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Lecture 14: Special interactions. What did we cover in the last lecture? Restricted motion of molecules near a surface results in a repulsive force which.

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Presentation on theme: "Lecture 14: Special interactions. What did we cover in the last lecture? Restricted motion of molecules near a surface results in a repulsive force which."— Presentation transcript:

1 Lecture 14: Special interactions

2 What did we cover in the last lecture? Restricted motion of molecules near a surface results in a repulsive force which acts to push surfaces apart and increase the entropy This method is used to stabilise colloids and surfaces when charges cannot be used effectively (i.e. in non polar organic solvents) For rod like molecules with a separation, d, the pressure between the surfaces at separation, D, is given by

3 In this lecture… 1) The properties of water as a solvent 2) Hydrogen bonding 3) Hydrophobic interactions Further Reading Intermolecular and surface forces, J. Israelachvili, Chapter 8, p122

4 Water is odd stuff It has high melting and boiling points for a low molecular weight liquid It also has a high latent heat of vapourisation and surface energy when compared to other liquids Water is arguably the most important liquid on earth All biological organisms have a high water content Nanoscience also exploits the dielectric properties of water to help stabilise particles and surfaces in solution (electrostatic colloidal interactions)

5 What gives water these properties? These properties are due to the special nature of the interactions between water molecules Atoms such as O, N, F and Cl are highly electronegative (they tend to pull electrons towards themselves) Covalent bonds with these atoms can be quite polar. Bonds between these atoms and hydrogen tend to have a greater polarity. Dipolar interactions between these bonds are referred as hydrogen bonds

6 How strong are hydrogen bonds? The energies associated with individual H-bonds are typically kJmol -1 (~ 4-16 k B T room ) Comparing this with the energies associated with other bonds we have Metallic and Covalent Bonds ~ 500 kJmol -1 (~200 k B T room ) Van der Waals (dispersion forces) bonds ~ 1 kJmol -1 (~0.4 k B T room ) They are easier to overcome than simple dispersion interactions but also relatively stable at room temperature

7 The structure of ice The directional nature of hydrogen bonds causes water molecules to pack into an open tetrahedral structure In doing so, the average number of nearest neighbour molecules decreases from ~5 in liquid to 4 in ice (resulting in a lower density) This is unusual as crystalline solids are usually more dense than the liquid because atoms/molecules become more closely spaced

8 The structure of water At low (~ room or body) temperature, the polar nature of the interactions between water molecules causes liquid water to retain some of this structure The hydrogen bonds are still present, but the molecules are more mobile. The interactions between H-bonds are dipole-dipole interactions (see lecture 3)

9 Hydrogen bonding 1) Dispersion forces act between all molecules and have no preferred direction 2) H bonds are dipole-dipole interactions between specific functional groups. As such they have some directionality 3) They also have energies of ~ 4-16 k B T room (dispersion energies ~ 1 k B T room ) 4) This makes them ‘sticky’ and more stable than simple dispersion forces but they can also be disrupted by thermal motion of the molecules

10 Hydrogen bonding in nanoscience Molecules which form many H bonds with neighbours have higher dissociation energies and are thus more sticky If the molecules are designed carefully, the directionality of H- bonds can be used to form supramolecular aggregates which self assemble in solution or on a surface PTCDI + melamine network Courtesy of Peter Beton

11 Hydrogen bonding in nature Biological molecules also exploit the directionality of H-bond formation to produce rigid structural elements that give these large molecules a well defined shape Proteins (  -helices and  -sheets) DNA (Base pair bonding)

12 Problem Calculate an estimate of the electrostatic energy of interaction associated with a single hydrogen bond in water if the intramolecular O-H separation is 0.1 nm, the intermolecular O-H separation is 0.176nm and the amount of charge separation along each bond is +/- 0.1e. Give your answer in units of k B T room

13 The interaction of water with non polar molecules When a nonpolar molecule comes into contact with polar liquids such as water it can only interact via dispersion forces (~0.4 k B T room ) These are much weaker than the hydrogen bonds (~4-16 k B T room ) As a result water molecules tend to reorient themselves near non polar molecules to form a more ordered structure which maximises H-bonding (called a Clathrate cage) These structures greatly reduce the entropy of the water molecules surrounding the non polar molecule

14 The hydrophobic effect There is a large entropic penalty (~10-30kJmol -1 ) associated with increasing the local order (decreasing disorder) of water molecules around a nonpolar molecule If the interfacial area between water and the nonpolar groups is reduced the system can offset this unfavourable entropy This is physical origin of the hydrophobic effect Note: The hydrophobic effect is not the result of a direct interaction between nonpolar molecules. + V2V2 V1V1 V 1 + V 2 Area, S 1 Area, S 2 Area < S 1 + S 2

15 Hydrophobic interaction energy There is no satisfactory theory to explain the details of the hydrophobic interaction. This is because the interaction between two surfaces involves many layers of intervening molecules Measurements show that the interaction energy, U, decays exponentially with the separation between surfaces where  1 is the interfacial energy with water, S, is the interfacial area and o is the range of the interaction o ~1-2nm

16 The hydrophobic force The hydrophobic force between two flat surfaces is given by where  1 is the interfacial energy with water, S, is the interfacial area and o is the range of the interaction o ~1-2 nm Area, S o

17 Problem Calculate the magnitude of a)The Van der Waals (dispersion) pressure and b)The pressure due to hydrophobic forces acting between two non polar interfaces at a distance of 0.5 nm in water if the Hamaker constant for the system is A=0.5x J and the interfacial energy between dodecane and water is 40 mJm -2. Assume that the range of the hydrophobic interaction is 1 nm

18 Hydrophobic interactions in nature The hydrophobic interaction is also used by protein molecules to ensure that different sections of the molecule fold up in a specific way. This is important because structure controls function As we will see in future lectures, the hydrophobic force is also important in the formation of biological (cell) membranes

19 Summary of key concepts Hydrogen bonds and hydrophobic interactions are attractive interactions. They are stronger than simple dispersion interactions (~4-30 kJmol -1 vs ~1kJmol -1 ) They have more specificity than dispersion forces and are used in both nature and nanoscience to build structures The hydrophobic effect arises due to changes in the entropy of water surrounding nonpolar molecules.


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