13. States of matter liquid and solids 1 Chem123/125 Spring 2003 Instructor: C. Chieh (sounds like Jay, Nickname peter) Cyberspace Chemistry (CaCt) science.uwaterloo.ca/~cchieh/cact

13. States of matterliquid and solids

1

Chem123/125Spring 2003

Instructor: C. Chieh (sounds like Jay, Nickname peter)

Cyberspace Chemistry (CaCt)science.uwaterloo.ca/~cchieh/cact/

Office: C2-263Phone: 888-4567 ext. 5816e-mail: [email protected]

You: Please provide your ID number when sending e-mails.


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Please sit in the front & center

The communication is easier and better for both you and the instructor. The better the communication, the better you learn.


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I (student)

learn how to learn,

learn the basics and techniques of science, and apply them to solve problems to survive,

make my plan to fit the class schedule; write CaCt quizzes to test my ability,

practice, practice and practice problems solving on suggested questions in part a and b of handout,

ask my professor or my TA for help whenever I need, will not wait till I got a failure mark,

read all instructions, and distinguish assumptions and facts

Check counseling services for study skills: www.adm.uwaterloo.ca/infocus


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AnnouncementsDid you fail CHEM 120 last term?

Then you cannot take CHEM 123 this term. You must select another course for this term.

CHEM 120 is offered this term from 5:30-6:50 pm in DC 1351 on Mon & Tues.

Did you miss the CHEM 120 exam for valid medical reasons?

The make-up exam will be held on Friday, January 9th from 2-5 pm. Bring your medical documentation with you to the exam. Contact the course coordinator to find out where the exam is being written. ([email protected])


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Announcements – cont.Tutorials start in week #2 for even-numbered tutorial sections (i.e. sections 102, 104, 106, etc.)

Tutorials start in week #3 for odd-numbered tutorial sections (i.e. sections 101, 103, 105, etc.)

CHEM 123L and CHEM 125L: Labs start this week!


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Need to LearnFeeling and longing motive all human endeavor

Everything that the human race has done and thought is concerned with the satisfaction of deeply felt needs and the assuagement of pain.

Albert Einstein


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13. States of Matter

Solid crystal, glass (amorphous)

Liquid types of liquid

GasABCD gas laws

Other terms related to states: plasma, solutions, mixtures, colloid, liquid crystals, super fluid, supercritical fluid, phase

Know properties of your material


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SolidsCrystals and amorphous (glass or frozen liquid)

Describe the difference between crystals and amorphous solids


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Crystals Common crystals:

Diamond, ice, dry ice, quartz, calcite, aragonite, sulfur, phosphorus etc.

Crystal properties:

Melting point, arrangement of atoms (crystal structures), electrical conduction, hardness, crystal habit etc.

Give some unique features of crystals


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Liquid Common liquids:

Water, alcohol, acetone, carbon tetrachloride, gasoline, nitrogen, helium, hydrogen, oil, etc.

Properties of liquid:

Freezing point, boiling point, vapor pressure, viscosity, color, surface tension, motion and arrangement of molecules, dielectric constant etc.

Describe the arrangement of molecules in liquid and some common properties of liquid


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PhasesA phase is a distinct and homogeneous state of a system with no visible boundary separating it into parts.

Phases: solid, liquid, gas, solution, different solid or liquid

Think of some systems and determine the number of phases in them. Milk, orange juice, coke, soup, wine, beer, absolute alcohol, gasoline, gas, natural gas, etc.

Ruby imbedded in rocks, many phases

Explain the concept of phase, define and exemplify


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Phase Transitions - Terms

solid

liquid gas

melt

vaporize

depositfreeze

sublimatecondense

Explain and name phase transitions


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Phase Transitions – in action

temperature

time

energy

thermometer

melting

boiling

E = C Tenergy = heat capacity * change in T

Explain changes as energy is transferred into a system


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Phase Transitions

- Notations & EnergyPlease consider how energy causes the changes

H2O (s) H2O (l) Hf 6 kJ/mol

H2O (s) H2O (g) Hs 47

H2O (l) H2O (g) Hv 41

H2O (l) H2O (s) -Hf -6

H2O (g) H2O (s) -Hs -47

H2O (g) H2O (l) -Hv -41

Please consider other phase transitions and their transition energies.Explain the meaning of each equation

18 g Ice at 273 K

Water at 298 K

1 mole H2O vapor

44.0 kJ

6.01 kJWater at 273 K

Energy level diagram given in Chem120


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Phase Transitions-Vapor Pressurean equilibrium depending on Temperature

Temperature

Vapor pressure mmHg or Pa

Vapor pressure of ice

Vapor pressure of water

Vapor pressures of ice and water are the same

Critical point

1.8: System on slide 8


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Phase Transitions-Vapor Pressure

Temperature




Triple point (ice, water and vapor) Difference between triple

point and melting point

Critical point


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Phase Transitions-Vapor Pressure

Temperature




Boiling point

101.3 kPa Variation of bp with altitude

Critical point


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Phase Transitions-Vapor Pressure Application

Vapor pressure of alcohol


Distillation

T

P


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Phase Transitions-Vapor PressureClausius-Clapeyron Equation

The Clausius-Clapeyron equation correlate vapor pressure and heat of phase transition:

– Hvap

ln P = ––––––– + B RT

note ln P = 2.303 log P

How to get straight-line plots?

ln P

1T

Explain the meaning of the Clausius-Clapeyron equation


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Phase Transitions-Vapor Pressure Clausius-Clapeyron Equation

The vapor pressure of ether is 400 and 760 mmHg at 18 and 35oC. What is the heat of vaporization?

P T400 291 (= 273+18)760 308 (= 273+35)

P1 – Hvap 1 1ln ––– = ––––––– (–– – –– ) P2 R T 1 T2

400 – Hvap 1 1ln ––– = –––––––––– (––– – –––) 760 8.31 J mol-1 291 308

- 0.642 = – Hvap*2.283e-5

Derived from the C-C equation

Hvap = 28721 J mol-1

dP H P–– = –––– (–––)dT R T2

H Ln P = – –––– + B R T


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Phase Transitions-Vapor Pressure Clausius-Clapeyron Equation

The heat of vaporization for water is 40.7 kJ mol-1. Calculate vapor pressure at 300 K.

P T/ KP1 300760 373expect U to know

R = 8.314 J mol-1

P1 - Hvap 1 1ln ----- = ---------- ( ---- - ---- ) P2 R T 1 T2

P1 -40700 J mol-1 1 1ln ------- = ------------------- ( ---- - ----- ) = - 3.914 760 8.31 J mol-1 300 373

P1 / 760 = e-3.914 = 0.020

P1 = 0.020*760 mmHg = 15.2 mmHg

when P=800, b.p.=?


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Regarding ln N = ln 10 log N

a b = N = x y

b (log a a) = log a N log x N = (log x x) y

b (log a a) = log a N = y log a x

(log a a) b = log a N = y log a x

= log x N log a x

log a N = log a x log x Nlog e N = log e x log x N

ln N = ln 10 log N

e a

10 x2.303


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Practice Problems

• Use a calculator to evaluate the vapor pressure of water at 272, 273, and 360 K. (Use data from lecture material)H / R = 41000/8.3142=4931P at 272 K = 748 Pa; P at 273 K = 799 Pa, P at 360K = 62846 Pa P at 373 = 101.3 k Pa

• Use a spreadsheet to plot the vapor pressure of ice for temperature between 253 and 275 K. H/R = 47000/8.3142 = 5653;

Facts about H2O;

m.p. = 272 Kb.p. = 373 KHsub = Hf + Hv

T P /Pa250 91253 119255 141273 610274 658


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Phase Transitions -Phase Diagrams of H2O

100oC

Temperature

Vapor pressure

1 atm



W

I

V

Critical point

Draw and explain the phase diagram of water using data on Cact

b.p.


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A Student’s Question

For water, the phase diagram says that at 20o C and 1 atm, the water exists as only one phase (liquid), but it still has a vapour pressure.

It seems to me that at one atm. water will be 2 phases (vapour and liquid) at any temperature between 0 and 100 degrees celcius.

I have the same question about ice too.


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Phase Transitions -Phase Diagrams of CO2

Temperature


Vapor pressure of solid CO2

Vapor pressure of liquid CO2

Liquid CO2

d

Gas CO2

Critical point

1 atm

216 K

304 K

73 atm

5.1 atm

266 K

Solid CO2

What are the differences between phase diagrams of H2O and CO2


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Phase transition - Phase Diagram of Sulfur

Rhombic Mono-clinic

liquid

vapor

Triple point


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Intermolecular Forces

Intermolecular forces: forces of interaction among molecules

Intermolecular forces strength examplevan der Waalsor London dispersion ~ 0.1 kJ mol-1 Ar, S8, CH4

Dipole-dipole ~ 10 kJ mol-1 CO, NO2

hydrogen bonding ~20-40 kJ mol-1 H2O, HF, NH3 ionic attraction ~100-1000 kJ mol-1 NaCl, KBr

Intramolecular: covalent bonding ~100-2000 kJ mol-1 diamond

Explain each term and correlate properties such as hardness, m.p., b.p., H, of materials with intermolecular forces in them.


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Intermolecular Forces - explained

Summarize types of intermolecular forces


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Intermolecular Forces – London (dispersion) ForceNon-symmetric distribution of electrons around nuclei resulting in temporally dipole-dipole interaction among molecules.

This force exists among all material, but they are very weak when compared to other interactions.

- - - -- 10+ * - - - - -

Temporary dipole of Ne (Z = 10)

- - - -- - 10+ * - - - -

- - - -- 10+ * - - - - -

What gives rise to London dispersion force?


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Dipole MomentThe product of magnitude of charge on a molecule and the distance between two charges of equal magnitude with opposite sign is equal to dipole moment; D (unit is debye, 1 D = 3.34E–30 C m (coulumb.metre); representation Cl+H, a vector )

Dipole moment = charge x distanceSymbol: µ = e– x d = q * dbond

For Cl+H, µ = 1.03 D, dH–Cl = 127.4 pmTwo ways of lookint at H+Cl, q = 1.03*3.34e–30 C m / 1.274e-12 m = 2.70e-20 C (charge separation by H–Cl )Ionic character = q / e– = 0.17 = 17%

d = 3.44e-30 C m / 1.60e –19 C (e– charge) = 2.15E–11 m = 0.215 pm (+e– by 0.215 pm)

H–Cl = 1.03 DH–F = 1.9 D, find d and % ionic character for them.

Identify molecules that have dipole moment?


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Dipole moment of H2O Please verify:The dipole moment of individual water molecules measured by Shostak, Ebenstein, and Muenter (1991) is 6.18710–30 C m (or 1.855 D). This quantity is a vector resultant of two dipole moments of due to O–H bonds. The bond angle H–O–H of water is 104.5o. Thus, the dipole moment of a O–H bond is 5.05310–30 C m. The bond length between H and O is 0.10 nm, and the partial charge at the O and the H is therefore q = 5.05310–20 C, 32 % of the charge of an electron (1.602210–19 C). Of course, the dipole moment may also be considered as separation of the electron and positive charge by a distance 0.031 nm. For the water molecule, a dipole moment of 6.18710–30 C m many be considered as separation of charge of electron by 0.039 nm.


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Intermolecular Forces – Hydrogen BondAttractions between H atom covalently bonded to small electronegative atom (N, O, F) and these electronegative atoms are called H-bond,

X-H - - - Y (X, Y = N, O, F)

Strange properties of water (high density at 277 K, high m.p. & b.p. to molecules of similar mass, high heat capacity, etc.

Lewis structure of H2O

H : O : H

H-bonds are responsible for many important phenomena.

Describe H-bonding


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Intermolecular Forces – H-bond and b.p.

Explain unusual properties due to the formation of hydrogen bonds.

Identify H-bonding molecules.

Apply the hydrogen bond formation ability of molecules to predict their properties.


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Intermolecular Forces – H-bond and DNA

The double-helix DNA has two strands of phosphoric-acid and sugar linked bases of Adenine, Guanine Cytosine or Thymine.

The A-T and G-C pairs stack on top of each other.

http://www.accessexcellence.org/AB/GG/dna_molecule.html


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Intermolecular Forces – Solids


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Arrangement of Atoms in Diamonds

Image from WebElements.com

http://www.webelements.com/webelements/elements/text/Ge/xtal.html

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/sili2.html


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Types of Solid - Covalent Crystals

Diamond (C ), Si, Ge, GaAs, CdS, SiO2 (quartz) etc. consist of network formed by covalent bonds. These are high melting, hard, crystalline crystals.


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Types of Solid - Molecular CrystalsMolecular crystals consist of molecules in their solids.

When only London dispersion force hold the molecules together such as solid Ne, Ar, CO2, C6H6, CCl4, I2 etc. they are soft with low m.p. and low heat of vaporization.

When the molecules have permanent dipole moment, the molecules are together more strongly. They have higher m.p. and heat of vaporization than those with only London dispersion force.Packing of I2 molecules

from Web Elements

Identify the molecules of I2


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Types of Solid - Sphere PackingSphere packing is one of the models used to illustrate packing of atoms in metallic crystals. There are two types of sphere packing:

Sequence Type

ABAB… hcp (hexagonal closest packing) ABCABC... ccp (cubic closest packing) or fcc (face centred cubic)

ABC

ABAB..ABCABC..


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Types of Solid -

Metallic Crystals

Three common types:ccp hcp bcc

A slight distortion of ccp leads to bcc.


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Types of Solid - Metallic Structures


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Types of Solid – H-bonded CrystalsIn water, some clusters of water molecules held by H-bonds exist.

Ice, solid NH3, and molecules with H-bond ability crystallize making the most number of H-bonds. These solids have higher m.p. than the molecular and dipole-moment molecules.

Hydrogen bonding leads to the peculiar properties of melting curve for ice, and highest density for water at 4oC.

Crystal structure of icelowtem.hokudai.ac.jp/~frkw/english/ss2.html


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Solids - Unit Cell of CrystalThe smallest convenient unit when repeatedly stacked together generate the entire crystal structure.

Two choices of unit cell in a 2-D space

How many disk does each unit have?Left 1 right one large one small

extend to 3-D space


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Solids - Unit Cell & Density

A metal X (atomic mass M) crystallizes in a simple cubic crystal structure with one atom per unit cell with the length of the unit cell a in cm, then its density d g cm-1 is.

M / NA

d = ------------ a 3

Mass of an atom / unit cell

Volume of a unit cell

If the cubic unit cell has n atoms, then n M d = (NA a 3 ).

NA is the Avogadro’s number


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Solids - Unit Cell & Radius of atomSimple cubic:Cubic edge a = 2 time radius of atoms,

a = 2 r

Body centre cubic:face diagonal df

2 = 2 a2

body diagonal, db2 = df

2 + a2 = 3 a2

= (4r)2

Face centre cubic:face diagonal df

2 = 2 a2 = (4r)2 a = 22 r

http://www.webelements.com/webelements/elements/text/Na/xtal.html

http://www.webelements.com/webelements/elements/text/Po/xtal.html


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Solids – Density and Radii of Atoms

Copper crystallize in ccp type structure with a density of 8.92 g mL-1. Calculate its atomic radius based on the hard-sphere packing model.

Let the cell edge be a and radius be r. Data NA = 6.023e23, atomic mass of Cu 63.5 and a = 22 r

4 * 63.5 gd = ------------------ = 8.92 g cm-3

6.02e23 * a3

Thus, a3 = 4*63.5 / (6.02e23 * 8.92) cm3 = 4.73e-23 cm3

And a = 3.617e-8 cm = 22 r

Thus, r = 1.279e-8 cm or 0.127 nm

volume

Work on part b problems Wk 2 & 3

http://www.webelements.com/webelements/elements/text/Cu/xtal.html


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Solids – Tetrahedral HolesFour balls of radii R touching each other forms a tetrahedral configuration, the center of which is called a tetrahedral hole.

To evaluate the size of this hole, imagine the largest ball of radius r placed in the hole touching all for large balls.

The 4 balls can be placed on alternate corners of a cube, whose edge = aSince the 4 balls touch each other face diagonal = 2R = 2aSince large balls touching small ballbody diagonal = 2(R+r) = 3 a

a = edge

2R = fd = 2 a

2(R + r) = bd = 3 aR＋ r 3 ------- = ----- R 2

r---- = 0.225 R

Details of deriving these are described in lecture.


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Another Td picture

What is the radius of the largest ball that can be placed in a tetrahedral hole without disturbing the packing of the spheres of radius R?

How do you go about to solve this problem?

Solve the same problem for an octahedral hole.


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Solids – Octahedral HolesThe equatorial plane of the octahedral is a square of edge a.= 2R

Place a small ball in the octahedral hole touching all 6 balls. Then

Diagonal = 2 (R + r) = 2 a = 2 (2R)

Therefore2(R+r) 2 --------- = ------ = 2 2R 1

r ---- = --------- = ________?

R


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Ionic Solids – cont.


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Chemical Energy and Hess’s LawC(graphite) + 0.5 O2 CO H° = - 110 kJ/mol.

2 C(graphite) + O2 2 CO H° = - 220 kJ/mol (multiplied by 2)

6 C(graphite) + 3 O2 6 CO H° = - 660 kJ/mol (multiplied by 6)

2 CO C(graphite) + O2 H° = 220 kJ/mol (+ve)

CO (g) + 0.5 O2 (g) CO2 (g) H° = - 283 kJ/mol. 1 mol C and O2

1 mol CO2

1 mol CO & 0.5 mol O2- 393 kJ

- 283 kJ

A quantity difficult to measure: - 110 kJ


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Lattice EnergyThe Lattice energy, U, is the amount of energy given off when an ionic solid is formed from its gassious ions

a Mb+(g) + b Xa- (g) MaXb(s) U kJ/mol

This quantity cannot be experimentally determined directly, but it can be estimated using Hess Law in the form of Born-Haber cycle. It can also be calculated from the electrostatic consideration of its crystal structure. Solid U kJ/mol Solid U kJ/mol Solid U kJ/mol Solid U kJ/mol

LiF –1036 LiCl – 853 LiBr – 807 LII – 757NaF – 923 NaCl – 786 NaBr – 747 NaI – 704

KF – 821 KCl – 715 KBr – 682 KI – 649

MgF2 – 2957 MgCl2 – 2526 MgBr2 – 2440 MgI2 – 2327


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Born-Haber Cycle for Lattice EnergyNa+ (g) + Cl(g) + e–

Na+ (g) + Cl– (g)

Na (g) + Cl(g)

Na (g) + ½ Cl2(g)

Na (s) + ½ Cl2(g)

Na Cl (s)

IPNa = +496 kJ

½ BECl-Cl = +122 kJ

s ubNa = +107 kJ

fNaCl = –4 11 kJ

Cl = –349 kJ

Lattice Energy NaCl

= – 7 87 kJ

Draw diagram and explain all steps


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Evaluating the Lattice Energy

Another approach to evaluate lattice energy

Na(s) + ½ Cl2(l) NaCl (s) - 411 Hf

Na(g) Na(s) - 107 -Hsub

Na+(g) + e– Na(g) - 496 -IP

Cl(g) ½ Cl2(g) - ½*244 - ½ *BE

Cl– (g) Cl (g) + e– 349 -EA

Add all the above equations leading to

Na+(g) + Cl-(g) NaCl (s) -787 kJ/mol Lattice Energy

Use these data to draw a diagram


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States of matter and intermolecular forces

Summary• Energy causes change of states (calculations)

• Kinds of intermolecular force and how they affect properties of material (qualitative and descriptive) London dispersion force, dipole-dipole interactionH-bonding, ionic interaction, covalent bonding metallic bond

• Vapor pressure, phase transitions, phase diagrams

• Sphere packing for structure model: simple, bcc, ccp, hcp etc.

• Density and unit cell (edge)

• Unit cell and radii of atoms

• Born-Haber Cycle

Documents

13. States of matter liquid and solids 1 Chem123/125 Spring 2003 Instructor: C. Chieh (sounds like Jay, Nickname peter) Cyberspace Chemistry (CaCt) science.uwaterloo.ca/~cchieh/cact