Question:

Consider steady-state diffusion in a binary A-B liquid at constant temperature and pressure. The mole-fraction of A at two different locations is 0.8 and 0.1. Let $N_{A1}$ be the diffusive flux of A calculated assuming B to be non-diffusing, and $N_{A2}$ be the diffusive flux of A calculated assuming equimolar counter-diffusion. The quantity $\frac{(N_{A1} - N_{A2})}{N_{A1}} \times 100$ is \(\underline{\hspace{2cm}}\) (rounded off to one decimal place).

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Fluxes differ because assuming B non-diffusing vs. equimolar counter-diffusion changes the molar-average velocity correction.
Updated On: Jan 13, 2026
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Correct Answer: 53 - 54

Solution and Explanation

Step 1: Case 1 - B is Non-Diffusing ($N_B = 0$)

When B is non-diffusing, we have unimolecular diffusion of A. The flux equation is:

$$N_{A1} = -\frac{cD_{AB}}{1-x_A} \frac{dx_A}{dz}$$

For steady-state diffusion between two points:

$$N_{A1} = \frac{cD_{AB}}{L} \ln\left(\frac{1-x_{A2}}{1-x_{A1}}\right)$$

where $L$ is the distance between the two locations.

Substituting values:

$$N_{A1} = \frac{cD_{AB}}{L} \ln\left(\frac{1-0.1}{1-0.8}\right) = \frac{cD_{AB}}{L} \ln\left(\frac{0.9}{0.2}\right)$$

$$N_{A1} = \frac{cD_{AB}}{L} \ln(4.5) = \frac{cD_{AB}}{L} \times 1.504$$

Step 2: Case 2 - Equimolar Counter-Diffusion ($N_A = -N_B$)

For equimolar counter-diffusion, Fick's first law applies directly:

$$N_{A2} = -cD_{AB} \frac{dx_A}{dz}$$

For steady-state between two points:

$$N_{A2} = \frac{cD_{AB}}{L}(x_{A1} - x_{A2})$$

Substituting values:

$$N_{A2} = \frac{cD_{AB}}{L}(0.8 - 0.1) = \frac{cD_{AB}}{L} \times 0.7$$

Step 3: Calculate the Ratio

$$\frac{N_{A1} - N_{A2}}{N_{A1}} = \frac{\frac{cD_{AB}}{L} \times 1.504 - \frac{cD_{AB}}{L} \times 0.7}{\frac{cD_{AB}}{L} \times 1.504}$$

$$= \frac{1.504 - 0.7}{1.504} = \frac{0.804}{1.504}$$

$$= 0.5346$$

Step 4: Calculate the Final Answer

$$\frac{N_{A1} - N_{A2}}{N_{A1}} \times 100 = 0.5346 \times 100 = 53.46$$

Rounded to one decimal place: 53.5

Answer

The quantity $\frac{(N_{A1} - N_{A2})}{N_{A1}} \times 100$ is 53.5 (rounded to one decimal place).

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