In my recent posting "Green Gold", I asserted that 6 grams of porous carbon contained an internal surface area equivalent to the area of the Melbourne Cricket Ground (arena). In arriving at this figure, I assumed a maximum value of 3,000 square metres (m*2) per gram of carbon, which was toned-down some from the statement in wikipedia on "Gold Mining": "The gold is adsorbed into the porous matrix of the carbon. Carbon has so much internal surface area, that a gram and a half has the equivalent surface area of the Melbourne Cricket Ground." I am not pedantic by nature, but given the importance of porous materials - activated carbon, zeolites, clays, silica gels etc. - I decided to look into the matter further.

Porous carbon is produced by "activating" (heating) wood, and it is the most dense types of wood (e.g. teak) that give the highest surface area materials. The procedure always results in a distribution of pore sizes, rather than the single pore dimension found in zeolites (molecular sieves). The latter feature can be adapted to produce types of porous carbon (called carbon molecular sieves) with a very narrow range of pore sizes by forming an organic polymer within the zeolite structure (which must fit that tight porous constraint), turning it to carbon by heating, and then dissolving (leaching) the zeolite template away usually using hydrofluoric acid (HF). The most commonly encountered forms of activated carbon have surface areas of around 1,000 or so square meters (m*2) per gram, and some with smaller pores offer internal surfaces amounting to 2,000 m*2. I am aware of porous carbons, formed by heating e.g. teak, and which contain a quite restricted range of pores in the low micropore (i.e under 20 Angstrom units or 2 nanometers nm) range, say under 10 A, which can reach 3,000 m*2/g and there is one claim of a material that has 5,000 m*2/g. However, if a gram and a half of some porous carbon really is enough to equal the MCG (at 18,100 m*2), it must have some particular properties to give it an internal surface area of around 12,000 m*2/g!

I shall first propose a highly simplified model of a porous solid. I shall assume that the extended material can be made up of a large number of small cubes, each of which contains a spherical pore. The solid is then "constructed" by placing a large number of cubes side by side, so that six cubes fit around each (one at each of the six sides - left, right, forward, backward, top, bottom). If we assume that a hypothetical solid has pores of 3 A (0.3 nm) in diameter, it would fit inside a cube of 3 x 3 x 3 = 27 A*3 (cubic Angstroms). The radius (r) is half that, i.e. 1.5 A, and so the internal surface area of the pore is obtained from:

4 x pi x r*2 = 4 x pi x (1.5)*2 = 28.27 A*2 (square Angstroms).

Assuming a density for the carbon of 0.6 g/cm*3 (cubic centimetre), we see that 1 gram of the porous carbon would occupy 1/0.6 = 1.67 cm*3

1 cm = 10*8 A, and so 1 cm*3 = 10*24 A*3. 1.67 cm*3 = 1.67 x 10*24 A*3

Therefore, you could fit 1.67 x 10*24 A*3/27 A*3 = 6.19 x 10*22 cubes into 1g of porous carbon.

This gives a total surface area of 6.19 x 10*22 x 28.27 A*2 = 1.75 x 10*24 A*2.

1 m = 10*10 A, and 1 m*2 = 10*20 A*2. Hence 1g of the carbon would contain 1.75 x 10*24 A*2/10*20 A*2 = 1.75 x 10*4 = 17,500 m*2.

----------------------------------------------------------------------------------------

We can work the calculation similarly for a 5 A pore, with the radius r = 2.5 A, and an internal surface area of 4 x pi x (2.5)*2 = 78.54 A*2 located in a cube of 5 x 5 x 5 = 125 A*3

Again, the volume of 1 g is 1.67 x 10*24 A*3, which could fit 1.67 x 10*24/125 = 1.336 x 10*22 cubes.

Hence, the total internal surface area amounts to: 1.336 x 10*22 x 78.54 = 1.049 x 10*24 A*3 = 10,493 m*2/g

-----------------------------------------------------------------------------------------

Now, our hypothetical "Australian" carbon would need an internal surface amounting to around 12,000 m*2/g. If the density is also 0.6 g/cm*3 (as assumed above), that gives an internal surface area of 12,000 x 0.6 = 7,200 m*2/cm*3 = 7.2 x 10*23 A*2. This has to be divided (fitted) into a volume of 1 cm*3 = 7.2 x 10*23 A*2/10*24 A*3 = 0.72/A

Our model of the solid is of a number (n) pores contained (obviously) in n cubes, and defining the geometry of the pore = 4 x pi x r*2 and the cube = (2r)*3 = 8r*3, and taking their ratio:

4n x pi x r*2/8n x r*3 we find that the term n cancels, leaving us with a simple geometrical factor:

4 x pi x r*2/8 x r*3 = 7.2 x 10*23 A*2/10*24 A*3 = 0.72/A = 1.57/r = 0.72/A

Therefore, r = 1.57 A/0.72 = 2.18 A, and so the pore diameter is 2r = 2 x 2.18 = 4.36 A.

To check the calculation, we can work out from 4 x pi x r*2 that the pore would have a surface are of 59.72 A*2 and fit into a cube of 4.36*3 = 82.88 A*3. In 1 cm*3 = 10*24 A*3/82.88 A*3 = 1.2 x 10*22 cubes, and so the surface area is 1.2 x 10*22 x 59.72 = 7.2 x 10*23 A*2 = 7,200 m*2, which is our correct starting value.

I am not aware of any such porous carbon, in fact with an area of 12,000 m*2/g and so I propose that the figure is nearer 15 grams of porous carbon (one decimal point place removed from the gram and a half that was supposed) to equal the area of the MCG. Even if the super material does exist,which I doubt, and can be synthesised, it is unlikely to be made on a large scale for gold mining, which would use the best but more common or garden variety. 18,100 m*2/15 grams = 1,207 m*2/g, which is reasonable for most carbons I have worked with.

---------------------------------------------------------------------------------------

Just to complete the exercise about porous solids, for 10 A pores, we end up with r = 5A, and so the pore has an internal surface area of 314.2 A*2 (from 4 x pi x r*2) in a cube of 10 x 10 x 10 = 1000 A*3, giving 1 x 10*21 cubes in a 1 cm*3 cube. In that case, the total surface area would be 1 x 10*21 x 314.2 = 3.142 x 10*23 A*2 = 3,142 m*2 or around 5,000 m*2/g for a density of 0.6 g/cm*3.

For a pore of 1 cm diameter in a 1 cm*3 cube, with r = 0.5 x 10*8 A, we arrive at a surface area of 3.142 x 10*16 A*2/1 x 10*20 = 3.142 x 10*-4 m*2 i.e. 3.142 cm*2! (just 5 square centimeters /g). This shows how the surface area drops as the size of the pores increases.

------------------------------------------------------------------------------------------

These are very rough and ready calculations, and the measured surface areas are always a good deal less than this, for the following reasons: (1) there is always a pore size distribution with a substantial fraction of larger pores (meso 20 - 500 A; and macro > 500 A) which reduces the intrinsic internal surface area of the material from that expected if it contained only micropores (< 20 A); (2) simple arguments about the geometry and number of pores tell nothing about the pathways by which molecules are actually absorbed into the porous particle, to which there are always barriers and restrictions, and hence the actual surface covered is less than the total surface area. As an example, zeolite X, where the pores are (almost) all small (10 A) and well defined, the actual adsorption capacity of nitrogen molecules on the surface accords with a value of around 700 m*2/g, rather than a value of around 3,000 m*2/g (assuming a density of 1 g/cm*3); (3) last but certainly not least - the structure of a porous solid is not a neatly packed arrangement of pores within cubes, but more complex, and with substantial inaccessible space occupied by its essential framework.

....this makes the original figure of one and a half grams of carbon to match the area of the MCG even less likely, however!

[For comparison: a tennis court has an area of 262 m*2; a football pitch (for international games) around 7,400 m*2; a cricket ground anywhere in the range 15,000 - 18,000 m*2. The MCG arena is reckoned at 18,100 m*2 and the entire Oval cricket ground site covers 24,279 m*2].

## 1 comment:

I enjoyed reading thiis

Post a Comment