Consider steady-state equimolar counterdiffusion through the converging-diverging tube. Given: r1=1cm; r2=2cm; NA=10-6 kmol/m2.s at the vessel 1 end of the tube. The rate of transport of B at the vessel 2 end of the tube is

5*10 -6 kmol/s, towards vessel 2

2*10-6 kmol/s, towards vessel 1

2.5*-7 kmol/s, towards vessel 1

5*10-6 kmol/s, towards vessel 2

9*10-7 kmol/s, towards vessel 2

Consider steady-state equimolar counterdiffusion through the convergin

1.	Consider steady-state equimolar counterdiffusion through the converging-diverging tube. Given: r₁=1cm; r₂=2cm; N_A=10^-6 kmol/m².s at the vessel 1 end of the tube. The rate of transport of B at the vessel 2 end of the tube is
A.	2*10<sup>-6</sup> kmol/s, towards vessel 1
B.	2.5*<sup>-7</sup> kmol/s, towards vessel 1
C.	5*10<sup>-6</sup> kmol/s, towards vessel 2
D.	9*10<sup>-7</sup> kmol/s, towards vessel 2
Answer» C. 5*10<sup>-6</sup> kmol/s, towards vessel 2

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Consider steady-state equimolar counterdiffusion through the converging-diverging tube. Given: r1=1cm; r2=2cm; NA=10-6 kmol/m2.s at the vessel 1 end of the tube. The rate of transport of B at the vessel 2 end of the tube is

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Consider steady-state equimolar counterdiffusion through the converging-diverging tube. Given: r₁=1cm; r₂=2cm; N_A=10^-6 kmol/m².s at the vessel 1 end of the tube. The rate of transport of B at the vessel 2 end of the tube is