Question

Consider interphase mass transfer of a species S between two immiscible liquids A and B. The interfacial mass transfer coefficient of S in liquid A is twice of that in liquid B. The equilibrium distribution of S between the liquids is given by $y_S^A = 0.5 y_S^B$, where $y_S^A$ and $y_S^B$ are the mole-fractions of S in A and B, respectively. The bulk phase mole-fraction of S in A and B is 0.10 and 0.02, respectively. If the steady-state flux of S is estimated to be 10 kmol h$^{-1}$ m$^{-2}$, the mass transfer coefficient of S in A is \(\underline{\hspace{2cm}}\) kmol h$^{-1}$ m$^{-2}$ (rounded off to one decimal place).

Hint:

Always apply flux equality and equilibrium relations together in interphase mass transfer problems.

Answer 1

Correct Answer: 221 - 223

Given:
$k_A = 2 k_B$
Equilibrium relation: $y_i^A = 0.5 y_i^B$
At steady state:
$N_A = k_A (y_{A} - y_i^A)$ and $N_B = k_B (y_i^B - y_{B})$
Since flux must be equal:
$k_A (0.10 - y_i^A) = k_B (y_i^B - 0.02)$
Using equilibrium: $y_i^A = 0.5 y_i^B$
Also: $k_A = 2k_B$
Substitute:
$2k_B (0.10 - 0.5 y_i^B) = k_B (y_i^B - 0.02)$
Divide by $k_B$:
$2(0.10 - 0.5y_i^B) = y_i^B - 0.02$
Solve:
$0.20 - y_i^B = y_i^B - 0.02$
$0.22 = 2 y_i^B$
$y_i^B = 0.11$
$y_i^A = 0.055$
Flux:
$N_A = k_A (0.10 - 0.055)$
$10 = k_A (0.045)$
Thus:
$k_A = \dfrac{10}{0.045} = 222.2$ kmol h$^{-1}$ m$^{-2}$
Rounded: $222.2$ kmol h$^{-1}$ m$^{-2}$