Lectures 6&7 POLYMER BLENDS Entropy of Binary Mixing Consider a mixing of two species A and B VB VA These two species originally occupied volumes VA and VB. After mixing the total volume occupied by a mixture is equal to VA+VB. The mixture is uniform and two components are randomly mixed filling the entire volume. The volume fractions of two components in the binary mixtures are VA+VB VA A = V A + VB VB B = = 1 A V A + VB Entropy of Binary Mixing If each molecule occupies a volume v0 the total number of molecules A in a mixture nA is equal to VA/v0 and there are nB=VB/v0 of molecules B. To calculate the entropy of the final (mixed) state we have to calculate the number of different ways of arranging nA and nB molecules between nS=nA+nB lattice sites. To place first A molecule we have nS empty sites. 1 nS Entropy of Binary Mixing For the second molecule there are only nS-1 available sites. 1 1 nS -1 2 Continue this process until all A molecules are used we will end up with 3 6 nS ! nS ! n S ( n S 1)( n S 2)....( n S n A + 1) = = 1 9 (nS n A )! n B ! 7 4 2 possible arrangements of A molecules. 5 8 Entropy of Binary Mixing However, all these arrangements of A molecules are not independent. For example, a distribution of A molecules will 3 6 not change if we exchange monomer 1 with 1 9 monomer 7, or 3 with 8. 7 4 2 The total number of permutations between nA 5 8 elements is equal to nA!. Thus, the number of different arrangements of nA molecules between nS sites is nS ! n A! n B ! After filling the lattice sites with A molecules there is only one independent configuration for B molecules. The number of different states for mixing A and B molecules is AB nS ! = n A! n B ! Entropy of Binary Mixing The entropy of mixing A and B molecules is S mix = S AB S A S B = k ln nS ! AB = k ln n A! n B ! A B This expression can be simplified if we use Stirling's approximation for factorial ln N ! N ln S mix = k ln = k n A ln where N e nS ! n n n = k n S ln S n A ln A n B ln B n A! n B ! e e e nA n n B ln B nS nS nA A = n A + nB and = kn S ( A ln A + B ln B ) nB B = n A + nB Placing Polymers on a Lattice Let us now calculate the entropy of mixing polymer molecules with solvent. As before we will assume that there are nA monomers and nB solvent molecules in a mixture. Each polymeric molecule has degree of polymerization NA and there are nM=nA/NA macromolecules in a system. To place first monomer on a lattice there ns available empty sites 1 nS Placing Polymers on a Lattice To place a second monomer we have to consider only z surrounding cell. 1 1 z We have also multiply this z factor by the probability that the selected cell is empty. This probability is equal to 1-1/ns . For the third monomer there are only z-1 surrounding cells, because one neighboring cell is already filled by a monomer. 1 1 z-1 The probability that selected cell is empty is equal to 1-2/ns. Placing Polymers on a Lattice Continue this process until all monomers are placed on the lattice we will have nM N A 1 1 2 3 nS z 1 ( z 1) 1 ( z 1) 1 ...( z 1) 1 = nS nS nS nS z ( z 1) n M ( N A 1) z ( z 1) n S ( n S 1)( n S 2)...( n S n M N A + 1) = n M ( N A 1) nS n M ( N A 1) nM nM nS ! nS n M ( N A 1) nB ! To obtain the number of different configurations for mixing polymers with solvent we have to divide it by nM!, because now all our polymeric molecules are identical. We have to divide n the final expression by the factor 2 M which is due to the fact that two ends of a chain are indistinguishable AB z = 2 nM ( z 1) nM ( N A 1) nS ! n ( N 1) nS M A nB ! nM ! Entropy of Polymer-Solvent Mixture The entropy of mixture of polymer A and solvent B is S AB = k ln AB n ! z nM = k ln nM ( z 1) nM ( N 1) nM ( N 1S ) 2 nS nB ! nM ! Once again using the Stirling's approximation for the factorial we can write S AB = k ln AB = k ln k n A ln (z 1) + n S ln z 2 nM ( z 1) n M ( N A 1) nS ! n M ( N A 1) nS nB ! nM ! nS n n n M ln M n B ln B n M (N A 1) ln n S e e e Entropy of Mixing Polymer and Solvent To obtain an entropy of mixing we have to subtract from entropy of AB mixture the entropy of pure polymer A. The entropy of pure polymer can be obtained from the expression for entropy of mixture SAB by setting nS=nA and nB=0. nM n A! S A = k ln A = k ln z 2 k n A ln (z 1) + n A ln nA n n M ln M n M ( N A 1) ln n A e e ( z 1) nM ( N A 1) nA n M ( N A 1) nM ! Entropy of mixing is equal to S mix = S AB S A k n S ln nS n n n n A ln A n B ln B + n M ( N A 1) ln A e e e nS k [n B ln B + n M ln A ] kn S B ln B + A NA ln A Free Energy of Mixing The free energy of mixing is equal to Fmix = TS mix B ln B + A ln A , regular solutions = kTn S B ln B + A ln A , polymer solutions B NB NA ln B + A NA ln A , polymer blends Dependence of free energy of ideal mixing on volume fraction -0.1 Fmix n s kT 0.2 0.4 0.6 0.8 1 A -0.2 N A = N B = 20 -0.3 N A = 100 N B = 1 -0.4 -0.5 -0.6 -0.7 NA = NB = 1 Energy of Binary Mixing The energy of mixing in the lattice theory assumes random distribution of particles on a lattice Let us choose a particle. The surrounding cells can be occupied either by the particle of the same sort A or by the B particle. The probability of this neighboring molecule to be A is equal to fA and probability of this molecule to be B is equal to fB=1-fA. If the interaction energy between A-B pair is uAB and interaction between A-A is uAA. The A z=4 average interaction energy between molecule A and its neighbor is U A = u AA A + u AB B For the site occupied by B molecule U B = u BB B + u AB A where uBB is the interaction energy between B-B pair Energy of Binary Mixing Each lattice site has z neighbors. Thus interaction energy between Selected molecule and all its neighbors is z times larger. The average energy per molecule is equal to a half of this energy to avoid a double counting of each contribution. Thus the average energy per molecule is UA z=4 A z = UA 2 z UB = UB 2 The total energy of pairwise interaction is equal to U AB = n A U A + n B U B = n S [ A U A + B U B ] = [ zn S 2 2 u AA A + 2u AB A B + u BB B 2 Energy of Binary Mixing The interaction energy in pure A sample is equal to zn S A zn A U AA = U AB ( B = 0, A = 1) = u AA = u AA 2 2 The same is true for pure B sample zn S B zn B U BB = U AB ( B = 1, A = 0) = u BB = u BB 2 2 The energy of binary mixing is equal to U mix = U AB U A U B = [ zn S zn S zn S 2 2 u AA A + 2u AB A B + u BB B u AA A u BB B = 2 2 2 zn S A B [2u AB u AA u BB ] 2 Energy of Binary Mixing per Site and Flory Interaction Parameter The energy of mixing per site is equal to U mix U mix z = = A (1 A )[2u AB u AA u BB ] nS 2 The Flory's interaction parameter is defined to characterize the differences in interaction energies of a mixture z (2u AB u AA u BB ) = 2 kT The energy of mixing per lattice site in terms of Flory interaction parameter is U mix = kT A (1 A ) Free Energy of Mixing The free energy of mixing per lattice site is Fmix = U mix T S mix nS B ln B + A ln A , regular solutions = kT A (1 A ) + kT B ln B + A ln A , polymer solutions B NB NA ln B + A NA ln A , polymer blends Flory Parameter For non-polar mixture the the interaction parameter c can be estimated from energies of vaporization of pure components of a mixtures. Energy to create \"surfaces\" of molecules zn A 2 E vap = U AA = u AA = n A E A = n A v0 A 2 The solubility parameter is defined as energy of vaporization per molecule divide by molecular volume A = E A vA Flory Parameter The same is true for B-B interaction energy z 2 u AA = v0 A 2 For non-polar molecules the interaction between A and B is z u AB = v0 A B 2 Using representation of the interaction energies in terms of solubility parameters we can write for the Flory's parameter the following expression z (2u AB u AA u BB ) v0 2 ( A + B2 2 A B ) = = 2 kT kT v0 ( A B )2 = kT General Expression for the Flory Parameter For polymers the Flory's interaction parameter is written as sum of two terms B A+ T Parameter A is entropic part and B/T is enthalpic part of interaction parameter Stability of Homogeneous Phase Dependence of the free energy of the system on volume fraction at different values of the interaction parameter =3 Fmix kT 0.2 -0.1 0.4 A 0.6 0.8 1 =2 -0.2 -0.3 -0.4 -0.5 =0.5 Fmix = (1 A ) ln (1 A ) + A ln A + A (1 A ) kT Stability of Homogeneous State Consider stability of homogeneous mixture of A and B monomers with volume fraction of A monomers f. Let us compare the free energy of homogeneous state -0.57 Fmix -0.58 with free energy of the kT -0.59 system with the same f1 f f2 0.7 average volume fraction f 0.6 0.5 0.4 -0.61 but consisting of two phases -0.62 A with polymer volume -0.63 fractions f1 and f2. -0.64 Because of coexistence of two phases the total volume V of the system is divided into two parts with volumes V1 and V2. Thus the average polymer volume fraction is equal to where fi=Vi/V 2 V1 V2 1 f2 = = 1 + 2 = f11 + (1 f1 ) 2 f1 = V V 2 1 2 1 Stability of Homogeneous State The free of two phase system is equal to Fmix 12 2 1 Fmix (1 ) + Fmix ( 2 ) ( ) = f1 Fmix (1 ) + f 2 Fmix ( 2 ) = 2 1 2 1 If the composition dependence of the free energy is locally concave (see figure) the free energy of the two phase state is higher than Fmix kT -0.57 -0.58 -0.59 -0.61 F ( ) -0.62 kT -0.63 Fmix ( )-0.64 kT 12 mix f1 0.4 f 0.5 f2 0.6 the free energy of homogeneous state. Thus the homogeneous 0.7 state is locally stable. A Stability of Homogeneous State However if the composition dependence of the system free energy is locally convex the free energy of homogeneous state is always higher than the free energy of the mixed state and system can lower its energy by separating into two phases. Fmix ( ) 0.055 kT 0.05 12 Fmix ( ) 0.045 0.04 kT 0.035 0.03 0.025 0.4 f1 0.5 0.6 0.7 f f2 fA Criterion of Local Stability The criterion of local stability can be written in terms of the second derivative of the system free energy Fmix kT -0.57 2 Fmix > 0, locally stable 2 F Tk 0.055 xim -0.58 0.05 -0.59 2 Fmix < 0, unstable 2 0.045 -0.61 -0.62 -0.63 0.4 0.5 0.6 0.7 0.04 0.035 0.03 0.025 -0.64 0.4 0.5 0.6 0.7 f 0.04 0.02 0.2 unstable 0.4 0.6 -0.02 metastable stable Fmix kT stable metastable Stable and Unstable Regions 0.8 1 A -0.04 -0.06 2 Fmix >0 2 2 Fmix <0 2 fmix>0 2 Phase Diagram Free energy of mixing of two polymers with degrees of polymerization NA and NB Fmix (1 ) ln (1 ) + ln + (1 ) = kT NB NA Fmix k NA =200; NB=100; =5K/T Phase Diagram Fmix kT N = 2.7 Fmix (1 ) = ln (1 ) + ln + (1 ) kT N N Chemical potential in two phases Fmix N (N )c = = ' Fmix =0 = " 1 Fmix 1 1 1 1 = ln (1 ) + ln + + (1 2 ) kT N N N N Solving for parameter b = ln[ (1 )] 1 1 1 ln (1 ) + ln = (1 2 )N 1 2 N N Coexistence curve Critical Point Critical point is a point where both spinodal and binodal lines intersect. ln[ (1 )] 1 1 1 b = ln (1 ) + ln = Binodal: (1 2 )N 1 2 N N Spinodal: 2 Fmix 1 1 = kT + 2 = 0 N N (1 ) 2 sp = 1 1 1 1 1 + = 2 N N (1 ) 2 N (1 ) For symmetric blends these two equations have a solution critical composition critical interaction parameter c = 1 / 2 and c = 2 / N Critical temperature: = A+ B T Tc = B B = c A 2 / N A Critical Point For asymmetric blends: 2 Fmix 1 1 = kT + 2 = 0 N A N B (1 ) 2 sp ( ) = 1 1 1 + 2 N A N B (1 ) Determine a minimum of spinodal sp ( ) = 1 1 1 1 1 1 = + + =0 2 N A N B (1 ) 2 N A 2 N B (1 ) 2 The solution of this equation gives the critical composition c = NB NA + NB Critical interaction parameter c = 1 2 1 NA + 1 NB 2 Examples of Phase Diagrams = A+ B<0 10 000 g mol-1 b t>0 133 000 g mol-1 20 000 g mol-1 spinodal 51 000 g mol-1 200 000 g mol-1 Polymer blend of poly(vinyl methyl ether) (M=51 500 g mol-1) and polystyrene of different molecular weights. Lower critical solution temperature - LCST 53 300 g mol-1 Solution of polyisoprene in dioxane. Upper critical solution temperature - UCST