Question

In an inter-phase mass transfer process, the lesser the solubility of a given solute in a liquid, the higher are the chances that the transfer process will be:

• A. liquid phase resistance controlled
• B. gas phase resistance controlled
• C. impossible
• D. driven by a non-linear driving force

Hint:

Controlling resistance rules in mass transfer: - Highly soluble gas (e.g., \(\text{NH}_3\) in water) \(\rightarrow m\) is very small \(\rightarrow \frac{1}{k_l}\) becomes negligible \(\rightarrow\) Gas-phase resistance controls. - Sparingly soluble Insoluble gas (e.g., \(\text{O}_2\) or \(\text{CO}_2\) in water) \(\rightarrow m\) is very large \(\rightarrow \frac{1}{m \cdot k_g}\) becomes negligible \(\rightarrow\) Liquid-phase resistance controls.

Answer 1

The Correct Option is A

Concept: According to the classical two-film theory of interphase mass transfer developed by Whitman, the overall mass transfer resistance for a solute moving between a gas phase and a liquid phase is distributed across two stationary film layers positioned on either side of the phase interface. The overall mass transfer coefficient based on the gas phase ($K_G$) and the liquid phase ($K_L$) can be related to the individual local film coefficients ($k_g$ and $k_l$) using the system equilibrium relationship. Assuming a linear equilibrium relationship modeled by Henry's Law: \[ y_i = m \cdot x_i \] Where:
• \( y_i \) and \( x_i \) represent the mole fractions of the solute at the gas and liquid sides of the interface, respectively.
• \( m \) represents the slope of the equilibrium curve. The parameter \( m \) is inversely proportional to the solubility of the gas solute in the liquid solvent (\( m \propto \frac{1}{\text{Solubility}} \)).

Step 1: Setting up the total mass transfer resistance equation.
Using the two-film model framework, the total mass transfer resistance expressed on an overall liquid-phase basis is given by the following mathematical formula: \[ \frac{1}{K_L} = \frac{1}{k_l} + \frac{1}{m \cdot k_g} \] Where:
• \( \frac{1}{K_L} \) represents the total, overall mass transfer resistance on a liquid-phase basis.
• \( \frac{1}{k_l} \) represents the local mass transfer resistance inside the liquid film boundary layer.
• \( \frac{1}{m \cdot k_g} \) represents the contribution of the local gas-film resistance to the liquid-phase basis.

Step 2: Evaluating the mathematical limit for low gas solubility.
The problem states that the solute has a very low solubility in the liquid solvent ("the lesser the solubility of a given solute"). A very low solubility implies that a high gas-phase partial pressure is required to dissolve even a small concentration of solute into the liquid phase. Therefore, the Henry's law equilibrium constant slope $m$ becomes extremely large: \[ \text{Solubility} \downarrow \quad \Rightarrow \quad m \rightarrow \text{very large} \, (m \uparrow\uparrow) \] Let us examine the effect of an extremely large value of $m$ on the term representing the gas-film resistance in our liquid-basis resistance equation: \[ \text{As } m \rightarrow \infty \, , \quad \frac{1}{m \cdot k_g} \rightarrow 0 \] Substituting this limit back into the total resistance equation yields: \[ \frac{1}{K_L} \approx \frac{1}{k_l} \]

Step 3: Determining the controlling resistance phase.
Because the term $\frac{1}{m \cdot k_g}$ becomes negligibly small, the gas-film boundary layer offers almost no relative resistance to the mass transfer process. Instead, nearly all of the mass transfer resistance is concentrated within the liquid film layer ($\frac{1}{k_l}$). Consequently, the rate of mass transfer is entirely limited by how quickly the solute can diffuse through the liquid film boundary. This condition is described as a liquid phase resistance controlled process.