Question

The individual mass transfer coefficients (\(\text{mol}/\text{m}^2\cdot\text{s}\)) for absorption of a solute from a gas mixture into a liquid solvent are \( k_L = 4.5 \) and \( k_G = 1.5 \). The slope of the equilibrium line is 3. Which one of the following resistance(s) is/are controlling?

• A. Liquid - side
• B. Gas - side
• C. Interfacial
• D. Both liquid and gas-sides

Hint:

To quickly find the controlling resistance, evaluate the ratio: \[ \frac{\text{Liquid Resistance}}{\text{Gas Resistance}} = \frac{m / k_L}{1 / k_G} = \frac{m \cdot k_G}{k_L} \] Substituting the values: \(\frac{3 \times 1.5}{4.5} = \frac{4.5}{4.5} = 1\). Since this ratio equals exactly 1, both resistances are identical in magnitude, meaning both phases share control over the mass transfer rate.

Answer 1

The Correct Option is D

Concept: The two-film theory relates the overall mass transfer resistances to the individual film resistances on the gas side and liquid side. To determine which phase resistance controls the rate of interphase mass transfer, we calculate the numerical values of both individual film resistances and evaluate their relative contributions to the total mass transfer resistance. The equations for total mass transfer resistance can be expressed on either a gas-phase basis ($1/K_G$) or a liquid-phase basis ($1/K_L$). Let us write the resistance balance equation on an overall gas-phase basis: \[ \frac{1}{K_G} = \frac{1}{k_g} + \frac{m}{k_l} \] Where:

• \( \frac{1}{K_G} \) is the total, overall mass transfer resistance on a gas-phase basis.

• \( R_G = \frac{1}{k_g} \) is the independent resistance contributed by the gas-side film layer.

• \( R_L = \frac{m}{k_l} \) is the independent resistance contributed by the liquid-side film layer, scaled by the equilibrium slope.

• \( m \) represents the slope of the linear equilibrium distribution line (\( y = m \cdot x \)).

Step 1: Extracting numerical parameters from the problem text.
The problem provides the following parameters:

• Individual liquid-film mass transfer coefficient: \( k_L = 4.5 \, \text{mol}/(\text{m}^2\cdot\text{s}) \)

• Individual gas-film mass transfer coefficient: \( k_G = 1.5 \, \text{mol}/(\text{m}^2\cdot\text{s}) \)

• Slope of the thermodynamic equilibrium line: \( m = 3 \)

Step 2: Calculating individual resistance values on a gas-phase basis.
Let us compute the numerical value of the resistance offered by the gas-side film: \[ R_G = \frac{1}{k_G} = \frac{1}{1.5} = \frac{2}{3} \approx 0.667 \, \frac{\text{m}^2\cdot\text{s}}{\text{mol}} \] Next, let us compute the numerical value of the resistance offered by the liquid-side film, converted to a gas-phase basis: \[ R_L = \frac{m}{k_L} = \frac{3}{4.5} = \frac{3}{9/2} = \frac{6}{9} = \frac{2}{3} \approx 0.667 \, \frac{\text{m}^2\cdot\text{s}}{\text{mol}} \]

Step 3: Evaluating the total overall resistance and resistance distribution.
Now, sum both film resistances to find the total overall mass transfer resistance ($1/K_G$): \[ \frac{1}{K_G} = R_G + R_L = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \approx 1.333 \, \frac{\text{m}^2\cdot\text{s}}{\text{mol}} \] Let us determine the percentage contribution of each film layer to the total resistance: \[ \text{Gas Film Resistance \%} = \frac{R_G}{1/K_G} \times 100\% = \frac{2/3}{4/3} \times 100\% = 50\% \] \[ \text{Liquid Film Resistance \%} = \frac{R_L}{1/K_G} \times 100\% = \frac{2/3}{4/3} \times 100\% = 50\% \] The calculations show that the gas-side film and the liquid-side film each contribute exactly 50% of the total mass transfer resistance. Because neither individual resistance is negligible and both contribute equally to limiting the mass transfer rate, the process is controlled by

both liquid and gas-sides resistances.