Thermodynamically Stable Emulsions Using Janus Dumbbells as Colloid Surfactants

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1 Thermodynamically Stable Emulsions Using Janus Dumbbells as Colloid Surfactants Fuquan Tu, Bum Jun Park and Daeyeon Lee*. Description of the term notionally swollen droplets When particles are adsorbed onto the interface of an emulsion droplet, particles will displace a small portion of the oil phase which leads to the swelling of the droplet as shown in Fig. S1. droplet with radius of R becomes R after the adsorption of Janus dumbbells. notionally swollen droplet can be regarded as a droplet with a cavity as shown in Fig. S1 (c) such that the droplet has a radius of R but still maintains a volume of oil phase that is equal to the droplet shown in Fig. S1 (a) and (b). This approach simplifies the calculation of free energy of droplet formation dg ow ow np ag where ow =4 R because we do not have to consider the adsorption of individual Janus dumbbells one at a time; instead, we can consider the attachment of n p Janus dumbbells at a time for the calculation of G d. The extra interface created by the cavity will eventually disappear when particles are adsorbed to the interface therefore there is no need to consider the free energy for creating this interface in the free energy of droplet formation. Figure S1. (a) bare oil droplet with a radius of R. (b) The droplet in (a) covered with close-packed Janus dumbbells with a slightly larger radius R. (c) Notionally swollen droplet that has a radius of R but has the same volume of oil phase as that in (a) and (b). S1

2 B. Effect of line tension of the free energy of emulsion formation Theoretically, line tension () is estimated to be ~ O(1-11 N). 1 The contribution of line tension to the free energy of creating a unit area of a hexagonal element for a symmetric Janus dumbbell of aspect ratio of 1.5 is / R / R / R expression gives a value of P P P. When R P = 1 nm and T = 9 K, the above k B T/m. s shown in the main text, the magnitude of free energy of creating a unit area of a hexagonal element is ~ 1 19 k B T/m, which is at least two orders of magnitude larger than the contribution due to line tension; thus, the effect of line tension on the free energy of emulsion formation is also negligible. C. Effect of S on G a d The adsorption entropy of particles ( the continuous phase ( demix S a S ) can be estimated by the entropy of demixing of particles from ). The entropy change due to demixing np dumbbells can be estimated using 4 nptdemixs npkt ln (S1) where is the volume fraction of np dumbbells in the continuous phase. s shown in Fig. S, G is d plotted as a function of the radius of oil droplet both with and without (note that S a is a part of a S G and a d ow ow p a G n G ). In this calculation, the continuous water phase is.5 m in volume and the radius of oil droplet ranges from 5 to 5 μm. The Janus dumbbells have an aspect ratio of 1.5 and the amphiphilicity of 4. s can be seen in Figure S, There is no observable difference whether S is considered or not. It is found that TdemixS a G a is well below 1-4 ; therefore S a is much smaller than G a thus negligible. S

3 Figure S. The free energy of forming a single droplet coated with Janus dumbbells without (black square) and with (red circle) the adsorption entropy of dumbbells. The continuous phase is an aqueous phase of.5 m in volume. The Janus dumbbells have an aspect ratio of 1.5 and amphiphilicity of 4. D. Consideration of colloidal interactions between dumbbells. We compare the magnitudes of potential energies associated with electrostatic interactions, van der Waals interactions and hydration interactions to the free energy of creating a hexagonal element illustrated in Fig. 1(b) of the main text. We incorporate the three interactions mentioned above using the following approach. The free energy for creating a hexagonal element while considering particle-particle interactions can be IN expressed as G G+ U + U + U as described in the main context. Notice that U ele hex hex ele vdw hyd = sphsph, U = vdw V vdw and Uhyd =V hyd where sphsph, V vdw and V hyd are pair potentials for electrostatic, van der Waals and hydration interactions, respectively. The factor comes in because every dumbbell has six nearest neighbors. s shown in Fig., the magnitude of G/ is hex hex k B T/m. Since the area of a hexagon is hex R where R 1 nm, the free energy for P P G G /.51 creating the hexagon excluding particle-particle interactions is 5 k B T. hex hex hex hex S

4 a. Electric double layer repulsion There could be electrostatic double layer repulsion between the two hydrophilic lobes. The potential energy can be calculated by considering the potential energy of electrostatic repulsion between two equal-sized spheres. The formula is approximation, where 1 64an kbt sphsph exp( S ) following the Derjaguin z e n ze is the Debye length and. tanh( ) and is Stern kt kt potential. a is the radius of the sphere. n is the number density of ions in the bulk solution. S is the shortest distance between these two spheres. z is the number of charge. 4 Thus, the pair potential due to electrostatic interactions can be represented by sphsph k T B B 4 B a exp( S ) kbt. For z 1 z e sphsph a mv (highly charged), z 1and S, kt 11 B with water 8 and T 9 K. k T e B Each hexagonal element will be connected with six other hexagonal elements thus one dumbbell will have six nearest neighbors. Therefore, if electrostatic interaction is included, every hexagonal element created will result in another sphsph energy penalty (because sphsph is pair potential energy). We see k B T is much less than 1% of G / that 66 sphsph. hex hex hex The calculation above most likely overestimates the electrostatic repulsion. First, stern potential larger than 1mV is not very common. Usually, a smaller potential is expected. For example, Levine et al. estimated the double layer interaction between silica particles of.86 μm in radius that are close to each other using the surface potential of 5 mv and 5 mv. The estimated values of sphsph for these particles were 65.5 k B T and k B T, respectively. 5 Moreover, in the case of dumbbells, the polar lobes are not completely spherical, thus the electrostatic potential is expected to be smaller than the spherical case. To consider the electrostatic interaction beyond the nearest neighbors, we need to consider the Debye length 1. Debye length can be calculated using z e n z e N VC where N V k T k T B B is the vogadro's number and C is the concentration of ions in bulk solution. ssuming C 4 1 M.1 S4

5 mol/m, Debye length in this case is kt nm, which is much smaller than the 1 B z e N VC radius of dumbbell lobes ( R 1 nm). Therefore, electrostatic double layer interactions beyond the nearest neighbors can be ignored. b. Dipole-dipole interaction It has also been shown that at short separation ( r 1 ) where r is the center to center separation distance between two interfacial particles, the potential energy due to electrostatic interaction is dominated by electric double layer interaction. 5, 6 Therefore dipole-dipole interaction is ignored. c. Van der Waals interaction Following the idea of Levine et al. 7, the potential energy due to Van der Waals interaction can be approximated by the sum of two terms: attractive interactions between two spherical particles in the aqueous phase and oil phases, respectively. Thus the pair potential for van der Waals interaction can be pwpa pdpa calculation using V vdw 1S 1S where pwp is the Hamaker constant for two polystyrene surfaces in water, and pdp is the Hamaker constant for two polystyrene surfaces in decane. S is the nearest separation between two surfaces. We estimate pdp using pdp d p (where d.97 1 J (decane) and and p 6.51 J. 8 ). The two Hamaker constants are found to be pwp 1.41 J and pdp.11 J. If we use.1 S nm, V vdw 5 k B T which is negligible G /.5 1 k B T. compared to 5 hex hex hex d. Hydration force When two hydrophilic surfaces are very close to each other (distance in a few nm range), there will be a strong repulsive hydration force that decays exponentially. 9 The hydration potential energy is expressed as V rp exp D / hyd o where is a decay length, o closest separation of particle surfaces. Using.6 nm and P is a hydration force constant and D is the 6 P 1 Nm -, 1 which are reasonable estimates for particles with β = 4, V hyd 51 k B T (D = and r = 1 nm) which is negligible G /.5 1 k B T. compared to 5 hex hex hex o S5

6 E. lgorithm to calculate droplet radius covered by Janus dumbbells with hexagonally close- packed structure The radius of droplet without any Janus dumbbell adsorbed at the surface is R. We will consider a case when R is 1 μm. The radius of droplet fully covered with Janus dumbbells is R. With Janus dumbbells of known aspect ratio ( Rd 1.5 ) and lobe radius ( R RP 1 nm), we need to calculate R as well as how many Janus dumbbells i.e. n p are needed to fully cover the droplet surface. We develop an iteration algorithm to enable this calculation. First, we assign an initial guess for R for example, R μm. s shown in Fig. S, 1 where R H H R cos cos 1 d, when the radius of droplet is much larger than the size of Janus dumbbells ( R 5R ). Thus the radius of curvature to the point where dumbbells contact with each other is Rpc R H as shown in Fig. S(a). Because Janus dumbbells form hexagonally closepacked structure, the number of Janus dumbbells can be determined as n p 4 Rpc R pc R R where.969 is the packing fraction for D hexagonally close-packed monolayer. Then the radius of droplet without any Janus dumbbell on the surface under the initial guess R μm is calculated as R R np VR VR VR " 4 R cos cos as shown in Fig. S(b). 1 and where V R cos cos R " R and R are compared with each other to obtain R R R " dif. If R dif 1 4 nm, we consider " R R such that R is the actual radius of droplet after a bare droplet with radius R is fully covered with Janus dumbbells otherwise R is updated to a new value satisfied. R R Rdif until R dif 1 4 nm is For R 1μm, R 1.5 and R R 1 nm, the calculated values for n p and R are.67x1 4 and 1.1 μm, respectively. d P S6

7 Figure S. (a) Schematic illustration of parameters and quantities used in the iteration algorithm. (b) Janus dumbbell adsorbed on a droplet of comparable size. The definitions of H, V R and V R are illustrated. F. Effect of curvature on the equilibrium position of Janus dumbbells When the radius of a droplet is comparable to that of a Janus dumbbell, the equilibrium configuration of the dumbbell may change due to the high internal pressure within the droplet. The equilibrium position of a Janus dumbbell on a curved interface is characterized by h, which is the displacement from the neck to the intersecting plane of dumbbells and oil droplet interface as shown in Fig. S4 (a). Note that h indicates that the interface is pinned at the neck (i.e., Janus boundary) of a Janus dumbbell. The equilibrium configuration of a Janus dumbbell is obtained by minimizing the adsorption energy of the dumbbell to the droplet interface. The equilibrium position is plotted as a function of the radius of the droplet as shown in Fig. S4(b). For a constant aspect ratio, the minimum droplet radius at which the interface stays pinned at the neck of a Janus dumbbell decreases as the extent of amphiphilicity (β) increases. The minimum droplet radius is around 5 nm for 9. Therefore with the highest amphiphilicity (i.e., β = 9 ), the assumption that a symmetric Janus dumbbell would maintain its upright configuration with its neck pinned at the interface is valid as long as the droplet radius is larger than 5 nm. S7

8 Figure S4. (a) Schematic illustration of the definition of h. (b) Equilibrium h normalized by the radius of a lobe ( R ) as a function of the droplet radius. G. Geometric consideration when the radius of emulsions droplets is small ( R 5R ) When R 5R, the assumption of hexagonal close-packing of Janus dumbbells at the droplet surface is no longer reasonable. To consider such a case, we rely on the tabulated values of the densest packing of circles on a spherical surface. 11 Two parameters and n p are used to define one specific packing. n p is the number of circles on the spherical surface and is the corresponding angular diameter of a circle. We implement a method to utilize these tabulated data to determine the densest packing of dumbbells on a droplet surface. To enable this, we first define several variables as shown in Fig. S5; R is the radius of a lobe and is related to the aspect ratio of the symmetric dumbbell so that cos 1 Rd where R d is the aspect ratio of a Janus dumbbell defined in the main text. The aim is to use to calculated R so that we can calculate the free energy of droplet formation G n G with corresponding n p where the area of droplet interface is d ow ow p a ow 4 R. Based on a geometric consideration, we find that R / R1 sin( / ) and R R R R R cos which 1 1 can be used to calculate R. Thus the free energy of droplet formation can be obtained. S8

9 Figure S5. Schematic illustration of the geometric consideration for utilizing tabulated data on the densest packing of circles on a spherical surface. The green circle represents the actual oil droplet. The dotted circle connects the centers of all the lobes. The black solid-line circle connects contact points of adjacent dumbbells. H. Equilibrium configuration of asymmetric Janus dumbbells The equilibrium configuration of asymmetric Janus dumbbells are obtained by calculating the adsorption energy as a function of the orientation angle. 1, 1 s shown in Fig. S6, the minimum adsorption energy is found at, which corresponds to the upright equilibrium configuration. We can see that the Janus dumbbell still has an upright equilibrium configuration when the asymmetry ( R / R ) is 1.. P S9

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