R. T. Sanderson Arizona State University Tempe 85281
Why Does Methane Burn?
The burning of methane is just one out of innumerable examples of chemical facts so familiar that they are commonly taken for granted. Of course methane burns! But why "of course?" If the burning of methane were not a well known phenomenon, would our students be able to predict it? In the teaching of chemistry, we must make a determined effort to persuade our students that no fact should be taken for granted. If they are to acquire some understanding of chemistry as a science, as an intellectual challenge rather than just a collection of facts, they must be brought face to face with the questions for which answers must be sought. They must be given, preferably through personal experience, a true appreciation of the extent and the limitations of human capacity to understand nature. Especially, they should be encouraged to ask why. Why does methane burn? I n an earlier paper1 I attempted to outline the general principles of chemical reaction. The relative importance of bond strength and entropy was discussed, and it was pointed out that if the heat of a reaction is relatively large, a t ordinary temperatures it will be the dominant factor in determining the free energy change. Furthermore, the entropy tends to depend more on the physical state than on the chemical composition. A relatively large entropy increase accompanies a change from more highly condensed, orderly phase (solid or liquid) to less condensed, more random phase (gas). Free energy changes for reactions tend to be more negative than the enthalpy change if the number of moles of gas in the products exceeds the number in the reactants, and vice versa, but not very different when the number of moles of gas is unchanged. The equation for the reaction we are studying here is: CHM
+ 20dd
-
COkd
+ 2H10(g)
Since 3 moles of gaseous reactants form 3 moles of easeous uroducts. the entronv c h a n ~ eASo for this reaction w d d probably be r~iativelysmall. It should not be surprising that the free energy change AGO = -191.4 kcal/mole and the enthalpy change A H o = -191.8 kcal/mole are very nearly the same, since TAS makes a verv small contribution in the relationshio. .. AGO = AHo - TAS". In other words, the enthalpy change AHo is the major contributor to the large negative AGO that tells us that methane will burn. SANDERSON, R. T., J . G E M Enuc., . 41, 13 (1964). SANDERSON, R. T., J. Inorg. Nucl. Chem., 28, 1553 (1966). SANDERSON, R. T., Advances in Chemistly Series No. 62, (Edzlor: GOULD,R. F.),1967, p. 187. SANDERSON, R. T.,"Inorganic Chemistry," Reinhold Publishing Corp., New York, 1967, p. 69, a
121. 41.5. ---, 3
See first reference in footnote 2.
When we inquire why methane will burn we are essentially asking why the reaction has a high negative enthalpy AH0. I n part this can be answered by point ing out the significance of the high negative AHo. It means that the bonds in COz and 2Hz0 are, on the average, stronger than the bonds in CHI and 202. But now we must ask, why are the bonds in COz and HzO stronger? The ultimate answer to this question is probably forever unattainable. Howevk, as the result of recent studies of bond energies,%we can now approach significantly closer toward a true understanding than was possible previously. For we can now calculate all the bond energies involved, from only a few fundamental properties of the atoms. Moreover, the calculation is much simpler than any wave mechanical calculation could be. The method consists of visualizing each bond as blending a covalent and an ionic nature. By first separating this blend into its component fractions, it is then possible to evaluate the individual contributions. The total bond energy is their sum. If the two bonded atoms are alike, their bond is called "homonuclear." One of the fundamental properties of atoms on which the new method of bond energy calculation is based is the homonuclear single covalent bond energy. Values obtained from a large amount of experimental work are available for many of the elements, including those involved in the present problem: C C , 83.1; H-H, 104.2; 0-0, 33.2 k~al/mole.~ The nonpolar heteronuclear single covalent bond energy E, is the geometric mean of the homonuclear single covalent bond energies of the separate elements, EL-., and EB-B, corrected for any deviation of the experimental bond length Ro from the sum of the nonpolar covalent radii, R,:
Since no two elements are identical in electronegativity, practically every heteronuclear covalent bond involves some degree of polarity. The energy if the bond were completely ionic is simply the Coulomb energy r;,:
-
E ,. - 3326% R,
(2)
where e is the charge in electrons and 332 the factor converting the energy to kcal/mole. The contributions made by the covalent and ionic energies can be apportioned by weighting coefficients t, t , = 1, t , being the average of the partial charges on the individual atoms:
+
E
=
t.E.
+ tiE
