Theories applied to Chemistry

Quantum Mechanics (Planck, Einstein, Schrödinger - ĤY=E ).YApplies to all chemical systems.

Although QM is the primary theory explaining the behavior of subatomic particles, atoms, and molecules it gives way to CM for much of chemistry due to its more complex mathematical foundations.

1

Wave-particle duality allows mathematical wave function solutions to predict chemical properties of systems.

Incorporates Heisenberg Uncertainty Principle.

The behavior of individual particles are not certain, but represented as outcome probabilities.

Statistical treatments make measurements on large systems essentially certain.

Theories applied to Chemistry

Classical Mechanics (Newton’s Laws of Motion?)

Applies only to larger (atoms /molecules), slower (relative to c) particles.

Although CM leads to a number of inconsistencies in chemical behavior it is still used for chemical topics where it provides accurate predictions with more conceptually accessible explanations. CM topics in chemistry include much of kinetics and thermodynamics.

Theories applied to Chemistry

Classical Mechanics (Newton - e.g. F = ma)

There are two distinct approaches to studying classical systems in chemistry ….

Macroscopic

Look at behavior of system as whole (P vs. V at cst T)Summarize behavior as a set of laws (Boyle’s Law etc.)Develop equations predicting the behavior of systems. (P1V1 = P2V2)

Microscopic (Classical)

Develop a physical model explaining molecular behavior.Apply Laws of motion to individual particles.Extrapolate to collection of particles (statistical mechanics).Derive equations predicting the behavior of systems (CM).

Kinetic Molecular Theory (KMT) see handout

Assume:

2. particles in constant, random motion

3. no attractive/repulsive forces

4. conservation of energy at every collision

x

y

z

1. gas particles have mass but no volume

x

y

z

Pressure of Ideal Gas

vx

vyvz

dpx = ? dt = ?

<v2> = 3nRT/nM = 3RT/M

v2 = vx2 + vy

2 + vz2 & <v2> = 3<vx

2 > & …. <vx

2> = <v2>/3

PA = 2mvx/(2x/vx) = mvx2/x

PAx = PV = mvx2

PV = nM<v2>/3

For N molecules ― PV = Nm<vx2> = nM<vx

2>

= nRT

<v2>1/2 = vrms = (3RT/M)1/2

P = F/A & PA = F = ma = m(dvx/dt) = dpx/dt

dpx = 2mvx dt = 2x/vx

x

y

z

Pressure of Ideal Gas

vx

vyvz

PV = Nm<v2>/3 = nM<v2>/3 = nRT

Derive expression for <v2>1/2 = vrms

P = F/A & F = ma = dpx/dt

Derive expression for KE (Etr) = ½nM<v2> in terms of T.

use KE (etr) = ½mv2 from Newton’s 2nd Law

Etr = ½Nm<v2> = ½nM<v2> for N particles

Etr = ½nM<v2> = 3nRT/2

etr = 3kT/2 (per molecule) since R = NAk or nR = Nk

rms-speed is a function of both T and M ……Kinetic energy is only a function of TIn the KMT all gases have an ‘equal’ impact on P at a fixed T, becausethe greater force of larger particles is offset by their slower speed

Or …. Etr,m = 3RT/2 (per mole)

…to derive an expression for KE (Etr) = in terms of T.

PV = Nm<v2>/3 = nM<v2>/3 = nRT x 3/2

<v2>1/2 = vrms = (3RT/M)1/2

The Barometric FormulaThis derivation is not in your text (see handout). However, it illustrates one practical application of statistical mechanics …The approximate atmospheric content as a function of altitude …

2

This model also serves as an example of the Boltzman Distribution Law, a law derived using statistical mechanics, that illustrates how particles distribute themselves over an energy gradient.

The Boltzman Distribution law is fundamental both to CM and QM. In the latter it is required to explain the characteristics of spectroscopy arising from the population distribution of quantum states.

We will apply the Boltzman Distribution law as a shortcut to avoid a more exact but tedious derivation of the speed distribution function.

dz = thickness of atmosphere layer

A = Surface area of layer

dm = mass of gas between z & dz dP = Pressure difference between z & z + dz.

Barometric Formula & Boltzmann Distribution

z = altitude

Fup = PA Fdown = g • dm + (P + dP)A

What is Fup and Fdown?

Barometric Formula

Let Fup = Fdown

PA = (dm)g + (P + dP)A

PV = nRT & n = ?

PA = MPVg/RT + PA + dP • A & dP = -MPVg/ART

integrate PºP dP/P = -Mg/RT 0

z dz

ln (P/P) = -Mgz/RT &...

Since V/A = dz …. dP/P = -Mgdz/RT

P = P exp(-Mgz/RT)

dm = MPV/RT dm/M

Boltzmann Distribution Lawby analogy to Barometric Formula

P/P = exp(-Mgz/RT)

P/P ~ N/No & DE (J/mol) = -Mgz (kg•m2•s-2/mol)

J = kg•m2•s-2

N/No = exp(-DE/RT) or....

N/No = exp(-De/kT)The Boltzmann Distribution Law gives the relative occupation for any two energy levels.e.g height in gravitational field, MO’s, vibrational energy states, translational energy of gases with varying speeds.

N/No = g/go •exp(-DE/RT)

Mountains ht (m) locationEverest 8848 HimalayasK2 8611Kilamanjaro 5895 AfricaEiger 3970 AlpsMatterhorn 4477 AlpsDenali 6194 AlaskaPikes Peak 4302 ColoradoMauna Kea 4205 HawaiiHarney Peak 2208 South Dakota (Black Hills)

StructuresBurj Dubai 818 United Arab Emirates Eiffel Tower 324 Paris, FranceSears Tower 527 ChicagoWashington Monument 169 Washington, D.CCN Tower 553 Toronto, Canada

P/P = exp(-Mgz/RT)

What is the atmospheric pressure at the top of ……..

CHM 3460Assignment #1Due Monday, 9/12/11Ball – chapter 19: #2 and #6

2. KE of Hg atom and 1 mole of Hg atoms that has speed = 200 m/s?

From CM etr = ½mv2 = 0.5 * 0.20059/6.022e23 * 2002 = 6.664 x 10-21 J

For 1 mole x by NA Etr = 4013 J

6. P of interstellar space containing 10 molecules of H2 per cm3 at 2.7K.

What is vrms? (The book is in error in calling this <v> and the # they give as an

answer is Vrms.)

Use PV = nRT (n = N/NA & R = NAk) P = NkT/V

P = 10/6.022 x 1023 * 1.38 x 10-23 • 2.7 0.013m3 = 3.73 x 10-16 Pa

vrms = (3RT/M)½ = (3 • 8.314 • 2.7/0.002016)0.5 = 183 m/s

if on earth vrms = 1920 m/s

alternately you could calculate vrms first and then use P = nM<v2>/3V to get P = 3.71 x 10-16 Pa

SPEED DISTRIBUTION

What is the composition of air?

Do all of the oxygen molecules in air have the same speed?

Are all of the oxygen molecules in air identical?

Do we want to include the difference between O2 molecules in our model?

<v2> = 3RT/M

16O 99.76% 15.9949117O 0.04%18O 0.20% 15.999

Are all of the identical oxygen molecules (e.g. 16O-16O = 99.52%) in air traveling at the same speed?

3

dNv/N = fraction of molecules with speed = v

SPEED DISTRIBUTION

G(v) = Speed Distribution Function = dNv/N

Fractional probability = ∫v1v2 G(v) dv

Distribution functions must be normalized such that…

∫0∞ G(v) dv = 1

Plot G(v) vs. vG(v) ordNv/N

v

What should the graph look like?

Molecules with speed = 0? Molecules with speed = ∞?

Expand to 3D velocity distribution function (v) (v) = g(vx) g(vy) g(vz)

Find 1D velocity distribution = g(vx)

DERIVATION STRATEGY

Change to speed distribution function G(v) removing dependence on specific direction of motion

What is the average vx for a gas molecule?

What should the function g(vx) look like?

→

→

→

→ → →→

→

Find the value of A (normalization constant)

see page 793 ….. ∫0∞ exp(-bx2) dx = ½(p/b)1/2

Find 1D velocity distribution = g(vx)

Find g(vx) = dN(vx)/N from Boltzmann Distribution….

Nvx/N = exp(-De/kT)

g(vx) = dN(vx)/N = A exp(-mvx2/2kT)

A = normalization constant

∫-∞∞ g(vx) dx = A ∫-∞

∞ exp(-mvx2/2kT) = 2A ∫0

∞ exp(-mvx2/2kT) = 1

(b = m/2kT)

A = 1/(2∫0∞ exp(-mvx

2/2kT)) A = 1/(p/b)1/2

g(vx) = (m/2pkT)½ exp(-mvx2/2kt)

KE = ½mvx2→

A = 1/(2pkT/m)1/2

0

5E-09

1E-08

1.5E-08

2E-08

2.5E-08

-1000 -800 -600 -400 -200 0 200 400 600 800 1000

g(v

x) (

s/m

)

v (m/s)

Velocity Distribution of Gases

He

CH4

O2

CO2

CO2

He

CH4

O2

g(vx) = (m/2pkT)1/2 exp(-mvx2/2kT)

Graph g(vx) vs. vx ….?

g(vx) = (M/2pRT)1/2 exp(-Mvx2/2RT)

Find (v) = g(vx) g(vy) g(vz): extrapolate to 3D ‘ray’

v2 = vx2 + vy

2 + vz2 = 3vx

2

g(vx) = (M/2pRT)½ exp(-Mvx2/2RT)

[(M/2pRT)½ ]3 = [exp(-Mvx

2/2RT)]3

F(v) = (M/2RT)3/2exp(-Mv2/2RT) dv

= [exp(-3Mvx2/2RT)]

<vx2> = <v2>/3

= [exp(-Mv2/2RT)]

For a large collection of particles …… vx

2 = vy2 = vz

2 and m/k = M/R

4

→→

Expand to 3D velocity distribution function (v) (v) = g(vx) g(vy) g(vz)

Find 1D velocity distribution = g(vx)

DERIVATION STRATEGY

Change to speed distribution function G(v) removing dependence on specific direction of motion

→

→ → →→

→

g(vx) = (M/2pRT)½ exp(-Mvx2/2RT)

F(v) = (M/2RT)3/2exp(-Mv2/2RT) dv→ →→

z

xdV

dV(cube) = dvxdvydvz = ? dv

G(v) = (M/2RT)3/2exp(-Mv2/2RT) 4v2 dv

(v) expanded to G(v) F(v) = (M/2RT)3/2exp(-Mv2/2RT) dv

dV = 4p(v+dv)3/3 – 4pv3/3 = 4v2dv

4pv3/3 + 4pv2dv + 4pvdv2 + 4pdv3/3

y

→→→

Scalar v represents radius of spherebut also represents speed

v = r

dv = dr

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0 500 1000 1500 2000

Speed Distribution of Gases

He

CH4

O2CO2

G(v)

v ms-1

G(v) = (M/2RT)3/2exp(-Mv2/2RT) 4v2

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0 200 400 600 800 1000 1200 1400 1600

G(v

) (

s/m

)

Speed (m/s)

CO2 Speed Distribution : T dependence

300

600

1000

300K

600K

1000K

G(v) for CO2

As T↑ G(v) becomes lower and broader.

Open Excel file – “speed graph”

0 250 500 750 10000

0.000500000000000005

0.00100000000000001

0.00150000000000001

0.00200000000000002

Neon: Speed Distribution (300 K)

<v> = ∫0∞ vG(v) dv

= (8RT/pM)½

= 561 m s-1

Vmp = v where dG(v)/dv = 0 = (2RT/M)½

= 497 ms-1

vrms = ∫0∞ (v2 G(v) dv)½

= (3RT/M)½

= 609 m s-1

Finding the average speed for any gas from ……

G(v) = (M/2RT)3/2exp(-Mv2/2RT) 4v2

<v> = ∫0∞ v • G(v) dv General formula

∫0∞ x2n+1 exp(-cx2) dx = n!/2cn+1 integral tables

Let n = 1 and c = M/2RT substitutions

v • G(v) dv = 4(M/2RT)3/2 • v3 • exp(-Mv2/2RT)

4p{M/(2pRT)}3/2 • ∫0∞ v3 exp(-cv2) dv = 4p{M/(2pRT)}3/2 • 1/(2c2)

= 4p{M/(2pRT)}3/2 • 4R2T2/(2M2)

= 8p R2T2M3/2 = (8RT/pM)1/2

23/2p3/2R3/2T3/2M2

0

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

0.0014

0 200 400 600 800 1000 1200 1400 1600

G(v

) (

s/m

)

v (m/s)

Neon: frac. with 400 < v < 440 T = 300 & M = 0.02018 kg/mol

fraction of molecules with speed between 400 ms-1 and 440 ms-1

= 400440 G(v) dv

G(v) = (M/2RT)3/2exp(-Mv2/2RT) 4v2

~ G(v) * dv (sub in v = 420 & dv = 40) better estimate if dv is small)

Speed Distribution19.17

Will they always have the same relative values? …. … or will variations in either T or M change their relative magnitudes?

vmp (2RT/M)½

<v> (8RT/pM)½

vrms (3RT/M)½

vmp (2RT/M)½ (2)½ 1

<v> (8RT/pM)½ (8/p) ½ 1.13

vrms (3RT/M)½ (3)½ 1.22

Assignment – 19.14 plus make graph of G(v) vs. v from 0 to 1000 m/s using increments of 100 m/s.

Show <vx> = 0 from g(vx) = (M/2pRT)1/2 exp(-Mvx2/2RT)

Collisions with Wall: dNw/dt … what factors?

1. Area of wall (A) = yz

2. average speed in direction of wall <vx> = 0

vx g(vx) dvx = <v>/4

3. The gas density (N/V) = PNA/RT

dNw/dt = A • <v>/4 • PNA/RT = A • (RT/2pM)½ • PNA/RT

Effusion Rate: dN/dt same as collisions with wall except replace area of wall with area of hole

x

z

y

vx

vyvz

5

2. average speed in direction of wall <vx> = 0

vx g(vx) dvx = <v>/4

0 x exp(-cx2) dx = 1/(2c) c = M/(2RT)

<vx> = (M/2pRT)1/2 • RT/M = {RT/(2pM)}½ x 8½/8½ = = {8RT/(16pM)}½ = ¼{8RT/(pM)}½ = <v>/4

g(vx) = (M/2pRT)1/2 exp(-Mvx2/2RT)

zb(b) (s-1) = how many collisions will one ‘b’ molecule make with all other ‘b’ molecules per unit of time?

Molecular collisions (like molecules) zb(b) (s-1) = function of ......

dN/dt = A • <v>/4 • N/VdNb(b) = the # of collisions of 1 ‘b’ molecule with other ‘b’ molecules.

How do these 3 factors change for molecular collisions?

<v>/4 was speed in direction of wall/hole (<vx>).How should we represent speed in direction of another molecule that is also moving?

N/V = density of molecules near hole.How should we represent N/V for collisions?

A = How do we represent the area of a collision?

Molecular collisions (like molecules) zb(b) (s-1) = function of ......

density of molecules N/V = PbNA/RT

Relative speed <vrel>

<vrel> (assume 90º∟) = (<v>2 + <v>2)1/2

<vrel><v>

<v><vrel> = 2½ • <v>

Collision area (p rcir)2 = pd2

<vrel> dt

2 • d

Molecular collisions (like molecules) zb(b) = A • <vrel> • Nb/V

zb(b) = pd2 • 2½<v> • PbNA/RT

<v> = (8RT/pM)½ <vrel> = 4(RT/pM)½

Zb(b) = the total # of collisions per unit volume

Zb(b) = zb(b) • ?Nb/V • ½

l (mean free path) = the average distance a molecule will travel before units (m) colliding with another ‘like’ molecule,

l = <vrel> zb(b) = ms-1/s-1 = 1/(A • Nb/V) = RT/(pd2PbNA)

or … zb(b) = ms-1/m = <vrel>/l

Collisions between like molecules –

Density - # molecules are available for collision (m-3): N/V = PNA/(RT)

relative speed – effective collision speed (ms-1): <vrel> = 4{RT/(pM)}1/2

Area of molecular displacement (m2): A = pd2

Collision frequency (s-1): z = <vrel>•A•(N/V) = 4p1/2PNAd2/{MRT}1/2

Total collisions (s-1 m-3): Z = ½•z•(N/V) = 2p1/2(PNAd)2/{M1/2(RT)3/2}

Mean free path (m): l = <vrel>/z = RT/{pd2PNA} = kT/(pd2P)

Collisions between like unlike molecules –

Collisions between like unlike molecules -

<vrel> (ms-1) = (8RT/p)1/2•(1/Mb2 + 1/Mc

2)1/2

A (m2) = p(rb + rc)2

zbc (s-1) = <vrel>•A•(Nc/V) = {8p•(1/Mb

2 + 1/Mc2)}1/2•(rb + rc)

2•PcNA/{RT}1/2

Zbc (s-1 m-3) = ½•zbc•(Nb/V) = {2p•(1/Mb

2 + 1/Mc2)}1/2•(rb + rc)

2•Pc•Pb•NA2/{RT}3/2

lbc (m) = <vrel>/zbc = p(rb + rc)2 • PcNA/RT

Collisions between like unlike molecules -

<vrel> (ms-1) = (8RT/p)1/2•(1/Mb2 + 1/Mc

2)1/2 A (m2) = p(rb + rc)2

zbc (s-1) = <vrel>•A•(Nc/V) = {8p•(1/Mb

2 + 1/Mc2)}1/2•(rb + rc)

2•PcNA/{RT}1/2

Zbc (s-1 m-3) = ½•zbc•(Nb/V) = {2p•(1/Mb

2 + 1/Mc2)}1/2•(rb + rc)

2•Pc•Pb•NA2/{RT}3/2

lbc (m) = <vrel>/zbc = p(rb + rc)2 • PcNA/RT

Zbc (s-1 m-3) = ½•zbc•(Nb/V)

zbc (s-1) = <vrel> • A • (Nc/V)

zbc (s-1) = (<vb>2 + <vc>2)1/2 • p(rb + rc)2 • (Nc/V)

6. P of interstellar space containing 10 molecules of H2 per cm3 at 2.7K.

What is vrms?

Use PV = nRT (n = N/NA & R = NAk) P = NkT/V

P = 10/6.022 x 1023 * 1.38 x 10-23 • 2.7 0.013m3 = 3.73 x 10-16 Pa

19.20 Diameter (H2) = 1.10Å = 1.10 x 10-10 m

determine z, Z, and l. If finished answer #21/22

Collisions between like molecules –

Density - # molecules are available for collision (m-3): N/V = PNA/(RT)

relative speed – effective collision speed (ms-1): <vrel> = 4{RT/(pM)}1/2

Area of molecular displacement (m2): A = pd2

Collision frequency (s-1): z = <vrel>•A•(N/V) = 4p1/2PNAd2/{MRT}1/2

Total collisions (s-1 m-3): Z = ½•z•(N/V) = 2p1/2(PNAd)2/{M1/2(RT)3/2}

Mean free path (m): l = <vrel>/z = 1/{A•(N/V)} = RT/{pd2PNA} = kT/(pd2P)

v G(v) v G(v) • dv

0 0

100 0.00034 15 8.24E-05

200 1.13E-03 105 0.003766

300 1.85E-03 1005 0.000568

400 2.10E-03 5005 1.5E-69

500 1.84E-03 10005 4.9E-278

600 1.31E-03

700 0.00078 vmp 395

800 0.00039 <v> 446

900 1.64E-04 vrms 484

1000 6.00E-05

Ball – 19.14 with speed distribution graph from 0 to 1000 m/s

0 100 200 300 400 500 600 700 800 900 10000

0.000500000000000001

0.001

0.0015

0.002

0.00250000000000001

v (m/s)

G(v)

Gas = Oxygen at 300K MW = 0.032 kg/mol

G(v) = 4p{M/(2pRT)}3/2 • v2 • exp {-Mv2/(2RT)}

19

zb(b) = A • <vrel> • Nb/V 3.8 x 10-20 238 1 x 107

<vrel> = 4(RT/pM)½ = 4 • {8.314 • 2.7/(p • 0.002)}½ = 238 m/s

Zb(b) = zb(b) • ?Nb/V • ½ = 4.53 x 10-4 m-3

l = <vrel> zb(b) = ms-1/s-1 = 1/(A • Nb/V) = RT/(pd2PbNA)

= 2.63 x 1012 m

20 H2MW

(kg/mol) T (K) d (m) <vrel> (m/s) A (m2) z (s-1) N/V l (m) book Z

6 0.002016 2.7 1.10E-10 238 3.80E-20 9.05E-11 10000000 2.63E+12 2.63E+12 4.53E-04

19.20

A = pd2 = p • (1.10 x 10-10)2

N/V = PNA/(RT) = 3.73 x 10-16 • 6.022 x 1023/(8.314 • 2.7) =

Brownian Motion The movement of a particle through a medium

l (mean free path) = the average distance a molecule will travel before units (m) colliding with another ‘like’ molecule,

l = <vrel> zb(b) = ms-1/s-1 = 1/(A • Nb/V) = RT/(pd2PbNA)

http://www.youtube.com/watch?v=ZAGloLXO9L0

Diffusion Demonstration

1D: (Dx) = (2Dt)½ 3D: (Dx) = (6Dt)½

Dgas = 3p/16 • l • <vrel>

Transport PropertiesdY/dt = -cst • Area • gradient (dy/dx)

Diffusion – matter over [ ] gradientJ = -D dc/dx or dni/dt = -Dik A (dci/dx)

name PropertyY

Gradientdy/dx

equation

thermal conductivity heatq

temperatureT

dq/dt = -k A (dT/dx)

diffusion molesni

concentrationni

dni/dt = -Dik A (dci/dx)

viscosity Momentumpx

Velocityvy

dpx/dt = -h A (dvy/dx)

Electricalconductivity

ChargeQ

PotentialF

dQ/dt = -k A (df/dx)

A theoretical model is developed to predict the constant value similar to barometric formula and effusion

Diffusion: change of concentration with timeDependent on concentration gradient and diffusion coefficient.

[ ]

t

dni/dt = -Dik A (dci/dx) Fick’s 1st Law

Fick’s 1st law: dni/dt = -Dik A (dci/dx)

Fick’s 1st law: the rate of change in the amount of solute present at a position in a solution is a function of the area observed, the concentration gradient, and the intrinsic ‘ease of movement’ of the solute in the given solvent.

Fick’s 2nd law: (dci/dt)x = -Dik A (d2ci/dx2)t

Fick’s 2nd law: the rate of change in concentration with change in time is a function of the diffusion coefficient of the solvent, the area, and the 2nd derivative of Dci with respect to position.

if ri > rs then Dis = kT/(6sri)

if ri ~ rs then Dis = kT/(4sri)

Diffusion – matter over [ ] gradient dni/dt = -Dis A (dci/dx)

In liquids ……Dis depends on the absolute and relative size of solute (i) vs. solvent (s), and the viscosity of the solvent.

Gases : ~ 3 cm2/minLiquids : ~ 0.03 cm2/minsolids : < 10-8 cm2/min

32 W MW (kg/mol) T (K) P (torr) P (Pa) A (m2) <v>/4 N/V dN/dt (s-1) dN/dt (g/hr)

0.18385 4500 4.98 664 0.000001 180 1.07E+22 1.923E+18 2.113

dN/dt = A • <v>/4 • N/VDensity - # molecules are available for collision (m-3): N/V = PNA/(RT)

relative speed – effective collision speed (ms-1): <v> = {8RT/(pM)}1/2

19.32 Tungsten effusion – MW = 0.18385 kg/molGiven: T = 4500 K - dN/dt = 2.113 g/hour - A = 0.10 mm2.

find PW at 4500K?Convert all units to SI and find <v>/4

Find N/V from effusion equation

Solve for P – which will represent Tungsten vapor pressure

For Friday do 19.28 and 19.31