: Gases are highly compressible.
* Also, recall that there are no attractive or repulsive forces between
gas particles.
a. Gases are free to move inside their containers.
b. Property: A gas expands until it takes the volume and shape of
its container.
* Next, recall that gas particles move rapidly in constant random
motion.
a. The particles travel in straight paths and move independently of
each other.
b. Only when a particle collides with a container wall or another gas
particle does it deviate from its straight-line path.
* In addition, recall that collisions between gas particle are elastic,
meaning that the total kinetic energy remains constant and that the
kinetic energy is transferred without loss from one particle to
another.
* Finally, recall that the average kinetic energy of a collection of gas
particles is directly proportional to the Kelvin temperature of the
gas.
3. Assumptions.
a. The particles are so small compared with the distance between
them that the volume of the individual particles can be assumed
to be negligible (zero).
b. The particles are in constant motion. The collisions of the
particles with the wall of the container are the cause of the
pressure exerted by the gas.
c. The particles are assumed to exert no forces on each other; the
are assumed to neither attract nor to repel each other.
d. The average kinetic energy of a collection of gas particles is
assumed to be directly proportional to the Kelvin temperature
of the gas.
4. Pressure is defined as the force exerted by particles colliding with a
a container wall. This is not the same as atmospheric pressure.
5. Real gases do not conform to these assumptions because they have
finite volumes and do exert forced on each other.
B. Justification of the gas laws using KMT.
1. Pressure and Volume (Boyle’s Law).
a. Law: For a given sample of gas at constant T and n, if the volume
of a gas is decreased, the pressure increase.
b. KMT: Since a decrease in volume means that the gas particles
will hit the container wall more often, the pressure should
increase.
2. Volume and Temperature (Charles’s Law).
a. Law: At constant P and n, the volume of a gas is directly
proportional to the Kelvin temperature.
b. KMT: When a gas is heated to a higher temperature, the speeds
of its molecules increase and thus hit the walls more often with
more force. The only way to keep the pressure constant in this
situation is to increase the volume of the container.
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3rd 9-weeks Notes (The end is near…in more ways than one.)
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3. Volume and Number of Moles (Avogadro’s Law).
a. Law: The volume of a gas at constant P and T depends directly
on the number of gas particles present.
b. KMT: An increase in the number of gas particles at the same
temperature would cause the pressure to increase if the volume
were held constant. The only way to return the pressure to its
original value is to increase the volume.
c. It is important to recognize that the volume of a gas depends
only on the number of gas particle present.
d. The individual volumes of the particles are not a factor because
the particle volumes are so small compared to the distances
between the particles.
6. The Meaning of Temperature.
a. Temperature is a measure of the energy of molecular motion; a
transfer of KE from collisions of higher energy objects to
collections of lower energy particles.
b. Equation:
7. Root Mean Square Velocity.
a. The root mean square velocity (urms) is the square root of the
average of the squares of the individual velocities of gas
particles.
b. Derivation:
4. Mixture of Gases (Dalton’s Law).
a. Law: The total pressure exerted by a mixture of gases is the
sum of the pressures of the individual gases.
b. KMT: All gases are independent of one another and the volumes
of the individual particles are unimportant. Thus the identities
of the gas particles do not matter.
5. Deriving the Ideal Gas Law.
a. Derivation:
c. Calculate the root mean square velocity for the atoms in a
sample of helium gas at 25 C.
b. The experimental and theoretical equations for the ideal gas law
are virtually identical, giving confidence in the KMT.
25
3rd 9-weeks Notes (The end is near…in more ways than one.)
HW # Handouts
VII. Effusion and Diffusion.
A. Effusion.
1. Effusion is the process by which a gas escapes from its container
through a tiny hole into an evacuated tube.
2. Thomas Graham (1805-1869), a Scottish chemist, found
experimentally that the rate of effusion of a gas is inversely
proportional to the square root of the mass of its particles.
3. Graham’s Law of Effusion:
VIII. Real Gases.
A. Deviations.
1. Molecules have finite molecular volumes.
2. Interactions exist between molecules.
3. When volume decreases, condensation often occurs.
4. Gases behave nearly ideally under ordinary conditions of low pressure
and high temperature.
B. Reasons for Deviations.
1. Graph:
4. Effusion Rates.
* Calculate the ratio of the effusion rates of hydrogen gas (H2)
and uranium hexafluoride (UF6), a gas used in the enrichment
process to produce fuel for nuclear reactors.
2. When Pext > 10 atm, deviations occur.
5. KMT: The effusion rate for a gas depends directly on the average
velocity of its particles.
a. The faster the gas particles are moving, the more likely they are
to pass through the effusion orifice.
b. Prediction:
3. Intermolecular attractions lower molecular speeds and lessen the
force of collision with the container wall.
a. Increasing the external pressure causes this to occur.
b. Lowering temperature yields the same effect.
c. Both cases may cause condensation.
d. Because the volume decreases in both cases, the PV/RT is
lowered.
4. Molecular Volume.
a. As external pressure increases, free volume decreases and
molecular volume becomes a significant factor, outweighing the
importance of the effects of intermolecular attractions.
B. Diffusion.
* Diffusion is the movement of one gas through another and uses the
same equation as effusion.
b. The “V” used in the ratio PV/RT still refers to the volume of the
container, not the volumes of the gas molecules.
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c. Because “V” is artificially too large, the ratio PV/RT increases.
5. The van der Waals Equation.
a. This equation adjusts pressure “up” and volume “down.”
b. The equation:
c. The a and b variables are the van der Waals constants.
* They are positive values specific to each gas.
* They are obtained from a table.
* a is related to molar mass and relate with the strength of the
intermolecular attractions.
* b is a measure of actual molecular volume.
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I. Types of Attractions.
A. Intramolecular Attractions.
1. These are forces within a molecule or polyatomic ion.
2. They are also referred to as bonding forces.
3. These influence chemical properties.
B. Intermolecular Attractions.
1. These are forces between molecules, ions, or atoms.
D. Ability of Flow.
1. For a gas, this ability is very high.
2. For a liquid, this ability is moderate.
3. For a solid, this ability is almost none.
III. A Look at Phase Changes for Water.
* Diagram:
2. They influence physical properties.
II. A Comparison of States of Matter.
A. KMT.
1. For a gas, the energy of attraction is less than the energy of motion.
The particles, therefore, are far apart.
2. For a liquid, the energy of attraction is stronger, but KE allows for
movement. The particles, therefore, are closer, but with motion.
3. For a solid, the energy of attraction is greater than the energy of
motion. The particles, therefore, as arranged in a fixed, organized
pattern.
B. Shape/Volume.
1. For a gas, the molecules conform to the container in both shape and
volume.
2. For a liquid, the molecules conform to the container shape; the
volume is limited by surface.
3. For a solid, the compound has a defined shape and volume.
C. Compressibility.
1. For a gas, the compressibility is high because of the great distances
between particles.
2. For a liquid, the compressibility is very low.
3. For a solid, the compressibility is virtually nonexistent.
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IV. Types of Intermolecular Attractions.
A. Introduction.
1. Bonding forces are relatively strong because they involve higher
charges that are closer together (covalent radius).
4. Drawing Hydrogen Bonds between Molecules.
2. Intermolecular forces are relatively weak because they typically
involve charges that are far apart (van der Waals radius).
B. Ion-Dipole Forces.
1. This occurs when an ion is attracted to a polar molecule.
2. Example:
C. Dipole-Dipole Forces.
1. This occurs between two polar molecules.
2. Example:
E. Ion-Induced Dipole Forces.
1. This occurs when an ion is attracted to a distorted nonpolar molecule.
D. Hydrogen Bonding.
1. This is a special, very strong type of dipole-dipole force.
2. This occurs only when a hydrogen is bonded to a nitrogen, oxygen, or
fluorine atom.
2. The electron density of the nonpolar molecule is shifted as a result
of a charged particle.
3. Example:
3. Example:
F. Dipole-Induced Dipole.
1. This occurs between a polar molecule and a distorted nonpolar
molecule.
2. Example:
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G. Dispersion(London) Forces.
1. These forces often occur when all particles collide.
2. This force is also related to molar mass because heavier molecules
tend to have more electrons capable of being polarized and distorted.
V. Properties of a Liquid.
A. Surface Tension.
1. Diagrams:
3. Shape is also a factor because more points of attraction means more
polarizability.
4. Example:
F. Predicting the Type and Relative Strength of Intermolecular Forces.
* In each of the following pairs, identify all the intermolecular forces
present and select the substance with the higher boiling point:
a. MgCl2 or PCl3
2. Surface tension is the energy required to increase the surface area
2
by a given amount ( in J/m ).
b. CH3NH2 or CH3F
3. To reduce the instability of the entire substance at the surface, a
liquid will reduce the number of particles exposed to the surface.
* The particles form a tight “skin” at the surface.
c. CH3OH or CH3CH2OH
4. The stronger the forces between the particles in a liquid, the
greater the surface tension.
d. Hexane or cyclohexane
e. CH3Br or CH3Cl
B. Capillarity.
1. Capillarity is the ability of a substance to naturally “rise up” a tube
against gravity.
2. Diagrams:
f. CH3CH2OH or CH3OCH3
g. C2H6 or C3H8
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C. Viscosity.
1. Viscosity is the resistance to flow.
b. Body-Centered Cubic.
* Diagram:
2. The viscosity of a liquid decreases with increasing temperature.
3. Molecular shape also plays a role in determining a liquid’s viscosity.
* Longer molecules have more contact points for attractive forces
and have higher viscosities.
VI. Properties of Solids.
A. Crystalline Solids.
1. Crystals have lattice structures made of unit cells.
a. A lattice is an array of points that forms a regular pattern.
b. A unit cell is the simplest arrangement of points that, when
repeated in all directions, gives the lattice.
c. Diagram:
c. Face-Centered Cubic.
* Diagram:
d. The coordination number of a particle in a crystal is the number
of nearest neighbors surrounding it.
2. Types of cubic unit cells.
a. Simple Cubic.
* Diagram:
3. Packing Efficiency.
a. Packing efficiency is the percentage of the available volume of
a crystal is occupied by “spheres.”
b. Simple cubic packing efficiency is 52% (48% air) and is very
rare.
c. Body-centered cubic packing efficiency is 68% and is more
common.
* Examples:
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d. Hexagonal closest packing efficiency is 74% (the most).
* Examples:
e. Cubic closest packing efficiency is 74% (the most) and is based
on the face-centered cubic unit cell.
* Examples:
* Diagram:
* Typical Properties: Wide range of hardness, wide range of
melting points, conductor.
* Examples:
4. Types and Properties of Solids.
a. Atomic Network.
* Structural Unit: Atom.
* Type of Bonding: Directional covalent bonds.
c. Atomic Group 8A.
* Structural Unit: Atom.
* Type of Bonding: London dispersion forces.
* Diagram:
* Diagram:
* Typical Properties: Very low melting points.
* Typical Properties: Hard, high melting point, insulator.
* Examples:
b. Atomic Metallic.
* Structural Unit: Atom.
* Type of Bonding: Nondirectional covalent bonds involving
electrons that are delocalized throughout
the crystal.
* Examples:
d. Molecular.
* Structural Unit: Molecule.
* Type of Bonding: Polar molecules- dipole/dipole interactions;
Nonpolar molecules- London dispersion
forces.
* Diagram:
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* Typical Properties: Soft, low melting, insulator.
* Examples:
e. Ionic.
* Structural Unit: Ion.
* Type of Bonding: Ionic.
* Diagram:
* Typical Properties: High melting point, insulator.
* Examples:
B. Amorphous Solids.
* They have an irregular lattice structure with many defects.
VII. Quantitative Aspects of Changes of State.
A. The Heating-Cooling Curve.
1. Diagram:
2. (*) indicates a change in heat accompanied by a change in
temperature that equals the change in KE as the most probable
speed changes.
3. (#) indicates a change in heat accompanied by a constant
temperature (constant KE), which is associated with a change in PE,
as the average distance between molecules changes.
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B. The Equilibrium Nature of Phase Changes.
1. Liquid-gas equilibria.
a. This is viewed for a CLOSED CONTAINER in a VACUUM.
* At 34.1 C, the vapor pressure of water is 40.1 torr. What is
the vapor pressure at 85.5 C? The Hvap of water is 40.7
kJ/mol.
b. The rate of condensation = rate of vaporization.
c. Vapor pressure is the pressure exerted by a vapor in a closed
system.
* Vapor pressure increases with temperature as the KE
increases.
* Vapor pressure increases with weaker intermolecular
attractions.
d. The Clausius-Clapeyron Equation relates vapor pressure and
temperature.
* Equation:
* The vapor pressure of ethanol is 115 torr at 34.9 C. If Hvap
of C2H5OH is 40.5 kJ/mol, calculate the temperature (in C)
when the vapor pressure is 760 torr.
e. Boiling can only occur in an open container and occurs when the
vapor pressure equals the external pressure.
* The normal boiling point of a substance is boiling at standard
pressure (1 atm).
2. Solid-liquid equilibria.
a. The rate of melting = rate of freezing.
b. The normal boiling point of a substance is melting at standard
pressure (1 atm).
3. Solid-gas equilibria.
* The rate of sublimation = rate of deposition.
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VIII. Phase Diagrams.
A. Diagram 1.
IX. The Properties of Mixtures.
A. Introduction.
1. A mixture is defined as two or more substances physically mixed
together but not chemically combined.
2. A solution is a homogeneous mixture with one phase.
B. Types of Solutions: Intermolecular Forces and the Prediction of
Solubility.
1. Definitions.
a. A solute is
b. A solvent is
c. Miscibility is
d. Solubility is
B. Diagram 2.
e. To be dilute means
f. To be concentrated means
2. The Solvation Process.
a. Solute particles must separate.
b. Some solvent particles must separate.
c. Solute and solvent particles must mix together.
d. Whether or not a solution occurs depends on the relative
strengths of the solute-solute, solvent-solvent, and solutesolvent intermolecular forces.
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3. Intermolecular attractions present in solutions.
*
4. Rule: “ Like dissolve like.”
* Molecular size may also be a factor.
5. Gas-liquid solutions.
a. Solute-solute attractions are very weak.
b. Solute-solvent forces are weak, but are the major factor in gas
solubility in a liquid.
6. Gas solutions and solid solutions.
a. Gas-gas: All gases are infinitely soluble in one another.
* Example:
b. Gas-solid: The gas occupies the spaces between the closely
packed particles.
* Example:
c. Solid-solid: These are called alloys.
* Substitutional: structural substitution.
** Examples:
* Interstitial: fill in the spaces within the structure.
** Example:
C. Energy Changes in the Solution Process.
1. Heats of Solution.
a. Explanation.
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b. Diagrams:
2. Heats of Hydration: Ionic Solids in Water.
a. Hsoln = Hsolute Hhydr
b. Hhydr is always exothermic (negative) because the ion-dipole
force is greater than the hydrogen bonding in water.
 
c. The charge density, the ratio of charge to volume, affects 
Hhydr.
* The trend down a group is to decrease Hhydr because a
larger ion means a larger radius between particles, therefore
less energy is required to break the attraction.
* The trend across a period is to increase Hhydr as charge
increases.
3. The Solution Process and the Tendency Toward Disorder.
a. Entropy is a measure of a system’s disorder.
b. Because the solution process occurs naturally, which creates
more disorder, entropy is increased.
D. Solubility as an Equilibrium Process.
1. Definitions.
a. An unsaturated solution means
b. A saturated solution means
c. A supersaturated solution means
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2. Factors Affecting Solubility.
a. Temperature.
* Solubility increases with temperature if the solution process
is endothermic (Hsoln > 0).
* Solubility decreases with temperature if the solution process
is exothermic (Hsoln < 0).
* Generally, ionic solids increase their solubility in water as
temperature increases.
* Generally, gases decrease their solubility in water as
temperature increases.
b. Pressure.
* Pressure changes have little effect on solid and liquid
solubility.
E. Quantitative Ways of Expressing Concentration.
1. Molarity (M)
2. Molality (m)
3. Parts by mass
4. Parts by volume
5. Mole fraction (X)
* Henry’s Law states that the solubility of a gas (Sgas) is
directly proportional to the partial pressure of the gas (Pgas)
above the solution.
6. What is the molality of a solution prepared by mixing 32.0 g CaCl2
with 271 g water?
* In a soft-drink plant, the partial pressure of CO2 gas inside a
bottle of cola is adjusted to 4 atm at 25 C. What is the
solubility of CO2 under these conditions? Henry’s law constant
7. Calculate the ppm by mass of calcium in a 3.50 g pill that contains
40.5 mg Ca.
for CO2 in water = 3 x 10-2 mol/L atm at 25 C.
8. The label on a 0.750-L bottle Italian Chianti indicates “11.5% alcohol
by volume.” What volume is alcohol (in L) does it contain?
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9. A sample of rubbing alcohol contains 142 g isopropanol (C3H7OH) and
58.0 g water. What are the mole fractions of alcohol and water?
F. Colligative (Collective) Properties of Solutions.
1. These properties of a solution are always compared against the pure
solvent.
2. Nonvolatile nonelectrolytic (ideal) solutions.
a. Vapor Pressure Lowering (P).
* This solution’s vapor pressure is always lower than the pure
solvent.
* Surface particles consist of some nonvolatile solute particles
that have replaced some solvent particles. Therefore, not as
many solvent particles are permitted to become a vapor.
10. Converting Units of Concentration.
* Hydrogen peroxide is a powerful oxidizing agent that is used in
concentrated solution in rocket fuel systems and in dilute solution
as a hair bleach. An aqueous solution H2O2 is 30% by mass and has
* Raoult’s Law for ideal solutions:
a density of 1.11 g/mL. Calculate its
a. Molality
* Calculate the expected vapor pressure at 25 C for a solution
prepared by dissolving 158.0 g of common table sugar (sucrose,
3
molar mass = 342.3 g/mol) in 643.5 cm of water. At 25 C, the
3
b. Mole fraction of H2O2
density of water is 0.9971 g/cm and the vapor pressure is
23.76 torr.
c. Molarity
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
* Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g water
at 25 C. The vapor pressure of pure water at 25 C is 23.76
torr.
* A solution was prepared by dissolving 18.00 g glucose in 150.0 g
water. The resulting solution was found to have a boiling point
of 100.34 C. Calculate the molar mass of glucose. Glucose is a
molecular solid that is present as individual molecules in
solution.
c. Freezing Point Depression (Tf).
* The freezing point of the solution is always lower than the
pure solvent under these conditions.
b. Boiling Point Elevation (Tb).
* This results from the increases pressure (P).
* Solute particles hinder the rate at which the solvent particles
freeze.
* Equations:
* The boiling point of the solution is always higher than the pure
solvent under these conditions.
* Equations:
* The boiling point of ethanol is 78.5 C. What is the boiling
point of a solution of 3.4 g vanillin (M = 152.14 g/mol) in 50.0 g
ethanol? (Kb of ethanol = 1.22 C/m).
* What mass of ethylene glycol (M = 62.1 g/mol), the main
component in antifreeze, must be added to 10.0 L water to
produce a solution for use in a car’s radiator that freezes at
-10.0 F (-23.3 C)? Assume the density of water is exactly 1
g/mL.
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* Diagrams:
* A chemist is trying to identify a human hormone, which
controls metabolism, by determining its molar mass. A sample
weighing 0.546 g was dissolved in 15.0 g benzene, and the
freezing-point depression was determined to be 0.240 C.
Calculate the molar mass of the hormone.
* Equations:
-3
* To determine the molar mass of a certain protein, 1.00 x 10 g
of it was dissolved in enough water to make 1.00 mL of
solution. The osmotic pressure of this solution was found to be
1.12 torr at 25 C. Calculate the molar mass of the protein.
d. Osmotic Pressure.
* Osmotic pressure () is the pressure that results from the
inability of solute particles to cross a semipermiable
membrane; the pressure required to prevent the net
movement of solvent across the membrane.
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* What concentration of sodium chloride in water is needed to
produce an aqueous solution isotonic with blood ( = 7.70 atm
at 25 C)?
3. Volatile nonelectrolytic solutions.
a. When a solution contains two volatile components, both
contribute to the total vapor pressure, but not necessarily in
equal amounts.
b. Equal amounts of liquid do not produce equal amounts of vapor.
4. Colligative Properties of Electrolytic Solutions.
a. Most electrolytic solutions are not ideal.
b. A vant Hoff factor (i) attempts to adjust the equations.
c. i =
c. A solution is prepared by mixing 5.81 g acetone (C3H6O, molar
mass = 58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass =
119.4 g/mol). At 35 C, this solution has a total vapor pressure of
260 torr. Is this an ideal solution? The vapor pressures of pure
acetone and pure chloroform at 35 C are 345 and 293 torr,
respectively.
d. Equations:
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G. Colloids.
1. A suspension contains very large particles that settle out of solution.
2. A solution contains very small solute particles that do not separate
from the solvent.
3. A colloid contains intermediate particles (solutelike) distributed
throughout a dispersing (solventlike) substance.
4. Evaluating colloids.
a. The Tyndall effect measures the scattering of light by a
colloidal suspension.
b. Brownian motion measures the change in speed and direction of
colloidal particles because of the action of the dispersing
medium.
5. Types of Colloids.
Colloid Type
Dispersed
Substance
Dispersing
Medium
Example
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I. Electromagnetic Radiation.
A. The Nature of Light.
1. The primary characteristics of a wave.
a. Wavelength () is the distance between two consecutive peaks
or troughs in a wave.
b. Frequency () is the number of waves (cycles) per second that
pass a given point in space.
* In the SI system, cycles is understood, and the unit per
-1
second becomes 1/s, or s , which is called hertz (Hz).
2. Frequency of Electromagnetic Radiation.
* The brilliant red colors seen in fireworks are because of the
emission of light with wavelengths around 650 nm when strontium
salts such as Sr(NO3)2 and SrCO3 are heated. Calculate the
2
frequency of red light of wavelength 6.50 x 10 nm.
c. Diagram:
B. Some Distinguishing Properties of Wave.
1. Refraction is the bending of a light wave (a change in angle) when it
strikes a boundary.
d. All types of electromagnetic radiation (waves), a breakdown of
light based on wavelength and frequency, travel at the speed of
light (c).
8
* The speed of light (c) = 2.9979 x 10 m/s.
* The electromagnetic spectrum:
e. Wavelength and frequency are inversely related.
*
 = c
where  is the wavelength in meters,  is the frequency in
cycles per second, and c is the speed of light.
2. Diffraction is the bending of a light wave around a boundary.
3. Light must be a wave because it possesses these properties.
II. The Nature of Matter.
A. Light vs. Matter.
1. Blackbody Radiation.
a. When a solid (specifically a particle) is heated, it emits light
waves.
* Examples:
b. German Max Planck (1858-1947) said that a hot, glowing object
emits or absorbs a specific amount of energy to produce this
light.
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c. This specific energy is said to be quantized, or specific, lost or
gained only in integer multiples of h.
d.
 
E = nh
-34
where h is Planck’s constant (6.626 x 10 J s),  is the
frequency of the electromagnetic spectrum absorbed or emitted, 
E is the change in energy for a system, and n is a whole-number
integer (1, 2, 3, ...).
e. This implies that light energy of matter is not continuous (like a
rainbow), but that it is absorbed or emitted only at specific
quantized energy states.
f. Is matter separate form light (energy)?
2. The Energy of a Photon (a packet of energy).
* The blue color in fireworks is often achieved by heating copper(I)
chloride (CuCl) to about 1200 C. Then the compound emits blue
light having a wavelength of 450 nm. What is the increment of
2
energy (the quantum) that is emitted at 4.50 x 10 nm by CuCl?
c. Confusing feature of the Photoelectric Effect.
* A threshold (minimum) frequency is required to knock an
electron free from a metal. Wave theory associates the light’s
energy with the wave amplitude (intensity), not its frequency
(color), so it predicts that an electron should be knocked free
when the metal absorbs enough energy from any color of light.
* Current flows the moment that light of high enough frequency
shines on the metal, regardless of its intensity. The wave
theory predicted that in dim light there should be a time lag
before current flowed, while the electrons absorbed enough
energy to break free.
4. Albert Einstein’s Photon Theory.
a. Einstein proposed that radiation is particulate, occurring as
quanta of electromagnetic energy (packets), later called
photons.
b. Einstein solved the mysteries of the Photoelectric Effect.
* A beam of light is composed of enormous numbers of photons.
Light intensity (brightness) is related to the number of
photons emitted per unit of time, not their energy. One
electron is freed from the metal when one photon of a certain
minimum energy is absorbed.
* An electron is freed the moment it absorbs a photon of enough
energy, not when it gradually accumulates energy from many
photons of lower energy.
5. In 1922, American Arthur Compton (1892-1962) performed
experiments involving collisions of X rays and electrons that showed
that photons do actually have a resting mass, a property of matter!
a. This is the Compton Effect.
3. The Photoelectric Effect.
a. When light strikes a monochromatic plate, an electrical current
flows (like a solar calculator).
b. Equations:
b. Light must transfer momentum to matter, like particles do; light
is behaving like a particle!
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6. Summary.
a. Energy is quantized. It can occur only in discrete units called
quanta.
b. Electromagnetic radiation, which was previously thought to
exhibit only wave properties, seems to show certain
characteristics of particulate matter as well.
* This is wave-particle duality or the dual nature of light.
3. In 1927, Davisson and Germer verified De Broglie’s concept when
an electron was seen to exhibit wave properties of interference and
diffraction.
III. The Atomic Spectrum of Hydrogen.
* The Atomic Spectrum.
1. The atomic spectrum (light emitted) was thought to be continuous.
a. Again, this is similar to a rainbow.
b. All of the wavelengths of visible light are present.
B. Wave-Particle Duality of Matter and Energy.
1. From Einstein, we have the following:
c. Diagram:
2. French physicist Louis de Broglie (1892-1987) proposed in 1923 that
if waves of energy have some properties of particles, perhaps
particles of matter have some properties of waves.
a. This is similar to a guitar string (a particle) producing sound (a
wave.)
2. When the emission spectrum of hydrogen in the visible region is
passed through a prism, only a few lines are seen.
a. Diagram:
b. De Broglie wave equation:
c. Calculations of Wavelengths.
* Compare the wavelength for an electron (mass = 9.11 x 10
7
-31
kg) traveling a speed of 1.0 x 10 m/s with the for a ball (mass
= 0.10 kg) traveling 35 m/s.
b. These lines correspond to discrete wavelengths of specific
(quantized) energy.
* Only certain energies are allowed for the electron in the
hydrogen atom.
c. The hydrogen emission spectrum is called a line spectrum.
IV. The Bohr Model.
A. The Quantum Model.
1. Niels Bohr (1885-1962) proposed in 1913 that the electron in a
hydrogen atom moves around the nucleus only in certain allowed
circular orbits.
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a. Diagram:
a. Bohr was able to calculate hydrogen atom energy levels that
exactly matched the values obtained by experiment.
b. The negative sign means that the energy of the electron bound
to the nucleus is lower than it would be if the electron were at
an infinite distance from the nucleus (n = infinity), where there
is no interaction and the energy is zero:
* Again, the closer to the nucleus, the more stable, and the less
energetic (less meaning a negative energy value).
c. Electrons moving from one energy level to another.
* Equation:
b. The atoms has stationary states, called energy levels, of specific
energy around the nucleus.
c. Electrons can move to other energy levels by absorbing (jumping
to higher energy levels) or emitting (jumping to lower energy
levels) photons of specific (quantized) energy.
* Energy levels farther from the nucleus more “unstable” and
therefore more “energetic.”
* As we will see, the closer to the nucleus an energy level is, the
more stable it is and the less energetic it is.
3. Energy Quantized in Hydrogen.
* Calculate the energy required to excite the hydrogen electron
from level n = 1 to level n = 2. Also calculate the wavelength of
light that must be absorbed by a hydrogen atom in its ground
state to reach this excited state.
d. The lowest (first) energy level of an atom is called ground state.
e. This model only works for one-electron atoms!
2. The most important equation to come from Bohr’s model is the
expression for the energy levels available to the electron in the
hydrogen atom:
E = -2.178 x 10
-18
2
2
J (Z /n )
in which n is an integer (the larger the value of n, the larger is the
orbit radius) and Z is the nuclear charge.
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4. Electron Energies.
* Calculate the energy required to remove the electron from a
hydrogen atom in its ground state.
B. The Heisenberg Uncertainty Principle.
1. The Heisenberg uncertainty principle states that it is impossible to
know simultaneously the exact position and velocity of a particle.
2. Stated mathematically:
5. Although Bohr’s model fits the energy levels for hydrogen, it is a
fundamentally incorrect model for the hydrogen atom, mainly
because electrons do not travel in circular orbits.
V. The Quantum-Mechanical Model of the Atom.
A. The Schrodinger Equation.
1. The Schrodinger equation is an extremely complex equation used to
describe the 3-D quantum mechanical model of the hydrogen atom.
C. The Physical Meaning of a Wave Function.
1. The square of the function indicates the probability of finding an
electron near a particular point in space.
2. The probability distribution, or electron density diagram, can be
plotted to give the most probable region where an electron may be
located.
* Examples:
2. The hydrogen electron is visualized as a standing wave (a stationary
wave as one found on a musical instrument) around the nucleus.
a. The circumference of a particular orbit would have to
correspond to a whole number of wavelengths.
b. Diagrams:
c. This is consistent with the fact that only certain electron
energies are allowed.
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3. The definition most often used by chemists to describe the size of
the hydrogen 1s orbital is the radius of the sphere that encloses
90% of the total electron distribution.
VI. Quantum Numbers.
A. Quantum Numbers.
1. A quantum number describes various properties of an atomic orbital
or probable electron location.
2. Types.
a. The principle quantum number (n) refers to the energy level or
integral values: 1, 2, 3, etc.
b. The angular momentum (azimuthal) quantum number (l) is
related to the shape of atomic orbitals.
* The integral values are from 0 to n - 1.
* l = 0 is called s; l = 1 is called p; l = 2 is called d; l = 2 is called
f.
c. The magnetic quantum number (ml) is related to the orientation
VII. Electron Spin and the Pauli Principle.
A. Electron Spin
1. Although the first three quantum numbers describe atomic orbitals,
one more is needed to describe a specific electron.
* This is called spin, and is not a property of the orbital.
2. The spin quantum number (ms) indicates the “direction” an electron
spins and can have one of two possible values, +1/2 or -1/2.
* Notation:
of the orbital in space.
* The integral values are between l and -l, including zero.
3. Quantum Numbers for the First Four Levels of Orbitals in the
Hydrogen Atom.
B. Pauli Exclusion Principle.
1. American Wolfgang Pauli stated that no two electrons in the same
atom can have the same set of four quantum numbers.
a. This is the Pauli exclusion principle.
b. Example:
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2. An atomic orbital can hold a maximum of two electrons and they must
have opposing spins.
3. The electrons of an atom in its ground state occupy the orbitals of
lowest energy.
VIII. Polyelectronic Atoms.
A. Electrostatic Effects.
1. In one electron systems, the only electrostatic force is the nucleuselectron attractive force.
2. Many-electron systems (polyelectronic atoms).
a. Nucleus-electron attractions.
b. Electron-electron repulsions raise orbital energy.
* Why?
b. Electron-electron repulsions.
3. For polyelectronic atoms, one major consequence of having
attractions and repulsions is the splitting of energy levels into
sublevels of differing energies.
a. The energy of an orbital in a many-electron atom depends on its
n value (size) and secondly on its l value (shape).
c. Electrons in outer orbitals (higher n) are higher in energy.
* Why?
b. This means that energy levels may split into s, p, d, and f
sublevels, depending on the n value.
B. The Effects of Electrostatic Interactions on Orbital Energies.
1. Recall that, by definition, an atom’s energy has a negative value.
a. A more stable orbital has a larger negative energy than a less
stable one.
b. Stronger attractions make the orbital lower in energy (larger
negative number).
c. Repulsions make the system higher in energy (smaller negative
number).
2. Major Conclusions.
a. A greater nuclear charge (Z) lowers orbital energies.
* Why?
d. Electrons that have a finite probability distribution near the
nucleus (penetration) have lower energy (more negative).
* Es < Ep < Ed < Ef.
* Why?
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XI. Periodic Trends in Atomic Properties.
A. Effective Nuclear Charge (Zeff)
C. Ionization Energy
1. What is ionization energy?
1.What is Zeff?
2.What type of ion is produced?
2. What is the group trend, with explanation?
3. Is the process endothermic or exothermic?
4. Electrons are easily removed until what occurs?
5. What is the group trend, with explanation?
3. What is the period trend, with explanation?
6. What is the period trend, with explanation?
B. Atomic Radius
1. What is an atomic radius?
2. What is the group trend, with explanation?
7. Why is each successive IE greater than the previous IE?
3. What is the period trend, with explanation?
8. Where are the exceptions and why do they occur?
4. Where are the exceptions and why do they occur?
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D. Electron Affinity
1. What is electron affinity?
3. What is the period trend, with explanation?
2. What is the group trend, with explanation?
3. What is the period trend, with explanation?
F. Problems.
1. Trends in Ionization Energy.
* The first ionization energy for phosphorus is 1060 kJ/mol, and
that for sulfur is 1005 kJ/mol. Why?
4. How does EA1 compare to EA2?
5. Where are the exceptions and why do they occur?
2. Ionization Energies.
* Consider atoms with the following electron configurations:
2
2
6
2
2
6
1
2
2
6
2
1s 2s 2p
1s 2s 2p 3s
1s 2s 2p 3s
E. Metallic Behavior
1. What defines metallic behavior?
2. What is the group trend, with explanation?
Which atom has the largest first ionization energy, and which
one has the smallest second ionization energy? Explain your
choices.
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