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van der Waals’ Interactions

van der Waals’ Interactions. Refers to all interactions between polar or nonpolar molecules, varying as r -6 . Includes Keesom, Debye and dispersive interactions. Values of interaction energy are usually only a few kT. Summary. Charge-charge. Coulombic. Dipole-charge. Dipole-dipole.

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van der Waals’ Interactions

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  1. van der Waals’ Interactions • Refers to all interactions between polar or nonpolar molecules, varying as r -6. • Includes Keesom, Debye and dispersive interactions. • Values of interaction energy are usually only a few kT.

  2. Summary Charge-charge Coulombic Dipole-charge Dipole-dipole Keesom Charge-nonpolar Dipole-nonpolar Debye Nonpolar-nonpolar Dispersive Type of InteractionInteraction Energy, w(r) In vacuum:e=1

  3. Interaction between ions and polar molecules • Interactions involving charged molecules (e.g. ions) tend to be stronger than polar-polar interactions. • For freely-rotating dipoles with a moment of u interacting with molecules with a charge of Q we saw: • One result of this interaction energy is the condensation of water (u = 1.85 D) caused by the presence of ions in the atmosphere. • During a thunderstorm, ions are created that nucleate rain drops in thunderclouds (ionic nucleation).

  4. Comparison of the Dependence of Interaction Potentials on r n = 1 Coulombic n = 2 n = 6 n = 3 van der Waals Dipole-dipole Not a comparison of the magnitudes of the energies!

  5. Cohesive Energy  • Def’n.: Energy required to separate all molecules in the condensed phase or energy holding molecules in the condensed phase. • In Lecture 1, we found that for single molecules with a potential w(r) = Cr -n, and with n>3: with r = number of molecules per unit volume s-3, where s is the molecular diameter. So for dispersive interactions, n = 6 and C is the London constant: 1/2 to avoid double counting! • For one mole, Esubstance = (1/2)NAE • Esubstance = sum of heats of melting + vaporisation. • Predictions agree well with experiment!

  6. Boiling Point • At the boiling point, TB, for a liquid, the thermal energy of a molecule, 3/2 kTB, will exactly equal the energy of attraction between molecules. • Of course, the strongest attraction will be between the “nearest neighbours”, rather than pairs of molecules that are farther away. • The interaction energy for van der Waals’ interactions is of the form, w(r) = -Cr-6. If molecules have a diameter of s, then the shortest centre-to-centre distance will likewise be s. • Thus the boiling point is approximately:

  7. C can be found experimentally from deviations from the ideal gas law: Evaluated at close contact where r =s. Comparison of Theory and Experiment Note that ao and C increase with s.

  8. H H H H C=O H H C-C H H H C-O-H H H Additivity of Interactions MoleculeMol. Wt.u (D) TB(°C) Dispersive only Ethane: CH3CH3 30 0 -89 Keesom + dispersive Formaldehyde: HCHO 30 2.3 -21 H-bonding + Keesom + dispersive Methanol: CH3OH 32 1.7 64

  9. ze q r Problem Set 1 1.Noble gases (e.g. Ar and Xe) condense to form crystals at very low temperatures. As the atoms do not undergo any chemical bonding, the crystals are held together by the London dispersion energy only. All noble gases form crystals having the face-centred cubic (FCC) structure and not the body-centred cubic (BCC) or simple cubic (SC) structures. Explain why the FCC structure is the most favourable in terms of energy, realising that the internal energy will be a minimum at the equilibrium spacing for a particular structure. Assume that the pairs have an interaction energy, u(r), described as where r is the centre-to-centre spacing between atoms. The so-called "lattice sums", An, are given below for each of the three cubic lattices. SC BCC FCC A6 8.40 14.45 12.25 A12 6.20 12.13 9.11 Then derive an expression for the maximum force required to move a pair of Ar atoms from their point of contact to an infinite separation. 2. (i) Starting with an expression for the Coulomb energy, derive an expression for the interaction energy between an ion of charge ze and a polar molecule at a distance of r from the ion. The dipole moment is tilted by an angle q with relation to r, as shown below. (ii) Evaluate your expression for a Mg2+ ion (radius of 0.065 nm) dissolved in water (radius of 0.14 nm) when the water dipole is oriented normal to the ion and when the water and ion are at the point of contact. Express your answer in units of kT. Is it a significant value? (The dipole moment of water is 1.85 Debye.)

  10. Molecular Crystals and Response of Condensed Matter to Mechanical Stress 3SCMP 2 February, 2006 Lecture 3 See Jones’ Soft Condensed Matter, Chapt. 2; Ashcroft and Mermin, Ch. 20

  11. Lennard-Jones Potential • The pair potential for isolated molecules affected by van der Waals’ interactions only can be described by a Lennard-Jones potential: w(r) = +B/r12 - C/r6 • The -ve r -6 term is the attractive v.d.W. contribution • The +ve r -12 term describes the hard-core repulsion stemming from the Pauli-exclusion. 12 is a mathematically-convenient exponent with no physical significance! • The two terms are additive.

  12. wmin -5 x 10-22 J L-J Potential for Ar (boiling point = 87 K) Actual s ~ 0.3 nm (Guess for B is too large!) Compare to: (3/2)kTB= 2 x 10-21 J (m) London Constant, C = 4.5 x 10-78 Jm6; Guessing B = 10-134 Jm12

  13. Intermolecular Force for Ar (boiling point = 87 K) F= dw/dr (m)

  14. Intermolecular Force for Ar (boiling point = 87 K) F= dw/dr (m) (m)

  15. The AFM probe is exceedingly sharp so that only a few atoms are at its tip! Weak Nano-scale Forces Can be Measured Sensitive to forces on the order of nano-Newtons.

  16. C A B C E D Measuring Attractive Forces at the Nano-Scale A = approach B = “jump” to contact C = contact D = adhesion E = pull-off Tip deflection  Force Vertical position

  17. Latex Particle Packing Tg = 20 °C

  18. w(r) = 4e[( )12 -( )6] L-J Potential in Molecular Crystals Noble gases, such as Ar and Xe, form crystalline solids (called molecular crystals) that are held together solely by the dispersive energy. In molecular crystals, the pair potential for neighbouring atoms (or molecules) is written as The molecular diameter in the gas state is s. Note that when r = s, then w = 0. eis a bond energy, such that w(r) = - ewhen r is at the equilibrium spacing of r = ro.

  19. [ - ] [( ) ] [ ] - ( - - ) L-J Potential in Molecular Crystals The minimum of the potential is found from the first derivative of the potential. Also corresponds to the point where F = 0. We can solve this expression for r to find the equilibrium spacing, ro: To find the minimum energy in the potential, we can evaluate it when r = ro:

  20. ro -e Lennard-Jones Potential + w(r) r s -

  21. Potential Energy of an Atom in a Molecular Crystal • For each atom/molecule in a molecular crystal, we need to sum up the interaction energies between all pairs (assuming additivity of the potential energies). • The total cohesive energy per atom is W = 1/2Srw(r) since each atom in a pair “owns” only 1/2 of the interaction energy. • The particular crystal structure (FCC, BCC, etc.) defines the distances between nearest neighbours, 2nd neighbours, 3rd neighbours, etc., and it defines the number of neighbours at each distance. • This geometric information that is determined by the crystal structure can be described by constants known as the lattice sums: A12 and A6. • For FCC crystals, A12 = 12.13 and A6= 14.45.

  22. So, for a pair we write the interaction potential as: For each atom in a molecular crystal, however, we write that the cohesive energy is: W = 2e[A12( )12 -A6( )6] w(r) = 4e[( )12 -( )6] From the first derivative, we can find the equilibrium spacing for an FCC crystal: ( ) Cohesive Energy of Atoms in a Molecular Crystal We notice that the molecules are slightly closer together in a crystal compared to when they are in an isolated pair (ro=1.12s).

  23. Cohesive Energy of Atoms in a Molecular Crystal - - We can evaluate W when r = ro to find for an FCC crystal: W This expression represents the energy holding an atom/molecule within the molecular crystal. Its value is only 8.6 times the interaction energy for an isolated pair. This result demonstrates that the dispersive energy is operative over fairly short distances, so that most of the interaction energy is contributed by the nearest neighbours. In an FCC crystal, each atom has 12 nearest neighbours!

  24. F ro The tensile stress st is defined as a force acting per unit area, so that: -  The tensile strainetis given as the change in length as a result of the stress: - Elastic Modulus of Molecular Crystals We can model the intermolecular force using a spring with a spring constant, k. The force, F, to separate two atoms in the crystal is:F = k(r - ro). ao ao

  25. st Y F A et L The Young’s modulus, Y, relates tensile stress and strain: Y can thus be expressed in terms of atomic interactions: - - Connection between the atomic and the macroscopic What is k?

  26. + W ro s r -8.6e - + F ro r - Elastic Modulus of Molecular Crystals F = 0 when r = ro

  27. Force to separate atoms is the derivative of the potential: [ ] - So, taking the derivative again: [ ] - ( ) But we already know that: ( ) So we see that: Elastic Modulus of Molecular Crystals We will therefore make a substitution forswhen finding k.

  28. Combining the constants to create new constants, C1 and C2, and setting r = ro, we can write: [ ] - - Finally, we find the Young’s modulus to be: - Elastic Modulus of Molecular Crystals [ ] - To find k, we now need to evaluate dF/dr when r = ro. As ro3 can be considered an atomic volume, we see that the modulus can be considered an energy density, directly related to the pair interaction energy.

  29. The definition of the bulk modulus, B, is: - This identity tells us that: - So B can be written as: - - After writing V in terms of s, and differentiating W, we obtain for an FCC molecular crystal: Bulk Modulus of Molecular Crystals We recall the thermodynamic identity: dU = TdS - PdV If we neglect the kinetic energy in a crystal, then U  W.

  30. Theory of Molecular Crystals Compares Well with Experiments w(ro)

  31. F Response of Condensed Matter to Shear Stress A y A How does soft matter respond to shear stress? When exposed to a shear stress, the response of condensed matter can fall between two extremes: Hookean (solid-like) or Newtonian (liquid-like)

  32. q The shear straingsis linearly related to the shear stress by the shear modulus, G: Elastic Response of Hookean Solids The shear strain gs is given by the angle q (in units of radians). A F Dx y A No time-dependence in the response to stress. Strain is instantaneous and constant over time.

  33. Viscous Response of Newtonian Liquids The top plane moves at a constant velocity, v, in response to a shear stress: A F Dx v y A There is a velocity gradient (v/y) normal to the area. The viscosityh relates the shear stress, ss, to the velocity gradient. The shear strain increases by a constant amount over a time interval, allowing us to define a strain rate: Units of s-1 The viscosity can thus be seen to relate the shear stress to the shear rate: hhas S.I. units of Pa s.

  34. Hookean Solids vs. Newtonian Liquids Hookean Solids: Newtonian Liquids: Many substances, i.e. “structured liquids”, display both type of behaviour, depending on the time scale. Examples include colloidal dispersions and melted polymers. This type of response is called “viscoelastic”.

  35. Response of Soft Matter to a Constant Shear Stress: Viscoelasticity Viscous response Elastic response t The shear strain, and hence the shear modulus, both change over time:gs(t) and When a constant stress is applied, the molecules initially bear the stress. Over time, they can re-arrange and flow to relieve the stress: (strain increases over time) (strain is constant over time)

  36. Response of Soft Matter to a Constant Shear Stress: Viscoelasticity Slope: tis the “relaxation time” t t We see that 1/Go (1/h)t Hence, viscosity can be approximated as h Got

  37. Example of Viscoelasticity

  38. g Constant strain applied s Stress relaxes over time as molecules re-arrange time Stress relaxation: Physical Meaning of the Relaxation Time time

  39. Typical Relaxation Times For solids,tis exceedingly large: t1012 s For simple liquids,tis very small: t10 -12 s For soft matter,ttakes intermediate values. For instance, for melted polymers,t1 s.

  40. h h ss Shear thinning h or thickening: Viscosity Sometimes Depends on the Shear Rate h ss Newtonian:

  41. An Example of Shear Thickening Future lectures will explain how polymers and colloids respond to shear stress.

  42. 1.Calculate the energy required to separate two atoms from their equilibrium spacing ro to a very large distance apart. Then calculate the maximum force required to separate the atoms. The pair potential is given as w(r) = - A/r6 + B/r12, where A = 10-77 Jm6 and B = 10-134 Jm12. Finally, provide a rough estimate of the modulus of a solid composed of these atoms. 2. The latent heat of vaporisation of water is given as 40.7 kJ mole-1. The temperature dependence of the viscosity of waterhis given in the table below. (i) Does the viscosity follow the expectations of an Arrhenius relationship with a reasonable activation energy? (ii) The shear modulus G of ice at 0 C is 2.5 x 109 Pa. Assume that this modulus is comparable to the instantaneous shear modulus of water Go and estimate the characteristic frequency of vibration for water, n. Temp (C)0 10 20 30 40 50 h(10-4 Pa s) 17.93 13.07 10.02 7.98 6.53 5.47 Temp (C)60 70 80 90 100 h(10-4 Pa s) 4.67 4.04 3.54 3.15 2.82 3. In poly(styrene) the relaxation time for configurational rearrangementstfollows a Vogel-Fulcher law given as t = toexp(B/T-To), where B = 710 C and To = 50 C. In an experiment with an effective timescale oftexp= 1000 s, the glass transition temperature Tg of poly(styrene) is found to be 101.4 C. If you carry out a second experiment withtexp = 105 s, what value of Tg would be obtained? Problem Set 2

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