Water vapor diffuses through an air film to a catalytic surface where it is converted to hydroxylradicals, very reactive oxidants, with the following reaction:H₂O → OH.The mole fraction of water vapor at z = 8 is 1500 ppm. The reaction at the surface may beassumed to be instantaneous. No reaction occurs in the gas film. With the assumption that thecatalyst surface is flat, determine the rate of water vapor diffusion per hour through one squaremeter of catalyst surface area.H₂OOH.z = dz = 0

It is given that water vapor diffuses through an air film to a catalytic surface where it is converted to hydroxyl radical, the following reaction takes place:thickness, ,mole fraction of water vaporNo reaction occurs at the gas film, the catalyst surface is flat.In the given case, molar flux will be due to diffusion and bulk flow , represented as:From Fick's Law of diffusion,The above expression can also be written as: here, is the molar flux , here represents , is the diffusion coefficient , is the mole fraction of A, is the concentration, and is the gradient of the mole fraction with respect to the distance.Now, the bulk flow can be written as:From stoichiometry, So, So, the flux for A is,On simplifying,Now write the mass balance, at a steady state,Now, divide the equation by and differentiate, to obtain,Now, apply boundary conditions as,Now, use and apply it on , to get,Integrate the above expression, to get,Separating the constants on one side, and integrating,Now, apply boundary conditions on at , so,So, Now, put the above value in ,The above is the concentration profile.Now, for flux, substitute the value of in Since the surface area of the catalyst is given as and it is flat, so, .So , equation becomes, Therefore the flux is expressed as: and the final value achieved is . If C, D are given, then substitute the values to find the more detailed answer.