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Here 1 mole O2, diffuses to the carbon particle, and 2 moles of CO diffuses back in the opposite direction. If we denote O, by A and CO by B, then N. = -2N, i.e. N, = - %N, DAn dpa (N, + N. RT dz DAn dp. NAPA RT dz 2P 2PDA2P - PA2 -In p-PAs RTL

Here 1 mole O2, diffuses to the carbon particle, and 2 moles of CO diffuses back in the opposite direction. If we denote O, by A and CO by B, then N. = -2N, i.e. N, = - %N, DAn dpa (N, + N. RT dz DAn dp. NAPA RT dz 2P 2PDA2P - PA2 -In p-PAs RTL

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Non-equimolar counter diffusion When a carbon particle is burnt in limited supply of 0, then the following reaction takes place, 20 + 0, = 200 Here 1 mole O, diffuses to the carbon particle, and 2 moles of CO diffuses back in the opposite direction. If we denote o, by A and CO by B, then N, = - 2N, L.e. N, = - %N. - 2N, i.e. N, = - % NA DAn dpa NA (Na + N RT dz DAn dpa NAPA NA RT dz 2P 2PDAn2P - PA2 In 2P-PA1 RTL
Transcribed Image Text:Non-equimolar counter diffusion When a carbon particle is burnt in limited supply of 0, then the following reaction takes place, 20 + 0, = 200 Here 1 mole O, diffuses to the carbon particle, and 2 moles of CO diffuses back in the opposite direction. If we denote o, by A and CO by B, then N, = - 2N, L.e. N, = - %N. - 2N, i.e. N, = - % NA DAn dpa NA (Na + N RT dz DAn dpa NAPA NA RT dz 2P 2PDAn2P - PA2 In 2P-PA1 RTL
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