STATES OF MATTER - properties of gases and liquids (fluids) and solids

23. Graham's Law of Diffusion and calculations using the formula to calculate the relative rates of diffusion of two gases - descriptions of diffusion rate experiments and exam practice questions

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 any medium e.g. open space, porous materials or a mixture of gases through a tiny hole (effusion) is inversely proportional to the square root of its density.

  • rate of diffusion 1/√d

  • and

  • rate of diffusion 1/√M_r

  • because the density of an ideal gas is proportional to its molecular mass,

  • so 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_rRT, M_r = mRT/PV

  • so, since d = m/V, then M_r 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 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'.

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 ²/₃rds 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.

• • NH_3(g) + HCl_(g) ==> NH₄Cl_(s)

• (b) M_r(NH₃) = 17, M_r(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.

• Diffusion demonstration 2 - porous pot experiments

• Experiments to qualitatively observe the relative rates of diffusion of air, hydrogen and carbon dioxide molecules

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

• 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		9.27
    –––––––––	=	–––––––––––––––––––	=	––––  = 1.09
    rC6H14		√MrC5H12 = √72		8.49

  • 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

• 

  • 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		18.76
      –––––––––	=	–––––––––––––––––––	=	–––––  = 1.004
      r238UF6		√Mr235UF6 = √349		18.68

  • (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?

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)

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