Chapter 18. Molecular interactions

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Atkins / Paula 《 Physical Chemistry, 8th Edition 》

Chapter 18. Molecular interactions In this chapter we examine molecular interactions in gases and liquids and interpret them in terms of electric properties of molecules, such as electric dipole moments and polarizabilities. 在本章我們將檢視氣體和液體分子的交互作用並且根據分子電性質解釋他們的電偶極矩和極化率。 Chap 18. Molecular interactions

Chapter 18. Molecular interactions Table of Contents 目錄內容

Electric properties of molecules 分子的靜電性質 18.1 Electric dipole moments 電偶極矩 18.2 Polarizabilities 極化性 18.3 Relative permittivities 相對電容率(介電常數) Interactions between molecules 分子間的作用力 18.4 Interactions between dipoles 偶極矩間的作用力 18.5 Repulsive and total interactions 斥力與分子總作用力 Gases and liquids 氣體與液體中的分子作用 18.6 Molecular interactions in gases 氣體中的分子作用 18.7 The liquid-vapor interface 氣液體介面中的分子作用 18.8 Condensation 凝結作用 Chap 18. Molecular interactions

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• All these electric properties of molecules, such as electric dipole moments and polarizabilities, reflect the degree to which the nuclei of atoms exert control over the electrons in molecule, either by causing electrons to accumulate in particular regions, or by permitting them to respond more or less strongly to the effects of the external electric fields. • 分子所有這些電性質，例如電偶極矩和極化率，反映出分子中原子核對電子的控制，造成電子在特定區域的累積程度，或是外在電場作用下電子容許有多少的反應。

Chap 18. Molecular interactions

Electric properties of molecules • Many of electric properties of molecules can be traced to 分子的許多電性的成因可以被追蹤到有如下:  the competing influences of nuclei with different charges, or 原子核對不同電荷影響力的競爭,或  the competition between the control exercised by a nucleus and the influence of an externally applied field. 原子核的控制力與外加電場影響力之間的競爭。 • The former competition may result in an electric dipole moment. The latter may result in properties such as refractive index and optical activity. 前者競爭可導致偶極矩。後者可導致例如折射率和光學活性等物性。


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18.1 Electric dipole moments • An electric dipole moment µ consists of two equal and opposite charges, q , separated by a distance R. • The moment of dipole, the dipole moment µ, is a vector of magnitude µ =q R, that points from the negative charge (-) to the positive charge (+). • Dipole moment are normally reported in the cgs unit Debye but their SI unit is coulomb-meter (C m). • The original definition of one Debye is the dipole moment of two equal and opposite charges of magnitude 1 e.s.u. separated by 1 Å. 1 D =3.33564x10-30 C m Chap 18. Molecular interactions

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• In elementary chemistry, an electric dipole moment is represented by the arrow ↛ added to the Lewis structure for the molecule, with the + marking the positive end. Note that the direction of the arrow is opposite to that of μ. • 在普通化學課本，由箭頭↛代表電偶極矩，附加於分子 Lewis結構，正端標記+。注意箭頭方向與 μ 定義不同。 Chap 18. Molecular interactions

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• The dipole moment of a pair of charges +e and -e separated by 100 pm is 1.6 x x10-19 C m, corresponding to 4.8 D. 一對偶極矩電荷若是由+e 和 -e分離的距離為100 pm為 1.6 x10-19 C m，對應到4.8 D。 • Dipole moments of small molecules are typically about 1 D. 典型的小分子的偶極矩是大約1 D。


Polar molecular • Polar molecular (極性分子) Molecules with permanent non-zero electric dipole moments are classified as polar molecules. The permanent dipole moment arise from differences in electronegativity which results in atoms possessing partial charges or imbalance in the distribution of electric charge or other features of bonding. (e.g. CO molecule)

分子有永久的非零電偶極矩則被分類為極性分子。當原子的陰電性差別，導致部份電荷不平衡；或從其他鍵結特性，結果均可出現永久偶極矩的分子。 (例如。 CO分子) • A dipole moment exists in a polar molecule can also be modified by an applied field.

一個極性分子存在的偶極矩，可能被外加電場修飾。 Chap 18. Molecular interactions

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Non-polar molecular

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• Non-polar molecule (非極性分子) • Nonpolar molecules acquire an induced dipole moment in an electric field on account of the distortion the field causes in their electronic distributions and nuclear positions; (一個非極性分子在一個電場獲取了感應電偶極矩，是由於電場在片刻造成的他們的電子分佈和核位置的畸變。) • However, this induced moment is only temporary, and disappears as soon as the perturbing field is removed. Polar molecules also have their existing dipole moments temporarily modified by an applied field. (然而，這感應電偶極矩只是臨時的，並且當去除干擾的外加電場時感應電偶極矩即消失。極性分子也可感應外加電場，而暫時地修改他們現有的永久偶極矩。) Chap 18. Molecular interactions

Stark effect • Stark effect is used to measure the electric dipole moments of molecule for which a rotational spectrum can be observed. In many cases microwave spectroscopy cannot be used because the sample is not volatile, decomposes on vaporization, or consists of molecules that are so complex that their rotational spectra cannot be interpreted. (當分子的旋轉光譜可以被觀察時，Stark 效應用於測量分子的電偶極矩。許多情況下因為樣品不揮發，或在汽化時分解，或分子包括很複雜的結構，使得旋轉光譜無法解釋。則不能使用微波光譜。) • In such cases the dipole moment may be obtained by measurement on a liquid or solid bulk sample using others method. (此時需以其他方法量測液態或固態樣品.) Chap 18. Molecular interactions

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• All heteronuclear diatomic molecules are polar, and typical values of μ include 1.08 D for HCl and 0.42 D for HI (Table 18.1). (所有異核雙原子分子皆有極性，並且μ 的典型值例如(表18.1): HCl為1.08 D ; HI為0.42 D) • Molecular symmetry is of the greatest importance in deciding whether a polyatomic molecule is polar or not. (在決定一個多原子分子是否有極性時，分子對稱才是最重要的因素。) • Indeed, molecular symmetry is more important than the question of whether or not the atoms in the molecule belong to the same element. (分子對稱的重要性的確超過在分子中的原子是否屬於同一個元素的問題。) Chap 18. Molecular interactions

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• Homonuclear polyatomic molecules may be polar if they have low symmetry and the atoms are in inequivalent positions. (同核多原子分子如果他們有低的對稱性且原子是在不對等的位置，則可能為極性的。) • For instance, the angular molecule ozone, O3 (2), is homonuclear; however, it is polar because the central O atom is different from the outer two (it is bonded to two atoms, they are bonded only to one); moreover, the dipole moments associated with each bond make an angle to each other and do not cancel. (舉例來說，角型的臭氧分子，臭氧O3（ 2 ）是同核的但也是極性的，因為中央的O原子是由不同的外二（它是鍵結兩個原子的Ｏ，其他的是只有鍵結一個原子的Ｏ）; 此外，每一個鍵結的偶極矩，形成的角度，彼此不抵消。) Chap 18. Molecular interactions

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• Heteronuclear polyatomic molecules may be nonpolar if they have high symmetry, because individual bond dipoles may then cancel. (異核多原子分子如果有高對稱性，可以是非極性的。因為個別鍵結的偶極矩，可能會相互抵消。) • The heteronuclear linear triatomic molecule CO2 for example, is nonpolar because, although there are partial charges on all three atoms, the dipole moment associated with the OC bond points in the opposite direction to the dipole moment associated with the CO bond, and the two cancel (3). (舉例來說，線性異核三原子分子的CO2為非極性，因為，雖然在所有三個原子上有部分電荷，但在OC鍵結上的偶極矩指向CO鍵結上偶極矩的相反方向，使偶極矩兩者間相抵消（見結構圖3）。) Chap 18. Molecular interactions

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Dipole moment of a polyatomic molecule • To a first approximation, it is possible to resolve the dipole moment of a polyatomic molecule into contributions from various groups of atoms in the molecule and the directions in which these individual contributions lie (Fig. 18.1). • Thus, 1,4-dichlorobenzene is nonpolar by symmetry on account of the cancellation of two equal but opposing C-C1 moments(exactly as in carbon dioxide). • 1,2-Dichlorobenzen, however, has a dipole moment which is approximately the resultant of two chlorobenzene dipole moments arranged at 60 to each other, Chap 18. Molecular interactions

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18.1 The resultant dipole moments (grey) of the dichlorobenzen e isomers (b to d) can be obtained approximately by vectorial addition of two chlorobenzene dipole moments (1.57 D).


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18.2 The resultant dipole moments (grey) of the dichlorobenzen e isomers (b to d) can be obtained approximately by vectorial addition of two chlorobenzene dipole moments (1.57 D).


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• This technique of ‘vector addition’ can be applied with fair success to other series of related molecules, and the resultant µ res of two dipole moments µ1 and µ2 that make an angle θ to each other (4) is approximately

µ res 2 = µ12 + µ22 + 2µ1µ2 cos θ • When the two dipole moments have the same magnitude (as in the dichlorobenzenes), this equation simplifies to

µ res 2 = 2µ12 + 2µ12cos θ = 2µ12 (1+cos θ) = 4µ12 cos2(θ/2)

µ res

= 2µ1 cos (θ/2) Chap 18. Molecular interactions

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• Self-test 18.1. • Estimate the ratio of the electric dipole moments of ortho (1,2-) and meta (1,3-) di-substituted benzenes. [µ (ortho)/ µ (meta) =1.7] Ans: Since µ = 2µ1 cos (θ/2) ;

µ (ortho)/µ (meta) = cos (60º/2) / cos (120º/2) 3 2   3  1.7 1 2


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• A better approach to the calculation of dipole moments is to take into account the locations and magnitudes of the partial charges on all the atoms. These partial charges are included in the output of many molecular structure software packages, • To calculate the x-component, for instance, we need to know the partial charge on each atom and the atom x-coordinate relative to a point in the molecule and form the sum

µx= ∑J qJ xJ • Here qJ is the partial charge of atom J, xJ is the xcoordinate of atom J, and the sum is over all the atoms in the molecule. Chap 18. Molecular interactions

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• When the two dipole moments have the same magnitude (as in the dichlorobenzenes), this equation simplifies to • µ = (µx2 + µy2 + µz2 ) 1/2


Example 18.1 Calculating a molecular dipole moment • Estimate the electric dipole moment of the amide group shown in (5) by using the partial charges (as multiples of e ) in Table 18.2 and the locations of the atoms shown. Method We use eqn 18.3a to calculate each of the components of the dipole moment and then eqn 18.3b to assemble the three components into the magnitude of the dipole moment. Note that the partial charges are multiples of the fundamental charge, e =1.609x10-19 C.


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-

+X

-

+Y


-

Answer

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• The expression for µx is :

µx = (-0.36 e) x (132 pm) + (0.45 e) x (0 pm) + (0.18 e) x (182 pm) + (-0.38 e) x (-62.0 pm) = 8.8 e pm = 8.8 e pm x (1.609x10-19 C/e)x(10-12 m/pm) = 1.4x10-30 C m • Corresponding to µx = 0.42 D. The expression for µy is

µy = (-0.36 e) x (0 pm) + (0.45 e) x (0 pm)

+ (0.18 e) x (-86.6 pm) + (-0.38 e) x (107 pm) = -56 e pm = -9.1x10-30 C m


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• We can find the orientation of the dipole moment by arranging an arrow of length 2.7 units of length to have x, y, and z components of 0.42, -2.7, and 0 units; the orientation is superimposed on (6).

µ = (µx2 + µy2 + µz2 ) 1/2 µ 2= (0.42)2+(-2.7)2+02 = 7.466 µ = 2.7 D


Self-test 18.2 • Calculate the electric dipole moment of formaldehyde, using the information in (7). • The expression for µx is :

µx = (0.02 e) x (0 pm) + (-0.38 e) x (0 pm) + (0.18 e) x (+94 pm) + (0.18 e) x (-94 pm) = 0.0 e pm = 0.0 e pm x (1.609x10-19 C/e)x(10-12 m/pm) =0Cm

µy = (0.02 e) x (0 pm) + (-0.38 e) x (118 pm) + (0.18 e) x (-61.0 pm) + (0.18 e) x (-61.0 pm) = -44.8 e pm = -66.80 e pm x (1.609x10-19 C/e)x(10-12 m/pm) = 1.075x10-30 C m

• [-3.2 D] Chap 18. Molecular interactions

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(b) Polarization • Polarization (P ) of a bulk sample is the electric dipole moment density, the mean electric dipole moment of the molecules, the mean electric dipole moment of the molecules, , multiplied by the number density, N=N/V • P = N

(18.4)

• In the following pages we refer to the sample as a dielectric, by which is meant a polarizable, nonconducting medium.


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• In the absence of an applied field, the individual molecule dipoles point in random direction, so the net dipole moment per unit volume is zero. An isotropic fluid sample has the polarization of zero since the molecules adopt any random orientations, so =0. • In the presence of a field, the dipoles become partially aligned because of thermal motion. Some orientations have lower energies than others. So the electric dipole moment density is nonzero. At temperature T, with the field E applied along z direction:  2E Z  3kT • We show in the Justification below that, at a temperature T Chap 18. Molecular interactions

Justification 18.1 • If the energy of the dipole in the field is U(θ) = - μ•Ε = -μΕcos θ • The number of molecule with potential energy U a dipole has an orientation in the range θto θ+dθ is given by the Boltzmann distribution, N    N 0 e E cos kT • When the exponent is small: N    N 0 1 E cos kT 

• Each molecule contributes a term μcos θ to the net dipole moment in the direction of the electric field. The polarization due to the orientation of dipole is estimated by multiplying N(θ) by μcos θ and integrating

dP   cos  e E cos kT d





0

 cos  e E cos kT d


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Justification • Energy barrier of the dipole reorientation in the field is E(θ) = -μΕcosθ • The probability dP that a dipole has an orientation in the range θto θ+dθ is given by the Boltzmann distribution,

dP  e -E   kT sin  d





0

e -E   kT sin  d

• The average value of the parallel dipole moments is 

Z    cos  dP   cos  dP 

  e -E   kT sin  cos d 0





e -E   kT sin  d

E -E   kT  E cos  kT  x cos   xy x , y  cos  kT  1   e x cos sin  cos  d   ye xy dy Z  0   11 x cos  sin  d xy dy e e   0

0

1


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Justification • Apply the integrals:

x  e x e xy dy  e 1 x 1

• then

Z

1



ye xy

1

e x  e x e x  e x dy   2 x x

x  e x x  e x e e   1   2 x  e x   ye xy dy   x x e 1   1  1    x   x  x  x xy dy e  e  x x e e e  1

• Use the Langevin function L (x ):  e x  e x 1  x L x    x       e  e x x  3 • we have

Z

with x  1

2  e x  e x 1  E E    x      x x 3kT 3kT  e e


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• Langevin function L (x) The Langevin function L (x) is the following mathematical function It occurs in the expression for the mean electric dipole moment of a system of polar molecules that are free to rotate in an electric field of strength E at a temperature T, < µ > = µ L (µ E /kT) When L (x) = µ This term contributes to the polarization of a medium.

2

E /3kT.

induced dipole moment µ * is the dipole induced by an applied electric field of strength E

µ *=αE.


Dielectric • Dielectric (介電材料) A dielectric is a polarizable non-conducting medium. • ferroelectric solid (鐵電物質) A ferroelectric solid has a permanent polarization because of a cooperative shift of some of its atoms in a specific direction.


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• When two charges q1 and q2 are separated by a distance r in a vacuum, the potential energy of their interaction is q1q2 V  4 or • When the same two charges are immersed in a medium (such as air or a liquid), the potential energy is reduced to q1q2 V  4r • where  is the permittivity of the medium. The permittivity is normally expressed in terms of the dimensionless relative permittivity, r (formerly and still widely called the dielectric constant) of the medium: r = / 0 Chap 18. Molecular interactions

Polarization of a dielectric

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• The capacitance C of a capacitor is the ratio of the charge Q on the conducting plates to the potential difference ΔΦ between the plates: Q C 



• On the parallel-plate capacitors, if the space is vacuum between the plates with area A and distance d, the capacitance Co is given by:  0A Co 

d


Fig. 18.A1

Fig. 18.A1 Parallel-plate capacitor filled with a dielectric. The plates are labeled A and A’; the dielectrics between the plates. Chap 18. Molecular interactions

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• The permittivity of vacuum ε0 • ε0=1/μ0c

2

=8.854187817…x10-12 C2N-1m-2 (exactly).

• the permeability of vacuum μ0=4πx10-7 N A-2 (exact). • If a dielectric is inserted between the plates, the capacitance is increased considering the polarization of the dielectric. • When a dielectric is placed in an electric field, the electrons in the molecules are pulled toward the positive plate and the nuclei are pulled toward the negative plate. The center of charge in molecules is shifted , the displacement produces a layer of charges on the surfaces of opposite charged plates. These surface charges reduce the electric field strength within the dielectric due to the opposite field. Chap 18. Molecular interactions

Fig. 18.A2 Fig. 18.A2 Representation of a capacitor containing dipolar molecules. The molecules are shown in their instantaneous positions indicating the partial orientation of the dipoles.


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• Dielectric in a capacitor is said to be polarized by the applied E -field. • Polarization P of dielectric is defined as the vector sum of dipole moments µ i of individual molecules per unit volume  P  i V (P )

i

• If N molecule per unit volume with charges ±q separated by δ, the magnitude of polarization vector (P ) is

P  Nq • The surface charge density σpol is the charge per area, which is equal to the dipole moment per unit volume :

P   pol  Q A Chap 18. Molecular interactions

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Electric field in the dielectric • Capacitor with a dielectric between plates will change its capacitance due to the change of E-field in dielectric . • The surface charge density on the plates σfree are driven by the external voltage on the wire, and will not change by inserting the dielectric. • Gauss’s law show that the electric field strength E in the dielectric due to polarization is

 free   pol E ε0

• If there were no polarization, σpol =0, we have E = σfree /ε0 Since σpol =P, the electric field strength E is

 free  P E ε0

E   free ε 0

as  pol  0

• As P is produced by E, it can not use to calculate E. Chap 18. Molecular interactions

Electric field in the dielectric • As P ∝ E, we need to know how P depends on E.

P  el ε 0E • The electric susceptibility (χel) of the dielectric between plates is the proportional constant of polarization.

  el ε 0E free • The electric field strength E now is E  ε0 E   free ε 0  ε 0 el  • solve as

• So the electric field strength E in dielectric is reduced by the surface charge of dielectric by a factor of 1 1 el  • Now the potential of a capacitor filled with a dielectric is

  Ed   freed ε 0  ε 0 el 

ε ε A r 0 • Capacitance is C   free A   ε 0  ε 0 el A d  d Chap 18. Molecular interactions

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Electric field in the dielectric

• The relative permittivity (εr) of the dielectric is

ε r  1 el   C C 0 • which is so called the dielectric constant. • Since the polarization is given by P  el ε 0E • Now the dielectric constant of a medium as

P εr  1  ε 0E • Measurements of the relative permittivity yield the polarization information of the dielectric. Chap 18. Molecular interactions

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18.2 Polarizabilities • A molecule under external electric field may redistribute its charge and forms a induced dipole moment μ *. • The dipole moment due to the distortion polarization is polarizability α :

*  E

where α is the polarizability of a molecule.

• The molecular polarizability is a proportional tensor since the dipole moment vector is not necessarily parallel to the electric field strength vector. In general α is symmetric • If the matrix is diagonalized, then αxy = αyz = αzx = 0 • In the principal axis system, the induced dipole moment is parallel to the electric field vector, the mean electric polarizability is (αxx + αyy + αzz)/3 • The substance is said to be isotropic if αxx = αyy = αzz and the unit for polarizability is C2 m2 J-1 = C m / J C-1 m-1 Chap 18. Molecular interactions

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Polarizability • hyperpolarizability If the response has nonlinear contribution, then the additional term are referred to as hyperpolarizability β of the molecule. 1 *  E  E 2   2 • (a) Polarizability volume Molecular polarizabilities are often expressed in terms of the polarizabilities volume, α＇, where α’= α / 4πεo. • The dimensions of polarizability volume are those of volume and its magnitude is similar to that of the volume of the molecule. Chap 18. Molecular interactions

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18.2 Polarizability of a dielectric

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• The electric polarizability are often in the form:

   ' E  E 4 0 • α’ has a unit of volume as a03 ~ (C2 m2 J-1)/ (J-1 C2 m-1) • The macroscopic polarizability of a bulk sample is the number per unit volume (N/V) multiplied by the dipole moment αE from the molecular property α.

N P  E V


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Justification 18.2

• The change in energy when the field increases from 0 to E is E

E

0

0

E     * dE   

1 2 EdE   E 2

• The contribution to the hamiltonian when a dipole is exposed to an electric field in the z-direction is

H 1   Z E

• Use second order perturbation theory

E 2   n

n

 * H 1

*

2

n

E 00   E n0 

2

 z ,0n 2  E  0  0  E  E n 0 n

• The polarizability of the molecule in the z-direction is 2

z ,0n 2e 2R 2    2 0  0   E E E n n o Chap 18. Molecular interactions

(b) Polarization at high frequency

• There are three kinds of polarization under an electric field: Orientation polarization: From the permanent dipoles. Distortion polarization: From the distortion of nucleus position. Electronic polarization: From the distortion of the electron distribution.


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Polarization • Orientation polarization : The polarizability of a molecule varies with frequency of the applied field. At low frequencies, there are orientation, distortion, and electronic contributions. The orientation polarization is the polarization that stems from the alignment of molecular dipoles in the applied field. This contribution is lost above about 1011 Hz. • Distortion polarization : The next contribution to disappear after polarization is the distortion polarization, the polarization that stems from the distortion of the locations of the nuclei. This contribution disappears when the frequency exceeds the vibrational frequency of the molecule. Chap 18. Molecular interactions

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• Electronic polarizability : At optical frequencies, only the electronic polarization, the polarization stemming from the shift in positions of the electrons, remains. This contribution also vanishes at frequencies in excess of electronic excitation frequencies.


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Justification 18.3

• The polarizability of a molecule in the presence of an electric field that is oscillating a frequency  in the zdirection is obtained using time-dependent perturbation theory and is 2

2 n 0 z ,0n  ω   2  n n 0  2

• provided that  is not close to 

n0,

• As  → o, the eq. reduced to a static polarizability. • As  → ∞, the polarizability becomes

 ω  

2

 2

n  n

2

0

z ,0n  0


as   

18.3 Relative permittivities • Permittivity The permittivity, ε, of a modified Coulomb potential in the medium from q/4πεo r to q/4πεrεo r; where ε0 is the vacuum permittivity, the latter is related to the vacuum permittivity by ε0 µ

0

c2 =1

The numerical value (which is based on the defined values of c and µ 0) is ε0 = 8.854187816 x 10-12 J-1 C2 m-1. relative permittivity, εr , of a medium (formerly dielectric constant) is defined as εr = ε / ε0 Chap 18. Molecular interactions

60

Reduction for electrolyte solutions • The relative permittivity can have a very significant effect on the strength of the interactions between ions in solution. For instance, water has a relative permittivity of 78 at 25°C, so the interionic Coulombic interaction energy is reduced by nearly two orders of magnitude from its vacuum value.


61

Debye equation • The relative permittivity of a substance is large if its molecules are polar or highly polarizable. The quantitative relation between the relative permittivity and the electric properties of the molecules is obtained by considering the polarization of a medium, and is expressed by the Debye equation:

 r  1  Pm  r  2 M


62

Debye equation • The Debye equation gives the relation between the relative permittivity and the electric properties of a molecule ε r  1 Pm

εr  2



M

• where M is the molar mass and ρ is the mass density of the sample. Pm is the molar polarization of the sample: NA  2      Pm  3εo  3kT  • The term μ 2/3kT from the thermal averaging of the electric dipole moment in the presence of the applied field. The corresponding expression without the contribution from the permanent dipole moment is called the Clausius-Mossotti equation. Chap 18. Molecular interactions

63

64

• Clausius-Mossotti equation • When there is no contribution to the polarization from permanent electric dipole moments the ClausiusMossotti equation is used

ε r  1 N A  ε r  2 3Mε 0 • The Clausius-Mossotti equation is used when there is no contribution from permanent electric dipole moments to the polarization, either because the molecules are nonpolar or because the frequency of the applied field is so high that the molecules cannot orientate quickly enough to follow the change in direction of the field. Chap 18. Molecular interactions

Example 18.1

65

• The relative permittivity of 1H35Cl(g) • Give the polarizability and dipole moment of HCl(g) in table 18.A2, calculate the relative permittivity at 1 atm and 273 K.

2 εr 1 N     ε r  2 3Vε 0  3kT

2   N AP      1.456x 103  3kT   3RTε 0 

1 21.456x 103  εr   1 . 0041 1 1.456x 103 


Orientation polarization of a dielectric • Polarization due to the orientation of dipole is given by

N  2E P  V 3kT

• The magnitude of the polarization of a dielectric in an electric field is the sum of the distortion polarization and the orientation polarization:

N P  V

2    3kT 

 E i 

E i : average local electric field

• The local filed is not the same as the applied field since the molecular dipole also produce a field.

E i  E ε r  2 / 3 • Under a low concentration, 2  N   ε r  2  • Now P     E V  3kT  3  Chap 18. Molecular interactions

66

67

• Rearrange:

N P  V

 2  ε r  2   E    3kT  3  

P and ε r  1 ε 0E

• yield the Clausius-Mossotti equation

2  εr 1 N      ε r  2 3Vε 0  3kT  • Replace the number density with molar mass M, (N/V = nNA/V=ρNA/M) the molar polarization (Pm) is

M Pm  

 εr  1  N A  2        3kT   ε r  2  3ε 0  • The unit of Pm is a molar volume

• The macroscopic Pm now connected to molecular properties αand μ, measure of molar polarization Pm at varies temperature gives α and μ. Chap 18. Molecular interactions

68

M Pm  

 εr  1  N A  2        3kT   ε r  2  3ε 0 

1/T Dependence of the molar polarization (Pm) of fluorobenzene on temperature. Chap 18. Molecular interactions

69

N A • The intercept is , gives the molecular 3ε 0 polarizability as α=1.146x10-39 C2 m2 J-1

NA • The slope is , gives the dipole moment as 9ε 0k 2

µ=5.26x10-30 C m.

• The dipole moment of molecules are often expressed in terms of Debye (D). 1 D=3.336x10-30 C m. Thus the dipole moment of fluorobenzene is 1.57 D. (Table 18.2) Chap 18. Molecular interactions

Example 18.2 • The relative permittivity of a substance is measured by comparing the capacitance of a capacitor with and without the sample present (C and C0 , respectively) and using εr = C / C0 . The relative permittivity of camphor (8) was measured at a series of temperatures with the results given below. Determine the dipole moment and the polarizability volume of the molecule.


70

Example 18.2 Method Equation 18.14 implies that the polarizability and permanent electric dipole moment of the molecules in a sample can be determined by measuring εr at a series of temperatures, calculating Pm, and plotting it against 1/T. The slope of the graph is NAμ2/9εok and its intercept at 1/T= 0 is NAα/3εok . We need to calculate (εr -I)/(εr + 2) at each temperature, and then multiply by M/ρ to form Pm.


71

72

Example 18.2 εr  1 Method According to eqn 15, we need to calculate εr  2

at each T, and then multiply by M/ρ to form Pm, θ/℃

ρ/(g cm-3)

εr

0

0.99

12.5

20

0.99

11.4

40

0.99

10.8

60

0.99

10.0

80

0.99

9.50

100

0.99

8.90

120

0.97

8.10

140

0.96

7.60

160

0.95

7.11

200

0.91

6.21


73

Example 18.2 • For Camphor, M =152.23 g mol-1, (εr-1)/(εr+2)

Pm/(cm3 mol-1)

θ/℃

(103K)/T

0

3.66

12.5

20

3.41

11.4

0.776

119

40

3.19

10.8

0.766

118

60

3.00

10.0

0.750

115

80

2.83

9.50

0.739

114

100

2.68

8.90

0.725

111

120

2.54

8.10

0.703

110

140

2.42

7.60

0.688

109

160

2.31

7.11

0.670

107

200

2.11

6.21

0.634

106

εr

0.793

122

The intercept lies at 83.52 and the slope is 10.55 Chap 18. Molecular interactions

74

The intercept lies at 83.5 and the slope is 10.6


75

N A • The intercept lies at 83.5 = , so α’=3.3x10-23 cm3 3ε 0

NA • The slope is 10.6 = , gives the dipole moment 9ε 0k 2

as µ = 4.39x10-30 C m, corresponding to 1.32 D.

• Because the Debye equation describes molecules that are free to rotate, the data show that camphor, which does not melt until 175°C, is rotating even in the solid. It is an approximately spherical molecule.


Self-test 18.3 • Determination of the polarizability volume • The relative permittivity of chlorobenzene is 5.71 at 20℃ and 5.62 at 25 ℃. Assuming a constant density (1.11 g cm3), estimate its polarizability volume and dipole moment. • [M = 112.5 g cm-3; slope=8.142,α’= 1.35 x 10-23 cm3; intercept=34.1404; µ = 1.16 D]


76

Maxwell equations • The Maxwell equations that describe the properties of electromagnetic radiation relate the refractive index at a (visible or ultraviolet) specified wavelength to the relative permittivity at that frequency: • nr = εr1/2 • Therefore, the molar polarization, Pm, and the molecular polarizability, α, can be measured at frequencies typical of visible light (about 1015 to 1016 Hz) by measuring the refractive index of the sample and using the Clausius-Mossotti equation.


77

Interactions between molecules • The van der Waals forces are the weak, attractive intermolecular forces arising from permanent or induced electric dipole moments. A van der Waals molecule is a cluster of closed-shell atoms or molecules that are held together by van der Waals forces that depends on the distance between the molecules as l/r 6. • In addition, there are interactions between ions and the partial charges of polar molecules and repulsive interactions that prevent the complete collapse of matter to nuclear densities. • The repulsive interactions arise from Coulombic repulsions and, indirectly, from the Pauli principle and the exclusion of electrons from regions of space where the orbitals of neighbouring species overlap. Chap 18. Molecular interactions

78

18.4 Interactions between dipoles • When two charges q1 and q2 are separated by a distance r in a vacuum, the potential energy of interaction is

q1 q 2 V r   4ε 0r

• When a dipole μ1 = q1 ℓ and a point charge q2 is close to each other, the potential energy of interaction is

1 q 2 V r   4ε 0r 2 • Sum of the repulsion of like charges and the attraction of opposite charges given the total potential energy of interaction. For a point dipole, l  r.


79

80

The potential energy of interaction between a dipole and a point charge is the sum of the repulsion of like charges and the attraction of opposite charges. For a point dipole, l  r.


81

Justification 18.4

The interaction between a point charge and a point dipole The sum of the potential energies of repulsion between like charges and attraction between opposite charges in the orientation shown in (9) is Answer The sum of two contributions with x = ℓ/2r is   1  q1 q 2 q1 q 2  q1 q 2   V r    4ε 0  r   r    4ε 0r   2 2 

1   1    1 x 1 x 

Because ℓ « r for a point dipole, this expression can be simplified by expanding the terms in x and retaining only the leading term: q1 q2 q1 q2 l ٛ 2xq1 q2 2 2  V r    1  x  x  ...  1  x  x  ...   4ε 0r 2ε 0r 2 4ε 0r






82


83

The potential energy of interaction between two dipoles is the sum of the repulsions of like charges and the attractions of opposite charges. This illustration shows a collinear arrangement of dipoles. Chap 18. Molecular interactions

84

Example 18.3

• Calculating the interaction energy of two dipoles • Calculate the potential energy of interaction of two dipoles in the arrangement shown in Fig 18.9 when their separation is r and θ =0 Answer The sum of four contributions with x = ℓ/r is 1  q1 q 2 q1 q 2 q1 q 2 q1 q 2  q1 q 2    V r     4ε 0  r   r r r   4ε 0r

1   1 2   1 x  1 x

The expansion of two terms in x and only the first surviving term retains, which is 2x 2. Now with μ1 = q1ℓ and μ2 = q2ℓ, the potential energy 2 2x q1 q 2 1 2  V r    3 4ε 0r 2ε 0r Chap 18. Molecular interactions

85

A parallel arrangement of electric dipoles as used for Self-test 18.4


86

Self-test 18.4 • Derive an expression for the potential energy when the interaction of two dipoles are in the arrangement shown in (11) when their separation is r and θ =90° Answer The sum of four contributions with x = ℓ/r is

q1 q 2 q1 q 2 q1 q 2 1  q1 q 2  V r      r 4ε 0  r r 2  2 r 2  2 d 2 = (r 2+ℓ2) = r 2(1+x2 ) ≈ r

2

 2q1 q 2  1   1  2 ε r  4  1 x 0  

(1+ x2/2 - x4/8 +… )2

Now with μ1 = q1ℓ and μ2 = q2ℓ, retains term x 2. theV (r ) is 2 4 1   1 2  r  1 2   x x    1   1  1  V r    2  2 2ε 0r   d  2ε 0r    2 8     2 4 4   1 2  x x x   1 2    1  1     2  2ε 0r    2 8 4   4ε 0r 3 Chap 18. Molecular interactions

  

87

For this arrangement, the expression should be multiplied by cosθ when the dipole lies at an angle θ to the axis of the dipole.


88


Multipole

89

• Monopole : is a point charge • Dipole : A point dipole is an electric dipole in the limit R → 0 • Multipole An electric multipole is an array of point charges, of which an electric dipole, two equal and opposite point charges, is a special case. • n-pole Specifically, an electric n-pole is an array of n point charges with an n-pole moment but no lower moment. A single point charges with zero overall charge and no dipole moment is called a quadrupole. Chap 18. Molecular interactions

90


91

• An octupole consists of an array of point charge that sum to zero and which has neither a dipole moment nor a quadrupole moment, (as for CH4 molecule). • The interaction energy falls off more rapidly the higher the order of the multipole. • The n-pole interacting with an m-pole, the potential energy varies with distance as

V r  

1

4ε 0  r

n m 1


92

Typical charge arrays corresponding to electric multipoles. The field arising from an arbitrary finite charge distribution can be expressed as the superposition of the fields arising from a superposition of multipoles. Chap 18. Molecular interactions

93

(b) Electric field

• The potential energy of a charge q1 in the presence of another charge q2 is V (r )= q1φ where φ is the Coulomb potential.

q1 q2 • V (r ) =q1  , and the electric field is E   4ε 0 r q2 • The electric field around a point charge q2 is E  2 4ε 0 r • For a point charge q1 close to a dipole μ2 = q2 ℓ, the q1 2 potential energyV r   2 4ε 0r • So the electric field around a dipole μ2 is E 

2 3 4ε 0 r

• The electric field of a multipole decreases more rapidly with distance than a monopole (a point charge).


94


Further Information 18.1 dipole-dipole interaction • Potential energy of interaction between two dipoles • The potential energy of μ2 in the electric field E1 that generated by μ 1 is given by the dot (scalar) product:

1  2 V r    E1  2 2 4ε 0r • Consider a distribution of charges qi located at xi yi zi from the origin. The Coulomb potential φ at r is qi qi  r      1 2 / 2 2 2 i 4ε 0 x  x i   y  y i   z  z i  i 4ε 0 r  ri  • Assume all charges are close to the origin (or ri 0), as expected on the basis of the increase in polarity of the solvent, but exothermic (∆transferH < 0). Therefore, it is a large decrease in the entropy of the system (∆transferS < 0) that accounts for the positive Gibbs energy of transfer. Chap 18. Molecular interactions

116


117

• For example, the process CH4(in CCl4 ) → CH4 (aq) has ∆transferG =+12 kJ mol-1, ∆transferH = -10 kJ mol-1, and ∆transferS = -75 J K-1mol-1 at 298 K. • Substances characterized by a positive Gibbs energy of transfer from a nonpolar to a polar solvent are called hydrophobic.


hydrophobicity constant, π

118

• It is possible to quantify the hydrophobicity of a small molecular group R by defining the hydrophobicity constant, π, as π = log (S/S₀)

where S is the ratio of the molar solubility of the compound R-A in octanol, a nonpolar solvent, to that in water, and S₀ is the ratio of the molar solubility of the compound H-A in octanol to that in water. • Positive values of π indicate hydrophobicity and negative values of π indicate hydrophilicity, the thermodynamic preference for water as a solvent. It is observed experimentally that the π values of most groups do not depend on the nature of A. Chap 18. Molecular interactions

119

• For example: • R = CH3, CH2CH3, (CH2)2CH3, (CH2)3CH3, (CH2)4CH3 • π = 0.5,

1.0,

1.5,

2.0,

2.5

and we conclude that acyclic saturated hydrocarbons become more hydrophobic as the carbon chain length increases. • This trend can be rationalized by ∆transferH becoming more positive and ∆transferS more negative as the number of carbon atoms in the chain increases.


At the molecular level

120

• Formation of a solvent cage around a hydrophobic molecule involves the formation of new hydrogen bonds among solvent molecules. This is an exothermic process and accounts for the negative values of ∆transferH . • On the other hand, the increase in order associated with formation of a very large number of small solvent cages decreases the entropy of the system and accounts for the negative values of ∆transferS. However, when many solute molecules cluster together, fewer (albeit larger) cages are required and more solvent molecules are free to move. The net effect of formation of large clusters of hydrophobic molecules is then a decrease in the organization of the solvent and therefore a net increase in entropy of the system. Chap 18. Molecular interactions

121

• This increase in entropy of the solvent is large enough to render spontaneous the association of hydrophobic molecules in a polar solvent. • The increase in entropy that results from fewer structural demands on the solvent placed by the clustering of nonpolar molecules is the origin of the hydrophobic interaction, which tends to stabilize groupings of hydrophobic groups in micelles and biopolymers (Chapter 19). The hydrophobic interaction is an example of an ordering process that is stabilized by a tendency toward greater disorder of the solvent.


(h) The total attractive interaction

122

• We shall consider molecules that are unable to participate in hydrogen bond formation. The total attractive interaction energy between rotating molecules is then the sum of the three van der Waals contributions discussed above. (Only the dispersion interaction contributes if both molecules are nonpolar). • In a fluid phase, all three contributions to the potential energy vary as the inverse sixth power of the separation of the molecules, so we may write

V = -C6 /r

6

• where C6 is a coefficient that depends on the identity of the molecules. Chap 18. Molecular interactions

123

• Although this equation represents the attractive interactions between molecules, it has only limited validity. • First, we have taken into account only dipolar interactions of various kinds, for they have the longest range and are dominant if the average separation of the molecules is large. However, in a complete treatment we should also consider quadrupolar and higher-order multipole interactions, particularly if the molecules do not have permanent electric dipole moments. • Secondly, the expressions have been derived by assuming that the molecules can rotate reasonably freely. That is not the case in most solids. and in rigid media the dipole-dipole interaction is proportional to 1/r 3 because the Boltzmann averaging procedure is irrelevant when the molecules are trapped into a fixed orientation. Chap 18. Molecular interactions

Axilrod-Teller formula • A different kind of limitation is that the eqn relates to the interactions of pairs of molecules. There is no reason to suppose that the energy of interaction of three (or more) molecules is the sum of the pairwise interaction energies alone. • The total dispersion energy of three closed-shell atoms, for instance, is given approximately by the Axilrod-Teller formula:

C6 C6 C6 C' V  6 6 6 3 rAB rBC rCA rAB rBC rCA  • where C’ = a (3 cos θA cos θB cos θC +1) Chap 18. Molecular interactions

124

The Axilrod-Teller formula


125

Ailrod-Teller formula

126

• Ailrod-teller formula : This formula allows the calculation of the interaction of three or more molecules • V = C6/rAB6 - C6/rBC6 – C6/rCA6 + C ’/(rAB rBC rBC)3 where C’ = a (3 cos θA cos θB cos θC +1) • The parameter a is approximately equal to (¾) α'C6; the angles θ are the internal angles of the triangle formed by the three atoms (14). The term in C' (which represents the non-additivity of the pairwise interactions) is negative for a linear arrangement of atoms (so that arrangement is stabilized) and positive for an equilateral triangular cluster. • It is found that the three-body term contributes about 10 per cent of the total interaction energy in liquid argon. Chap 18. Molecular interactions

18.5 Repulsive and total interactions

127

• When molecules are squeezed together, the nuclear and electronic repulsions and the rising electronic kinetic energy begin to dominate the attractive forces. • The repulsions increase steeply with decreasing separation in a way that can be deduced only by very extensive, complicated molecular structure calculations. • However, in many cases, progress can be made by using a greatly simplified representation of the potential energy, where the details are ignored and the general features expressed by a few adjustable parameters. Chap 18. Molecular interactions

128

18.9a In a two-dimensional molecular dynamics simulation that uses periodic boundary conditions when one particle leaves the cell its mirror image enters through the opposite face.


18.5 Repulsive and total interactions • Hard-sphere potential • One such approximation is the hard sphere potential, in which it is assumed that the potential energy rises abruptly to infinity as soon as the particles come within a separation d: • A hard-sphere potential is a model intermolecular potential in which there is no interaction between two particles until they are at a specified distance, when the potential energy climb abruptly to infinity • V = ∞ if r ≤ d • V = 0 if r > d Chap 18. Molecular interactions

129

22.4 Repulsive and total interactions

130

• Mie potential • This is an approximation used for the intermolecular potential n m energy where n>m C C

    V r        r  r 

• Lennard-Jones potential

• The Lennard-Jones potential is a model intermolecular potential energy that expresses the potential energy of two molecules as a function of their separation r in terms of two parameters. The n 6 Lennard-Jones (n,6)-potential is C  C 

V r   

   r  r 

• The first term (when n >6) expresses the repulsive contribution to the potential energy, which is operative at short distances (essentially contact) and the second term expressed the longer range attractive interactions. Several attractive intermolecular interactions are inversely proportional to r 6. Chap 18. Molecular interactions

131


The Lennard-Jones potential the relation of the parameters to the features of the curve, and the two contributions. Note that 21/6 = 1.122 . . . .

  ro 12  ro  6  V r   4         r   r     ε : well depth ro: closest distance when V (r )=0


132

Lennard-Jones (12,6)-potential

133

• For mathematical reasons it is sometimes convenient to use the Lennard-Jones (12,6)-potential, with n =12. which is normally parametrized as follows:   ro 12  ro 6  V r   4        r    r  • The parameter ε is the depth of the potential well and ro is the separation at which the curve cuts through V (r )=0. The well minimum occurs at r = 21/6 ro • The exp-6 potential: V = 4ε{e–r/r0 – (r0/r)6}. A potential with an exponential repulsive term and a 1/r 6 attractive term. Chap 18. Molecular interactions

134


135


Atomic force microscopy

136

• The potential energy of interaction of molecules was the primary interest in the molecular-size probes and the surface.. • Atomic force microscopy (AFM): a technique in which the force between a molecular sized probe and a surface is monitored. • As force, F, is the negative slope of potential, 13 7  dV r  24   ro   ro    F  2     dr ro   r   r  


The greatest attractive force

137

• For a Lennard Jones potential, the net attractive force is greatest at a distance of r = 1.244σand with a force F = -2.397ε/r o:   ro 12  ro 6  V r   4        r    r  2  V r   4 12 13 6 7   12ro r  6ro r 2 r r 12 14 6 8  4 156ro r  42ro r



 4r



14



156r 156r

 4r 29 6 6 r  ro 7 14

12 o

 42ro r 6 6

 42ro r 29 r  6 ro 7

12 o


6 6

   0

Molecular beam

138

• Molecular beam A molecular beam is a narrow stream of molecules with a narrow spread of velocities and in some cases, in specific internal states or orientations. Molecular beam studies of non-reactive collisions are used to explore the details of intermolecular interations with a view to determining the shape of the intermolecular potential. Molecular beam studies of reactive collisions are used to study the details of collisions that lead to the exchange of atoms and hence to explore the shape and features of the molecular potential energy surface. Surfaces, particularly the ability of particular features to act as sites of catalytic activity, are also studied using molecular beams. (Figure) Chap 18. Molecular interactions

139


140

Fig.18.13 A 3D QSAR analysis of the binding of steroids, molecules with the carbon skeleton shown, to human corticosteroid-binding globulin (CBG). The ellipses indicate areas in the protein's binding site with positive or negative electrostatic potentials and with little or much steric crowding. It follows from the calculations that addition of large substituents near the left-hand side of the molecule (as it is drawn on the page) leads to poor affinity of the drug to the binding site. Also, substituents that lead to the accumulation of negative electrostatic potential at either end of the drug are likely to show enhanced affinity for the binding site. (Adapted from P. Krogsgaard-Larsen, T. Liljefors, U. Madsen (ed.), Textbook of drug design and discovery, Taylor & Francis, London (2002).) Chap 18. Molecular interactions

141


Effects of Molecular Interactions on Collisions

142

• The extent of scattering may depend on the relative speed of approach as well as the impact parameter b. Two paths leading to the same destination will interfere quantum mechanically; in this case they give rise to quantum oscillations in the forward direction. Interference between the numerous paths at that angle modifies the scattering intensity markedly.



143

Potential Energy The basic arrangement of a molecular beam apparatus. The atoms or molecules emerge from a heated source, and pass through the velocity selector, a train of rotating disks. The scattering occurs from the target gas (which might take the form of another beam), and the flux of particles entering the detector set at some angle is recorded. (a) The definition of the solid angle, d, for scattering.

(b) The definition of the impact parameter, b, as the perpendicular separation of the initial paths of the particles. Chap 18. Molecular interactions

144


145


The impact parameter • impact parameter • The impact parameter, b, is the initial perpendicular distance between a target molecule and the line of approach of a projectile molecule. (Figure)


146

147



Potential Energy Three typical cases for the collisions of two hard spheres:

(a) b = 0, giving backward scattering; (b) b  RA + RB, giving forward scattering; (c)

0  b  RA + RB, leading to scattering into one direction on a ring of possibilities.

(The target molecule is taken to be so heavy that it remains virtually stationary.)


148


• Figure Collision of two spherically symmetrical molecules with an impact parameter b. The diagram is drawn so that the center of gravity of the system does not shift during the collision. The angle of deflection is χ.


149

150

18.17 The extent of scattering

may depend on the relative speed of approach as well as the impact parameter. The dark central zone represents the repulsive core; the fuzzy outer zone represents the long-range attractive potential.


151


152


153

18.18 Two paths leading to the

same destination will interfere quantum mechanically; in this case they give rise to quantum oscillations in the forward direction.


154

• Quantum oscillation • A particle with a certain impact parameter might approach the attractive region of the potential in such a way that the particle is deflected towards the repulsive core, which then repels it out through the attractive region to continue its flight in the forward direction. Some molecules, however, also travel in the forward direction because they have impact parameters so large that they are un-deflected. The wave functions of the particles that take the two types of path interfere, and the intensity in the forward direction is modified. This effect is called quantum oscillation. (Figure) Chap 18. Molecular interactions


155

Fig. 18.19 The dark central zone represents the repulsive core; the fuzzy outer zone represents the long-range attractive potential. The interference of paths leading to rainbow scattering. The rainbow angle, r, is the maximum scattering angle reached as b is decreased. Chap 18. Molecular interactions

156

• Rainbow scattering • As the impact parameter b for a molecular collision is decreased the angle of scattering increases, passes through a maximum, and then decreases again. There is strong constructive interference between neighboring paths at the turning point, and hence a pronounced peak in scattered intensity. The phenomenon is termed rainbow scattering, and the angle at which it occurs is the rainbow angle, θr, (Figure)


157



• Collision trajectories for a pair of molecules interacting by a Lennard-Jones 6-12 potential.


158

Surface Tension

159

• 表面張力（surface tension）在兩相(特別是氣-液)界面上，處處存在著一種張力，它垂直與表面的邊界，指向液體方向並與表面相切。 • 把作用於單位邊界線上的這種力稱為表面張力，用γ表示，單位是 N/m. • 氣-液界面的表面張力γ是製造單位界面(dAS) ，與所需功 dw 的比例常數 dw = γ dAS. • 廣義的表面自由能定義：保持對應量的特徵自然變數不變，每增加單位表面積時，對應量熱力學函數的增加值。 • 狹義的表面自由能定義：保持溫度、壓力和組成不變，每增加單位表面積時，Gibbs自由能(以功的形式)的增加值，稱為表面 Gibbs自由能，或簡稱表面自由能或表面能，單位為 J/m2 。 Chap 18. Molecular interactions

160

F  2γl; AS  2h l F dh 2γl dw   γ dAS 2l dh 2l


Table 6.2 Surface Tensions of Some Liquids

Substance

t/℃

γ/mN m-1

Inorganic Aluminum

30

840

Gold

40

1102

Lead

60

453

Mercury

100

472

Lead chloride

140

138

Sodium chloride

180

106

Oxygen

-183

13.1

Nitrogen

-183

6.2


161

Table 6.2 Surface Tensions of Some Liquids

Substance

t/℃

γ/mN m-1

Organic Acetone

20

23.7

Benzene

20

28.88

Ethanol

20

22.8

Diethyl ether

20

16.96

Glycerol

20

63.4

Water

0

75.7

15

73.5

25

72.0

50

67.9

100

58.8


162

163


164


Curved surfaces

165

• 凸面(Convex)的曲率半徑取正值，凹面(Concave)的曲率半徑取負值。 • A curved surface of a liquid, or a curved interface between phases, exerts a pressure so that the pressure is higher in the phase on the concave side of the interface. (在液體屈面或是兩相的界面兩邊,施加於凹面的內部壓力較外部壓力大) • The pressure inside of a rubber balloon is higher than the atmospheric pressure. (橡皮氣球的內部壓力比大氣壓力較高) • Bubble is a region in which vapor is trapped by a thin film of liquid. (液膜泡是平衡蒸氣包在液體薄膜中) • Cavity is a vapor-filled hole in a liquid. (氣泡是在液體中的洞,填充了平衡蒸氣壓) • Droplet is a small volume of liquid at equilibrium with and surrounded by its vapor. (液滴是小體積的液體被周圍的平衡蒸氣壓包圍)


166

Figure A

• Spherical droplet of a pure liquid in contact with its vapor in a closed container at temperature T. Solid line for the position of the surface, and the dashed line shows the effect of an infinitesimal expansion of the droplet. Chap 18. Molecular interactions

167

Surface Energy of droplet

• The surface area AS is related to the volume VL of the liquid and to the amount nL of the liquid.

4 3 VL  π r 3

AS  4 π r 2

dVL  4 π r 2dr

dAS  8 π r dr

• At constant V and T, the work of surface formation is equal to the change in Helmholtz energy (dA) of the system. • The fundamental equations at constant T for two phases are: 2 dAL T  -PL dVL  μL dnL  γ dAS dAS  dVL

r

dAV T  -PV dVV  μV dnV

• At equilibrium:

dA T

 0  PV -PL  dVL 

2γ

dVL

or PL-PV  

2γ

r r where dA = dAL + dAV ; dVV = - dVL ; dnL = - dnV ; V =  L ; Chap 18. Molecular interactions

Phase Equilibrium

168

• The Laplace equation for the vapor pressure at a curved surface show that, as bubble or cavity becomes larger, the pressure difference across its surface decreases.

Pin - Pout = 2 γ/ r where γ is the surface tension and r is the radius, Thus the pressure inside the curvature is higher than the pressure outside the boundary, Pin > Pout .


Surface Energy of cavity

169

• For a cavity, the outward force = Pressure x area = 4πr 2Pin where the pressure inside is Pin and its radius is r. • The inward force from the external pressure = 4 π r 2Pout • The change in surface area when the radius of sphere change from r to r + dr is 2 dAS  4 π r  dr  - 4 π r 2  8 π r dr • As work = force x distance. The work done when the surface is stretched by dr is γdAS dw  F x dr  8 πγr dr

• The force opposing stretching through a distance dr when the radius is r. At equilibrium, the outward and inward forces are balanced. The force inwards arises from both external pressure and the surface tension:4 π r 2Pin = 4 π r 2Pout + 8 π γr • Now we have

Pin - Pout = 2 γ/ r


170


171

The balance of forces that results in a contact angle, θc. γsg = γsl + γlg cos θc.


Capillary rise and surface tension

172

• Capillary action is the tendency of liquids to rise up capillary tubes of narrow bore. The height h to which a liquid of mass density ρ and surface tension γ will rise is

2γ h ρ gr

where g is the acceleration of free fall. • There may be a nonzero angle between the edge of the meniscus and the wall called the contact angle θc, • If the contact angle between the tube and the liquid is θc rather than zero, then the right hand side will multiply by cos(θc). 1 γ  ρ g hr cosC  2 Chap 18. Molecular interactions

Illustration

Illustration • If water at 25℃ rises through 7.36 cm in a capillary of radius 0.20 mm, its surface tension at that temperature is 1 γ  ρ g hr 2

= ½ x(997.1 kg m-3)x(9.81 m s-2)x(7.36x10-2 m)x(2.0x10-4 m) =72 mN m-1 where we have used 1 kg m s-2 = 1 N


173

174


Phase Equilibrium

175

• 溫度升高，表面張力下降，當達到臨界溫度TC 時，表面張力趨向於零。這可用熱力學公式說明：

dG  S dT V dP   dA   B dnB 運用全微分的性質，可得：

B

S  ( )T ,P ,nB  ( ) A ,P ,nB A T 等式左方為正值，因為表面積增加，熵總是增加的。所以 γ 隨T 的增加而下降。


176

S  ( )T ,P ,nB  ( ) A ,P ,nB A T


Contact Angle and Interfacial Tension

177

• The contact angle θc , is related to the surface tensions of the solid/gas; solid/liquid; and liquid/gas interfacial surface tension: γsg; γsl; γlg; by cos θc = (γsg- γsl) / γlg ; • Capillary rise occurs for 0° < θc< 90° • Capillary depress occurs for 90° < θc< 180°

An insect walking on the water


178

= 0° -

< 90° >0

= 90° > 90° =0


= 180° 0° corresponds to the wetting of a surface, and it is possible when wad > 2 γlg • equation can be written as:

cos C  

Wad

 lg

-1


181

Surface Wetting • the liquid 'wets' (spreads over) the surface, corresponding to 0 < θc < 90°, when 1 < wad /γlg < 2 • The liquid does not wet the surface, corresponding to 90° < θc < 180°, when 0 < wad /γlg < 1. • For mercury in contact with glass, θc =140°, which corresponds to wad /γlg =0.23, indicating a relatively low work of adhesion of the mercury to glass on account of the strong cohesive forces within mercury.


182

Surface Wetting

• The criteria for surface wetting is θc >0° corresponds to the wetting of a surface, and it is possible when wad > 2 γlg

(Source: Fonds der Chemischen Industrie, Germany; imageseries "Tenside") Chap 18. Molecular interactions

183

18.8 Condensation

184

• The vapor pressure of a liquid depends on the pressure applied to the liquid. Because curving a surface gives rise to a pressure differential of 2 γ /r, we can expect the vapor pressure above a curved surface to be different from that above a flat surface. By substituting this value of the pressure difference into eqn 4.3 (P = P * exp(2γV̿ L /rRT) here P is the vapor pressure when the pressure difference is zero) we obtain the Kelvin equation for the vapor pressure of a liquid when it is dispersed as droplets of radius r: • P = P * exp(2γV̿ L /rRT)


Chap 18 Phase Equilibrium

• For droplet in a vapor: For cavity in a liquid: 2γ 2γ PV -PL  PL-PV  r r • When the radius of curvature is increased to infinity, this pressure difference disappears, as it must for a planar surface. • The pressure is always greater on the concave side(凹面) of the surface. • The volume of droplet VL equals to the molar volume V̿L times the amount nL :  2  2 dAS    dVL    V L dnL r  r  Chap 18. Molecular interactions

185

186

Equilibrium Vapor Pressure

• The fundamental equation for G for the liquid phase is

dG L T,P  μ *L dnL  2/r V L dnL





 μL*  2V L/r dnL  μL dnL

• The fundamental equation for G for the vapor phase is dG V T,P  μ V dn V • At equilibrium dG T,P  0  dG L T,P  dG V T,P  μL*  2V L/r dnL  μ V dn V



dnL  -dn V ; μ V  μL*  2V L/r  μL Chap 18. Molecular interactions



Kelvin equation

187

• Assuming that the vapor is an ideal gas,

μV   RT lnP/P   μL*  2 γV L /r • For a flat surface, (r →∞) μV   RT lnP * /P   μL*

• The difference gives the Kelvin equation:

RT lnP/P *   2 γV L /r

• For the vapor pressure of liquid that is dispersed as droplet of radius r , P = P * exp(2γV̿ L /rRT) • P* is the vapor pressure of the bulk liquid of molar volume ̿V̿ and γ is the surface tension of the liquid. • For the vapor pressure inside a spherical cavity of radius r change the sign in the exponent: P = P * exp(-2γV̿ V /rRT) Chap 18. Molecular interactions

L

188

Fig 6.7

Fig 6.7 Effect of radius of curvature of a surface on the vapor pressure of water at 25 ℃ . (a) for droplet and (b) for cavity. Droplet (a) (b)

 2V in P P *  exp   rRT

Bubble


   

189

• The analogous expression for the vapor pressure inside a cavity can be written at once. The pressure of the liquid outside the cavity is less than the pressure inside, so the only change is in the sign of the exponent in the last expression. • For droplets of water of radius I μm and I nm the ratios p/p* at 25℃ are about 1.001 and 3, respectively. • The second figure, although quite large, is unreliable because at that radius the droplet is less than about 10 molecules in diameter and the basis of the calculation is suspect. The first figure shows that the effect is usually small; nevertheless it may have important consequences. Chap 18. Molecular interactions

190

Phase Equilibrium Example The size of a water droplet when condensation starts Water vapor is rapidly cooled to 25 ℃ to find the degree of supersaturation required to nucleate water droplets spontaneously. It is found that the vapor pressure of water must be four times its equilibrium vapor pressure at 25 ℃. (a) Calculate the radius of a stable water droplet formed at this degree of supersaturation. (b) How many water molecules are there in the droplet?

Ans: Assumed that the surface tension is independent of the radius of curvature, 2V L 2 0.07197 N m-1  18x10 -6 m3 mol -1  a)r    0.75 nm 8.3145 J K -1 mol -1 298 K  ln 4 RT ln P/P sat  3 4 4 -9 3 πr ρ  0.75x10 m 1x10 3 kg m -3 3  b) N  3 M/N A 18x10 -3 kg mol -1 6.022x10 23 mol










 -1



 59

Phase Equilibrium

191

• A metastable phase (亞穩定相): A thermodynamically unstable phase but prevented by kinetic reason from a transition to a stable phase. • A superheated liquid is a liquid above its boiling temperature. • A supercooled liquid is a liquid below its freezing point. • When a vapor is cooled such that its chemical potential lies above the chemical potential of its liquid. This thermodynamically unstable vapor phase is said to be supersaturated (過飽和). • It is only unstable with respect to the bulk liquid but not small droplets. • The nucleate (起成核作用) means to provide a center at which a condensation process can occur at spontaneous nucleation centers (自發凝結核核中心 ). • This is the chance formation of droplets that are large enough to initiate condensation in a supersaturated phase.

人工造霧機: Chap 18. Molecular interactions

192

• Consider the formation of a cloud: Warm, moist air rises into the cooler regions higher in the atmosphere. At some altitude the temperature is so low that the vapor becomes thermodynamically unstable with respect to the liquid and we expect it to condense into a cloud of liquid droplets. • The initial step can be imagined as a swarm of water molecules congregating into a microscopic droplet. Because the initial droplet is so small it has an enhanced vapor pressure. Therefore, instead of growing it evaporates. This effect stabilizes the vapor because an initial tendency to condense is overcome by a heightened tendency to evaporate. Chap 18. Molecular interactions

193

• The vapor phase is then said to be supersaturated. It is thermodynamically unstable with respect to the liquid but not unstable with respect to the small droplets that need to form before the bulk liquid phase can appear, so the formation of the latter by a simple, direct mechanism is hindered.


194

• Clouds do form, so there must be a mechanism. Two processes are responsible. • The first is that a sufficiently large number of molecules might congregate into a droplet so big that the enhanced evaporative effect is unimportant. The chance of one of these sponetaneous nucleation centers forming is low, and in rain formation it is not a dominant mechanism. • The more important process depends on the presence of minute dust particles or other kinds of foreign matter. These nucleate the condensation (that is, provide centers at which it can occur) by providing surfaces to which the water molecules can attach. Chap 18. Molecular interactions

195

• Liquids may be superheated above their boiling temperatures and supercooled below their freezing temperatures. In each case the thermodynamically stable phase is not achieved on account of the kinetic stabilization that occurs in the absence of nucleation centers. For example, superheating occurs because the vapor pressure inside a cavity is artificially low, so any cavity that does form tends to collapse. This instability is encountered when an unstirred beaker of water is heated, for its temperature may be raised above its boiling point. Violent bumping often ensues as spontaneous nucleation leads to bubbles big enough to survive. To ensure smooth boiling at the true boiling temperature, nucleation centers, such as small pieces of sharp-edged glass or bubbles (cavities) of air, should be introduced. Chap 18. Molecular interactions

雲霧的成因 • 霧和雲是由浮游在空中的小水滴或冰晶組成的水汽凝結物，只是霧生成在大氣的近地面層中，而雲生成在大氣的較高層而已。霧既然是水汽凝結物，因此應從造成水汽凝結的條件中尋找它的成因。大氣中水汽達到飽和的原因不外兩個：一是由於蒸發，增加了大氣中的水汽；另一是由於空氣自身的冷卻。對於霧來說冷卻更重要。當空氣中有凝結核時，飽和空氣如繼續有水汽增加或繼續冷卻，便會發生凝結。凝結的水滴如使水準能見度降低到1千米以內時，霧就形成了。 • 另外，過大的風速和強烈的擾動不利於霧的生成。 • 因此，凡是在有利於空氣低層冷卻的地區，如果水汽充分，風力微和，大氣層結穩定，並有大量的凝結核存在，便最容易生成霧。一般在工業區和城市中心形成霧的機會更多，因為那裏有豐富的凝結核存在。


196

Further Information 18.2 • Crossed beam technique • When an incident molecular beam is directed in to a second molecular beam in order to study the collisions of these highly selected states, the method is called a crossed beam technique.


197

198

• Incident beam flux • The incident beam flux I is the number of particles from the incident molecular beam passing through a given target area. • Differential scattering cross-section • In molecular beam work, the differential scattering cross-section, σ, is the constant of proportionality between the number of molecules that are scattered per unit time into a cone that represents the area covered by the eye of the detector, dI , and the intensity, I, of the incident beam, the number density of target molecules, and the path length, dx, through the sample: dI = σ I N dx Chap 18. Molecular interactions

199


200


201


state-to-state cross section

202

• A more detailed version of this quantity is the stateto-state cross section σnn’ , for scattering from an initial state n to a final state n’, which depends among other things, on the speed of approach υ.


203


204


205


206


207


208


homopolar contribution

209

• homopolar contribution (同極的貢獻 ) One of the contribution to the dipole moment of a molecule is the imbalance in atomic radii result in an imbalance of charge density because the enhanced charge density associated with the overlap region close to the nucleus of smaller atom. • The diagram shows how the charge accumulation leads to a region of negative charge closer to the smaller atom. • Such a "homopolar dipole" contribution to the total dipole moment can arise even in the absence of a difference in electronegativity between the two atoms. Chap 18. Molecular interactions

210

18.1 One of the contributions to the dipole moment of a molecule is the imbalance of charge arising from the overlap of orbitals of different radii. This diagram shows how the charge accumulation leads to a region of negative charge closer to the smaller atom. Chap 18. Molecular interactions

211

18.27 The radial distribution function of the oxygen atoms in liquid water at three temperatures. Note the expansion as the temperature is raised.

Source: A.H. Narten, M.D. Danford, and H.A. Levy, Discuss. Faraday Soc. 43, 97 (1967) Chap 18. Molecular interactions

212

18.28 The radial distribution function for a simulation of a liquid using impenetrable hard spheres (ball-bearings)
