Tarnishing with simultaneous dissolution of gas in the metal

In this section, an important special case of high-temperature oxidation will be discussed. Up to now we have assumed that there is no appreciable dissolution of the electronegative component X in the metal during the oxidation process. However, in many important practical systems this is by no means the case, as one can easily appreciate by looking for example at the phase diagram for the Zr-0 system. What rate law should we then expect for the growth of the product layer when dissolution of the element X in the metal occurs simultaneously with the growth of the oxidation product, if local equilibrium can be assumed The situation for a onedimensional experiment when the compound MeX is formed is illustrated in Fig. 8-4. 

Analogous situations for the case of classical solid state reactions have already been mentioned in sections 6.2 and 7.2. An example is the reaction between CoO and Cr2 03 to form CoCr2 04 with the simultaneous dissolution of Cr2 03 in CoO. However, the dissolution of gases in metals during oxidation is a technologically important and, at the same time, a very illustrative example, and so the quantitative treatment of this problem will be outlined here. An important practical example is the oxidation of zirconium. 

The second term in square brackets accounts for the transport of X into the metal according to Pick s first law. The simplifying assumption has been made that Cx, the equilibrium solubility of X in the.metal, is small, so that the mole fraction of metal in the metal phase at x = if 

Continuing with the solution of (8-24), we must now solve Pick s second law for the diffusion of X into the metal with the boundary conditions Cx 0 for. x— oo, and Cx Cx for X = (J. According to eq. (5-42), this solution is  

By taking the derivative (dcx/dx) at x = from eq. (8-25), and substituting this into eq. (8-24), we obtain the following transcendental equation for the growth rate constant a  

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