Post on 30-May-2015
description
14.2 The Boltzmann factor e(-ε/kT)
April 12, 2023
Connections to kT
kT
multiply byNA
1000
6.02 × 1023
103particles mol–1
energy per mole of particles
kJ mol–1
Average energy per particle
J
divide by e
1.6 × 10–19 C
average energy per particle in eV
divide by k
1.38 × 10–23 J K–1
absolute temperature
K
Aims
• Examine one derivation of the Boltmann factor
• Use the Boltzmann factor• State what is proportional to the rate of a
reaction
Getting extra energy by chance
• Processes happen when the energy ε per particle is a multiple of the average kinetic energy kT per particle
• The multiple can be as high as 20 or 30• To achieve this it must get lucky
Think about playing pool…
How many will get lucky?
• A small fraction f will get this extra energy• This come from the other particles• With a large number of particles the fraction f
who get more energy will remain the same• The chances of getting lucky decreases with
the number of collision required• A particle needs a run of luck…
How often you get lucky?
• The chances of hitting ten heads in a row is 1 in 210 which means every 1000 ish turns it will happen
• Particles at room temperature interact billions of times per second so they get lucky often (10, 20 even 30 kT)
Climbing a ladder of energy by chance
One particle may acquire energy several timesin succession
With many particles, a fraction f acquire energy at each step of ladder
fraction f3
fraction f2
fraction f
gets lucky three times
gets lucky twice
gets lucky onceenergy
energy
energy
By chance, particles may get extra energy from the random thermal motion of surroundingparticles.
Flying high
kTef /
The Boltzmann factor and the atmosphere
Column of air in the atmosphere
pressure must be largerlower down because ofextra weight ofatmosphere above
number per unit volume npressure p
mass of extra layer = nmA dhweight of extra layer = nmgA dh
number per unit volume n + dnpressure p + dp
assume: temperature same at all heights
m = mass of moleculen = number of molecules per unit volumek = Boltzmann constantT = temperature
area A
lowerpressure here
extra weight
higherpressurehere
height h + dh
height hdh
Gas laws
pV = NkTp = (N/V) kTp = nkT(pressure increases with density)
difference in pressuredp = kT dn
Extra pressure due to weight of extra layer
pressure difference between layersdp = weight of extra layer/area Adp = – nmgA dh/A(pressure decreases with height)
difference in pressuredp = –nmg dh
kT dn = –nmg dh
dn/dh = – (mg/kT )n
rate of change of number with height proportional to number
n/n0 = exp(–mgh/kT)
n/n0 = exp(–/kT ) = mgh = difference in potential energy
Ratio of numbers of particles in states differing by energy is equal to theBoltzmann factor exp(–/kT)
nkTp
nkTp
kTV
Np
NkTpV
dd
hnmgpA
hnmgAp
A
Fp
dd
dd
d
)/()/(
0
d
d
dd
kTkTmgh een
n
nkT
mg
h
n
hnmgnkT
Activation processes
An energy hill has to be climbedbefore the process can happen
The energy needed has to beacquired by chance from randomthermal agitation of the surroundings
energy needed forprocess to be possible
Examples of activation processes
Viscosity of liquids
particles needenough energy tobreak out of ‘cage’formed by theirneighbours, so liquidcan flow
atoms needenough energy tobecome ionised,freeing electronswhich can conduct
Ionisation of semiconductor
Evaporation of liquids Ionisation in a flame
Many chemical reactions Nuclear fusion in the Sun
particles need enoughenergy to break awayfrom attractions ofparticles at surface,and leave the liquid
molecules needenough energy todissociate into ions,which can conduct acurrent
particles must collidewith enough energyto get close enoughto bond to make anew molecule
protons must haveenough energy to getclose enough to fuseinto deuterium
activationenergy
+_
_ +
+
++
_
_
_
+ +
The energy needed for a process to happen has to be acquired by chancefrom the random thermal agitation of the surroundings
Soft matter
Soap, cell membranes and wristwatches
The molecules don’t care
Quick check questions (page 132)
Questions 1-7 (pg 134)
Test