Interphase mass transfer and overall coefficients

Two-resistance theory of interphase transfer, interface compositions, overall gas- and liquid-phase coefficients from individual ones and equilibrium slope, and how to tell which phase controls.

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Why it matters

In absorption, stripping, distillation and extraction the solute must leave one phase and enter another, so it meets two resistances in series. Interface compositions cannot be measured, so designers use overall coefficients based on bulk compositions and equilibrium data. Knowing which phase controls tells you whether to spend money on gas-side turbulence, liquid-side agitation or a better solvent.

Key ideas

Two-resistance (two-film) theory (Lewis–Whitman).

  1. Each phase has its own resistance, described by an individual coefficient (k_y or k_G on the gas side, k_x or k_L on the liquid side).
  2. The interface itself offers no resistance: the two phases are in equilibrium at the interface, so (x_i, y_i) lies on the equilibrium curve.
  3. At steady state the flux leaving one phase equals the flux entering the other.

Interface composition. Writing N_A = k_y (y_b − y_i) = k_x (x_i − x_b) shows that the interface point lies where a straight line of slope −k_x/k_y drawn from the bulk point (x_b, y_b) meets the equilibrium curve. (For concentrated solutions use the F-type or k'-type coefficients; this lesson treats dilute systems.)

Overall driving forces. Equilibrium links the phases, so the bulk liquid composition can be expressed as an equivalent gas composition y* = m·x_b (gas in equilibrium with the bulk liquid), and the bulk gas as x* = y_b/m. Overall driving forces are (y_b − y*) or (x* − x_b). With a linear (or locally linear) equilibrium of slope m, the resistances add like resistors in series.

Controlling resistance.

  • Very soluble gas (small m, e.g. NH₃ or HCl in water): m/k_x is small, so gas-film control, K_y ≈ k_y.
  • Sparingly soluble gas (large m, e.g. O₂, N₂, CO₂ in water): 1/(m·k_y) is small, so liquid-film control, K_x ≈ k_x.
  • Fractional resistance of the gas phase = (1/k_y)/(1/K_y).

When overall coefficients are awkward. If the equilibrium curve is strongly curved, m changes with composition and K_y varies through the column. Then work with individual coefficients and interface compositions, or use the local slope between the relevant points (m' between interface and bulk liquid for K_y, m'' between bulk gas and interface for K_x).

Henry's-law form. If equilibrium is written p_A = H·C_A, the same algebra gives K_G and K_L with H in place of m.

Formulas

Individual coefficients (dilute): N_A = k_y (y_b − y_i) = k_x (x_i − x_b) (y_b − y_i)/(x_b − x_i) = −k_x/k_y

Overall coefficients with equilibrium y* = m·x: N_A = K_y (y_b − y*) = K_x (x* − x_b) 1/K_y = 1/k_y + m/k_x 1/K_x = 1/k_x + 1/(m·k_y) K_x = m·K_y

  • k_y, k_x, K_y, K_x in mol/(m²·s) (or kmol/(m²·s)); m – slope of equilibrium line, dimensionless (mole-fraction basis); y* – gas mole fraction in equilibrium with bulk liquid; x* – liquid mole fraction in equilibrium with bulk gas.

Henry's-law form, p_A = H·C_A: 1/K_G = 1/k_G + H/k_L 1/K_L = 1/k_L + 1/(H·k_G) N_A = K_G (p_A,b − p*) = K_L (C* − C_A,b), with p* = H·C_A,b and C* = p_A,b/H

  • k_G, K_G in mol/(m²·s·Pa); k_L, K_L in m/s; H in Pa·m³/mol.

Fractional resistances: gas phase = (1/k_y)/(1/K_y), liquid phase = (m/k_x)/(1/K_y)

Worked examples

Example 1 – interface composition and overall coefficient (standard). Given: at one point in an absorber y_b = 0.04, x_b = 0.01; equilibrium y = 1.5x; k_y = 1.2 × 10⁻³ kmol/(m²·s), k_x = 3.0 × 10⁻³ kmol/(m²·s).

  1. Flux continuity: k_y (y_b − 1.5x_i) = k_x (x_i − x_b).
  2. x_i = (k_y·y_b + k_x·x_b)/(1.5k_y + k_x) = (4.8 × 10⁻⁵ + 3.0 × 10⁻⁵)/(1.8 × 10⁻³ + 3.0 × 10⁻³) = 0.01625; y_i = 1.5 × 0.01625 = 0.02438.
  3. N_A = k_x (x_i − x_b) = 3.0 × 10⁻³ × 0.00625 = 1.875 × 10⁻⁵ kmol/(m²·s). Check: k_y(y_b − y_i) = 1.2 × 10⁻³ × 0.01562 = 1.875 × 10⁻⁵. ✓
  4. 1/K_y = 1/k_y + m/k_x = 833.3 + 500 = 1333.3, so K_y = 7.50 × 10⁻⁴ kmol/(m²·s).
  5. y* = 1.5 × 0.01 = 0.015; N_A = K_y (y_b − y*) = 7.50 × 10⁻⁴ × 0.025 = 1.875 × 10⁻⁵. ✓
  6. Gas-phase resistance fraction = 833.3/1333.3 = 0.625.

x_i = 0.0163, y_i = 0.0244, N_A = 1.88 × 10⁻⁵ kmol/(m²·s), K_y = 7.50 × 10⁻⁴ kmol/(m²·s); gas phase offers 62.5 % of the resistance.

Example 2 – liquid-film control for a sparingly soluble gas (GATE level). Given: CO₂ absorbed into CO₂-free water; H = 3000 Pa·m³/mol (p = H·C); k_L = 1.0 × 10⁻⁴ m/s; k_G = 1.0 × 10⁻⁶ mol/(m²·s·Pa); bulk gas partial pressure of CO₂ = 10 kPa.

  1. 1/K_L = 1/k_L + 1/(H·k_G) = 1/(1.0 × 10⁻⁴) + 1/(3000 × 1.0 × 10⁻⁶) = 10 000 + 333 = 10 333 s/m.
  2. K_L = 9.68 × 10⁻⁵ m/s; liquid-side resistance = 10 000/10 333 = 96.8 %.
  3. C* = p_b/H = 10 000/3000 = 3.33 mol/m³; C_b = 0.
  4. N_A = K_L (C* − C_b) = 9.68 × 10⁻⁵ × 3.33 = 3.23 × 10⁻⁴ mol/(m²·s).

K_L = 9.68 × 10⁻⁵ m/s; N_A = 3.23 × 10⁻⁴ mol/(m²·s); the process is liquid-film controlled, so increasing gas velocity would barely help.

Common mistakes

  • Adding coefficients instead of resistances (1/K = Σ 1/k).
  • Putting m in the wrong place: it multiplies the liquid resistance in 1/K_y and divides the gas resistance in 1/K_x.
  • Mixing Henry's constant definitions (p = H·C, p = H·x, y = m·x, C = H·p) – check units before substituting.
  • Using y* = m·y or forgetting that y* is in equilibrium with the bulk liquid.
  • Assuming interface resistance – the interface is at equilibrium.
  • Using a constant K_y when the equilibrium line is strongly curved.

For GATE CH

Typical questions: compute K_y or K_L from individual coefficients and m or H; find the fraction of resistance in each phase; decide which phase controls for a given solubility; locate interface compositions from the slope −k_x/k_y; compute flux from an overall driving force. Practise unit consistency between different forms of Henry's law.

Quick check

  1. Why can the interface be treated as being at equilibrium?
  2. Write 1/K_y in terms of k_y, k_x and m.
  3. For NH₃ absorbed in water, which phase usually controls?
  4. If k_y = k_x and m = 1, what fraction of the resistance lies in the gas phase?

Answers: 1. interfacial resistance is negligible compared with the film resistances; 2. 1/K_y = 1/k_y + m/k_x; 3. gas phase; 4. 50 %.

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is interphase mass transfer?Concept

    Interphase mass transfer refers to the movement of a chemical species from one phase to another, such as from a liquid to a gas or vice versa. This process is crucial in operations like distillation, absorption, and extraction, where the transfer of mass across phase boundaries is necessary for separation and purification.

  2. 2.Explain the concept of overall mass transfer coefficients.Concept

    Interface compositions cannot be measured, so the flux is written with an overall driving force based on bulk compositions and equilibrium: N_A = K_y(y_b − y*) = K_x(x* − x_b), where y* is the gas composition in equilibrium with the bulk liquid. With a linear equilibrium y = m·x, the two film resistances add in series: 1/K_y = 1/k_y + m/k_x and 1/K_x = 1/k_x + 1/(m·k_y). The overall coefficient is constant only if the equilibrium line is straight over the range considered.

  3. 3.Why are overall mass transfer coefficients used in chemical engineering processes?Application

    Overall mass transfer coefficients are used because they simplify the analysis of mass transfer processes involving multiple phases. By combining the resistances of each phase into a single coefficient, engineers can more easily design and optimize equipment like absorbers and distillation columns.

  4. 4.What happens if the resistance to mass transfer is higher in one phase compared to the other?Application

    If the resistance to mass transfer is higher in one phase, it becomes the controlling resistance, meaning it limits the overall rate of mass transfer. In such cases, efforts to enhance mass transfer should focus on reducing the resistance in the controlling phase, such as by increasing agitation or using a more efficient packing material.

  5. 5.Explain how the two-film theory is related to interphase mass transfer.Concept

    The Lewis–Whitman two-film theory places one resistance on each side of the interface, each modelled as a stagnant film, with well-mixed bulk phases beyond. The interface itself is assumed to offer no resistance, so the compositions there are in equilibrium. At steady state the flux through the gas film equals that through the liquid film, which fixes the interface compositions (a line of slope −k_x/k_y from the bulk point meets the equilibrium curve). Adding the two resistances gives the overall coefficient.

  6. 6.How does temperature affect interphase mass transfer rates?Application

    Temperature can significantly affect interphase mass transfer rates. Generally, increasing the temperature increases the kinetic energy of molecules, which can enhance diffusion rates and reduce the viscosity of liquids, thereby increasing mass transfer rates. However, it can also affect solubility and vapor pressure, which must be considered.

  7. 7.What is the role of diffusion in interphase mass transfer?Concept

    Diffusion is a key mechanism in interphase mass transfer, as it describes the movement of molecules from regions of high concentration to low concentration. In mass transfer operations, diffusion across phase boundaries is often the rate-limiting step, especially in systems with low solubility or high viscosity.

  8. 8.For a dilute system whose equilibrium distribution coefficient is m = 1 (in consistent concentration units), the individual gas- and liquid-side coefficients are 0.1 m/s and 0.05 m/s. Calculate the overall coefficient.Numerical

    With m = 1 the resistances simply add: 1/K = 1/k_G + m/k_L = 1/0.1 + 1/0.05 = 10 + 20 = 30 s/m, so K = 0.0333 m/s. Two-thirds of the resistance is in the liquid phase. For any other m the liquid resistance would be multiplied by m (for a gas-based overall coefficient), so m must never be dropped silently.

  9. 9.For a system with an equilibrium distribution coefficient m = 1 (consistent units), the liquid-film coefficient is 0.02 m/s and the overall coefficient is 0.015 m/s. What is the gas-film coefficient?Numerical

    With m = 1, 1/K = 1/k_G + 1/k_L, so 1/k_G = 1/0.015 − 1/0.02 = 66.67 − 50 = 16.67 s/m and k_G = 0.06 m/s. The liquid phase then carries 50/66.67 = 75 % of the resistance, so it controls the rate.

  10. 10.Describe how interphase mass transfer is applied in a distillation column.Application

    On each tray or packing section, rising vapour contacts descending liquid. The more volatile component transfers from the liquid into the vapour and the less volatile component condenses from the vapour into the liquid, often close to equimolar counter-diffusion when molar latent heats are similar. The rate is governed by vapour- and liquid-side resistances combined through the local slope of the equilibrium curve. Repeated contacting enriches the vapour in the light component going up and the liquid in the heavy component going down.

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