In equimolar counter diffusion for every mole of one component diffusing in one direction there is$,$ by definition, one mole of the other component diffusing in the opposite direction. In this problem we will consider mass transfer by diffusion in the vertical column of constant crosssection $($shape of the cross$-$section is not significant$)$ shown in Figure Q$2$ below. The column is open at both ends. Inside the column there are two components A and B$.$ Gas streams containing both components A and B flow across each of the two open ends of the column at the top and the bottom. The concentration of A in the lower stream is greater than the concentration of A in the upper stream giving rise to a concentration difference, which drives the mass transfer of A from the bottom to the top of the column. For equimolar counter diffusion, we have N$\_($Ay$)=-$N$\_($By$)$ in the column, where N$\_($Ay$)$ and N$\_($By$)$ are the molar fluxes of the components A and B respectively in the vertical direction in the column, relative to stationary coordinates. The geometry has been chosen so that there is mass transfer in one direction only, the y $-$direction and Cartesian coordinates can be used. Use a differential control volume inside the column and a$)$ develop the mass balance equation in molar terms for components A and B ; b$)$ derive an expression for the molar fluxes N$\_($Ay$)$ and N$\_($By$),$ relative to stationary coordinates; c$)$ derive the concentration profiles of components A and B along the column. Plot the profiles in a graph with concentration in the x $-$axis and vertical distance in the y $-$axis; $12$ d$)$ compare the molar flux of A in this case to that for diffusion through a stagnant film. P$.$ Anxrli February $2024$ You can assume that the gases are ideal. Figure Q$2$: Schematic of the vertical column where the equimolar counter diffusion of gases A and B takes place.