Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2015

Lecture Notes 38: 1 May

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Equilibrium Vapor Pressure, cont.

Last time we found the Clausius-Clapeyron Equation described the plot for the variation of vapor pressure with temperature. We can use this equation to find useful information such as the boiling points of liquids at different elevations (and thus pressures).

Clausius-Clapeyron equation

Example: Find the boiling point of water at 10,000 ft elevation if the atmospheric pressure is 508.4 mmHg. deltaHvap = 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 = -deltaHvap/R (1/T1 - 1/T2) + b - b

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

ln P1/ P2 = deltaHvap/R (1/T2 - 1/T1)

and

1/T2 - 1/T1 = (R/deltaHvap) (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 deltaHvap!

Heating & Cooling Curves

A consideration of vapor pressure etc. leads to the behaviors of substances with increasing (or decreasing) temperature (see Fig 10.42, p 488 of Zumdahl 9th ed):

plot of heating curve for water

Note that this curve is completely reversible, that is you can trace it from left to right as a heating curve, or right to left as a cooling curve. Or you can flip the curve around the y-axis to get the cooling curve.

Example calculations

How much heat is needed to melt 100.0 g of ice at –10.0°C and get water at 10.0°C?

Cpice = 2.03 J g–1 °C–1, Cpwater = 4.2 J g–1 °C–1, Cpsteam = 2.0 J g–1 °C–1, deltaHvap = 40.7 kJ/mol, deltaHfus = 6.02 kJ/mol

This is a four part problem, first need to calculate the heat to raise the temperature of the ice to the melting point, then need to calculate the heat of melting the ice, then calculate the heat needed to raise the liquid to 10.0°C, and finally add the three parts together.

1. (Cpice)(mass)(deltaT) = (2.03 J g–1 °C–1)(100.0 g)(10.0°C) = 2.03 x 103 J = 2.03 kJ

2. (deltaHfus)(moles water) = (6.02 kJ/mol)(100.0g/18.01g/mol) = 33.45 kJ

3. (Cpwater)(mass)(deltaT) = (4.2 J g–1 °C–1)(100.0 g)(10.0°C) = 4.2 x 103 J = 4.2 kJ

4. Adding the three parts, deltaHprocess = 2.03 kJ + 33.45 kJ + 4.2 kJ = 39.68 kJ = 39.7 kJ

How much heat is released when 100.0g of 110.0°C steam is condensed to give water at 90.0°C?

For this problem we follow a similar strategy to that above, but backwards.

1. (Cpsteam)(mass)(deltaT) = (2.0 J g–1 °C–1)(100.0 g)(10.0°C) = 2.0 x 103 J = 2.0 kJ

2. (deltaHvap)(moles water) = (40.7 kJ/mol)(100.0g/18.01g/mol) = 225.99 kJ

3. (Cpwater)(mass)(deltaT) = (4.2 J g–1 °C–1)(100.0 g)(10.0°C) = 4.2 x 103 J = 4.2 kJ

4. Adding the three parts, deltaHprocess = 2.0 kJ + 225.99 kJ + 4.2 kJ = 232.19 kJ = 232.2 kJ

Solids and Crystals

Solids: Recall earlier definition - solids have fixed or definite shapes and volumes. By this definition solids are strictly limited to the crystalline solids. (The amorphous (noncrystalline) solids discussed in our text are what we have discussed as supercooled liquids.)

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© R A Paselk

Last modified 1 May 2015