Fixed motor

3.3 Microstructural deformation by chemical reaction

3.3.1 Fixed motor

We first consider the fixed motor scenario as it is easiest to describe. Since the reaction only takes place at the motor surface (heterogenous reaction), the reactants and products diffuse in the fluid surrounding the motor with relative translational diffusivities D_R and D_P and number densities n_R and n_P, respectively. (In this case, the motor does not move, thus the relative diffusivities are just the diffusivities of the reactants and products.) In a frame fixed on the osmotic motor, each reactant and product particle moves diffusively.

Therefore, the probability flux for the reactant is

j_R=−D_R∇g_R. (3.5)

Conservation of reactant particles at steady state requires that∇ ·j_R= 0. The probability of finding a reactant particle relative to the motor,gR, satisfies a pair-level diffusion equation

∇²g_R= 0. (3.6)

To fully determine the pair-distribution function gR, Eq. (3.6) must be accompanied by appropriate boundary conditions. The reactant microstructure is undisturbed far from the osmotic motor particle, giving

gR∼1 as r → ∞. (3.7)

Collisions at the motor/reactant boundary give rise to the heterogeneous reaction, mathe- matically stated as

n· ∇g_R= R_R h(n)

n_RD_R at r =a+bR. (3.8)

The products satisfy a similar equation with the subscript R replaced by P. However, the osmotic force in (3.3) is proportional to the total probability density of bath particles g= (n_Rg_R+n_Pg_P)/n_T, wheren_T =n_R+n_P is the total number density of bath particles, which satisfies

∇²g= 0, (3.9)

g∼1 as r → ∞, (3.10)

n· ∇g= R_R h(n) n_RD_R

nR

n_T

1−sDR

D_P

at r=a+bR. (3.11)

The distribution of reactions on the surface is determined by the dimensionless function h(n), which we take to be 1 on the reactive half and 0 on the passive half. The model

can be easily extended for particles with different reactive surfaces (different forms forh(n) see Chapter 4). The nonuniform reaction causes an anisotropic environment of reactant particles around the motor. The surface reaction rate is modeled as a first-order reaction:

R_R =κn_Rg_R, whereκ is the speed of reaction. Defining the microstructural deformation functionsfR=gR−1 and f = (g−1)/(nR/nT(1−sDR/DP)) that represents the deviation from equilibrium for the reactant and the bath particle distributions, respectively, and all lengths nondimensionalized by the sum of the motor and reactant radii: a+b_R, it is easy to see thatfRand f satisfy the same Laplace’s equation and boundary conditions. For the total probability density of bath particles,

∇²f = 0, (3.12)

f ∼0 as r→ ∞, (3.13)

n· ∇f =Da h(n) (1 +f_R) at r= 1, (3.14)

which also describes the reactant distribution if f is substituted byf_R. We have defined a Damk¨ohler number

Da= κ(a+bR)

D_R , (3.15)

that describes the ratio of the speed of reaction κ to Brownian motion, DR/(a+bR).

In general, increasing Damk¨ohler number corresponds to driving the system away from equilibrium. Thus one need only to determine the reactant probability density to completely

solve the problem. The net osmotic force (3.3) becomes

F^osm = −n_RkT I

r=a+bR

ng_R(r)dS−n_PkT I

r=a+bP

ng_P(r)dS

= −n_RkT(a+b_R)²

1−sD_R DP

I

r=1

nf(r)dΩ, (3.16)

where dΩ is the solid angle of integration and the total microstructural deformation func- tion f depends on the Damk¨ohler number. For spherical reactants and products of dif- ferent radii, the integral should be over the “contact” surface at a+b_R and at a+b_P. However, this introduces a negligible error, especially in the limit a b_R,P. The net os- motic force is proportional to the thermal energy, kT, times the area of contact, (a+b)², times the jump in concentration of the reactant across the motor, n_RR

nf(r)dΩ, times the stoichiometric/diffusivity factor (1−sDR/DP), which tells how many product particles are produced per reactant, s, and how fast the products diffuse relative to the reactants, D_R/D_P. Thus, the osmotic force is larger for relatively large motor particles and scales asF^osm ∼nRkBT a²(1−sDR/DP). For bath particles that are much larger than the mo- tor, the available free space of self-propulsion is reduced, and consequently the entropic force of the bath particles is reduced. When bath particle collisions with the motor are uniform around the surface, as in the case of an inert or an uniformly reactive particle, the motor/bath boundary gives rise to the no-flux condition from the impenetrability of the motor particle and the Neumann boundary condition, respectively. In both cases, the microstructural deformation function f(r) is symmetric and uniform everywhere, and thus the osmotic force (3.16) reduces to zero.

The correctness of the factor (1−sD_R/D_P) can be appreciated by considering some special cases. If we had simple R → P (or s = 1) and the product had the same dif-

fusivity as the reactant, the net osmotic force would be zero. This would correspond, if you will, to simply changing the “color” of the reactant which cannot produce any net force. However, if the reactants and products have different diffusivities, say because the reaction changes the “shape” of the particle, or perhaps its interaction with the solvent (hydrophilic/hydrophobic), then even if only one product is produced for each reactant there will still be a net osmotic force on the motor. The sign of the force will depend on which diffuses faster. As another example, suppose that s = 2, but the products and reactants have the same diffusivity. There will be now a net increase of bath particles on the reactive side and the osmotic force will be to the left in Figure 3.1 rather than to the right. If we restrict ourselves to strictly hard-sphere particles of the same density, then D_R/D_P =b_P/b_R=s^−1/3 from conservation of mass, and thus the stoichiometric/diffusivity factor becomes (1−sD_R/D_P) = (1−s^2/3). Since one needs only to solve (3.14) in any situation, all cases can be treated at the same time and the precise details of stoichiometry and diffusivity, an be addressed later in the final scaling factor, (1−sD_R/D_P), for the net osmotic force. And, of course, Eq. (3.16) correctly gives zero osmotic force in the absence of any reactive particles: nR= 0.