Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 31: 11 April

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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

public domain image via Wikipedia Creative Commons

 

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!

Heating & Cooling Curves

A consideration of vapor pressure etc leads to the behaviors of substances with increasing (or decreasing) temperture (see Fig 10.44, p 464 of Zumdahl 5th ed):

• 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.

    

    public domain image via Wikipedia Creative Commons

    • 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.

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© R A Paselk

Last modified 11 April 2011