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Week 9: Lectures 25 – 27
Lecture 25: W 10/19 Lecture 26: F 10/21 Lecture 27: M 10/24 Reading:
BLB Ch 18.1 – 18.4, 5.5, 11.3 – 11.6 Homework:
BLB 18: 9, 11, 15, 29, 69; Supp 18: 1 – 3; BLB 11:33, 37, 39, 31, 43, 47, 50; Supp 11:5 - 13
Reminder:
• Angel Quiz 8 due on Thur 10/20 • ALEKS Objective 9 due on Tue 10/25
• Jensen Office Hour: 501 Chemistry Building
Tuesdays & Thursdays, 10:30 – 11:30 am
• TA office hours, evening tutor sessions, guided study group info posted on Angel ->
Help Available
• Exam 3: Monday Nov 7 6:30 – 7:45 pm
Jensen Chem 110 Chap 10 Page: 2
Real gases deviate from ideal behavior
The 5 key postulates of KMT
! Straight-line motion in random directions
! Molecules are small & have “no” volume compared to the total volume
! No intermolecular forces
! Elastic collisions
! Mean kinetic energy ! " T (in K)
For a non-ideal gas (real gas) this is not true for two types of conditions:
high P low T
Reasons:
• Molecules (or atoms) have a finite size molecules occupy space
• Molecules have attractive forces these forces become stronger when they are close together
Jensen Chem 110 Chap 10 Page: 3
Real Gases – Pressure
For 1 mol of any ideal gas: PV/(RT) = 1
For 1 mol of different gases at same T & different P:
At low P, the ideal-gas equation is valid: PV/(RT) = 1
In general, deviation from ideal behavior increases as P increases:
• At high P, attractive forces between molecules lead to the appearance of a smaller n
PV/(RT) < 1 Example: CO2 at 200 atm
• At very high P, finite molecular volume leads to repulsion and the appearance of a larger n
PV/(RT) > 1 Example: CO2 at 800 atm
Low P PV/(RT) ~ 1 High P: PV/(RT) < 1
Very high P PV/(RT) > 1
Jensen Chem 110 Chap 10 Page: 4
Real Gases – Temperature
For 1 mol of any ideal gas: PV/(RT) = 1
For 1 mol of N2 gas at different T & different P
• At low T, molecules stick together:
PV/(RT) ! 1 • As temperature increases, molecules move
faster, the behavior of real gases becomes more ideal.
Jensen Chem 110 Chap 10 Page: 5
Summary
Example: Under which conditions is Ar(g) most likely to approach ideal behavior?
A. 10 atm and 100°C
B. 1.0 atm and -200°C C. 0.10 atm and –100°C D. 20.0 atm and 100°C E. 0.10 atm and 200°C
High P PV/RT < 1
Attractive forces lead to the appearance of a smaller n
Very high P PV/RT > 1
Finite molecular volume leads to repulsion and the appearance of a larger n
Low P PV/RT ! 1
Attractive forces and finite molecular volume have minimal impact
As T #, real
gas behavior becomes more ideal
At high T, Kinetic energy overcomes attractive forces
Jensen Chem 110 Chap 10 Page: 6
Correcting the Ideal Gas Law
• For any gas, we can measure P, V, T
• At higher P, measured P is too small due to attractive forces between molecules
The amount of “missing” P is
proportional to
1. size of attractive forces (a):
2. frequency of collisions: (n2/V2)
To compensate, use: • At higher P, measured V is too
large due to finite molecular volume or excluded volume per mole (b)
Actual volume:
Vactual = Vmeasured – Vexcluded
To compensate, use: V – nb
P +
Jensen Chem 110 Chap 10 Page: 7
Real Gases: the van der Waals Equation
The Equation of state for IDEAL gases:
PV = nRT
The Equation of state for REAL gases:
• NOTE: a & b are specific for the particular gas we are talking about (BLB Table 10.3).
a (L2-atm/mol2) b (L/mol) He 0.0341 0.0237 Xe 4.19 0.0510
H2 0.244 0.0266 Cl2 6.49 0.0562
CH4 2.25 0.0428 CCl4 20.4 0.1383
correction for
intermolecular forces
between gas molecules
correction for
excluded volume (size)
of gas molecules
Chapter 18: Atmospheric Chemistry
Outline:
• Composition of Atmosphere
• Regions of Atmosphere
• Photochemical reactions in the upper atmosphere
$ Chemistry in upper atmosphere protects us from UV radiation
Photoionization Photodissociation(natural ozone cycle)
$ Human activity has impact on the chemistry of the upper atmosphere
Ozone depletion by CFCs
• Photochemical reactions in the lower atmosphere
$ Water Vapor, CO2 and the Climate
$ Sulfur Compounds and Acid Rain
$ Nitrogen Oxides and photochemical Smog
Jensen Chem 110 Chap 18 Page: 2
Composition of the Atmosphere
Composition of Dry Air Near Sea Level Component Content (mole fraction) Molar Mass
Nitrogen 0.78084 28.013
Oxygen 0.20948 31.998
… … …
Xenon 0.000000087 131.30
Mole fraction:
Express mole fraction in parts per million (ppm):
1 ppm = xi ! 106
Example: Neon xNe = 0.00001818, what is the mole
fraction of Ne in ppm?
xi =moles of i
total moles
Jensen Chem 110 Chap 18 Page: 3
Partial Pressures of Components
Example: The mole fraction of NO on a
smoggy day is measured at 20 ppm. If the barometric pressure is 732 torr, what is the
partial pressure of NO in the atmosphere?
A. 146.4 torr B. 0.01464 torr
C. 0.0366 torr
D. 15.2 torr
E. 36.6 torr
Jensen Chem 110 Chap 18 Page: 4
Regions of the Atmosphere
Based on temperature profile 1. Troposphere:
T % as altitude # 2. Stratosphere:
T # as altitude #
3. Mesosphere: T % as altitude #
4. Thermosphere: T # as altitude #
Jensen Chem 110 Chap 18 Page: 5
Pressure Profile
• Pressure at any altitude depends on the weight of gas above it.
• Pressure decreases
exponentially with altitude.
At high altitude, pressure is low . . . • How does low pressure
affect density? • How does low pressure
affect molecular collisions? • How does low collision
frequency affect the frequency of chemical reactions?
Jensen Chem 110 Chap 18 Page: 6
Important Chemistry in the Atmosphere
Photochemical Reactions: reactions that involve the absorption or emission of a photon.
• Photoexcitation: electronic excitation
NO2 + h& ' NO2* (* = excited state)
• Photodissociation: bond broken by absorption of a photon (light)
O2 + h& ' O + O
Requires energy " the bond energy
• Photoionization: removal of a valence e– from a molecule by absorption of a photon
N2 + h& ' N2+ + e–
Requires energy " the ionization potential
Jensen Chem 110 Chap 18 Page: 7
Example: What is the maximum wavelength of
a photon needed to photoionize O2 in the upper atmosphere?
O2 + h& ' O2+ + e– IE = 1205 kJ/mol
A. 99.4 nm
B. 165 nm
C. 274 nm D. 0.399nm
E. 39.9 nm
Jensen Chem 110 Chap 18 Page: 8
Importance of Atmospheric Ozone (O3)
• EDG = ___________________, MG = ________, bond angle = 117 ° , bond length = 1.28Å
• Light blue gas
• Pungent odor (smell near electrical discharges)
• (Hf°= 142kJ/mol (reactive… less stable than O2)
Atmospheric Concentration: • In the troposphere: O3 is an irritant (smog) • in the stratosphere: O3 is essential; peak
concentration at~25 km; [O3] ~ 10 ppm The (small) amount of O3 in the stratosphere reflects the delicate balance between creation and destruction of O3.
Jensen Chem 110 Chap 18 Page: 9
The Natural Ozone Cycle
• Formation of O3
O2 (g) + h& ' 2 O (g)
O (g) + O2 (g) 'O3 (g)
• Destruction of O3
O3 (g) + O (g) '2 O2 (g)
• Protection of Earth#s surface from UV-radiation:
O3 (g) + h& ' O (g) + O2 (g)
Ozone Depletion
Chlorofluorocarbons (CFCs), CFCl3, CF2Cl2, CF3Cl, destroy ozone cycle CF2Cl2 + h& ' CF2Cl• + Cl• () < 240 nm) 2Cl• + 2O3 ' 2ClO• + 2O2 2ClO• + h& ' 2Cl• + O2
NET: 2 O3 + h& ' 3 O2
• Cl atom acts as a catalyst to destroy O3
• 1 Cl atom can destroy > 100,000 O3 molecules!?!!
Jensen Chem 110 Chap 18 Page: 10
The Greenhouse Effect
• Energy coming to Earth is mostly solar $ Some is reflected by atmosphere (~30%) $ Some UV is absorbed by atmosphere $ Most hits Earth#s surface, where it is
absorbed and causes warming
• Energy radiated by Earth#s surface is mainly infrared (IR) heat
• The atmosphere is transparent to UV and Visible light, but NOT to IR
• The H2O, CO2, CH4 and O3 in atmosphere absorbs IR light
Thermal regulation by these gases (mostly CO2 and H2O) in the atmosphere leads to the Greenhouse Effect
Jensen Chem 110 Chap 18 Page: 11
Sulfur Compounds and Acid Rain
Natural rain: pH ~ 6 (slightly acidic due to dissolved CO2)
Acid rain: pH of 4 to 4.5 (more acidic due to dissolved pollutants like H2SO4, HNO3)
Sources: • bacterial decay of organic matter • volcanic gases and forest fires • fossil fuels combustion and industrial processes
Example: Coal burning (contaminated by sulfur): S8 (s) + 8 O2 (g) ' 8 SO2 (g)
SO2 reacts in the atmosphere:
2 SO2 (g) + O2 (g) ' 2SO3 (g)
SO3 (g) + H2O (g) ' H2SO4
sulfuric acid – dissolves in water and falls to earth as acid rain
• affects pH of soil and water • corrodes metals (Fe) • dissolves stone (marble, limestone)
CaCO3(s) + H2SO4(aq) ' CaSO4(aq) + CO2(g) + H2O(!)
Jensen Chem 110 Chap 18 Page: 12
Acid Rain Natural rain: pH ~ 6
slightly acidic due to dissolved CO2
Acid rain: pH of 4 to 4.5 more acidic due to dissolved pollutants like H2SO4, HNO3
Jensen Chem 110 Chap 18 Page: 13
Photochemical smog
primary h& secondary pollutants pollutants NO, NO2, CO O3 hydrocarbons
In auto engine: N2 +O2 ' 2NO (H =181 kJ
In air: 1. 2NO + O2 ' 2NO2
NO2 + h& ' 2NO + O ) = *400 nm
O + O2 + M ' O3 + M (M=another molecule)
2. NO2 + H2O' HNO3 (burns eyes)
Ozone: good in the stratosphere, but not in the troposphere
• Diminishes respiratory capabilities • Reacts with NO to form NO2 and O2
• Photodissociates to form reactive oxygen radicals, which react with hydrocarbons, etc.
Jensen Chem 110 Chap 11 Page: 1
Kinetic Molecular Description of Matter Gas: Kinetic energy >> intermolecular forces Liquid: Kinetic energy + intermolecular forces Solid: Kinetic energy << intermolecular forces Kinetic Energy " T: Heating: T # , KE # : solid ' liquid ' gas
Cooling: T % , KE % : gas ' liquid ' solid
Jensen Chem 110 Chap 11 Page: 2
Structure Affects Function
Functional group
Structure MW g/mol
BP °C
hydrocarbon
72 36
aldehyde
72 75
ketone
72 79
amine
73 77
alcohol
74 117
carboxylic
acid 74 141
Jensen Chem 110 Chap 11 Page: 3
Viscosity and Surface Tension
• Viscosity: resistance to flow; higher viscosity: slower flowing
$ stronger IM forces, more ___________ to flow
$ IM forces increases, viscosity ___________
• Surface tension: energy needed to increase surface area;
$ Intermolecular interactions are favorable
(heat is required to break them)
$ Surface molecules have fewer interactions.
$ IM forces increases, surface tension__________
Example: Which of the following has the highest viscosity?
A. B. C.
Jensen Chem 110 Chap 11 Page: 4
Vapor Pressure
Vapor pressure (v.p.): pressure exerted by a vapor in equilibrium with its liquid or solid phase;
• Dynamic equilibrium: no NET change, but change is occurring on molecular level
forward rate = backward rate evaporation = condensation
Jensen Chem 110 Chap 11 Page: 5
Vapor Pressure and Temperature
$ At higher temperature, more molecules have energies sufficient to escape the liquid, so the vapor pressure is higher
$ For different liquids, as IM forces increases,
vapor pressure (at a given T) decreases
Example: Which of the following has the highest vapor pressure?
A. B. C.
Jensen Chem 110 Chap 11 Page: 6
Vapor Pressure and Boiling Point
Boiling point (BP): T at which v. p. = Pext
• As Pext increases, BP __________
• Normal boiling point: T at which v. p. = ____
Example: Why does it take longer to hard boil eggs at higher altitudes than at lower altitudes?
Jensen Chem 110 Chap 11 Page: 7
Phase Diagrams
Plot of pressure vs. temperature of the system showing the boundaries between the phases.
Be able to identify the following;
$ normal melting point
$ normal boiling point
$ critical point
$ triple point
$ coexistence curves
Jensen Chem 110 Chap 11 Page: 8
Comparison of H2O and CO2 Phase Diagrams
Differences: 1) Triple Point: 2) Direction of the solid-liquid coexistence line:
CO2 slants to the ____ as P increases: MP __ as P# H2O slants to the ____ as P increases: MP__ as P#
Example: Using the phase diagram for water, how many phases are encountered if the water sample is heated from -10ºC to 100 ºC? The pressure is kept at 780 torr through the heating process.
Jensen Chem 110 Chap 11 Page: 9
Phase Changes Endothermic Process: Requires energy to disrupt intermolecular forces.
Melting (fusion) Vaporization Sublimation
Exothermic Process: Energy is released when intermolecular interactions are formed
Freezing Condensation Deposition
Jensen Chem 110 Chap 11 Page: 10
Heating Curve and Calorimetry
Two types of changes take place as heat is added:
1. Heating within a single phase (in blue):
q = m Cp (T (p , constant pressure)
T # , average KE # , total energy # ,molecular separation # , molecular attractions % , molecular order % .
2. phase transition (in red): changes from one physical state to another
q = n (Hx (x = fusion, vaporization)
T is constant, average kinetic energy is constant, but total energy # , separation between molecules # , molecular attractions % , molecular order % .
Jensen Chem 110 Chap 11 Page: 11
Heating within a Single Phase:
q = C m (T
C = heat capacity (J/g-K or J/mol-K) m = amount of substance (g or mol) (T = Tfinal - Tinitial
Note: (T (in K) = (T (in°C) Heat Capacity (C): Amount of heat required to raise an object#s temperature by 1K (or 1°C).
Usually C is given for a specified amount of pure substance Example: liquid H2O
specific heat , C = 4.184 J/g-K
molar heat capacity , Cm = 75.2 J/mol-K
$ Specific heat is different for each phase
$ Specific heat is different per gram than per mole
Jensen Chem 110 Chap 11 Page: 12
Example: The specific heat of toluene (C7H8)
is 1.13 J/g-K. How many Joules of heat are needed to raise the temperature of 0.5 mole
of toluene from 0°C to 25°C?
Jensen Chem 110 Chap 11 Page: 13
Practice Example: What is the specific heat
of gold, if raising the temperature of 985 g of Au from 30°C to 45°C requires 1.891 kJ
of energy?
A. 0.000128 J/g-K B. 4.81 J/g-K
C. 0.128 J/g-K
D. 0.256 J/g-K
E. 0.385 J/g-K
Jensen Chem 110 Chap 11 Page: 14
Heating during a Phase Transition:
q = m (Hx (x = fusion, vaporization)
(H = heat (kJ/g or kJ/mol) given off (-) or absorbed (+) when a change occurs
m = amount of substance (g or mol)
(Hfusion = heat needed to melt a mole of substance
(Hvap = heat needed to vaporize a mole of substance
Jensen Chem 110 Chap 11 Page: 15
Calculation of Energy Change
Measure the heat required for each segment of heating curve, then sum all individual
steps.
Example: Ethanol (C2H5OH) melts at –114 °C.
The enthalpy of fusion is 5.02 kJ/mol. The
specific heats of solid and liquid ethanol are 0.97 J/g–K and 2.3 J/g–K, respectively. How
much heat is needed to convert 25.0 g of
solid ethanol at –135 °C to liquid ethanol at
–50 °C?
A. 207.3 kJ
B. –12.7 kJ
C. 6.91 kJ D. 4192 kJ
E. 9.21 kJ
Jensen Chem 110 Chap 11 Page: 16
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