STATES OF MATTER 
properties of gases and liquids (fluids) and solids
23.
Graham's Law of Diffusion and calculations using the formula for relative rates
of diffusion of two gases
Helpful for UK
advanced level chemistry students aged ~1618, IB courses and US grades 1112 K12 honors.
Subindex for this page on diffusion
23a Graham's Law
of diffusion and formula
23b Molecular
mass, speeds and kinetic energy
23c
Demonstrations of diffusion and relative rate of diffusion calculations
23d
Examples of using diffusion to separate gaseous molecules
23e
Fick's
1st Law of Diffusion and links to diffusion contexts in biology
23a. Graham's Law of Diffusion
and calculation formula

Diffusion, or the
'selfspreading' of molecules, naturally arises out of their constant chaotic
movement of particles in all directions, though on a time average basis, more molecules
will move in the direction of a region of lower concentration down a diffusion
gradient if such a
situation exists e.g. initially after 'pouring bromine vapour into air' in gas jars.

See part 4.
What is diffusion?
Particle model and examples
of demonstrating diffusion in gases

Molecules of differing
molecular mass diffuse at different rates.

The smaller the molecular mass, the
greater the average speed of the molecules at constant temperature.

The greater the average speed of the
particles the greater their rate of diffusion.

See notes on the
MaxwellBoltzmann distribution of molecular velocities.

This conceptually explains Graham's Law
of diffusion, explained below.

It has been shown that,
assuming ideal gas behaviour and constant temperature, the relative rate of
diffusion of a gas through porous materials or a mixture of gases or a tiny hole
(effusion) is inversely
proportional to the square root of its density.

Since the density of an
ideal gas is proportional to its molecular mass, the relative rate of
diffusion of a gas is also inversely proportional to the square root of its
molecular mass*.

* PV = nRT, PV = m/M_{r}RT,
M_{r} = mRT/PV

so,
since d = m/V, then M_{r} is proportional to
density.

r_{1} 

√d_{2} 

√M_{2} 

= 

= 

r_{2} 

√d_{1} 

√M_{1} 
 Which is the mathematical ratio
representation of Graham's law of diffusion for comparing two gases of
different molecular masses.

Graham's Law arises from
the fact that the average kinetic energy** of gas particles is a constant for
all gases at the same temperature.

**The formula for kinetic
energy is KE = ^{1}/_{2}mu^{2}, where m = mass
of particle, u = velocity. This means the average mu^{2}
is a constant for constant kinetic energy, so u is proportional to 1/√m and the m can be shown
via the Avogadro Constant to
be proportional to M_{r}, the molecular mass of the gas.

You have to think of the molecules
'hitting' the space of the pore or tiny hole and passing through the. The
greater the speed the more chance the particle has of passing through this
'porous space'.
23b. Molecular mass and average speeds of molecules
Atomic masses: H = 1, C = 12, N = 14, O =
16, Cl = average 35.5, Br average 80
Relative molecular
masses = M_{r} : H_{2}
= 2, He = 4,
CH_{4} = 16,
NH_{3}
= 17, N_{2}
=
28, O_{2}
= 32, HCl = average 36.5,
CO_{2} = 44,
Br_{2} = 160
(molecular mass) Average speeds of particles
at 0^{o}C
(2) hydrogen 1698 m/s
(4) helium 1202 m/s (note that helium
balloons deflate far too quickly!)
(16) methane 600 m/s
(17) ammonia 661 m/s
(28) nitrogen 454 m/s (981 m/s at 1000^{o}C)
(32) oxygen 424 m/s
(44) carbon dioxide 362 m/s
All
substances have the same average kinetic energies at the same
temperature, despite the fact that there is a wide range of
speeds and kinetic energies due to the random motion and collisions.
A little bit of kinetic energy theory
and Graham's Law of Diffusion
Assuming you are dealing with average values of KE and v
KE =
½mv^{2}
2KE = mv^{2}
v^{2} = 2KE/m therefore
v =
√(2KE/m)
BUT think from the above equation:
v_{av}
√(1/M_{r})
THEN
think average rate of diffusion (ROD) is proportional to the average speed of
molecules, so,
ROD
√(1/M_{r}) which is a form of Graham's Law
of diffusion/effusion. Actually
in 1831, Graham expressed his law in terms of density:
r_{diff}
√(1/d)
derived from experimental results, but, at the same temperature
pressure:
M_{r}
d (ref. Avogadro)
Graham's Law of Diffusion
expressed in terms of two gases of different molecular masses
r_{diff1} 

√M_{r2} 

= 

r_{diff2} 

√M_{r1} 
and relating the relative diffusion rates of
two different molecules 1 and 2 
23c.
Demonstrations of diffusion and
relative rate of diffusion calculations

Diffusion demonstration
1

The reaction
between hydrogen chloride and ammonia gases


Two cotton wool plugs are
separately soaked in concentrated aqueous ammonia and hydrochloric acid
solutions respectively and sealed in a long tube with rubber bungs.

Using a simple chemical
equation and Graham's Law of diffusion, account for (a) the appearance of a
'white smoke ring' and (b) the fact the smoke ring occurs about ^{2}/_{3}rds
along from the ammonia end of the tube.

(a) The aqueous
ammonia will give off ammonia fumes and the conc. hydrochloric acid gives off
hydrogen chloride fumes which will diffuse down the tube towards each other.

When they meet an acid base reaction gives fine crystals of the salt ammonium
chloride.


(b) M_{r}(NH_{3})
= 17, M_{r}(HCl) = 36.5

If
r is the relative rate
of diffusion the following ratio applies,

r_{NH3} 

√M_{r}HCl
= √36.5 

6.04 

= 

= 
= 1.47 
r_{HCl} 

√M_{r}NH_{3}
= √17 

4.12 

and this shows that
ammonia will diffuse about 50% faster than hydrogen chloride so the smoke
ring will appear much nearer the HCl end of the tube the NH_{3} end of
the tube.
An
alternative maths approach to the experiment above
NH_{3(g)} + HCl_{(g)}
===> NH_{4}Cl_{(s)}
fine crystals of ammonium
chloride salt formed
Molecular masses: HCl 36.5, NH_{3} 17
Prediction of relative rates of diffusion from
Graham's Law
Ammonia: 1/√(17)
= 0.243
Hydrogen chloride: 1/√(36.5)
= 0.167 0.243/0.167 = ~1.5 and observation matches prediction!
The smoke ring is about 2/3rds of the
way along from the ammonia end of the tube. 

Diffusion demonstration 2  porous
pot experiments

The relative rates of
diffusion of air, hydrogen and carbon dioxide molecules

The porous pot is inverted and connected to a U tube
manometer system and allowed to settle down so bother coloured fluid
levels are at the same height. This means there is no pressure
differential in the U tube.

A large inverted beaker is placed in position over
the porous pot and the levels should still be equal. You then pass
hydrogen gas up into the beaker where it will accumulate, being less
dense than air. Within a short time the liquid moves away from the
porous pot because the faster moving lower molecular mass molecules
of hydrogen diffuse into the pot faster than the air molecules of
nitrogen and oxygen can diffuse out.

Here the porous pot is inverted, but still attached
to a simple U tube manometer system. A large beaker is held in
position surrounding the porous pot and the system allowed to settle
so the liquid levels are equal. Carbon dioxide gas is passed into
the beaker where it will accumulate, being denser than air. Within a
short time the coloured liquid moves towards the porous pot because
the faster lower molecular mass molecules of air (mainly nitrogen
and oxygen) diffuse out the porous pot faster than the slower higher
molecular mass molecules of carbon dioxide can move in.

Experiments 2. and 3. illustrates Graham's Law of
diffusion and the different rates of diffusion create a small
(temporary) difference in pressure between the gases in the porous
pot and the gases outside of the porous pot.
See also States
of Matter Parts
4. and 5. for more on diffusion demonstrations
4.
What is diffusion?
Particle model and examples
of demonstrating diffusion in gases
6.
Particle model of a liquid and diffusion experiments and associated phenomena
23d. Examples of using diffusion to separate gaseous
molecules

Zeolites are silicate
minerals that are porous at the molecular level and they are used as catalysts
and 'molecular sieves' in the petrochemical industry in processes such as
cracking and subsequent molecule separation.

(a) Calculate the relative
rates of diffusion of pentane CH_{3}CH_{2}CH_{2}CH_{2}CH_{3},
hexane CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3}
or 2methylpentane (CH_{3})_{2}CH_{2}CH_{2}CH_{2}CH_{3}
into a zeolite mineral.

Atomic masses: C = 12, H =
1

Hexane and 2methylpentane
are structural isomers of C_{6}H_{14} with the same molecular
mass.

r_{C5H12} 

√M_{r}C_{6}H_{14}
= √86 

9.27 

= 

= 
= 1.09 
r_{C6H14} 

√M_{r}C_{5}H_{12}
= √72 

8.49 
 Relative rate of diffusion is
1.09 : 1.00 for
pentane : hexane/2methylpentane.
 So, e.g. pentane should
diffuse into the zeolite mineral faster than hexane.

(b) In practice
2methylpentane does not diffuse into the zeolite as fast as hexane or maybe
not at all.
A
very brief description of the uranium enrichment
(a) Separation by diffusion
The UF_{6} gas is forced through semipermeable porous
membrane and uses lots of energy.
Its a sort of
multiple porous pot system!
The slightly 'lighter' ^{235}UF_{6}
molecules diffuse through the surface of the porous tubes on average
a tiny bit faster than the heaver' ^{238}UF_{6}
molecules and gradually the ratio changes in favour of U235
hexafluoride in the enriched stream/
(b) Gas
centrifuge
This enrichment process now employs huge gas
centrifuge system to separate the 'lighter' ^{235}UF_{6}
molecules from the heavier ^{238}UF_{6} molecules
(less energy).
The heavier molecules of ^{238}UF_{6} are spun out
in a rotating cylinder to the outer region of the chamber,
and retained, with the help of slower diffusion in reverse (left
diagram)
The lighter molecules of ^{235}UF_{6}
tend to concentrate a little bit
more towards the centre of the drum, again helped by their natural
process of a higher rate of diffusion. The very slightly enriched 'central' mixture is then fed
into the next rotating drum.
This is repeated many times in
many cylinders (diagram above right) to
complete the uranium enrichment process.
The enriched UF_{6} is chemically
processed back to uranium metal and used for fuel rods in nuclear
reactors, in which plutonium is produced, this is then extracted for
use as superior nuclear fuel rods or converted to weapon grade
plutonium for nuclear armed missiles.

23e. Fick's
1st Law of Diffusion
ROD =
rate of diffusion of specific substance
ROD
difference in
concentration x surface area /
diffusion distance
ROD
∆c x a / d
ROD = net rate of particle transfer down a
diffusion gradient.
∆c =
difference in concentration from the top to the
bottom of the diffusion gradient (i.e. from the higher region to the
lower region in terms of concentration.
a = area over which this net rate of diffusion is
taking place
d = 'horizontal' distance over which the diffusion
takes place (width of gradient)
This means the net rate of material
transfer is increased by:
 increasing the difference
in concentrations
 increasing the surface are
through which diffusion takes place
 decreasing the distance
over which the concentration gradient is active i.e. the
distance the diffusing particles (molecules or ions) have to
travel.
These diffusion factors are
VERY important for various biological situations (links
below)

Biology links to diffusion situations
Transport
links e.g. movement of water and nutrients by diffusion
(1)
Introduction to diffusion and demonstration experiments
and
its importance in biology
(2)
A particle model and factors
affecting the rate of diffusion and Fick's Law of diffusion
(3)
The action of
partially permeable cell
membranes  selective diffusion and examples
(4)
Osmosis  examples and explanation
(7)
Active transport  explanation of why needed and examples
explained
(8)
A quick comparison of diffusion, osmosis and active transport
Gas exchange
by diffusion links
2.
Introduction to exchange surfaces in cells and
organs
3.
Gas exchange in
the human lungs by diffusion, comments on breathing, COPD and ventilators
4.
Gas exchange and the structure of fish gills
5.
The function of villi in the exchange
surface of the small
intestine
Learning objectives for diffusion
Be able to explain how
diffusion takes place.
Be able to quote Graham's
Law of diffusion.
Be able to do relative
rates of diffusion calculations using the Graham's Law formula in terms
of gas density or molecular mass of the molecules.
Using Graham's law of
diffusion be able to explain the observations in experiments like the
porous pot demonstrations.
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UK GCSE level (~US grade 810) school chemistry revision
notes
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preuniversity chemistry revision notes
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