Lecture: P1_Wk1_L5 Inter-Molecular Forces: Keesom Force. Ron Reifenberger Birck Nanotechnology Center Purdue University 2012

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1 Lecture: P_Wk_L5 Inter-Molecular Forces: Keesom Force Ron Reifenberger irck Nanotechnology Center Purdue University

2 Last Lecture: Electrostatic Intermolecular Interactions ion () fixed angle () ion olar molecule angle averaged (3) non-olar molecule olariation induced diole (4) P_Wk_L5

3 This Lecture: Electrostatic Intermolecular Interactions olar molecule interacts with interacts with olar molecule non-olar molecule fixed angle angle averaged olariation (5) Keesom (6) induced diole (7) Debye non-olar molecule P_Wk_L5 interacts with non-olar molecule Keesom:Debye:London :: fluctuating induced dioles (8) London 3

4 5. Interaction between olar molecules: fixed orientation already four arameters: searation distance () + 3 angles (Θ,Θ,φ) P_Wk_L5 -q +q E U( θ, θ, ϕ, ) = U ( ) = E = = 3 E, diole 4κε o U ( ) = 3 or ˆ ˆ ( ) 3 ( ˆ)( ˆ) [ θ θ θ θ ϕ ] = f ( θ ) ( ) < 3, θ, ϕ ; f θ, θ, ϕ 4κε κε relative twist cos( )cos( ) sin( )sin( )cos( ) lowest energy when θ = θ = U(,, ϕ, ) = 3 4κε +q Θ angle φ Θ -q 4

5 Diole-Diole Interactions The magnitude and sign of the diole-diole energy deends strongly on orientation U() is minimum U() is maximum U() is ero P_Wk_L5 5

6 Reresentative lots of f(θ,θ,φ) 36 o f(θ,θ,φ) color bar:. 36 o θ. -. θ o θ o 36 o φ=; in lane configuration o o 36 o θ φ=9 o ; erendicular configuration P_Wk_L5 f U( θ, θ, ϕ, ) = 4κε 3 ( θ, θ, ϕ) ( θ, θ, ϕ) = [ cos( θ )cos( θ ) sin( θ )sin( θ )cos( ϕ) ] f 6

7 Φ diagram for d/s.75 + x q Symbol q q q 3 q 4 q Interaction otential energy between in-lane (φ= o ) ermanent dioles - + x - q 3 q 4 Actual Charge + q - q + q - q d s Fixed, NO rotation U total (in -8 J) q =x -8 C; s= nm d/s=.75 d/s=.5 d/s= Φ (degrees) Note: = P_Wk_L5 7

8 6. Interaction between olar molecules: angle averaged diole-diole interaction must be roerly angle-averaged by allowing the orientation of dioles and to fluctuate. Must calculate U ( ) U (, θ, θ, φ) e U(, θ θ φ,, ) kt where here U(, θ, θ, φ) = [ θ θ θ θ φ] cos cos sin sin cos 3 4κε... reresents the roer thermal average over all angles The comlete derivation is given in Aendix A : U Keesom C ( ) = = 3 kt 4κε K 6 6 (also known as the Keesom interaction; W.H. Keesom, Physik. Z., 9 (9) and, 643 (9)) P_Wk_L5 8

9 U The Keesom interaction between two ethanol molecules from P_Wk_L3 Keesom Material =.7D Chemical Formula 3 ( 5.66 ) relative twist angle <Θ > <φ> <Θ > = nm =.7D Cm (. 7D) D ( ) = = 3 kt 4κε 3.38 / 3 4 ( ) 8.85 C Nm 6 3 ( 6 9 J K)( K) 4 m = Nm 3.4 J κ, dielectric constant m, diole moment (in Debye),3 α o /4ε o olariability volume (in -3 m 3 ) Ethanol (CH 3 )CH OH ( ) 9 5 ( 5.38 ) (.44 ) (.56 ) = J ev.8 J J P_Wk_L5 9 = = 9 ev 9

10 7. Interaction otential energy between a olar and non-olar molecule olar molecule +q, α o, -q θ secify location of non-olar molecule E ˆ (, θ) = cosθ ˆ + sinθθ diole 3 4κε o E (, θ) 4cos θ sin θ diole = + 3 4κε o α o, δ - = 3cos θ + 3 4κε o δ + P (aligned with E diole ) non-olar molecule When a olar molecule is in close roximity to a non-olar molecule, the olariing electric field is generated by the fixed diole ( ) of the olar molecule. This induces a diole in the electron cloud of the non-olar molecule. The resulting interaction energy is called the Debye interaction energy. U Debye αo, ( ) = (4 κε ) 6 P_Wk_L5

11 ( E) = α E induced o, In this case, E is due to a diole U (, θ ) = ( E) E = E ( αo, ) Debye induced diole θ α o, 3 6 4κε o ( 4κε o ) 3cos + 3cos θ + o, = α = if an angle average over θ is erformed, cos U The induced diole moment is a function of E Debye ( ) = α o, ( 4κε ) o 6 θ = 3 (See Aendix, P_Wk_L4) P_Wk_L5

12 If the non-olar molecule is relaced by a olar one, then the second molecule could induce a diole in the first: +q olar molecule +q, α o, θ, α o, -q olar molecule -q The net effect is additive, so we must have U Debye ( ) = α + α o, o, 6 (4 κε ) If the two olar molecules are identical, then U Debye ( ) α o = (4 κε ) 6 P_Wk_L5

13 Aendix A: Derivation of the angle-averaged Keesom interaction The derivation of the angle-averaged Keesom interaction is usually not given in most textbooks. Rather, the final result is often written down after some statement like After a lengthy calculation, we find.. In what follows the derivation is given using the same logic followed when discussing the ion-diole interaction in the revious lecture (C_Wk_L3). The starting oint is the diole-diole interaction energy for two dioles and searated by a distance U (, θ, θ, φ) [ θ θ θ θ φ] cos cos sin sin cos = 4κε 3 P_Wk_L5 3

14 At first, you might be temted to erform a straightforward angle average of U(, θ, θ, φ) as follows: U (, θ...) = [ θ θ θ θ φ] 3 4κε [ ] Let U(, θ...) = U ( ) cosθ cosθ sinθ sinθ cos φ = U ( ) f ( Ω) where Uo( ) and f ( Ω ) = cosθcosθ + sinθsinθcosφ 3 4κε U ( ) cos cos sin sin cos o θ= θ = θ= θ= o ϕ= θ = θ = dϕ U (, θ,...)sinθ dθ sin θ dθ dϕ f( Ω)sinθ dθ sinθ dθ I = U ( ) = Uo( ) I dϕ sinθ dθ sinθ dθ dϕ sinθ dθ sinθ dθ θ = θ = θ = θ = [ ] I = cosθ cosθ + sinθ sinθ cosφ sinθ dθ sinθ dθ dϕ A dϕ sinθ θdθ sinθ θdθ ϕdϕ θdθ θdθ ϕ= θ = θ = ϕ= θ = θ = = cos cos + cos sin sin = I = dϕ sinθ dθ sinθ dθ = 8 θ = θ = IA U ( ) = Uo( ) = I The result is ero because the angle average is erformed for a free diole in which all angles are equally likely. Instead, a weighted angle average must be calculated that referentially weights those angles having lowest energy. This involves a correct treatment of the oltmann weighting factor and requires some understanding of statistical thermodynamics. P_Wk_L5 o A 4

15 Weighting the angles using the oltmann factor gives U ( ) U (, θ,...) e let Uo( ) =, kt U(, θ,...) kt ϕ θ θ θ θ θ θ d U (,,...) e sin d sin d U(,,...) kt θ= θ= U(, θ,...) kt dϕ e sinθdθsinθdθ θ = θ = U(, θ,...) kt U (, θ,...) e = U( ) then o f ( Ω) dϕ f( Ω) e sinθ dθ sinθ dθ θ = θ = f ( Ω) dϕ e sinθdθsinθdθ θ = θ = Evaluating such a quantity is simlified by recogniing that the numerator and denominator are related by a derivative. Using the well known roerties of the natural logarithm: P_Wk_L5 U(, θ,...) kt d f ( Ω) U (, θ,...) e = Uo ( ) ln ϕ θ θ θ θ d e sin d sin d d θ= θ= d since ln d e sin d sin d d f ( Ω) ϕ θ θ θ θ θ = θ = = f ( Ω) ϕ θ θ θ θ d e sin d sin d θ= θ= dϕ θ = θ = f e d d f ( Ω) ( Ω) sinθ θsinθ θ 5

16 P_Wk_L5 ( ) if f Ω <<, we can write [ ] f ( Ω) ϕ sinθ θsinθ θ θ = θ = This means we must evaluate the integral d e d d f ( Ω) dϕ e sinθdθsinθdθ dϕ + f( Ω ) + f ( Ω ) +... sinθdθsinθdθ θ = θ = θ = θ = = dϕ θ = θ = θ = θ = sinθ dθ sinθ dθ [ ] + dϕ f( Ω) sinθ dθ sinθ dθ = I + I + I +... I = I = 8 I = 3 + dϕ f ( Ω ) sinθdθsin θdθ +... θ = θ = I3 = dϕ [ cosθcosθ + sinθsinθcosφ] sinθdθsinθdθ = θ = θ = = I + I + I dϕ 4 cos θ θ cos sinθdθ sinθdθ θ = θ = a b c + dϕ [ 4 cosθ cosθ sinθ sinθ cosφ] sinθdθ sinθdθ θ = θ = + dϕ sin θ θ φ sin cos sinθdθsinθdθ θ = θ = 6

17 I = ϕ θ θ a 4 d cos cos sinθdθsinθdθ I b θ = θ = θ= θ= [ ] θ θ [ ] ( ) ( ) 4 6 = 4 cos cos = 4 = 4 = θ = θ = = 4 dϕ [ cosθcosθsinθsinθcosφ] sinθ θ θ θ d sin d θ = θ = = 4 cosϕdϕ cosθ θ θ θ sin cos sin dθ dθ θ = θ = θ= θ= ϕ= 3 3 [ ϕ] θ θ ϕ = 3 θ = 3 θ = = 4 sin sin sin = I = ϕ θ θ φ c d sin sin cos sinθdθsinθdθ = θ = θ = 3 3 cos φdϕ sin θ θ sin dθ dθ θ = θ = ϕ= θ = θ = = ( ϕ + sinϕcosϕ) ( cos3θ 9cosθ) ( cos3θ 9cosθ) ϕ= θ = θ = = [ ] ( ) = ( ) 9() + = = 3 9 P_Wk_L5 7

18 I3 = Ia + Ib + Ic = + + = = So I + I + I = U(, θ ) kt d f ( Ω) U ( ) U (, θ) e = Uo( ) ln ϕ θ θ θ θ d e sin d sin d d θ= θ= d = Uo( ) ln ( I + I + I3 +...) d d d = Uo( ) ln = Uo( ) ln( 8 ) + ln d 3 d 3 = + Uo( ) Uo( ) 3 P_Wk_L5 8

19 Uo( ) now since = and Uo( ), 3 kt 4κε we finally obtain U ( ) U(, θ,...) kt U (, θ,...) e = Uo( ) 3 = 3 3 4κε 3 kt 4κε = 6 3kT 4κε P_Wk_L5 9