Is the ethanol boom about to bust?

Yes

7 of 14

by Douglas Roger Sandy Dympnus Christopher Messer

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What are the actual numbers comparing ethanol versus conventional gasoline as fuel and as pollutants? Some time back I came upon a letter on a bulletin board asking how it was possible for a gallon of gasoline weighing around seven pounds to generate almost thirty pounds of carbon dioxide. With a degree in chemistry, I knew the answer, but refrained from supplying a response because I was more interested in what some of the other posters had to say about it. I could have gone ahead and wrote something to the puzzled poster, because no one attempted to answer his question. But it made me realize that nowhere had I seen the simple but revealing calculations that compared and contrasted the pollution potential and efficiency as fuels of conventional gasoline versus ethanol.

Gasoline is a complex blend of aliphatic (straight- and branched-chain) hydrocarbons, cycloalkanes (ring compounds in which each carbon atom has four single-bonded carbon and/or hydrogen atoms attached to each carbon atom), and aromatic (compounds based the benzene ring) hydrocarbon compounds, and may contain five to twelve carbon atoms per molecule, depending on the type of hydrocarbon. About five hundred different compounds can be used in gasoline blends, the choice of which can and does change seasonally, geographically, and in order to tailor the octane rating. To further complicate the picture, there is now a special class of components known as oxygenates that can be added to reduce the amounts of carbon monoxide produced by gasoline engines. In the United States, as a practical matter the two chemicals that have been used for this purpose are ethanol and the now-discredited MTBE; that is, methyltert-butyl ether.

With this in mind, it is evident that some simplification is needed for an explanation that is even possibly comprehensible to the average person. I propose to effect this simplification by considering the carbon dioxide generational potential of three compounds found in gasolineisooctane, benzene, and ethanoland comparing the three results. The balanced equations for complete combustion look like this:

Isooctane: 2C8H18 + 25O2 = 16CO2 + 18H2O Benzene: 2C6H6 + 15O2 = 12CO2 + 6H2O Ethanol: CH3CH2OH + 3O2 = 2CO2 + 3H2O

We can obtain a rough estimate of the amount of weight hydrocarbons, and specifically the carbon atoms they contain, gain upon being oxidized to water and carbon dioxide. What is important to notice here is that each carbon atom, which in the unburned material is (except at one spot in the case of ethanol) connected to one or more carbon atoms and/or one to three hydrogen atoms is now connected to two oxygen atoms of about 16 amu (atomic mass units) each. Thus, every carbon unit, which would initially be joined to from zero to three hydrogen atoms of about 1 amu each, after combustion would have its weight increased by twenty-nine to thirty-two amu. Since an atom of carbon is about 12 amu, this as a first approximation represents a weight increase of 242% to 266%. All the extra weight comes from atmospheric oxygen, less the weight of the attached hydrogen atoms, which reappears on the right of the balanced equations as water.

However, we are not restricted to rough approximation methods to determine the weight of carbon dioxide generated by the combustion of various components of gasoline. We can calculate exact theoretical yields that reflect familiar units; i.e., gallons and pounds, To begin, molecular weights are calculated from the above empirical formulas by adding up the gram-atomic weights of carbon, hydrogen, and oxygen each compound contains, thus:

Isooctane: 8(12.0111 g/mole)+18(1.00797 g/mole)=114.23 g/mole Benzene: 6(12.0111 g/mole)+ 6(1.00797 g/mole)=78.114 g/mole Ethanol: 2(12.0111 g/mole)+6(1.00797 g/mole)+15.9994 g/mole=46.069 g/mole Carbon dioxide: 12.0111 g/mole+2(15.9994 g/mole)=44.010 g/mole

We note that each mole of isooctane weighs 114.23 g. and produces eight moles of carbon dioxide, weighing 44.010 grams each, for a total of 352.08 g., representing a weight increase of 3.082 times (308%).

Each mole of benzene weighs 78.114 grams and produces six moles of carbon dioxide, weighing 44.010 grams each, for a total of 264.06 grams, representing a weight increase of 3.380 times (338%).

Each mole of ethanol weighs 46.069 grams, and produces two moles of carbon dioxide, weighing 44.010 grams each, for a total of 88.020 grams, representing a weight increase of 1.911 times (191%).

Finally, we need the densities of the first three compounds in pounds per gallon. Carbon dioxide is a gas, for which we can obtain a volume by making use of the ideal gas molar volume solution to the effect that 1 mole of any gas will occupy about 22.4 liters (or about 5.92 gallons) at standard temperature and pressure. The following densities were taken from sites on the web: Isooctane: .69 g/ml = .69 kg/l = .69 kg/l X 3.785 l/gal X 2.2046 lb/kg = 5.76 lb/gal Benzene: .887 g/ml = .887 kg/l = .887 kg/l X 3.785 l/gal X 2.2046 lb/kg = 7.40 lb/gal Ethanol: .785 g/ml = .785 kg/l = .785 kg/l X 3.785 l/gal X 2.2046 lb/kg = 6.55 lb/gal

For comparison, we have the density for a typical gasoline formulation of .75 g/ml, which by the process shown above works out to 6.26 pounds per gallon.

In order to determine the weight of carbon dioxide produced by the complete combustion of each of these, we need only multiply the number of pounds of the original material by the weight increase of the carbon dioxide, which is simply the ratio of the total gram molecular weight of the gasoline component to the total gram molecular weight of carbon dioxide produced by combustion. These are given above, and the calculations are:

Isooctane: 5.76 lb X 3.082 = 22.0 lb Benzene: 7.40 lb X 3.380 = 28.12 lb Ethanol: 6.55 lb X 1.911 =12.51 lb

The above calculations seem to lead us to conclude that by a universal change to ethanol, we could cut the carbon dioxide emissions of automobiles by a factor of about half, based on the weight of carbon dioxide produced per gallon of the above components. But this is not necessarily the case. The usefulness of ethanol as a direct substitute for blended gasoline, rather than as a material to reduce carbon monoxide emissions is actually called to question by the calculations we have done here. The big question is, is a gallon of ethanol sufficient to produce the same amount of work as a gallon of gasoline? This is unlikely, since the aim of an efficiently operating reciprocating engine is to convert fuel from a liquid to gaseous carbon dioxide and water with the biggest increase in total volume possible. To put it another way, when you want to obtain the greatest possible volume of carbon dioxide in the first place, it is counterproductive to replace it with a fuel that will produce less. Inspection of the balanced chemical reactions shows that a mole of isooctane produces seventeen moles of gaseous carbon dioxide and water, that a mole of benzene produces nine moles of these gases, but that a mole of ethanol produces only two and a half moles of gaseous products. Keep in mind that each mole of these gases will occupy the same volume at the same temperature and pressure, whether at STP (where each mole occupies 22.4 liters or 5.92 gallons), or at the high temperatures and pressures of a cylinder head.

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