Weak Bonds

Hydrogen bonds are a special case of weak bonds. Note that they are significantly stronger (>100 fold) than the other weak bonds at about 4-10% as strong as a covalent bond. Hydrogen bonds only occur when a hydrogen bound to a small, very electronegative atom is brought close to another small, very electronegative atom. Essentially this means that we only see hydrogen bonds between hydrogens bound to N, O, or F (second Period electronegative elements) and N, O, or F. So we can have O-H O, O-H N, O-H F, N-H O, N-H N etc. hydrogen bonds. This is because hydrogen bonds involve dipole-dipole interactions, but they also have covalent character (about 10% of the sharing we see in true covalent bonds) which requires that the participating atoms be small enough to get close enough to allow such partial sharing. (Grp IVA-VIIA bp plot, text Fig 10.4, p 427:

 

Hydrogen bonding accounts for much of the special properties of water, such as its very high boiling point (261°C higher than methane with only a 10% increase in MW), high viscosity, high heat capacity etc. which in turn are due to the strong bonds between the individual molecules so they stick together.

Examples of water excluding non-polar substances to force the formation of biomembranes, separate out oils etc.

Equilibrium Vapor Pressure

Occurs when rate of evaporation = rate of condensation. Must have some liquid (or solid for sublimation) present. (Figure 10.39, p 460)

Recall that:

• Equilibrium vapor pressure depends only on T and substance.
• Boiling occurs when vapor pressure = atmospheric pressure.
• Heat of Vaporization is the energy required to convert a mole of liquid to a gas.

Quantitative variation of vapor pressure with temperature:

Plot (P_vap vs. T; upward curve) Figure 10.42, p 462

 

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Plot (lnP_vap vs 1/T; linear with negative slope, T = K) Figure 10.42, p 4462

 

For the linear plot can find the equation (y = ax + b):

where a = the slope = -H_vap/R and R = 8.315 JK^-1mol^-1. So

This expression is known as the Clausius-Clapeyron Equation.We can use this equation to find useful information such as the boiling points of liquids at different elevations (and thus pressures).

Example: Find the boiling point of water at 10,000 ft elevation if the atmospheric pressure is 508.4 mmHg. H_vap = 4.39 x 10⁴ J/mol.

How do we solve this? If we take the difference between the two situations we get:

ln P₁ - ln P₂ = -H_vap/R (1/T₁ - 1/T₂) + b - b

reaarranging and recalling that log a - log b = loga/b

ln P₁/ P₂ = H_vap/R (1/T₂ - 1/T₁)

and

1/T₂ - 1/T₁ = (R/H_vap) (ln P₁/ P₂)

putting in numbers

1/T₂ = (8.315 JK^-1mol^-1/ 4.39 x 10⁴ J/mol) ln (760 mmHg / 508.4 mmHg) + 1/373.15 K

1/T₂ = 7.62 x 10^-5 + 2.68 x10^-3 = 2.76 x 10^-3

T₂ = 362.7 K = 89.7 °C

Notice that we can also use the data from vapor pressures (or boiling points) at two pressures to calculate a value for H_vap!

• Properties:
  • Hard and rigid - they have virtually no tendency to flow or diffuse.
  • Nearly incompressible - need to increase pressure about 1,000,000 times to decrease volume by half.
  • Very low thermal coefficients of expansion.
  • Crystal lattice
  • Melting and freezing points are sharp - all units in the interior of a perfect crystal have the same relationships, and therefore the same bonds. Thus when enough energy is added to break the bonds for one unit, there is enough to break bonds with all, so melting is sudden as all the particles break bonds with each other at same temperature and thus same energy.
    • heat of fusion/crystallization (saw last time with liquids)
  • Nearly all solids expand when they melt (after all the particles are moving faster). As a consequence, nearly all solids will sink in their liquid forms (water is of course a notable exception - we'll look at why later).
• Structure Determination: So how do we know how atoms are arranged in crystals?
  • X-ray diffraction is THE tool for crystal structure determination. It gives a full 3-D picture of how atoms are arranged via the interpretation of diffraction patterns. So what kind of information is obtained, and how do we use it to reconstruct a crystal?
    • Xray scattering (Figure 10.10, Zumdahl p 433) and diffraction patterns.

    

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    • Bragg Equation: we can determine the distances between layers in a repeating structure using x-rays. The relevant distances can be seen in Figure 10.11, Zumdahl p 433, resulting in the Bragg equation:

    n = 2d sin

  • STEM: a newer method with equal resolution is Scanning Tunneling Electron Microscopy. This technique give a highly detailed view of the surface of a crystal (or other object). Distances between surface atoms can be calculated etc.

Crystal (Solid) Structure

Crystal Structure (overheads)

• Lattice
  • Unit Cell
• Crystal lattice
  • 3 kinds of cubic lattice: simple cubic, body-centered cubic, and face-centered cubic. (Figure 10.9, p 432 of Zumdahl)

  

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  • For identical spheres can get two kinds of close packing: (Figure 10.13, p 437 of Zumdahl)

  

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  • cubic close packing, which turns out to be a face-centered cubic lattice (Figure 10.15, p 437 of Zumdahl)
  • hexagonal close packing (Figure 10.14, p 437 of Zumdahl)

  The images below show the so-called cannon ball stacking in close-packing. Note that the stack is a direct result of the HCP lattice shown in the image above, where just the top ball and the next layer of three balls are darkened. Can you discern the next (triangular) layer in the diagram which cooresponds to the third layer down in the pictures below?

  

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Last modified 10 April 2009