Fig. 7.1 - Industrial Process with Dispersion / Dilution / Reaction

The Langmuir Model for Adsorption

Fig. 7.2 - Langmuir's Model for Competitive Adsorption

The chemical equilibrium expression for this is
        Sol-bound + Ads-in-solution == Ads-bound + Sol-in-solution

Assumptions in the Langmuir Model

• all surface sites have the same adsorption energy for the adsorbate (and a similar statemt for the solvent)

• adsorption (of either solvent or adsorbate) at one site doesn’t affect the availability of (block) the next site to adsorb solvent or adsorbate

• adsorption (of either solvent or adsorbate) at one site doesn’t affect the energy of adsorption of the neighboring sites (as they adsorb either solvent or adsorbate)

• the activity of the adsorbate is directly proportional to its concentration (and a similar statemt for the solvent)

The Langmuir Isotherm Equation

The equilibrium constant K_EQ derived from the above chemical equation and assumptions is

[Eq. 1]           K_EQ = K_RP (C_S / C_A) [ / (1 - )]

Where
    C_A [mol/m³] is the concentration of adsorbate
    C_S is the concentration of solvent molecules
    _A is the fraction of surface sites covered by adsorbate
    K_RP is a lumped constant including the activity coefficients of the reactants and the products.

Since C_S doesn't change significantly as the adsorption of solvent varies, it can be considered a constant and we can combine the three constants into a single Langmuir adsorption constant K_ADS = K_EQ / (K_RP C_S).

When the Langmuir equation is solved for C_A we get

[Eq. 2a]           C_A = (1 / K_ADS) _A / (1 - _A)

In what follows we shall drop the subscript A, so C and will refer to adsorbate in solution and adsorbed, respectively.

Note that the concentration in equilibrium with a surface that is half-saturated is simply C_50% = 1 / K_ADS, so that

[Eq. 2b]           C = C_50% / (1 - )

Since C_90% = 9 C_50% and C_30% = 0.43 C_50%, if we wish to raise the fractional surface coverage from 30% to 90% (a factor of 3) we must raise the solution concentration by a factor of 21. And to raise the fractional surface coverage from 0.90 to 0.95 we must raise the concentration in solution by a factor of 19. Using the Langmuir model it is impossible to get 100% coverage.

When the Langmuir equation is solved for we get Eq 2c and Fig 7-3. This curve is called an adsorption isotherm because we typically measure and plot the points from an experiment run at a fixed temperature.

[Eq. 2c]           = C K_ADS / (1 + C K_ADS)

Fig. 7-3 - A Langmuir Adsorption Isotherm

Since Sol-site and Ads-site interactions are temperature dependent, K_ADS changes with temperature, usually decreasing as the temperature increases. Since K_ADS appears in the Langmuir equation is a product with C it serves as a scaling factor for the C axis. If theta is plotted as a function of C K_ADS, we get a universal curve that is independent of temperature, adsorbate, liquid, and solid (over the range for which the assumptions of the model are satisfied).

Total Adsorbate in Solution and on the Solid

The amount (in moles) of adsorbate in the total liquid volume (V, excluding the volume of solids) is

[Eq. 3a]           n_soln = C V

A calculation of the moles of adsorbate adsorbed on the powder requires that we first determine the specific surface area (area per unit mass) of the powder. If (as a simplified case) the powder is a mixture of several components each of which consists of equal-size cubes then since area of cube / (volume of cube) = 1 / (length of side)

[Eq. 3b]     A_sp = sum over all components [ mass fraction of component
                    * area of cube / (volume of cube * density) ]
                    = (6 / ) (f_i / D_i)

where D = length of an edge and = particle density

The area covered by a mole of surface sites is

[Eq. 3c]           A_site = N_Avo x_site²

The moles of dispersant adsorbed at fractional coverage is

[Eq. 3d]           n_ads = (A_sp / A_site) m_powder

Calculations when Total Dispersant is Fixed

The amounts of powder and dispersant remain constant when a system is diluted with fresh solvent or heated. These changes cause dispersant to desorb from the powder until solid coverage and solution concentration are again in equilibrium. If we add the left and right sides of Eq. 3a and Eq. 3d we get

[Eq. 4a]     n_total = C V + (A_sp / A_site) m_powder

We can use Eq. 2a to substitute for C and get an equation that we can solve for as

[Eq. 4b]     n_total = (V / K_ADS) [ / (1 - )] + A_sp m_powder

If we multiply all terms by (K_ADS / V) (1 - ) we get a quadratic equation

[Eq. 4c]     a ² + b + c = 0

where the lumped (dimensionless) parameters are
    a = (K_ADS / V) (A_sp / A_site) m_powder
    c = (K_ADS / V) n_total
    b = - 1 - a - c
The quadratic can be solved to give

[Eq. 4d]     = [-b - SQRT(b² - 4 a c)] / (2 a)

This is the equation to use if we know the total mass of dispersant from initial conditions and we need to estimate the fractional coverage of the surface after conditions change.

Estimating the Adequacy of Dispersant Dosage

We often imagine that optimum dispersion stabilization occurs when the particles are fully coated with dispersant. The Langmuir model indicates that this is impossible and that approaching full coverage requires having a lot of dispersant in solution -- and this woiuld be costly for a commercial process. So let us set some reasonable, but arbitrary, criteria for judging whether a given surface coverage is adequate to maintain a dispersion (that is, to prevent flocculation):
-- below 30% = likely inadequate
-- 30% to 59% = possibly adequate, but more likely inadequate
-- 60% to 90% = probably adequate
-- over 90% = likely requires too much dispersant in solition

Using these criteria we can estimate whether increased fines, dilution, or heating might reduce the surface coverage enough to make the dispersion unstable.
-- If the Langmuir calculation indicates a coverage lower than 60% we should consider increasing the dosage of dispersant so as to raise the coverage to 75% or more. We will spend more money to ensure a good dispersion.
-- If the Langmuir calculation indicates a coverage higher than 90% we should consider reducing the dosage of dispersant so as to drop the coverage to 80%. We can save money and still maintain a good dispersion.

SAMPLE CALCULATIONS

-- Given: Tank C contains 18 m³ of water at 30^oC, to which is added 2,000 kg of powder. [See the "more fines" section below for a discussion of why this high solids loading might be used.] For our initial calculations, let us assume that the powder is made up of uniform cubes 2.1 m on a side with a particle density of 2,000 kg/m³.

For this example using Eq. 3b
A_sp = 6 / (2 x 10^-6 m * 2 x 10³ kg/m³) = 1,500 m²/kg

-- Given: Each cube is covered a square array of surface sites, each of which is 1.0 nm from its neighbor.

For this example using Eq. 3c
A_site = 6.02 x 10²³ (1.0 x 10^-9 m)² = 602,000 m²/mol.

-- Given: The dispersant has M_A = 0.200 kg/mol (200 g/mol). The dispersant molecules are small enough that one molecule can adsorb on each site with no interference from adsorption at other sites. The agent adsorbs from aqueous solution onto the powder with an adsorption constant K_ADS = 1,000 m³/mol at 30^oC and K_ADS = 100 m³/mol at 50^oC.

-- Given: We wish to achieve 90% surface coverage. The solution concentration in equilibrium with this, using Eq. 2a, is
C = [0.9 / (1 - 0.9)] / (1,000 m³/mol) = 0.009 mol/m³

The amount in solution, using Eq. 3a, is
n_soln = 0.009 mol/m³ * 18 m³ = 0.16 mol
The amount adsorbed on the powder, using Eq. 3d, is
n_ads = 0.9 * (1,500 m²/kg / 602,000 m²/mol) * 2,000 kg = 4.27 mol
Thus the total dispersant required is 4.43 moles and the percent of total dispersant in solution is 3.6%

CALCULATION WITH A SIZE DISTRIBUTION: Real systems have a distribution of particle sizes. If we assume that large and small particles have the same K_ads, the distribution influences the adsorption only through A_sp.
-- Given: A distribution with some fines and some agglomerates:
20% fines with size 0.7 m contribute 857 m²/kg
60% medium with size 2.1 m contribute 857 m²/kg
20% coarse with size 6.3 m contribute 95 m²/kg
for a total of 1810 m²/kg. This larger specific area requires more dispersant than we computed the monosize case above, namely 5.41 adsorbed, so the total dispersant required is 5.57 moles, of which 2.9% is in solution.

DEAGGLOMERATION AND ATTRITION IN TANK C: A high solids loading is often desirable in a wet-in tank because the many particle-particle collisions induced by a high-shear mixer will help break up clumps (and may also produce fines). Particle-particle collisions increase as the square of the number density in suspension, and for a given particle size distribution the number density increases linearly with the mass density. The high energy input required to stir a high-solids suspension can cause undesired heating, so dispersion tanks often have cooling jackets. As fines are produced A_sp will increase, and if we don't add more dispersant the available agent will be spread over a larger surface area, and the fractional coverage will decrease. This will result in some of the dispersant in solution adsorbing on the additional surface to bring the adsorption back into equilibrium.
-- Given: A shift in the above distribution so that there are 30% fines and only 10% coarse. The specific surface area increases (21%) to 2,190 m²/kg, the fractional surface coverage drops to 0.758, and the solution concentration drops to 0.00313 mol/m³, with only 1.0% in solution.

DILUTION IN PIPE D: The dispersion is pumped out of the tank and diluted so that it will be easier to pump.
-- Given: Four times the original liquid is added in-line. The total amount of dispersant remains constant, so Eq. 4d is used to solve for fractional coverage and then Eq. 2a used to solve for solution concentration. These are used in Eq 3a and 3d to get total amounts. We find that the dilution reduces the fractional coverage to 0.732, the solution concentration drops slightly to 0.00273 mol/m³, with 4.4% of the agent in solution.

HEATING IN TANK E: The dispersion is then heated in a reactor in which the particles are used as a catalyst.
-- Given: K_ADS decreases rapidly with temperature to 100 m³/mol at 50^oC. The computation protocol used for the last case is used here, yielding a fractional coverage of 0.589 and a solution concentration of 0.0143 mol/m³, with 23% of the agent in solution.

In Conclusion

Symbol List

      a, b, c [dimensionless] = parameters in the quadratic equation
      m_powder [kg] = mass of powder
      n [mol] = moles of adsorbate
      x_site [m] = distance between surface adsorption sites

      [dimensionless] = fraction of surface sites covered by adsorbate

Subscripts:
      A for adsorbate
      S for solvent
      ads = adsorbed on the powder surface
      soln = dissolved in solution
      total = total of adsorbed plus dissolved