Jensen Chem 110 Chap 10 Page: 2 - Chemistry Coursescourses.chem.psu.edu/Chem110fall/lecture...

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


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


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.


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


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 +


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


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


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


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 #


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?


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


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


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.


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!?!!


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


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(!)


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


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.


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


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


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.


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


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.


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?


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


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.


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


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 % .


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


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?


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


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


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


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Jensen Chem 110 Chap 10 Page: 2 - Chemistry Coursescourses.chem.psu.edu/Chem110fall/lecture...

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