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Homework Statement
Problem 18B.13 from Transport Phenomena, BSL.
Tarnishing of metal surfaces. In the oxidation of most metals the volume of oxide produced is greater than that of the metal consumed. This oxide thus tends to form a compact film, effectively insulating the oxygen and metal from each other. For the derivations that follow, it may be assumed that
a) For oxidation to proceed, oxygen must diffuse through the oxide film and that this diffusion follows Fick's law.
b) The free surface of the oxide film is saturated with oxygen from the surrounding air.
c) Once the film of oxide has become reasonably thick, the oxidation becomes diffusion controlled; that is, the dissolved oxygen concentration is essentially zero at the oxide-metal surface.
d) The rate of change of dissolved oxygen content of the film is small compared to the rate of reaction. That is, quasi-steady state conditions may be assumed.
e) The reaction involved is [itex]\frac{1}{2} x \textrm{O}_2 + \textrm{M} \rightarrow \textrm{MO}_x[/itex]
We wish to develop an expression for rate of tarnishing in terms of oxygen diffusivity through the oxide film, the densities of the metal and its oxide, and the stoichiometry of the reaction. Let [itex]c_O[/itex] be the solubility of oxygen in the film, [itex]c_f[/itex] the molar density of the film, and [itex]z_f[/itex] the thickness of the film. Show that the film thickness is
[tex]z_f = \sqrt{\frac{2 D_{O_2 - MO_x} t}{x} \frac{c_O}{c_f}}[/tex]
This result, the so-called "quadratic law," gives a satisfactory empirical correlation for a number of oxidation and other tarnishing reactions. Most such reactions are, however, much more complex than the mechanism given above.
Homework Equations
Fick's first law of diffusion
Steady state molar balance
Unsteady state molar balance
The Attempt at a Solution
First, since this is a quasi-steady state problem, I started by doing a steady state mole balance for oxygen in the oxide film, which leads to the following D.E.
[tex]\frac{dN_{O_2 z}}{dz} = 0[/tex]
Fick's law of diffusion is given by [itex]N_{O_2 z} = -D_{O_2 - MO_x} \frac{d C_{O_2}}{dz}[/itex]. Since diffusivity is constant in this problem, the D.E. can be simplified to
[tex]\frac{d^2 C_{O_2}}{dz^2} = 0[/tex]
Integrating twice we get
[tex]C_{O_2} = c_1 z + c_2[/tex]
Our boundary conditions are
z = 0 → CO2 = cO
z = zf → CO2 = 0
So, the concentration profile and its derivative are given by
[tex]C_{O_2} = c_O \left(1 - \frac{z}{z_f} \right)[/tex]
[tex]\frac{d C_{O_2}}{dz} = - \frac{c_O}{z_f}[/tex]
Therefore, the molar flux of O2 is given by
[tex]N_{O_2 z} = \frac{D_{O_2 - MO_x} c_O}{z_f}[/tex]
Now we can proceed with the unsteady state part of the problem, using the expressions derived from the steady state analysis (quasi-steady state). This is where my solution differs from the one shown in the statement. We make an unsteady state molar balance for the oxide film
[tex]\frac{dM_{MO_x}}{dt} = \frac{2}{x} S N_{O_2 z}[/tex]
Where S is the surface area of the film and 2/x is the stoichiometric coefficient (for every mole of O2 that diffuses, 2/x moles of oxide are formed). We also know [itex]M_{MO_x} = S z_f c_f[/itex]. So the molar balance becomes
[tex]z_f \frac{d z_f}{dt} = \frac{2 D_{O_2 - MO_x}}{x} \frac{c_O}{c_f}[/tex]
Integrating and using the initial condition that when
t = 0 → zf = 0
We arrive at the expression for the film thickness as a function of time
[tex]z_f = \sqrt{\frac{4 D_{O_2 - MO_x} t}{x} \frac{c_O}{c_f}}[/tex]
However, my result differs by a factor of 2 from the answer provided by the book. I used 2/x as the stoichiometric coefficient, but the authors seem to have used only 1/x in order to arrive at the given result. At first, I thought only the x was significant and I could neglect the 2, but then I saw a similar problem using the same stoichiometry and it included the 2/x term. Maybe I missed a small detail I could've used to cancel the 2. I'm not sure, but this is really bugging me.
Thanks in advance for any input!
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