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# 510H9_2 - 193.6 Diffusion and chemical reaction in a liquid...

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Unformatted text preview: 193.6 Diffusion and chemical reaction in a liquid a. The differential equation for the steady-state diffusion from a sphere is ch __ 2 III __ 2 2 = JSABrzdrlr d—r—J— kc CA -0 or g—zd—EIE[§ Z—é]_ bI‘ 0 According to Eq. C.1-6b, the solution to this equation is C1 6—115 C2613; F =——— +— 6 5 Application of the boundary conditions that I‘ (1) = 1 and F(oo) = 0 determines the constants, and the ﬁnal result is (cf. Eq. 19B.6-1) _ JSABCAO d__I‘_ r=R R JSABCAO ll =+ (1+b) and the total loss of A from the sphere in moles per unit time is I! 2 ‘ WA =47zR2(‘®ABCA°) 1+ k‘R b. The unsteady-state mass balance on the dissolving sphere of species A is II 2 -£(% nR3psph) = 47tR2(———°®ABC£°MA )[1 + k1 R J l9"? Since we have used here the steady—state expression for the molar ﬂux at the surface of the sphere, this is a quasi-steady-state treatment. . The integration of this equation can be accomplished as follows: First we divide through by 4% 1081C,h to get dt: Apsph AB Next, put all the factors containing R on the left side R dR= JSABCAOMA 1+(]ch2 MAB) dt psph Next we introduce a new dimensionless variable Y = R kf”/J9AB and write 1 + Y dt= p\$th Then we integrate 1Y0 —dY——k”éA°MA 1 dt 1+Y psph and finally 1+Y k’tf M ’ (Y—Y0)—1n1er =——1—4°——A(t—to) or o psph °®AB 1 + ki7 °® AB R0 psph From this one can get the dependence of the sphere radius on the time. ...
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510H9_2 - 193.6 Diffusion and chemical reaction in a liquid...

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