## The element conservation method

Element conservation is expressed as ratios of the number of moles of one element to the number of moles of another element. For a reaction involving N molecular species, N-2 of these mole ratios are required. This method is best explained by applying it to the methane combustion reaction of Eq (9.1). For this reaction, the mass-action law is:

To solve for the 4 mole fractions in this equation requires three other equations. The first is the summation of the mole fractions:

A four-species reaction such as Eq (9.1) must involve three elements. This reaction interchanges C, H, and O among four molecular entities. Element conservation requires that two ratios of the number of moles of the elements, irrespective of their molecular states, be fixed and independent of the extent of reaction. Any two ratios of the three elements can be selected. Choosing the C/H and C/O ratios as specified yields the equations:

The coefficients of the mole fractions in the above ratios are the number of element moles per mole of a molecular species.

The initial molecular forms in the mixture do not affect the equilibrium composition as long as the element ratios are preserved. For example, the element ratios (C/H) = (C/O) = % apply to an initial (nonequilibrium) mixture of 1 mole of CH4 and 2 moles of O2. Alternatively, the same element ratios could be obtained from an initial mixture of 1 mole of CO2 and 2 moles of H2O. The same equilibrium composition would result from either of these initial states.

The general approach is to first solve Eqs (B) and (C) for three of the mole fractions in terms of the fourth, then substitute these results into Eq (A).

Example

From the (C/H) equation, x^o = 2xCo2 ; from the (C/O) equation, x^ = 2xCh4

Substituting these into Eq (B) gives xCh4 = Yi — xco2 , so that x^ = ~ 2xco2 . With three mole fractions expressed in terms of xqo2 , Eq (A) becomes: