A 50 mm (diameter) sphere of solid nickel is oxidized in a gas mixture containing 60% argon and 40% oxygen by volume. The rate of oxidation of nickel is controlled by the rate of transport of oxygen through the concentration boundary layer. The rate of oxidation (in moles of nickel per minute (mol/min), rounded off to two decimal places) is _________.

Given: Total pressure = 1atm.; Temperature = 1173 K; Concentration of oxygen at the surface of the solid = 0; Boundary layer mass transfer coefficient = 0.03 m/s; Universal gas constant, R = 8.205 x10^-5 m^3.atm. K^-1, mol^-1, Assume ideal gas behavior.

# Calculate the rate of oxidation.

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This is mass transfer case, which can be solved by using FICK's 1st law. Mass flux rate (mol/sec.m2) = D*(C/L) Hence, rate of oxidation (mol/""min"") = area*(D/L)*C*""60"" Given D/L = 0.03 m/s C = n/V = P/RT = 0.4/1173*8.205*.00001 = 0.4156 mol/m3 (theRead more

This is mass transfer case, which can be solved by using FICK’s 1^{st}law.Mass flux rate (mol/sec.m^{2}) = D*(C/L)Hence, rate of oxidation (mol/””min””) = area*(D/L)*C*””60″”Given D/L = 0.03 m/sC = n/V = P/RT = 0.4/1173*8.205*.00001 = 0.4156 mol/m^{3}(these are moles of oxygen in air)Area = 4*pie*(0.025)^{2}Rate of oxidation (mol/min) = 4*pie*(0.025)^{2}*0.03*0.4156*60 = 0.05875 mols of oxygen/minNow, this we have calculated in terms of moles of oxygen, but requirement is in terms of moles of Nickel.2Ni + O_{2}= 2NiO1 mol of O_{2}= 2 mol of Ni

See lessHence, rate of oxidation in terms of moles of nickel per min = 2*0.05875 = 0.11 mol/minIs there any other method of solving this question

Is there any other method of solving this question

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