Supporting Information
J. Marguerite Hughes, Damian Aherne and Jonathan N Coleman*
School of Physics and CRANN, Trinity College Dublin, Dublin 2, Ireland.
*colemaj@tcd.ie
S1 Testing the accuracy of the lattice model
The lattice model is the simplest possible way to calculate the energetics of liquids.
However, its simplicity is only an advantage if it is reasonably accurate. It has a number of elements which might be considered too simple. For example, the idea of placing molecules
on lattice sites was disliked by Hildebrand
1,2
as he considered it a poor reflection of reality. In
addition, lattice models consider only nearest neighbour interactions which may result in loss of precision. We can assess the accuracy of a lattice model by comparing its predictions to those obtained using a slightly more sophisticated method of calculation. Here we calculate the cohesive energy density for a liquid using a lattice model and compare it to the result found by integrating over the interaction between all pairs of molecules.
The lattice model
In a lattice model, the cohesive energy density can be found from the sum of the binding energies between all pairs of nearest neighbour lattice sites divided by the volume of the system:
E
C

Nz
S

2
1
Nv
S
 z
S

2 v
S
(S1)
Here N is the total number of lattice sites (molecules), z is the number of nearest
S neighbours per liquid lattice site (=6 in a cubic lattice model),

S
is the volume per lattice site (molecular volume) and

is the intersite binding energy. This binding energy can be thought of as the energy minimum of the pairwise intermolecular potential energy function
(see below).
Integration over all atom pairs
We can calculate the cohesive energy using a method first described by Hildebrand.
2
We will study a system of N molecules in a liquid of volume V . Consider a reference molecule. The potential energy of this molecule is the sum of the potential energies due to all
1

of the pairwise interactions with all other atoms in the liquid. We will assume the molecules are small and we can model them as hard spheres. The number of other molecules with centres in a spherical shell with inner radius r and outer radius r dr is dN

4

N r dr ( )
V
(S2) where ( ) is the radial distribution function. This function approaches 1 for large r but deviates from 1 at small r to describe the local structure of the liquid near the reference molecule. We can work out the potential energy of the reference molecule by multiplying dN by the potential energy of interaction,

, of the reference molecule with each molecule between r and r dr and integrating over all space (all atoms). Furthermore, we can extend this to the potential energy of all atoms by multiplying by N / 2 where the factor of 2 is to avoid double counting. Dividing by V gives the total interaction energy per volume i.e. the cohesive energy density:
E
C
 
 dN
 N
2 V
(S3)
Here the lower limit of integration is d , the molecular diameter. This is because, if we treat the molecules as hard spheres, the shortest possible distance between the centre of any molecule and the centre of the reference molecule is d .
We can approximate the intermolecular potential as the van der Waals potential
(ignoring the repulsive part):
  d
6
(S4)
In this scenario, for two nearest neighbour molecules (separated by a molecular diameter, d ), the potential is
 
, i.e. the intermolecular separation is defined by the bottom of the potential well.
This means we can write the cohesive energy density as
E
C
  


 
6
4

N r dr ( )
V
N
2 V
 
2
 d 6

N
 2

V


 d r
4
(S5)
We can make the approximation that the liquid volume is given by V

N
4
3

 
3
leading to
E
C
 
72



 d r
4
(S6)
2

For simplicity we can approximate ( )

1 , and approximating v
S

4
 d
( )
3 2
3 we get
E
C
 
4 v
S

(S7) which is within a numerical factor (of order 1) of the lattice model result in equation S1.
This result clearly shows that while the lattice model is crude, it gives results that are very close to those found using more sophisticated models. Thus we believe that its simplicity is a significant advantage and more than justifies its use here.
S2 Applying Hildebrand parameters to 2-dimensional platelets
We can use the lattice model to calculate the enthalpy of mixing of platelets in a solvent. We consider the initial state of the system as N
2D
identical platelets stacked to form a single crystal and a large volume of solvent. We assume N
2D
is very large, and that the platelets are big enough that the interactions between edges and solvent can be ignored. We consider each platelet as divided into n
× n
×1 sites arranged in a cubic lattice. Thus the total number of platelet lattice sites is
2 n N
2 D
. Within the crystal, each platelet has z
2 D
=2 neighbours. The solvent, meanwhile, consists of N
S
molecules arranged in a cubic lattice of
N
S
sites.
The final state is a uniform mixture of the two components consisting of
N
 2 n N
2 D

N
S
lattice sites in a cubic arrangement. The overall volume of the mixture is expressed as
 
(
2
V v N v n N
S S 2 D

N
S
) where v is the lattice unit (or solvent molecule)
S volume. From this, we can also see that the volume fraction of platelets in the mixture is given by

2 D

S
2 v n N
2 D
V
, and the volume fraction of solvent by
 
S v N
S
V
.
The enthalpy of mixing is defined as the enthalpy change going from the initial state of the system to the mixed state. We can compute

H
Mix
as the sum of 4 distinct terms:

H
Mix

H

H f (2 D )

H

H i (2 D )
(S8) each of which we will consider in turn.
H i (2 D )
represents the energy required to separate all of the platelets to infinite separation and can be thought of as the cohesive energy of the crystal:
H i (2 D )
 n N z
2 D 2 D

2 D

2 D
2
(S9)
3

where the factor of 2 is to avoid double counting, z
2 D
=2 is the number of van der Waals bonded nearest neighbours per platelet lattice site and

2 D

2 D
is the binding energy at the interface between 2 platelet lattice sites. We take

2 D

2 D
(and all other intersite energies) to be positive.
H represents the energy required to separate all of the solvent molecules to infinite separation and can be thought of as the cohesive energy of the solvent:
H

N z
S S
2
 where z =6 is the number of nearest solvent neighbours per lattice site,
S

(S10)
is the binding energy at the interface between 2 solvent lattice sites and the factor of 2 accounts for double counting.
Next we consider the energy released on condensation of solvent and platelets from infinity to form the final mixture of platelets within the lattice. We split this into two terms relating to platelets ( H f (2 D )
) and solvent molecules ( H ). To calculate the first term, we consider that each platelet will be surrounded by a combination of other platelets and solvent molecules, the average proportions of which are determined by their respective volume fractions

and
2 D

, where
S

2 D
 
S

1 :
H f (2 )
  n N z
2 D 2 D
2
(

2 D

2 D

2 D
 
S

2

) (S11)
Similarly, the environment experienced by an average solvent molecule is also dependent on both

and
S

. However, in this case there is an additional constraint. Whilst
2 D the number of nearest neighbours any solvent molecule has is z
S
(=6), only a maximum of z
2 D
(=2) of those may ever be occupied by platelets, due to the platelet dimensions. This is illustrated in the schematic S1.
In other words, ( z
S
 z
2 D
) neighbouring sites will always be occupied by solvent molecules, and z
2 D
neighbouring sites by a combination of platelets and solvent molecules
(the average proportions of which are given by

and
S

). Hence we can write
2 D
H
 
N ( z
S S
 z
2 D
)

2

N z
S 2 D
2
(
 
S
 
S

2 D

2 D
) (S12)
Summing all of these terms gives us
4


H mix

N z
S S
2
 

N
S
( z
S
 z
2 D
)
2
 
N z
S 2 D
2
(
    
S S S S

2 D 2 D
)

2 n N z
2 D 2 D
2
(

2 D

2 D

2 D
 
S

2

D S
)

2 n N z
2 D 2 D

2 D

2 D
2 which rearranges to

H mix

N z
S 2 D (

2
  
S
 
S

2 D

2 D
)

2 n N z
2 D 2 D
2
(

2 D

2 D

2 D
 
S

2
  
2 D

2 D
) and may be further simplified to

H mix

N z
2 D (

(1
 
S
)
 
S

2 D

2 D
)

2 n N z
2 D 2 D
2 2
(

2 D

2 D
(1
 
2 D
)
 
S

2

D S
)
Letting
 
2 D
(1

S
) yields

H mix

N z
2 D
2
 
S

2 D
)

2 n N z
2 D 2 D
2
(1
  
2 D

2 D
 
S

2 D
)
Finally, if we remember that
 
2 D

S
V
2 D , and 1
 
S
 v N
S
V
, we find that

H mix
 z
2 D
2
V v
S
(1
     
S

2 D
)
 z
2 D
2
V v
S
   
2 D

2 D
 
S

2 D
)
This means the enthalpy of mixing per unit volume is given by:



H
Mix
V
 

2 D z
2 D

(1
 
2 v
S
 
2 D

2 D

2

S

2 D

(S13)
We note that this is very similar to the standard expression for enthalpy of mixing for small molecule mixtures:

H
Mix
V
 z
0 D
2 v
S

(1
    
2
 
(S14)
We can express equation S13 in terms of Hildebrand solubility parameters by noting that this parameter is defined as the square root of the cohesive energy density. This allows us to write the cohesive energy density of the solvent as
E
 z
S
2
 v
S
 
2
(S15) where the factor of 2 is to avoid double counting. The cohesive energy of the initial platelet crystal can be written as
E
C ,2 D

2 z n N
2 D 2 D

2 D

2 D
2
2 D S
 z
2 D
2

2 D

2 D v
S
(S16)
5

We can use this expression to define the Hildebrand solubility parameter of platelets to be
E
C ,2 D
 z
2 D
2

2 D

2 D v
S
 z
2 D z
S
 2
T ,2 D
(S17)
We note that this new definition is allowed because the significant difference in geometry between a platelet and a solvent molecule which occupies a single lattice site makes using the same definition in each case inappropriate.
If we assume that the interactions between all lattice sites are dominated by the
London interaction, we can use the geometric mean approximation (see below) to estimate

S

2 D
:

S

2 D
  
2 D

2 D

2 v
S z
S
 
T ,2 D
(S18)
Inserting these expressions into equation S14 gives

H
Mix
V
 

2 D z
2 D

(1


 
T ,2 D

2 z
S

1
3

(1


 
T ,2 D

2
(S19)
This is very similar to the standard expression

H
Mix
V
 
(1


 

2
.
(S20) for solute N .
S3 Applying Hildebrand parameters to 1-dimensional rods
We can follow a similar procedure to derive the enthalpy of mixing for rigid rods in a solvent.
We consider the initial state of the system as N
1 D
identical rods cubic close packed to form a single crystal and a separate large volume of solvent. We assume N
1 D
is very large.
We consider each rod as divided into n
×1×1 sites arranged linearly. Thus the total number of platelet lattice sites is nN
1 D
. Within the crystal, each rod has z
1 D
(=4) nearest neighbours. The solvent consists of N
S
molecules arranged in a cubic lattice of N
S
sites. The final state is a uniform mixture of the two components consisting of nN
1 D

N
S
lattice sites in a cubic arrangement.
As before, we can divide

H
Mix
into 4 distinct terms, using equation S8:

H
Mix

H

H f (1 )

H

H i (1 )
6

H i (1 )
represents the energy required to separate all of the rods to infinite separation and can be thought of as the cohesive energy of the crystal:
H i (2 D )
 nN z
1 D 1 D

1 D

1 D
2
(S21) where

1 D

1 D
is the binding energy at the interface between two rod lattice sites and the 2 is to avoid double counting.
As before, H represents the energy required to separate all of the solvent molecules to infinite separation and can be thought of as the cohesive energy of the solvent:
H

N z
S
2
 where z (=6) is the number of nearest neighbours per solvent lattice site,
S

(S10)
is the binding energy at the interface between two solvent lattice sites and the factor of 2 accounts for double counting.
The energy released going from the state of infinite separation to the final state of rods and solvent mixed within the lattice is expressed the same way as for the platelets, with volume fractions

1 D
and

S
in this instance:
H f (1 )
  nN z
1 D 1 D
2
(

1 D

1 D

1 D
 
S

1

) (S22) and
H
 
N
S
( z
S
 z
1 D
)

2

N z
S 1 D
2
(
 
S
 
S

1 D

1 D
) . (S23)
Similarly to the platelet case, only a maximum of z
1 D
(=4) of a solvent molecule’s nearest neighbour sites may ever be occupied by rods, due to the rod length, as shown in schematic
S2.
Summing as before, we find the enthalpy of mixing of rods and solvent molecules to be

H
Mix
V
 

1 D z
1 D
2 v
S

(1
   
1 D

1 D

2

S

1 D

Defining the Hildebrand solubility parameter of rods to be
E
 z
1 D
2

1 D

1 D v
S
 z
1 D
 2 z
S
(S24)
(S25)
7

and assuming that the interactions between all lattice sites are dominated by the London interaction, we can use the geometric mean approximation (see below) to estimate

S

1 D
:

S

1 D
  
1 D

1 D

2 v
S z
S
 
1 D
(S26)
Inserting these expressions into equation S24 gives



H
Mix
V
 

1 D z
1 D

(1


  z
S

2 
2
3

(1


 

2
(S27)
This is again very similar to the standard expression

H
Mix
V
 
(1
 
S
 
N

2
. (S20)
Hence, for any nanomaterial N with dimension d

3 , we can extract two general definitions:


H
Mix
V
 d


1 d
3 

(1


 

2
(S28) and
E
 z
N
2
 v
S


1 d
3


2 (S29)
Figure S1: A) Initial platelet crystal; each platelet has z
2 D

2 nearest neighbours B) Initial solvent crystal ( z
S

6 nearest neighbours) C) A platelet in the solvent-platelet mixture may be entirely surrounded by solvent molecules D) A solvent molecule in the final mixture must always have z
S
 z
2 D

4 solvent nearest neighbours
8

Figure S2: A) and B) Initial rod and solvent crystals C) A rod may be surrounded by solvent molecules on z
1 D

4 sides in a mixture D) A solvent molecule must always have z
S
 z
1 D

2 nearest neighbour solvent molecules.
S4 Deriving Mixing Enthalpy from First Principles as a function of Dispersive and Polar
Solubility Parameters
Derivation of an expression for the enthalpy of mixing within a lattice model involves the calculation of the bracketed energetic term in equation S14. For simplicity we will label this as
    
2
 
To calculate

E , it is necessary to calculate

,

and

(S30)
. We will assume that each of these interaction energies is the sum of a dispersive and a dipolar term. For two small molecules (or lattice sites) A and B , we write this in the general form
    
where D and P represent dispersion and polar respectively.

For the dispersion interaction, the energy of interaction is given by 3
 
3
 
A B
2 (4

0
)
2
I I
1 2
1
I
1

I r
2 0
6
 k
1
 
B
(S31)
Here,

is the polarisability of a molecule of type A ,
A
I its ionization potential,
A r
0
the equilibrium molecular separation (lattice site size) and k is a constant.
1
Similarly, for two small molecules (or lattice sites) interacting by dipolar interactions, the energy of interaction is given by
  
2
 
2
3 (4
A B

0
)
2
1 1 kT r
0
6
 k
2
 
2
B
(S32)
9

where

is the dipole moment of a molecule of type A and
A k
2
is a constant. We can see that dispersive interactions are controlled by respective molecular polarisabilities, but dipoledipole interactions by the square of the dipole moment.
Inserting these expressions into equation S30 we get:

2 k
1
 
A B
 k
1

A
2  k
1

B
2 
2 k
2
 
2
A B
 k
2

A
4  k
2

B
4
 which may be simplified as:
  k
1
(

A
 
B
)
2  k
2
(

2
A
 
B
2 2
) (S33)
This is effectively a solubility parameter equation for

E , where the

A
and
 2
A terms may be thought of as solubility parameters in their own right. However, not only do they have different units from each other, but it is also desirable to recast them into new variables which have the same units as existing (Hildebrand or Hansen) solubility parameters, in order that comparisons may be easily made. Therefore, by combining equations S15 and
S33 we can write
3-5
 
2 v
S z
S

2  k
1

A
2 (S34) and
 
2 v
S z
S

2  k
2

A
4 (S35) for A, and similar for B. Here

and

are Hansen’s dispersive and polar parameters for material A and

is the lattice site (or solvent molecule) volume. Substituting these values
S into the equation for

changes it to
2

S z
S
 (
  
)
2 
(
  
)
2
(S36)
Converting this to enthalpy of mixing as described above and writing this specifically in the case of a nanomaterial-solvent system, this becomes



H
Mix
V
Mix

 d


1 d
3

(1
    
)
2 
(
  
)
2
(S37)
Comparing it to Hansen's expression for the enthalpy of mixing

H
Mix
V
 
(1
 
)



  

2 
1
4

  

2 
1
4

  

2


(S38) we see that the form is identical, except for the presence in Hansen's equation of an empirically obtained hydrogen - bonding term, and the factor of 1/4 in the P and H terms.
10

S5 The effect of dipole-induced dipole effects
Of course, this model is still quite limited; for example, it ignores at least one potentially important interaction: the dipole-induced dipole interaction.
These interactions may be described by an intermolecular ( i.e.
intersite) interaction energy:
3
  
1
(4
A B

0
)
2 6 r
0
 k
3 A B
(S41) where the various constants are defined as in the previous formulation. We see that such an interaction is controlled by the permanent dipole moment of one molecule and the molecular polarisability of the other, and that the full description of such an interaction must then include a term describing molecule A acting on B and molecule B acting on A .
Adding a term of type
  k
3
 
B
 k
3
 
A
into the expression for intersite interactions to represent the dipole-induced dipole interaction, the resulting

E is:
 
 k
1
(

A
 
B
)
2  k
2
(

2

A

B
)
 k
3

A
 
B
)(

2
A
 
B
2
)

(S42)
Defining

and

as before, it is evident that the dipole-induced dipole term can be written as the product of these terms. This yields
2

S z
S



(

  
)
2 
(
  
)
2 
2 k
3 k k
1 2
(
  
)(
  
Letting k
3 k k
1 2

A produces
2

S z
S
 (
  
)
2 
(
  
)
2 
A
  
)(
  
)


)
(S43)
(S44)
As before, we can use this to express the enthalpy of mixing for a solute of dimension d:




H
Mix
V
Mix


 d


1 d
3

(1
    
)
2 
(
  
)
2 
A
  
)(
  
)
(S45)
S6 Relating the dispersed concentration to enthalpy of mixing
11

We can use the expressions derived above to find the dispersed concentration of solute using recent work which has shown that for small molecules and rod-like solutes the maximum dispersed volume fraction is given by:
6
  exp 


 v
RT

H / V
Mix
 
 


(S46) where v is the volume per mole of the dispersed phase. Assuming we can model a platelet as a very low aspect ratio rod, we can apply this to 0D, 1D and 2D solutes. For simplicity, we will illustrate this first in the case where we consider only dispersive and dipole-dipole interactions. Thus, inserting S37 into S46, and making the approximation of low volume fraction ( 1 1 ), we obtain an expression for the maximum dispersed solute volume fraction:

N

 exp 1
 d

N
3

RT

(
  
)
2 
(
  
)
2
 

(S47)
This predicts that the dispersed volume fraction will behave as a two-dimensional
Gaussian function in

and

space, i.e. the product of two individual Gaussians of equal width , one a function of

and the other a function of

. We illustrate this for both rods and platelets in figures S3 and S4.
12

Figure S3: Calculated dispersed volume fraction for rods. (A) Dispersed volume fraction (in arbitrary units) in the case where only dispersive and dipole-dipole interactions are present
(equation 25). (B) A contour plot of the surface illustrated in A. (C) A contour plot plotted from equation 26 where dispersive, dipole-dipole and dipole-induced dipole interactions are present. Here, a nonphysical value of A = 0.8 is used in the dipole-induced dipole term. (D) A contour plot plotted from equation 26 with a realistic estimate of A = 0.1. Note its similarity to the case where only dispersive and dipolar interactions are considered (B). In all four graphs, d

1 ,

=50 L/mol (i.e. 50,000 g/mol assuming a density of 1000 kg/m
3
),
N

=18
MPa
1/2
and

=9.3 MPa
1/2
were used.
13

Figure S4 Calculated dispersed volume fraction for platelets. (A) Dispersed volume fraction
(in arbitrary units) in the case where only dispersive and dipole-dipole interactions are present (equation 25). (B) A contour plot of the surface illustrated in A. (C) A contour plot plotted from equation 26 where dispersive, dipole-dipole and dipole-induced dipole interactions are present. Here, a nonphysical value of A = 0.8 is used in the dipole-induced dipole term. (D) A contour plot plotted from equation 26 with a realistic estimate of A = 0.1.
Note its similarity to the case where only dispersive and dipolar interactions are considered
(B). In all four graphs, d

2 ,

=500 L/mol (i.e. 500,000 g/mol assuming a density of 1000
N kg/m 3 ),

=18 MPa 1/2 and

=9.3 MPa 1/2 were used.
14

We can follow the same procedure for the case where we also consider dipole induced dipoles. Using equations S45 and S46 the dispersed concentration is given by

N

 exp 1
 d

N
3

RT
 (
  
)
2 
(
  
)
2 
A
  
)(
  
)


(S48)
At first glance it would seem as though the last term, a product of the two different parameters, would significantly change the behavior compared to equation S47. In fact, this expression simply describes a specific case of a two-dimensional elliptical Gaussian function, i.e.
a Gaussian broadened in one dimension, narrowed in the other and rotated by a particular angle (in this instance

4
) in the
  
plane. The relative narrowing can be described by the ratio of widths: w narrow w broad

1

1

A
A
(S49)
We note that A is determined by a set of physical constants, allowing us to write:
A
 k
3 k k
1 2



3 1
2 (4

0
)
2


1 1
(4

0
)
2 6 r
0


I
1
I I
1 2

I r
1
2 0
6
 
2
3 (4
1

0
)
2
1 1 kT r
0
6



Approximating I
1

I
2

I gives kT
I
1

I
2
I I
1 2
(S50)
A

2 kT
I
(S51)
As all real molecules have I>2kT, this means A<1. A crude estimate of the magnitude of ionization potentials tells us that A

0.1
, giving w
Narrow

0.9
w
Broad
. This narrowing is very small, meaning that the incorporation of dipole-induced dipole forces hardly changes the dependence of concentration on

and

compared to that predicted by equation S47.
S7 Universally positive enthalpy of mixing
The sign of our general expression for the enthalpy of mixing states (S45) depends on the sign of

E. The expression for

E (S44) can be written as z
S

E
2

S

(
  
)
2 
(
  
)
2 
A
  
)(
  
) (S52) which can be abbreviated to
15

z
S

E
2

S

X
2 
Y
2 
2 AXY
Let us assume that

E<0
X
2 
Y
2 
2 AXY

0
(S53)
(S54)
This can only be true if either X or Y are negative (A is positive by definition). This expression can be rearranged to give
X

Y
 
2 A
Y X
Here both X/Y and Y/X must be negative. Thus
1
2


X

Y
Y X



A
(S55)
(S56)
The minimum value of the left hand side of this expression is 1 which means that the enthalpy of mixing can only be negative if A>1.
(1) Hildebrand, J. H. Annual Review of Physical Chemistry 1981 , 32 , 1.
(2) Hildebrand, J. H.; Wood, S. E. The Journal of Chemical Physics 1933 , 1 , 817.
(3) Israelachvili, J. Intermolecular and Surface Forces , Second Edition ed.;
Academic press, 1991.
(4) Miller-Chou, B. A.; Koenig, J. L. Progress in Polymer Science 2003 , 28 ,
1223.
(5) Hansen, C. M.; Skaarup, K. J. Paint Techn.
1967 , 39 , 511.
(6) Hughes, J. M.; Aherne, D.; Bergin, S. D.; Streich, P. V.; Hamilton, J. P.;
Coleman, J. N. Nanotechnology 2011 , submitted .
16