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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 ~16-18, IB courses and US grades 11-12 K12 honors.

Sub-index 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 'self–spreading' 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 Maxwell–Boltzmann 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/MrRT, Mr = mRT/PV

    • so, since d = m/V, then Mr is proportional to density.

  • r1   √d2   √M2
    ––– = –––– = ––––
    r2   √d1   √M1
  • 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 = 1/2mu2, where m = mass of particle, u = velocity. This means the average mu2 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 Mr, 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 = Mr : H2 = 2, He = 4, CH4 = 16

NH3 = 17N2 = 28O2 = 32, HCl = average 36.5CO2 = 44Br2 = 160

(molecular mass) Average speeds of particles at 0oC

(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 1000oC)

(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 = ½mv2    2KE = mv2    v2 = 2KE/m therefore    v = (2KE/m)

BUT think from the above equation: vav √(1/Mr)

THEN think average rate of diffusion (ROD) is proportional to the average speed of molecules, so, ROD √(1/Mr) which is a form of Graham's Law of diffusion/effusion.

Actually in 1831, Graham expressed his law in terms of density: rdiff √(1/d) derived from experimental results, but, at the same temperature pressure: Mr d (ref. Avogadro)

Graham's Law of Diffusion expressed in terms of two gases of different molecular masses
rdiff1   √Mr2
–––– = –––––
rdiff2   √Mr1

 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

  • HCl - NH3 diffusion expt.

  • 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 2/3rds along from the ammonia end of the tube.

    • Atomic masses: N = 14, H = 1, Cl = 35.5

  • (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.

    • NH3(g) + HCl(g) ==> NH4Cl(s)

  • (b) Mr(NH3) = 17, Mr(HCl) = 36.5

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

  • rNH3   √MrHCl = √36.5   6.04
    –––––––– = –––––––––––––––––––– =          –––––––– = 1.47
    rHCl   √MrNH3 = √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 NH3 end of the tube.

An alternative maths approach to the experiment above

NH3(g) + HCl(g) ===> NH4Cl(s) fine crystals of ammonium chloride salt formed

Molecular masses: HCl 36.5, NH3 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

porous pot diffusion experiments demonstration of relative rates of diffusion of air, hydrogen carbon dioxide molecules calculations of relative rates explained using Graham's Law of diffusion

  1. 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.

  2. 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.

  3. 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.

  4. 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

  • Example 1. Using zeolite minerals to separate hydrocarbon molecules

zeolite structure aluminiosilicate catalytic properties processing hydrocarbons in petrochemical industry

  • 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 CH3CH2CH2CH2CH3, hexane CH3CH2CH2CH2CH2CH3 or 2–methylpentane (CH3)2CH2CH2CH2CH3 into a zeolite mineral.

    • Atomic masses: C = 12, H = 1

    • Hexane and 2–methylpentane are structural isomers of C6H14 with the same molecular mass.

    • rC5H12   √MrC6H14 = √86  


      ––––––––– = ––––––––––––––––––– =

      ––––  = 1.09

      rC6H14   √MrC5H12 = √72  


    • Relative rate of diffusion is 1.09 : 1.00 for pentane : hexane/2–methylpentane.
    • So, e.g. pentane should diffuse into the zeolite mineral faster than hexane.
  • (b) In practice 2–methylpentane does not diffuse into the zeolite as fast as hexane or maybe not at all.

    • Suggest a reason for this behaviour?

    • The 'methyl branching' in 2–methylpentane makes it a more bulky molecule that has greater difficulty fitting into zeolite minerals.


  • Example 2. Uranium enrichment

  • (c) doc b

    • Enriching uranium means to increase the relative ratio of 235U/238U to produce uranium metal suitable for use as fuel rods in nuclear reactors.

    • It is the 235U isotope that is very fissile (readily undergoes fission e.g. nuclear equation above) but only occurs as a small % in uranium ores in which most uranium is the non–fissile 238U isotope.

    • There is only 0.72% of fissionable U-235 present in extracted uranium metal.

    • Grades of uranium-235 need to be 25% to 75% for use as a nuclear fuel.
    • U ore ==> U2O3 (yellow cake) ==> 235UF6 (enrichment) ==> 235U metal concentrate
    • To produce 'enriched' uranium metal it is first extracted by reduction from uranium oxide and then converted into gaseous uranium(VI) fluoride (uranium hexafluoride).

    • Atomic mass of F = 19, isotopic masses of U are 235 and 238, so the two molecular masses for isotopic variants of UF6 are 349 and 352.

    • (a) Calculate the relative rates of diffusion of the hexafluorides of the two uranium isotopes.

      • r235UF6   √Mr238UF6 = √352  


        ––––––––– = ––––––––––––––––––– =

        –––––  = 1.004

        r238UF6   √Mr235UF6 = √349  


    • (b) Suggest why the process must be repeated many times before enough enrichment has occurred.

      • For each diffusion 'run' only a very small enrichment occurs because of the similarity of the molecular masses of 235UF6 and  238UF6 and hence the very similar rates of diffusion.

      • The enrichment process now employs huge gas centrifuge systems to separate the 'lighter' 235UF6 molecules from the heavier 238UF6 molecules (see next section).

    • How is the separation done?

A very brief description of the uranium enrichment

(a) Separation by diffusion

The UF6 gas is forced through semi-permeable porous membrane and uses lots of energy.

Its a sort of multiple porous pot system!

The slightly 'lighter' 235UF6 molecules diffuse through the surface of the porous tubes on average a tiny bit faster than the heaver' 238UF6 molecules and gradually the ratio changes in favour of U-235 hexafluoride in the enriched stream/


(b) Gas centrifuge

This enrichment process now employs huge gas centrifuge system to separate the 'lighter' 235UF6 molecules from the heavier 238UF6 molecules (less energy).

The heavier molecules of 238UF6 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 235UF6 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 UF6 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:

  1. increasing the difference in concentrations
  2. increasing the surface are through which diffusion takes place
  3. 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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