Below are two attached powerpoint documents for Chemistry about : Thermochemistry Electromagnetic Radiation and Electronic Structure of Atom. Complete ALL the practice problems appear on the slides
c6_rfs.pdf
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Chapter 6
1
Electromagnetic Radiation
and the Electronic Structure of
the Atom
General
Chemistry
Vining
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6.1a Wavelength & Frequency
´ Wavelength (λ): distance
between peaks (or troughs)
´ Frequency (ν): usually
expressed in hertz (Hz), cycles
per second
´ Amplitude: height from
middle to top (or bottom) of
wave
´ Speed of light:
´ c = λν = 3.0 108 m/s
´ λ and ν are Inversely related
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6.1a Practice
´ What is the frequency of light that has a λ of 435 nm?
´ A radio station broadcasts at 91.7 MHz. What is the λ?
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6.1b The Electromagnetic Spectrum
´ The different types of
electromagnetic radiation
arranged by wavelength
(Fig 6.1.2)
´ Short: gamma and X-rays
´ Long: TV and radio
´ High/low energy?
´ Red vs blue light (note range
of visible spectrum)
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6.2 Photons and Photon Energy
´ Photoelectric effect: light of sufficient energy (long or
short λ?) can eject electrons from a metal’s surface
´ Ephoton = hν
´ h is Planck’s constant: 6.626 × 10-34 Js
´ For a single photon
´ Photon energy is directly related to frequency
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6.2 Practice
´ What is E for a photon with a ν of 8.66 × 1014 Hz?
´ What is E for a photon with a λ of 582 nm?
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6.3a Atomic Line Spectra
´ Continuous spectrum: rainbow;
white light separated into colors
using a prism
´ Line spectrum: specific
wavelengths from excited gas
phase elements (electric current
passed through them) after
being passed through a prism
´ Different for each element
´ Suggests that energy of atoms is
quantized
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6.3b Bohr Model
´ Electron energy is quantized
´ Principle quantum number (n): number assigned to these energy levels
´ Electrons in low energies can be promoted to higher energies by absorbing
energy
´ Electrons can release energy in the form of photons (light) to return to lower
energy levels
´ Ground state: n = 1
´ Excited state: n > 1
´ Explains hydrogen line spectrum
´ Energy change between energy levels:
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6.3b Example
´ Calculate the wavelength (nm) when an electron
moves from the n = 5 to the n = 3 energy level. Is this
radiation visible?
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6.3b Practice
´ Calculate the wavelength (nm) when an electron
moves from the n = 4 to the n = 1 energy level. Is this
radiation visible?
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6.4a Wave Properties of Matter
´ Bohr’s model fails when
applied to species with >1
electron
´ Still useful with energy transitions
´ Only considers particle nature
of electrons
´ Heisenberg Uncertainty
Principle
´ Matter, including electrons, has
wave-particle duality
´ de Broglie equation
´ h is Planck’s constant, m is
mass (kg), v is velocity (m/s)
´ Remember: 1 J = 1 kg
m2/s2
´ Not possible to know with great
certainty both the position and
momentum of an electron at
the same time
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6.4a Practice
´ Calculate the wavelength of an electron (9.11 x 10-31 kg) moving at a
speed of 2.2 x 106 m/s
´ Calculate the wavelength of a golf ball (45 g) moving at a speed of 72
m/s.
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6.4b Schrödinger Equation
´ Equation used to describe
wavelike behavior of
electrons
´ Solutions produce wave
functions (Ψ) that predict
the energy of an electron
and region of space most
likely for it to be found
´ Orbitals: Ψ2 = probability
density of finding an
electron in a given region
around a nucleus
´ Dot map vs boundary surface:
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6.5a Quantum Numbers
´ Organization of periodic table is related to organization
of electrons in atoms!
´ Principal quantum number (n): describes size and energy of the
shell in which orbital resides (period)
´ Angular momentum quantum number ( ): shape of orbital,
varies from 0 to n-1 (subshell/orbital type)
´ 0=s, 1=p, 2=d, 3=f
´ Magnetic quantum number ( ): orbital orientation in space,
values ranging from – to + (specific orbital)
´ Connect to the periodic table…
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Periodic Table of the Elements
1
1
2
H
H
He
1.00794*
4.002602
1.00794*
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.941*
9.012182
10.811*
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
39.948
22.98976928 24.3050
26.9815386 28.0855* 30.973762
19
20
21
22
23
K
Ca
Sc
Ti
V
39.0983
40.078
44.955912
47.867
50.9415
51.9961
54.938045
37
38
39
40
41
42
43
Rb
Sr
Y
Zr
85.4678
87.62
88.90585
91.224
92.90638
95.96
(97.9072)
101.07
102.90550
55
56
57
72
73
74
75
76
77
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
132.9054519 137.327
12.0107* 14.00674* 15.9994* 18.9984032 20.1797
24
25
26
35.4527*
34
35
36
As
Se
Br
Kr
72.63
74.92160
78.96
79.904
83.798
50
51
52
53
54
In
Sn
Sb
Te
I
Xe
112.411
114.818
118.710
121.760
127.60
126.90447
131.293
80
81
82
83
84
85
86
Tl
Pb
Bi
Po
At
Rn
207.2
28
29
30
Co
Ni
Cu
Zn
Ga Ge
55.845
58.933195
58.6934
63.546
65.38
69.723
44
45
46
47
48
49
Ag
Cd
106.42
107.8682
78
79
Cr Mn Fe
Nb Mo Tc
32.066*
33
27
Ru Rh Pd
Au Hg
32
138.90547
178.49
180.94788
183.84
186.207
190.23
192.217
200.59
204.3833*
87
88
89
104
105
106
107
108
109
110
111
112
113
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs Mt Ds
Rg
Cn Uut Fl
(271.133)
(270)
(223.0197) (226.0254) (227.0278) (265.1167) (268.125)
58
59
Ce
Pr
140.116
90
60
61
62
(277.150)
63
(276.151)
64
195.084 196.966569
31
(281.162)
65
(280.164)
66
(285.174)
67
(284.178)
68
114
**
(289.187)
69
208.98040 (208.9824) (209.9871) (222.0176)
115
116
117
118
(288.192)
(293)
(294)
(294)
Uup Lv** Uus UUo
70
71
Nd Pm Sm Eu Gd Tb
Dy Ho
140.90765
144.242
(144.9127)
150.36
151.964
157.25
158.92535
162.500
164.93032
167.259
168.93421
173.054
91
92
93
94
95
96
97
98
99
100
101
102
103
Es Fm Md No
Lr
Th Pa
U
Np Pu Am Cm Bk
Cf
Er Tm Yb Lu
174.9668
232.03806 231.03588 238.02891 (237.0482) (244.0642) (243.0614) (247.0704) (247.0703) (251.0796) (252.0830) (257.0951) (258.0984) (259.1010) (262.1096)
S.E. Van Bramer 8/29/2012
* 1995 IUPAC Values from Pure Appl. Chem., Vol 68, No. 12, pp. 2339-2359, 1996. doi: 10.1351/pac199668122339, http://pac.iupac.org/publications/pac/pdf/1996/pdf/6812×2339.pdf
**Names for elements 114 and 116 are from Pure Appl. Chem., Vol. 84, No. 7, pp. 1669–1672, 2012. doi: 10.1351/PAC-REC-11-12-03
All other values from: 2009 IUPAC Values from Pure Appl. Chem., Vol. 83, No. 2, pp. 359–396, 2011. doi:10.1351/PAC-REP-10-09-14, http://pac.iupac.org/publications/pac/pdf/2011/pdf/8302×0359.pdf
-Elements with one weight have uncertainty in the last digit.
-Elements with the weight in parenthesis, weight is given for the longest lived isotope.
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6.5a Practice
´ List all values for
´ List all values for
in n = 4
for an f orbital
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6.5b-c Orbital Shapes and Nodes
´ Designations, see
below images,
give info about
quantum numbers
´ Nodes: areas of
no electron
density; # planar
nodes =
´ What are the
quantum numbers
associated with
these diagrams?
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18
Summary of Subshells and Orbitals…
Jacqueline Bennett • Chemistry 111 • Spring 2019 • SUNY Oneonta
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…and the Periodic Table (again)
´ 2 electrons per orbital:
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6.5 Practice
´ Provide two reasonable sets of quantum numbers for
electrons for the following elements:
´ Lithium
´ Oxygen
´ Iron
´ Lead
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Chapter 5
Thermochemistry
1
General
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5.1a Kinetic and Potential Energy
´ Energy: ability to do work; cause
change
´ Work: force involved in
moving an object some
distance
´ Thermochemistry: study of how
heat energy is involved in
chemical change
´ Kinetic energy: E associated
with motion (e.g., mechanical
E and thermal E)
´ Potential energy: E associated
with position (e.g., chemical
energy)
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5.1a Practice
´ Be able to identify types
of energy, sources, and
whether kinetic and/or
potential (refer to table
on previous slide)
´ dropping a rock
´ using a flashlight
´ driving a car
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5.1b Energy Units
´ Energy of chemical
processes usually
reported in:
Know metric conversions, cal to J conversion
´ J: E required to accelerate
1 kg with a force of 1 N
over a distance of 1 m (1 J
= 1 kg•m2/s2)
´ cal: E needed to raise the
temp of 1 g pure water by
1 °C.
´ one food Calorie = 1 kcal
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5.1b Practice
´ Convert Calories to kWh
´ A bag of chips contains 150
Calories. How many kWh is
this?
´ A 60 W light bulb consumes
0.060 kWh of energy per
hour. How many of these
bulbs could “run” off of a
bag of chips?
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5.1c Principles of Thermodynamics
´ Thermodynamics: study of relationships
between heat, energy, and work and
their interconversions
´ System: item or reaction of interest
´ Surroundings: everything else
´ Isolated system: matter and energy NOT
passed to/from surroundings
´ Closed system: energy only CAN be
passed to/from surroundings (most
cases in chemistry)
´ Internal energy, Esystem, changes
when heat (q) is added/lost and
work (w) is done by/on the system
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5.1c Practice
´ A gas is compressed, and
during this process the
surroundings do 150 J of
work on the gas. At the
same time, the gas loses
280 J of energy to the
surroundings as heat.
What is the change in
internal energy of the
gas?
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5.2a Enthalpy
´ Enthalpy = internal energy + pressure
x volume
´ H = E + PV
´∆H = heat exchanged under
constant pressure
´enthalpy cannot be
measured directly but
enthalpy change, and
therefore, relative enthalpy
can be measured
Endothermic process: ∆H is (+)
heat is transferred into the system
Exothermic process: ∆H is (–)
heat is transferred out from
the system
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5.2b Representing Energy Changes
´ Enthalpy diagrams: help
chemists to visualize
chemical and physical
changes
´ ∆Hrxn = ∆Hprod – ∆Hreact
´ Endothermic:
´ Exothermic:
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5.3a Heat Transfer and Temperature
Changes: Specific Heat Capacity
´ Specific heat capacity:
energy required to raise
the temperature of 1 g
sample by 1 °C
´ Unit: (J/g•°C)
´ Why does 150 J heat
affect copper and glass
so differently?
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5.3a Example
´ Using the following data, determine the
specific heat capacity of silver.
q = 150 J
m = 5.0 g Ag
Tfinal = 145.0 °C
Tinitial = 20.0 °C
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Trends in Specific Heat Capacity
´ Can you spot any trends
here?
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5.3a Practice
Calculate the final temperature
reached when 324 J of heat is
added to a 24.5-g iron bar
initially at 20.0 °C. Use the
specific heat for iron from the
previous table.
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5.3b Heat Transfer Between Substances
´ Thermal equilibrium:
objects in contact at
different temperatures will
transfer heat from the
warmer to the cooler until
both objects are at the
same temperature
´ Warmer object cools
´ Cooler object warms
´ -heat lost = heat gained
´ -qwarmer = qcooler
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5.3b Example: Calculate heat lost/gained
´ A 12.00-g block of
copper at 12.0 °C is
immersed in a 5.00-g
pool of EtOH with a
temperature of 68.0 °C.
What is the final
(equilibrium) temperature
of the copper and EtOH?
´ qCu + qEtOH = 0
´ Specific heats: Cu = 0.385 J/g•°C
EtOH = 2.44 J/g•°C
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5.3b Practice: Final temp, no phase
changes
´ Calculate the final
temperature when a
10.0-g block of metal
(see below) at 95.0 °C is
dropped into a 50.0-g
pool of water at 20.0 °C
(4.184 J/g°C)?
´ Copper: 0.385 J/g°C
´ Gold: 0.129 J/g°C
´ Aluminum 0.897 J/g°C
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5.3c Energy, Changes of State, and Heating
Curves
What happens as you
add heat to ice starting
at -10 °C?
1.
2.
3.
4.
5.
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Phase changes vs warming
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What’s really happening?
´ Warming:
´ Molecules move more rapidly
´ Kinetic energy increases
´ Temperature increases
´ Melting/Boiling:
´ Molecules do NOT move more
rapidly
´ Temperature remains constant
´ Intermolecular bonds are broken
´ Chemical potential energy
(enthalpy) increases
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5.3c Example
´ How much energy is required to convert 40.0 g of ice at
–30 °C to steam at 125 °C? cice = 2.06 J/g•°C, cwater =
4.18 J/g•°C, csteam = 1.92 J/g•°C, ∆Hfus = 333 J/g, ∆Hvap =
2256 J/g.
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5.3c Practice
´ How much energy is required to convert 10.0 g of ice at
–10 °C to steam at 110 °C? cice = 2.06 J/g•°C, cwater =
4.18 J/g•°C, csteam = 1.92 J/g•°C, ∆Hfus = 333 J/g, ∆Hvap =
2256 J/g.
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5.4a Enthalpy Change for a Reaction
´ Breaking or forming chemical bonds changes enthalpy
´ ∆Hrxn: enthalpy change for a reaction
´ Zn(s) + S(s) → ZnS(s) + 205.98 kJ
´ ∆Hrxn= –205.98 kJ
´ Why negative?
´ N2(g) + O2(g) +66.36 kJ → 2NO2(g)
´ ∆Hrxn = ?
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5.4b Enthalpy Change and Chemical Equations
´ ∆Hrxn is a function of the C12H22O11(s) + 12 O2(g) à12 CO2(g) + 11 H2O(l) DH = -5645 kJ
reaction as written
´ enthalpy is an extensive
variable
C12H22O11(s) + 12 O2(g) à
´ if the moles are affected by
some amount, enthalpy is
affected by that same
amount
´ if the reaction is reversed,
the sign is changed
12 CO2(g) + 11 H2O(l) + 5645 kJ
2 C(s) + 2 H2(g) à
2 C(s) + 2 H2(g) + 52 kJ à
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C2H4(g) DH = +52 kJ
C2H4(g)
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5.4b Example
´ According to the
following reaction, 373 kJ
of energy is evolved for
each mole of CO that
reacts. Calculate the
enthalpy change for the
reaction.
´ 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)
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5.4b Practice
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = -2220 kJ
If we burn 0.25 mol propane, what quantity of heat is produced?
If 1.60 mol of CO2 is produced, what quantity of heat is produced?
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5.4b More Practice
4HCl(g) + O2(g) → 2H2O(l) + 2Cl2(g)
ΔH = -202.4 kJ
Calculate the enthalpy change for:
H2O(l) + Cl2(g) → 2HCl(g) + 1/2 O2(g)
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Where does enthalpy change come from?
´ Bond Energies
´ DH = energy needed to break bonds – energy released
forming bonds
Example: formation of water:
´ DH = [498 + (2 x 436)] – [4 x 436] kJ = -482 kJ
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5.4c Determining Enthalpy Change:
Calorimetry
´ Calorimetry: experiments that
measure heat exchange
(calorimeter)
´ Bond energies can predict
ΔH for gas phase reactions
only.
´ΔH for reactions not in
the gas phase is more
complicated, due to
solvent and solid
interactions.
´ So, we measure ΔH
experimentally.
´ Perform reaction in a way that measures
heat gained or lost by the system.
´ Two types:
´ Constant pressure: “coffee cup
calorimetry” measures DH
´ Constant volume: “bomb calorimetry”
measures DE
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5.4c Calculate enthalpy from constant P
calorimetry (coffee-cup, ∆H)
When 4.50 g NH4Cl is dissolved in 53.00 g of water in a styrofoam cup, the
temperature of the solution decreases from 20.40 °C to 15.20 °C. Assume
that the specific heat of the solution is 4.18 J/g • °C. Calculate DH for the
dissolution (kJ/mol). Assum …
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