The simplest scheme that will explain the lower-temperature results of methane oxidation is the following:

There is no H₂O₂ dissociation to OH radicals at low temperatures. H₂O₂ dissociation does not become effective until temperature reaches about 900 K.

As before, reaction (3.76) is slow. Reactions (3.77) and (3.78) are faster, since they involve a radical and one of the initial reactants. The same is true for reactions (3.80)–(3.82). Reaction (3.80) represents the necessary chain branching step. Reactions (3.79) and (3.83) introduce the formyl radical known to exist in the low-temperature combustion scheme. Carbon monoxide is formed by reaction (3.81), and water by reaction (3.78) and the subsequent decay of the peroxides formed. A conversion step of CO to CO₂ is not considered because the rate of conversion by reaction (3.49) is too slow at the temperatures of concern here.

It is important to examine more closely reaction (3.77), which proceeds [30,31] through a metastable intermediate complex—the methyl peroxy radical—in the following manner:

At lower temperatures, the equilibrium step is shifted strongly toward the complex, allowing the formaldehyde and hydroxyl radical formation. The structure of the complex represented in reaction (3.77) is well established. Recall that when O₂ adds to the carbon atom in a hydrocarbon radical, it forms about a 90° bond angle. Perhaps more important, however, is the suggestion [30] that at temperatures of the order of 1000 K and above, the equilibrium step in reaction (3.77) shifts strongly toward the reactants so that the overall reaction to form formaldehyde and hydroxyl cannot proceed. This condition would therefore pose a restriction on the rapid oxidation of methane at high temperatures. This possibility should come as no surprise, as one knows that a particular reaction mechanism can change substantially as the temperature and pressure change. There now appears to be evidence that another route to the aldehydes and OH formation by reaction (3.77) may be possible at high temperatures [9,31]; this route is discussed in the next section.

In contrast to reaction (3.76), at high temperatures the thermal decomposition of the methane provides the chain initiation step, namely

With the presence of H atoms at high temperature, the endothermic-initiated H₂–O₂ branching and propagating scheme proceeds, and a pool of OH, O, and H radicals develops. These radicals, together with HO₂ (which would form if the temperature range were to permit reaction (3.76) as an initiating step), abstract hydrogen from CH₄ according to

The fact that reaction (3.77) may not proceed as written at high temperatures may explain why methane oxidation is slow relative to that of other hydrocarbon fuels and why substantial concentrations of ethane are found [4] during the methane oxidation process. The processes consuming methyl radicals are apparently slow, so the methyl concentration builds up, and ethane forms through simple recombination:

Thus methyl radicals are consumed by other methyl radicals to form ethane, which must then be oxidized. The characteristics of the oxidation of ethane and the higher-order aliphatics are substantially different from those of methane (see Section H1). For this reason, methane should not be used to typify hydrocarbon oxidation processes in combustion experiments. Generally, a third body is not written for reaction (3.89), since the ethane molecule's numerous internal degrees of freedom can redistribute the energy created by the formation of the new bond.

Brabbs and Brokaw [36] were among the first who suggested the main oxidation destruction path of methyl radicals to be

These are radical–radical reactions or reactions of methyl radicals with a product of a radical–radical reaction (owing to concentration effects) and are considered less important than reactions (3.77) and (3.90). However, reactions (3.77) and (3.90) are slow, and reaction (3.96) can become competitive to form the important methoxy radical, particularly at high pressures and in the lower-temperature region of flames (see Chapter 4).

The methoxy radical formed by reaction (3.90) decomposes primarily and rapidly via

Although reactions with radicals to give formaldehyde and another product could be included, they would have only a very minor role. They have large rate constants, but concentration factors in reacting systems keep these rates slow.

Reaction (3.90) is relatively slow for a chain branching step; nevertheless, it is followed by the very rapid decay reaction for the methoxy (reaction (3.97)), and the products of this two-step process are formaldehyde and two very reactive radicals, O and H. Similarly, reaction (3.96) may be equally important and can contribute a reactive OH radical. These radicals provide more chain branching than the low-temperature step represented by reaction (3.77), which produces formaldehyde and a single hydroxyl radical. The added chain branching from the reaction path (reactions (3.90) and (3.97)) may be what produces a reasonable overall oxidation rate for methane at high temperatures. In summary, the major reaction paths for the high-temperature oxidation of methane are

Of course, all the appropriate higher-temperature reaction paths for H₂ and CO discussed in the previous sections must be included. Again, note that when X is an H atom or OH radical, molecular hydrogen (H₂) or water forms from reaction (3.88). As previously stated, the system is not complete because sufficient ethane forms so that its oxidation path must be a consideration. For example, in atmospheric-pressure methane–air flames, Warnatz [38,39] has estimated that for lean stoichiometric systems about 30% of methyl radicals recombine to form ethane, and for fuel-rich systems the percentage can rise as high as 80. Essentially, then, there are two parallel oxidation paths in the methane system: one via the oxidation of methyl radicals and the other via the oxidation of ethane. Again, it is worthy of note that reaction (3.88) with hydroxyl is faster than reaction (3.49) so that early in the methane system CO accumulates; later, when the CO concentration rises, it effectively competes with methane for hydroxyl radicals and the fuel consumption rate is slowed.

The mechanisms of CH₄ oxidation covered in this section appear to be most appropriate, but are not necessarily definitive. Rate constants for various individual reactions could vary as the individual steps in the mechanism are studied further.