3rd Year Physical Chemistry PDF

3rd Year Physical Chemistry Director: Prof Henry Curran (1) Dr Chong-Wen Zhou (2, 6) Prof Dónal Leech (3) Dr David Cheung (4, 5) 6 Topics: Exam: 2 hours; 4 × Sections; 2 × questions in each Section; attempt one question per section; => Answer 4 × questions Section A 1. Molecular Interactions (4 L...

3rd Year Physical Chemistry Director: Prof Henry Curran (1) Dr Chong-Wen Zhou (2, 6) Prof Dónal Leech (3) Dr David Cheung (4, 5) 6 Topics: Exam: 2 hours; 4 × Sections; 2 × questions in each Section; attempt one question per section; => Answer 4 × questions Section A 1. Molecular Interactions (4 Lectures/1 Tutorial) 2. Quantum Chemistry (4 Lectures/1 Tutorial) Section B 3. Electrochemistry (8 lectures/2 Tutorials) Section C 4. Macromolecules (4 Lectures/1 Tutorial) 5. Self-Assembly (4 Lectures/1 Tutorial) Section D 6. Spectroscopy (8 lectures/2 Tutorials) Notes available on Canvas  Third Year Physical Chemistry Practicals (Part of CH334 Module) Starting: Week of 17th February for 6 weeks (Tuesday or Wednesday) 1 × 3 hr (2–5 pm) session per week (Organic chemistry lab.) Oral Exam: Week 6 (to be decided) Practical is continuous assessment NO LATE HAND-UPS/SWOPPING DAYS Molecular Interactions Prof Henry Curran University of Galway Physical Chemistry (CH313) Atkins and de Paula, Focus 10 3 4 lectures plus one tutorial leading to one exam question Main Text: “Elements of Physical Chemistry” Atkins & de Paula, 7th Edition OTHERS. “Physical Chemistry” Atkins & de Paula, 8th Edition or any other PChem textbook Notes available on Canvas Focus 10 1  Background Atoms and molecules with complete valence shells can still interact with one another even though all of their valences are satisfied. They attract one another over a range of several atomic diameters and repel one another when pressed together. 5 Molecular interactions account for: – condensation of gases to liquids – structures of molecular solids – structural organisation of biological macromolecules as they pin molecular building blocks (polypeptides, polynucleotides, and lipids) together in the arrangement essential to their proper physiological function. 6  Van der Waals interactions Sum of the attractive and repulsive forces between partial charges in polar molecules other than those due to covalent bonds, hydrogen bonds or the electrostatic interaction of ions with one another or with other molecules Force between two permanent dipoles (Keesom) Force between permanent and induced dipole (Debye) Force between two instantaneously induced dipoles (London dispersion) – Interaction between species with neither a net charge nor a permanent electric dipole moment (e.g. two Xe atoms) 7  The total interaction Hydrogen bonding The hydrophobic effect Modelling the total interaction Molecules in motion 8  Van der Waals interactions Interactions between molecules include the attractive and repulsive interactions between the partial electric charges of polar molecules and the repulsive interactions that prevent the complete collapse of matter to densities as high as those characteristic of atomic nuclei. 9  Van der Waals interactions Repulsive interactions arise from the exclusion of electrons from regions of space where the orbitals of closed-shell species overlap. Those interactions proportional to the inverse sixth power of the separation are called Van der Waals interactions. 10  Van der Waals interactions Typically, one discusses the potential energy arising from the interaction. If the potential energy is denoted V, then the force is –dV/dr. If V = –C/r6 the magnitude of the force is: d  C  6C − 6  = 7 dr  r  r d n n −1   x = nx   dx  11  Interactions between partial charges Atoms in molecules generally have partial charges. Atom Partial charge/e C(=O) +0.45 C(–CO) +0.06 H(–C) +0.02 H(–N) +0.18 H(–O) +0.42 N −0.36 O −0.38 12  Interactions between partial charges If these charges were separated by a vacuum, they would attract or repel one another according to Coulomb’s Law: q1q2 V= 40 r where q1 and q2 are the partial charges and r is their separation 13  Interactions between partial charges However, other parts of the molecule, or other molecules, lie between the charges, and decrease the strength of the interaction. Thus, we view the medium as a uniform continuum and we write: q1q2 V= 4r where  is the permittivity of the medium lying between the charges. 14 The permittivity is usually expressed as a multiple of the vacuum permittivity by writing  = r0, where er is the relative permittivity (dielectric constant). The effect of the medium can be very large, for water at 25 ⁰C, r = 78. The PE (V) of two charges separated by bulk water is reduced by nearly two orders of magnitude compared to that if the charges were separated by a vacuum. 15  Coulomb potential for two charges vacuum fluid 16 Ion-Ion interaction/Lattice Enthalpy Consider two ions in a lattice 17  Ion-Ion interaction/Lattice Enthalpy two ions in a lattice of charge numbers z1 and z2 with centres separated by a distance r12: ( z1e) x ( z 2 e) V12 = 40 r12 where 0 is the vacuum permittivity. 18  Ion-Ion interaction/Lattice Enthalpy To calculate the total potential energy of all the ions in the crystal, we have to sum this expression over all the ions. Nearest neighbours attract, while second-nearest repel and contribute a slightly weaker negative term to the overall energy. Overall, there is a net attraction resulting in a negative contribution to the energy of the solid. 19 For instance, for a uniformly spaced line of alternating cations and anions for which z1 = +z and z2 = –z, with r the distance between the centres of adjacent ions, we find: z 2e 2 V =− x 2 ln 2 4 0 r 20  1  z 2e 2 z 2e 2 z 2e 2 z 2e 2  V= x− + − + −.....  4 0  r 2 r 3 r 4 r     1 1 1  2 2 z e V =− x 1 − + − +.....  4 0 r  2 3 4  2 2 z e V =−. ln 2 4 0 r 21 Born-Haber cycle for lattice enthalpy 22 Lattice Enthalpies, HL0 / (kJ mol−1) Lattice Enthalpy (HL0) is the standard enthalpy change accompanying the separation of the species that compose the solid per mole of formula units. e.g. MX (s) = M+(g) + X− (g) 23 Calculate the lattice enthalpy of KCl (s) using a Born-Haber cycle and the following information at 25 ⁰C: Process H0 (kJ mol−1) Sublimation of K (s) +89 Ionization of K (g) +418 Dissociation of Cl2 (g) +244 Electron attachment to Cl (g) –349 Formation of KCl (s) –437 24  Calculation of lattice enthalpy Process H0 (kJ mol−1) KCl (s) K (s) + ½Cl2 (g) +437 K (s) K (g) +89 K (g) K+ (g) + e– (g) +418 ½Cl2 (g) Cl (g) +122 Cl (g) + e– (g) Cl– (g) –349 KCl (s) K+ (g) + Cl– (g) +717 kJ mol−1 25  Lattice Enthalpies, HL0 / (kJ mol−1) LiF 1 037 LiCl 852 LiBr 815 LiI 761 NaF 926 NaCl 786 NaBr 752 NaI 705 KF 821 KCl 717 KBr 689 KI 649 MgO 3 850 CaO 3461 SrO 3283 BaO 3114 MgS 3 406 CaS 3119 SrS 2974 BaS 2832 Al2O3 15 900 Lattice Enthalpy (HL0) is the standard enthalpy change accompanying the separation of the species that compose the solid per mole of formula units. e.g. MX (s) = M+(g) + X− (g) 26  Electric dipole moments When molecules are widely separated it is simpler to express the principal features of their interaction in terms of the dipole moments associated with the charge distributions rather than with each individual partial charge. An electric dipole consists of two charges +q and –q separated by a distance l. The product ql is called the electric dipole moment, . 27  Electric dipole moments We represent dipole moments by an arrow with a length proportional to  and pointing from the negative charge to the positive charge: +  − Because a dipole moment is the product of a charge and a length the SI unit of dipole moment is the coulomb-metre (C m) 28  Electric dipole moments It is often much more convenient to report a dipole moment in debye, D, where: 1D = 3.335 64 x 10−30 C m because the experimental values for molecules are close to 1 D. The dipole moment of charges +e and –e separated by 100 pm is 1.6 x 10−29 C m, corresponding to 4.8 D. 29 Electric dipole moments: diatomic molecules A polar molecule has a permanent electric dipole moment arising from the partial charges on its atoms. All hetero-nuclear diatomic molecules are polar because the difference in electronegativities of their two atoms results in non-zero partial charges. 30  Electric dipole moments /D  /(10−30 m3) Ar 0 1.66 CCl4 0 10.5 C6H6 0 10.4 H2 0 0.819 H2O 1.85 1.48 NH3 1.47 2.22 HCl 1.08 2.63 HBr 0.80 3.61 HI 0.42 5.45 31 Electric dipole moments: diatomic molecules More electronegative atom is usually the negative end of the dipole. There are exceptions, particularly when anti-bonding orbitals are occupied. +  − – CO dipole moment is small (0.12 D) but negative end is on C atom. Anti-bonding orbitals are occupied in CO and electrons in anti-bonding orbitals are closer to the less electronegative atom, contributing a negative partial charge to that atom. If this contribution is larger than the opposite contribution from the electrons in bonding orbitals, there is a small negative charge on the less electronegative atom. 32 Electric dipole moments: polyatomic molecules Molecular symmetry is of the greatest importance in deciding whether a polyatomic molecule is polar or not. Homo-nuclear polyatomic molecules may be polar if they have low symmetry In ozone, dipole moments associated with each bond make an angle with one another and do not cancel.  + +  - - Ozone, O3 33 Electric dipole moments: polyatomic molecules Molecular symmetry is of the greatest importance in deciding whether a polyatomic molecule is polar or not. In carbon dioxide, dipole moments associated with each bond oppose one another and the two cancel.  + +  - - Carbon dioxide, CO2 34 Electric dipole moments: polyatomic molecules It is possible to resolve the dipole moment of a polyatomic molecule into contributions from various groups of atoms in the molecule and the direction in which each of these contributions lie. 35 Electric dipole moments: polyatomic molecules 1,2-dichlorobenzene: two chlorobenzene dipole moments arranged at 60⁰ to each other. Using vector addition the resultant dipole moment (res) of two dipole moments 1 and 2 that make an angle q with one another is approximately: res  res  ( +  + 21 2 cos  ) 2 2 1/ 2 1 2 1  2 36 Electric dipole moments: polyatomic molecules 37 Electric dipole moments: polyatomic molecules Better to consider 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. Dipole moments are calculated considering a vector, , with three components, x, y, and z. The direction of  shows the orientation of the dipole in the molecule and the length of the vector is the magnitude, , of the dipole moment.  = ( +  +  2 x 2 y z) 2 1/ 2 38 Electric dipole moments: polyatomic molecules To calculate the x-component we need to know the partial charge on each atom and the atom’s x- coordinate relative to a point in the molecule and from the sum:  x =  qJ xJ J z  where qJ is the partial charge of atom J, xJ is the x coordinate of atom J, and x y the sum is over all atoms in molecule 39 Partial charges in polypeptides Atom Partial charge/e C(=O) +0.45 C(–CO) +0.06 H(–C) +0.02 H(–N) +0.18 H(–O) +0.42 N −0.36 O −0.38 40  Calculating a Molecular dipole moment (182,-87,0) H +0.18 (0,0,0)  C -0.36 (132,0,0) N +0.45 O -0.38 (-62,107,0) x = (–0.36e) x (132 pm) + (0.45e) x (0 pm) +(0.18e) x (182 pm) + (–0.38e) x (–62 pm) = 8.8e pm = 8.8 x (1.602 x 10−19 C) x (10−12 m) = 1.4 x 10−30 C m = 0.42 D 41 Calculating a Molecular dipole moment y = (–0.36e) x (0 pm) + (0.45e) x (0 pm) +(0.18e) x (–86.6 pm) + (–0.38e) x (107 pm) = –56e pm = –9.1 x 10−30 C m = –2.7 D  = ( +  +  ) z = 0 2 2 2 1/ 2 x y z  = [(0.42 D)2 + (–2.7 D)2]1/2 = 2.7 D 42  Calculating a Molecular dipole moment (182,-87,0) H +0.18 (0,0,0)  C -0.36 (132,0,0) N +0.45 O -0.38 (-62,107,0) Thus, we can find the orientation of the dipole moment by arranging an arrow 2.7 units of length (magnitude) to have x, y, and z components of 0.42, –2.7, 0 units (Exercise: calculate  for formaldehyde) 43 Interactions between charge and dipole The potential energy of a dipole 1 in the presence of a charge q2 is calculated taking into account the interaction of the charge with the two partial charges of the dipole, one a repulsion the other an attraction. 1q2 l -q1 q2 V =− 40 r 2 q1 r 44 Interactions between a charge and a dipole A similar calculation for the more general orientation is given as: q 2 1q2 cos  l r V =−  40 r 2 q1 -q1 If q2 is positive, the energy is lowest when  = 0 (and cos  = 1), as the partial negative charge of the dipole lies closer than the partial positive charge to the point charge and the attraction outweighs the repulsion. 45  Interactions between dipoles-fixed The interaction energy decreases more rapidly with distance than that between two point charges (as 1/r2 rather than 1/r), because from the viewpoint of the point charge, the partial charges on the dipole seem to merge and cancel as the distance r increases. 46  Interactions between dipoles-fixed Interaction energy between two dipoles 1 and 2: l2 q2 -q2 1 2 (1 − 3 cos 2  ) l1 r V=  4 0 r 3 q1 -q1 For dipole-dipole interaction the potential energy decreases as 1/r3 (instead of 1/r2 for point-dipole) because the charges of both dipoles seem to merge as the separation of the dipoles increases. 47  Interactions between dipoles-fixed The angular factor takes into account how the like or opposite charges come closer to one another as the relative orientations of the dipoles is changed. – The energy is lowest when  = 0 or 180⁰ (when 1 – 3 cos2 = –2), because opposite partial charges then lie closer together than like partial charges. – The energy is negative (attractive) when  < 54.7⁰ (the angle when 1 – 3 cos2 = 0) because opposite charges are closer than like charges. – The energy is positive (repulsive) when  > 54.7⁰ because like charges are then closer than opposite charges. – The energy is zero on the lines at 54.7⁰ and (180 – 54.7) = 123.3⁰ because at those angles the two attractions and repulsions cancel. 48  Interactions between dipoles-fixed Calculate the molar potential energy of the dipolar interaction between two peptide links separated by 3.0 nm in different regions of a polypeptide chain with  = 180⁰, 1 = 2 = 2.7 D, corresponding to 9.1 x 10−30 C m. 1 2 (1 − 3 cos  ) 2 V= 4 0 r 3 (9.1𝑥10−30 𝐶𝑚)2 𝑥(−2) 𝑉= 4𝜋(8.854𝑥10−12 𝐽−1 𝐶 2 𝑚−1 )𝑥(3.0𝑥10−9 𝑚)3 𝑉 = −5.6𝑥10−23 J ≡ −33.72 J mol−1 49 Interactions between dipoles-moving The average energy of interaction between polar molecules that are freely rotating in a fluid (gas or liquid) is zero (attractions and repulsions cancel). However, because the potential energy for dipole-dipole interaction depends on their relative orientations, the molecules exert forces on one another, and do not rotate completely freely, even in a gas. Thus, the lower energy orientations are marginally favoured so there is a non-zero interaction between rotating polar molecules. 50 Interactions between dipoles-moving When a pair of molecules can adopt all relative orientations with equal probability, the favourable orientations (a) and the unfavourable ones (b) cancel, and the average interaction is zero. In an actual fluid (a) predominates slightly. 51 Interactions between dipoles-moving 2  2 2 212  22 V =− 3(40 ) 2 kTr 6 V =− 1 2 3(40 ) kTr 2 6 E  1/r6 => van der Waals interaction E  1/T => greater thermal motion overcomes the mutual orientating effects of the dipoles at higher T 52 Interactions between dipoles At 25 ⁰C the average interaction energy for pairs of molecules with  = 1 D is about –1.4 kJ mol−1 when the separation is 0.3 nm. This energy is comparable to average molar kinetic energy of 3/2RT = 3.7 kJ mol−1 at 25 ⁰C. These are similar but much less than the energies involved in the making and breaking of chemical bonds. 53 Induced dipole moments-Debye Interaction A non-polar molecule may acquire a temporary induced dipole moment * as a result of the influence of an electric field generated by a nearby ion or polar molecule. 54  Induced dipole moments The field distorts the electron distribution of the molecule and gives rise to an electric dipole. The molecule is said to be polarizable. The magnitude of the induced dipole moment is proportional to the strength of the electric field, E, giving: * = E where  is the polarizability of the molecule. 55  Induced dipole moments The larger the polarizability of the molecule the greater is the distortion caused by a given strength of electric field. If a molecule has few electrons (N2) they are tightly controlled by the nuclear charges and the polarizability is low. If the molecule contains large atoms with electrons some distance from the nucleus (I2) nuclear control is low and polarizability is high. 56  Induced dipole moments Polarizability also depends on the orientation of the molecule wrt the electric field unless the molecule is tetrahedral (CCl4), octahedral (SF6), or icosahedral (C60). – Atoms and tetrahedral, octahedral, and icosahedral molecules have isotropic (orientation-independent) polarizabilities – All other molecules have anisotropic (orientation- dependent) polarizabilities 57  Polarizability volume The polarizability volume has the dimensions of volume and is comparable in magnitude to the volume of the molecule   =' 40 58  Polarizability volume /D  /(10−30 m3) Ar 0 1.66 CCl4 0 10.5 C6H6 0 10.4 H2 0 0.819 H2O 1.85 1.48 NH3 1.47 2.22 HCl 1.08 2.63 HBr 0.80 3.61 HI 0.42 5.45 59  Polarizability volume What strength of electric field is required to induce an electric dipole moment of 1 D in a molecule of polarizability volume 1.1 x 10−31 m3?   = ' 40  −31 1.1 x 10 =   =1.224 x 10 J C m − 41 -1 2 2 40 1.0 x 10 −6 (3.335 64 x 10 −30 )  * = E  E = 1.224 x 10 − 41 = 2.725 x 10 5 JC −1m −1 = 2.725 x 10 5 Vm −1 = 2.725 x 10 2 kVm −1 = 2.725 kV cm −1 60 Dipole-induced dipole moments-Debye Interaction A polar molecule with dipole moment 1 can induce a dipole moment in a polarizable molecule the induced dipole interacts with the permanent dipole of the first molecule and the two are attracted together  2 2 V =− 1 the induced dipole (cyan arrows) follows the changing orientation 0 r 6 of the permanent dipole (yellow arrows) 61 Dipole-induced dipole moments For a molecule with  = 1D (HCl) near a molecule of polarizability volume ’ = 1.0 x 10−31 m3 (benzene), the average interaction energy is about –0.8 kJ mol−1 when the separation is 0.3 nm. E  1/r6 => van der Waals interaction 62  London Dispersion Interactions The dispersion interaction (London Force) between non-polar species arises from transient dipoles which result from fluctuations in the instantaneous positions of their electrons The dispersion interaction (London Force) between non-polar species arises from transient dipoles which result from fluctuations in the instantaneous positions of their electrons 63  Dispersion Interactions Electrons from one molecule may flicker into an arrangement that results in partial positive and negative charges and thus gives an instantaneous dipole moment 1. This dipole can polarize another molecule and induce in it an instantaneous dipole moment 2. Although the first dipole will go on to change the size and direction of its dipole (≈ 10−16 s) the second dipole will follow it; the two dipoles An instantaneous dipole on are correlated in direction, with one molecule induces a the positive charge on one molecule dipole on another molecule, close to a negative partial charge on and the two dipoles attract the other molecule and vice versa. thus lowering the energy. 64  Dispersion Interactions Overall, net attractive interaction Polar molecules interact by: – dispersion interactions and dipole-dipole interactions – dispersion interactions often dominant Dispersion interaction strength depends on: – polarizability of first molecule which is decided by nulcear control loose => large fluctuations in e- distribution – polarizability of second molecule V 1 2 65  Dispersion Interactions 2  ' I1 I 2 ' V=− x 6 x 1 2 London 3 r I1 + I 2 formula I1, I2 are the ionization energies of the two molecules The potential energy of interaction is proportional to 1/r6 so this too is a contribution to the van der Waals interaction. For two CH4 molecules, V = –5 kJ mol−1 (r = 0.3 nm) 66  Total interaction-Hydrogen bonding The Coulombic interaction between the partly exposed positive charge of a proton bound to an electron withdrawing X atom (in X—H) and the negative charge of a lone pair on the second atom Y, as in: -X—H+ …… Y- Strongest intermolecular interaction Denoted X—H……Y, with X and Y being N, O, or F – only molecules with these atoms ‘Contact’ interaction – turns on when X—H group is in contact with Y atom 67  Hydrogen bonding Leads to: – rigidity of molecular solids (sucrose, ice) – low vapour pressure (water) – high viscosity (water) – high surface tension (water) – secondary structure of proteins (helices) – attachment of drugs to receptor sites in proteins 68  Interaction potential energies Interaction type Potential Distance Energy Distance Typical dependence Typicalenergy Energy Comment Interaction Comment Equation dependence (kJ mol-1) q1q2 −1 ofVpotential = energy (kJ mol ) Ion-ion 40 r 1/r 250 Only between ions Ion–ion 1/r 1q2 cos  250 Only between ions Ion-dipole V =− 1/r 2 15 2 40 r 2 Ion–dipole 1/r 15   2 (1 − 3 cos  ) 2 Between stationary polar Dipole-dipole (fixed) V = 13 1/r3 2 Dipole–dipole 1/r 4 0 r 3 2 Between stationary molecules polar 212  22 molecules Between rotating polar Dipole-dipole (moving) V =− 1/r6 0.3 1/r3(40 ) kTr 2 6 6 molecules 0.3 Between rotating polar mo 26   ' ' I I Between all types of (dispersion) V =1/r − x 1 2 London London (dispersion) x 1 2 1/r6 2 2 Between all types of mole 3 r 6 I1 + I 2 molecules and ions and ions The energy of a hydrogen bond X–H  Y is typically 20 kJ mol−1 and occurs on contact for X, Y = N, O, or F. 69  The Hydrophobic effect An apparent force that influences the shape of a macromolecule mediated by the properties of the solvent, water. Why don’t HC molecules dissolve appreciably in water? Experiments show that the transfer of a hydrocarbon molecule from a non-polar solvent into water is often exothermic (H < 0) The fact that dissolving is not spontaneous must mean that entropy change is negative (S < 0). 70  The Hydrophobic effect For the process: CH4 (in CCl4) = CH4 (aq) H = −10 kJ mol−1, S = −75 J K−1 mol−1, and G = + 12 kJ mol−1 at 298 K. G = H − TS Substances characterized by a positive Gibbs energy of transfer from a non-polar to a polar solvent are classified as hydrophobic. 71  The Hydrophobic effect When a HC molecule is surrounded by water, the water molecules form a clathrate cage. As a result of this acquisition of structure, the entropy of the water decreases, so the dispersal of the HC into water is entropy-opposed. The coalescence of the HC into a single large blob is entropy-favoured. 72  The Hydrophobic effect The formation of the clathrate cage decreases the entropy of the system because the water molecules adopt a more ordered arrangement than in the bulk liquid. However, when many solute molecules cluster together fewer (but larger) cages are required and more solvent molecules are free to move. This leads to a net decrease in the organization of the solvent and thus a net increase in the entropy of the system. 73  The Hydrophobic effect 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 in the decrease in structural demands on the solvent is the origin of the hydrophobic effect. The presence of hydrophobic groups in polypeptides results in an increase in structure of the surrounding water molecules and a decrease in entropy. 74  Polymer solvation Some polymers has a negative entropy of solution, due to the decrease in volume as the polymer dissolves. Since the enthalpy of solution does not decrease too much with temperature, and the entropy of solution is negative and does not vary appreciably with temperature, these polymers are less soluble at higher temperatures. 75 LCT-Liquid Crystal Templating 76  Modelling the total interaction The total attractive interaction energy between rotating molecules that cannot participate in hydrogen bonding is the sum of the contributions from the dipole-dipole, dipole-induced-dipole, and dispersion interactions. Only the dispersion interaction contributes if both molecules are non-polar. 77  Modelling the total interaction All three interactions vary as the inverse sixth power of the separation. Thus, the total van der Waals interaction energy is: C V =− 6 r where C is a coefficient that depends on the identity of the molecules and the type of interaction between them. 78  Modelling the total interaction The attractive (negative) contribution has a long range, but the repulsive (positive) interaction increases more sharply once the molecules come into contact. Repulsive terms become important and begin to dominate the attractive forces when molecules are squeezed The potential energy of two together. closed-shell species as the distance between them is changed 79  Modelling the total interaction These repulsive interactions arise primarily from the Pauli exclusion principle, which forbids pairs of electrons being in the same region of space. The repulsions increase steeply in a way that can be deduced only by very extensive, complicated, molecular structure calculations. 80  Modelling the total interaction In many cases one may use a greatly simplified representation of the potential energy. – details ignored – general features expressed using a few adjustable parameters Hard-Sphere potential (approximation) – Assume potential energy rises abruptly to infinity as soon as the particles come within some separation  81  Modelling the total interaction V = ∞ for r ≤  V = 0 for r >  There is no potential energy of interaction until the two molecules are  separated by a distance  when the potential This very simple assumption is energy rises abruptly to surprisingly useful in assessing infinity a number of properties. 82  Modelling the total interaction Another approximation is to express the short-range repulsive potential energy as inversely proportional to a high power of r: * C V =+ n r where C* is another constant (the star signifies repulsion). Typically, n is set to 12, in which case the repulsion dominates the 1/r6 attractions strongly at short separations as: C*/r12 >> C/r6 83  Modelling the total interaction The sum of the repulsive interaction with n = 12 and the attractive interaction given by: C V =− 6 r is called the Lennard-Jones (12,6)-potential. It is normally written in the form:   12     6 V = 4   −     r   r   84  Modelling the total interaction The two parameters are  (epsilon), the depth of the well, and , the separation at which V = 0. The Lennard-Jones potential models the attractive component by a contribution that is proportional to 1/r6, and a repulsive component by a contribution proportional to 1/r12 85 Modelling the total interaction Species / kJ mol−1  / pm Ar 128 342 Br2 536 427 C6H6 454 527 Cl2 368 412 H2 34 297 He 11 258 Xe 236 406 86