# Difference between revisions of "The Electrostatic Component of Disjoining Pressure"

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− | <math> \upsilon = arcsin(1/cosh(z\phi_d/2)) | + | <math> \upsilon = arcsin(1/cosh(z\phi_d/2))</math> |

− | \varphi = arccos(sinh(z\phi_d/2)/sinh(z\phi_1/2))</math> | + | |

+ | <math>\varphi = arccos(sinh(z\phi_d/2)/sinh(z\phi_1/2))</math> | ||

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− | <math> \Pi_e/4nkT = cot^2\upsilon | + | <math> \Pi_e/4nkT = cot^2(\upsilon)</math> |

+ | |||

− | \kappa h = 2 sin\upsilon \cdot F(\varphi,\upsilon)</math> | + | <math>\kappa h = 2 sin(\upsilon) \cdot F(\varphi,\upsilon)</math> |

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− | <math> cos\varphi = cot\upsilon / sinh(z\phi_1/2)</math> | + | <math> cos(\varphi) = cot(\upsilon) / sinh(z\phi_1/2)</math> |

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− | <math> tan\varphi = tan\upsilon\cdot sinh(z\phi_\infty /2)</math> | + | <math> tan(\varphi) = tan(\upsilon)\cdot sinh(z\phi_\infty /2)</math> |

## Revision as of 17:04, 8 December 2011

By Bryan Hassell: AP 255 Fall 11

Keywords: disjoining pressure

## Contents

## Introduction

Investigating the mechanisms of spontaneous charging of interfaces between two phases, one being an electrolyte solution, leads to an equilibrium diffuse atmosphere. If two interfaces are brought together then their ionic atmospheres will overlap and this gives rise to the electrostatic component of the disjoining pressure.

## Methods of Calculation of the Disjoining Pressure <math> \Pi_e </math>

There are several approaches for calculating <math> \Pi_e </math>. One by Derjaguin [1] directly determined the electrostatic disjoining pressure acting on the surfaces of parallel plates in an electrolyte solution. He assumed that at equal potentials of the external and internal surfaces of the plate the hydrodynamic pressure is the same and so the disjoining pressure is just the difference between the Maxwell stresses at the external and internal surfaces of the plates:

<math> \Pi_e = \epsilon E_{ex}^2/8\pi - \epsilon E_{in}^2/8\pi </math>

<math> dG_{T,P,\mu_i} = S\Pi_e dh + S(\psi_1 d\sigma_1 + \psi_2 d\sigma_2) </math>

where <math> \mu_i </math> are the chemical potentials, T is the temperature, P pressure, S is the surface area of the plates and <math> \psi_1 </math> and <math> \psi_2 </math> are the surface potentials as seen in Figure 1. If you then subtract from both sides of the equation the differential of the last term from the previous equation you get:

<math> dG_{T,P,\mu_i} - Sd(\psi_1 d\sigma_1 + \psi_2 d\sigma_2) = -S\Pi_e dh - S(\sigma_1 d\psi_1 + \sigma_2 d\psi_2).</math>

The left hand side of this is a total differential, or

<math>\partial\Pi_e/\partial\psi_{h,\mu_I,\psi_2} = \partial\sigma_1/\partial h|_{\mu_i,\psi_1,\psi_2}.</math>

To get <math> \Pi_e </math> from this we need to integrate the left hand side so we need to transform the right hand side. Referring to the Poisson equation,

<math>{d^2\psi \over dx^2} = \left (\frac{4\pi}{\epsilon} \right) \rho </math>

where the potential <math>\psi</math> is a function of x as seen in Figure 1. The volume charge density is found as

<math> \rho = \sum_{i} ez_i \tilde{n}_i</math>

where the local concentrations of ions are determined by the concentrations of ions in the bulk and the work of their transfer from bulk into the interlayer, <math> w_i = z_i e\psi</math>, therefore the local volume density is a function of local potential, or <math>\rho = \rho(\psi)</math>. In the bulk of the solution where <math>\psi = 0, \tilde{n}_i = n_i</math> the solution is locally electroneutral (volume charge density = 0 and <math>\rho(0) = 0</math>). Multiplying both sides of the Poisson equation by <math> 2{d\psi \over dx}dx</math> and integrating gives:

<math> {\epsilon \over 8\pi} {\left (\frac{d\psi}{dx} \right) }^2 = -\int_{0}^{\psi} \rho(\psi)d\psi + C </math>

where the integration constant C is a function of interlayer thickness h; and as h <math>\rightarrow \infty </math> meaning <math>d\psi/dx \rightarrow 0</math> and <math>\psi \rightarrow 0</math>, then C also goes to 0. For the left plate, the previous equation can be replaced by

<math> {2\pi \sigma_{1}^{2} \over \epsilon } = C - \int_{0}^{\psi_1} \rho d\psi</math>

and by Gauss's theorem,

<math> \sigma_1 = - {\epsilon \over 4\pi} {\left (\frac{d\psi}{dx} \right)}_{x=0}</math>

and <math>\psi(0) = \psi_1</math>. If the potential of the right hand plate is considered constant or <math>\psi_2 =</math> const. then by differentiating the equation for the left hand plate we get

<math> {4\pi\sigma_1 \over \epsilon} {\left (\frac{\partial\sigma_1}{\partial h} \right)}_{\psi_1} = {\left (\frac{\partial\C}{\partial h} \right)}_{\psi_1}</math>

and using the identity

<math> {\left (\frac{\partial C}{\partial h} \right)}_{\psi_1} {\left (\frac{\partial h}{\partial \psi_1} \right)}_{\C} {\left (\frac{\partial\psi_1}{\partial C} \right)}_{h} = -1 </math>

and the relation

<math> \sigma_1 = {\epsilon \over 4\pi} {\left (\frac{\partial\psi_1}{\partial h} \right)}_{C,\psi_2}</math>

we get finally something we can use

<math> {\left (\frac{\partial\sigma_1}{\partial h} \right)}_{\psi_1,\psi_2} = -{\left (\frac{\partial\C}{\partial \psi_1} \right)}_{h,\psi_2} </math>

Substituting this into the original equation we were trying to solve for <math>\Pi_e</math> and integrating we get

<math> \Pi_e(h) = -C </math>

where we've taken into account that <math>\Pi_e(\infty) = 0</math> and <math> C(\infty) = 0 </math>. Taking the integration constant we found before, we finally get the key equation:

<math> \Pi_e(h) = -\int_{0}^{\psi}\rho(\psi)d\psi - {\epsilon E^2 \over 8\pi} </math>

This is the most general formula for <math>\Pi_e</math> which holds for both symmetrical and asymmetrical double layers in electrolyte solutions of arbitrary composition and doesn't have any assumptions on the relationship between local values of volume charge density to potential. This relationship depends on conditions of electrochemical equilibrium for all the charged dissolved components in the layer and is a complicated problem.

Another derivation of <math>\Pi_e</math> may be found by using hydrostatics and electrostatics. Without going into too much detail we first start with the equality which holds in equilibrium electrolyte layers:

<math> \nabla p + \rho \nabla \psi + {1 \over 8\pi} \nabla \left (E^2 \frac{\partial\epsilon}{\partial\rho_l}\rho_l \right) = 0 </math>

As it is seen, this is a balance of the hydrostatic pressure, electric force acting on the volume charge and the last term which is electrostriction. Writing this in differential form, then noting that in the bulk of solution <math>p\equiv p_o</math>, <math>\psi\equiv 0</math> and <math>E\equiv 0</math> and in the interlayer <math>p\not\equiv p_o</math>, <math>\psi\not\equiv 0</math> and <math>E\not\equiv 0</math> we can get the following equation:

<math> p_o = p(x,h) + \int_{0}^{\psi_{x,h}}\rho d\psi + \frac{1}{8\pi}E^2(x,h)\frac{\partial\epsilon}{\partial\rho_l}\rho_l</math>

Substituting the Poisson equation into the equilibrium equation we see that the pressure normal to the surfaces in any transverse cross section of the film are:

<math>p_1 \equiv p(x,h) - \frac{\epsilon E^2(x,h)}{8\pi}+\frac{E^2(x,h)}{8\pi}\frac{\partial\epsilon}{\partial\rho_l}\rho_l</math>

which is constant, only a function of h independent of x. This is compose of hydrostatic pressure, Maxwell stress, and striction pressure. The difference between <math>p_1(h)</math> and <math>p_o</math> is exactly the disjoining pressure so by subtracting the last two equations from each other, you get the equation derived before with the electrostriction term gone. This result shows that the effect of electrostriction is just that hydrodynamic pressure will be a function of local electrostatic potential and local field. Integration of <math>\Pi_e</math> over a distance h between the interfaces of diffuse layers makes it possible to obtain an important characteristic of overlapping diffuse atmospheres: their Gibbs free energy of interaction, <math> V_e \equiv \Delta G_e</math>.

Another method of calculating <math>\Pi_e</math> is based on calculating the free energy of overlapping electrical double layers. The energy of interaction Ve is found as the difference between the sums of the free energies of the overlapping and far away double layers, so the disjoining pressure is found by differentiating Ve with respect to distance. Calculating free energies is difficult computationally and in fundamental principles (as seen briefly in my other wiki The Free Energy of Charged Particles in an Electrolyte Solution) but can be seen in detail in references at the end of this wiki. The general stat-mech derivation of the integral relations for the free energy of double layers is given by Ikeda who analyzed two different boundary conditions for overlapping boundary layers: constant potential and constant surface charge density. Each of these conditions gives a distinct integral expression for free energy, which is a shortcoming of this method of calculating the interaction between double layers.

## Interaction Between Identical Layers. Boundary Conditions

If overlapping diffuse layers are identical, the system will have a symmetry plane at x = d = h/2 where <math>\psi = \psi_d</math> and E = 0. Now our key equation for disjoining pressure takes the form:

<math>\Pi_e(\psi_d) = -\int_{0}^{\psi_d}\rho d\psi > 0 </math>

where this formula allows us to derive the function <math>\Pi_e(h)</math> after dependence of <math>\rho</math> on <math>\psi</math> has been established and boundary conditions formulated. Using the Guoy-Chapman approach for volume charge density of a binary asymmetric electrolyte and integrating the let equation you get the isotherm

<math> \Pi_e(\phi_d)= \frac{1}{2}(z_1n_1 + z_2n_2)\theta f(\phi_d)</math>

where z1 and z2 are the valences of the counterions and secondary ions, n1 and n2 are their volume concentrations, <math>\phi_d = e\psi_d/\theta</math> and <math>\theta = kT</math>. The function <math>f(\theta)</math> depends on the value of potential and the valences of the ions:

<math> f(\theta) = \frac{1}{z_1}e^{z_1\phi} + \frac{1}{z_2}e^{-z_2\phi} - \frac{1}{z_1} - \frac{1}{z_2}</math>

and f(0) = 0. Using the last equation, for identical layers in a symmetric electrolyte, our equation becomes

<math> \Pi_e = 2n\theta(cosh (n\phi_d) - 1)</math>

This equation was shown by Langmuir when he considered the excess osmotic pressure of ions in the symmetry plane of an electrolyte film where E=0. But in strong electrolytes the osmotic pressure is given as

<math> P_{osmotic}=\theta(n_1 + n_2) - \frac{e^2\kappa}{6\epsilon}(z_{1}^{2}n_1 + z_{2}^{2}n_2)</math>

Really the second to last equation may be derived from the key equation given before if we neglect the second term in the last equation in comparison to the first term. But, **the relationship between disjoining pressure and osmotic pressure is in fact an arbitrary assumption that cannot be proved!** The practical important result is not <math>\Pi_e(\psi_d)</math> but <math>\Pi_e</math> as a function of h. This is not invariant w.r.t. boundary conditions which reflect the features of the charging mechanism. This mechanism determines, by electroneutrality condition, the changes in potential <math>\psi_1</math> of the interfaces of these layers. In a planar gap the electroneutrality condition is

<math> \sigma_s(\phi_1) + \sigma_d(\phi_1,\phi_d) = 0</math>

where <math>\sigma_s</math> is the surface charge density including that of the Stern monolayer and is determined by the equation that corresponds to a specific mechanism of surface charging, and <math>\sigma_d)</math> is the density of the diffuse charge between surface and symmetry plane and is found by the relation representing the first integral of the Poisson-Boltzmann equation. This integral is:

<math>\frac{d\phi}{dx} = -\kappa \sqrt{\frac{2}{z_1+z_2} [f(\phi)-f(\phi_d)]}</math>

where <math>1/\kappa</math> is the Debye screening length. The surface charge of the diffuse atmosphere is then:

<math> \sigma_d = -\frac{\epsilon}{4\pi}\frac{d\psi}{dx} \bigg|_{x=0} = \frac{\epsilon\theta\kappa}{4\pi e}\sqrt{\frac{2}{z_1+z_2} [f(\phi)-f(\phi_d)]}</math>

Finally, to get <math>\Pi_e(h)</math> we need to relate <math>\phi_1</math> and <math>\phi_d</math> to the distance h. This is done with the second integral of the Poisson-Boltzmann equation:

<math> \frac{\kappa h}{2} = \sqrt{ \frac{z_1+z_2}{2}} \int_{\phi_d}^{\phi_1}d\phi/\sqrt{f(\phi)-f(\phi_d) }</math>

If <math> \sigma_s = \sigma_s(\phi_1)</math> is known then using all these equations, you can define <math>\Pi_e(h)</math> for a given composition of the solution.

## Interaction at Constant Charge or at Constant Surface Potential

It is seen that the <math>\Pi_e</math> curves are different for systems which have the same potential of the isolated double layer <math>\psi_\infty = \psi_1(h\rightarrow \infty)</math> but differ in the mechanism of charging the interface. One factor may be used to see how large these differences can be. All possible curves fall within a region formed by the <math>\Pi_e</math> curves calculated for constant potential of diffuse layer interfaces and for constant charge density where repulsion is minimum in the first case but maximum in the latter. These two cases allow for the parameterization of the problem for symmetrical and asymmetrical electrolytes. Introducing new variables:

<math> \upsilon = arcsin(1/cosh(z\phi_d/2))</math>

<math>\varphi = arccos(sinh(z\phi_d/2)/sinh(z\phi_1/2))</math>

for symmetrical electrolytes transforms the last equation for disjoining pressure and the second integral o the Poisson-Boltzmann equation into the following forms:

<math> \Pi_e/4nkT = cot^2(\upsilon)</math>

<math>\kappa h = 2 sin(\upsilon) \cdot F(\varphi,\upsilon)</math>

And the relationship between <math>\varphi</math> and <math>\upsilon</math> is determined by the boundary conditions. For example, if <math>\phi_1</math> = const,

<math> cos(\varphi) = cot(\upsilon) / sinh(z\phi_1/2)</math>

and if <math>\sigma</math> = const,

<math> tan(\varphi) = tan(\upsilon)\cdot sinh(z\phi_\infty /2)</math>

The function F above is an elliptic integral of the first kind with modulus <math>sin\upsilon</math> and amplitude <math>\varphi</math>. Figure 6.2 shows some calculated results of <math>\Pi_e(\kappa h)</math> to show the difference between the two different boundary conditions. It is seen that this difference gets especially large at small values of h between the surfaces. It is noticed that for constant potential the disjoining pressure tends to a finite limit or

<math> \lim_{h \to 0} \Pi_e^{\psi} = 4nkTsinh^2(\phi_1/2)</math>

When h gets smaller at constant surface charge density on the other hand, it results in not only increasing the potential <math>\phi_d</math> but also theoretically unlimited growth in <math>\phi_1</math>. As a result, <math>\upsilon</math> becomes very small, <math>\varphi</math> goes to <math>\pi</math>/2 and so does the elliptic integral F therefore <math>\Pi_e/4nkT \approx 1/\upsilon^2</math> and <math>\kappa h = \pi\upsilon</math>. When <math>\upsilon</math> is eliminated the relation holds:

<math> \Pi_e/4nkT \approx \pi^2/(\kappa h)^2</math>

and replacing the reciprocal Debye length, we get the well known Langmuir formula:

<math> \Pi_e \approx \pi\epsilon (kT)^2/2z^2e^2h^2</math>

If surface charge density is constant, this formula holds at least for h << <math>1/\kappa</math>. It means that the disjoining pressure increases indefinitely when the interlayer thins out, being independent of the potential <math>\phi_\infty = \phi_1(\infty)</math> of the double layers prior to the onset of interaction… actually as h goes to 0 the dependence of <math>\Pi_e</math> on distance is modified to <math>\Pi_e\sim\sigma/h</math>.