27.1

3rd Law of ThermodynamicsThe 3rd Law of Thermodynamics states that the entropy of pure perfect crystalline substances approach the same value as absolute zero is approached and, since these entropies have the same value, it is convenient to define the entropies of substances in this limiting state as zero:

So

0 K [pure perfect crystal] 0

Would the following crystal of carbon monoxide, C O, (represented by the two-dimensional sketch) have an absolute entropy of zero at 0 K:

C O C O C O O C C O C O

C O O C C O C O C O C O

C O C O C O C O C O C O

C O C O O C C O C O O C

If not, why not?

The calorimetrically measured entropy of carbon monoxide differs from that predicted by the 3rd Law by 4.70 J / mole K. What fraction of carbon monoxide molecules were disordered in the imperfect crystal on which the calorimetric measurements were based?

In a rigorous application of the term pure in the 3rd Law, a substance would consist of single stable isotopes and not the mixtures that occur naturally. What is the contribution of the natural stable isotopic composition of lead (1.4 atom % 204Pb, 24.1 atom % 206Pb, 22.1 atom % 207Pb, 52.4 atom % 208Pb) to its entropy? Why is this contribution to the entropy usually ignored? Can you think of a situation when this contribution should not be ignored?

27.2

The 3rd Law allows us to calculate standard absolute or 3rd Law entropies by starting with a pure perfect crystalline substance at absolute zero and summing all the entropy changes that occur as that substance is brought to some other temperature, usually 298.15 K, at constant pressure:

So298.15 K =

So

0 K [pure perfect crystal] 0

298.15 K

dS

= 0 + 0 K

the lowest temperature for which data is available

Cp dT / T

+

temperature range over which the phase is stable

Cp dT / T

all phases

+

Ho phase change at To / To

all phase changes

Standard data for most thermodynamic variables are tabulated as differences or relative values, e.g., Eo, Ho, Go, etc. Entropies, however, are tabluated as absolute values, e.g., So. Comment on whether these entropies are really absolute values or are they too relative values.

27.3

We will illustrate the calculation of a standard absolute 3rd law entropy for molecular oxygen, O2, at 298.15 K.

The contribution to the entropy from 0 K to the lowest temperature for which experimental data is available is evaluated using the Debye T3 Law, which states that at very low temperatures the constant pressure heat capacity of a substance is proportional to the cube of the absolute temperature:

Cp, at low T = a T3

For example for O2 the lowest heat capacity measurement is for the alpha solid phase at 14.0 K:

Cp, 14.0 K [ O2 ()] = 0.8 cal / K = a (14.0 K)3

which allows the proportionality constant, a, to be determined as:

a = 3 x 10-4 cal / K-4

Once the proportionality constant is known the Debye T3 Law can be used to evaluate the contribution to the entropy from 0 K to 14.0 K:

0

14.0 K

C p, O2 dT / T =

0 K

14.0 K

a T3 dT / T

= (3 x 10-4 cal / K-4) 0 K

14.0 K

T2 dT

= (3 x 10-4 cal / K-4) [(14.0 K)3 - (0 K)3] / 3

= 0.3 cal / K

27.4

Peter Debye

Petrus (Peter) Josephus Wilhelmus Debye was born March 24, 1884, at Maastricht, the Netherlands.. His university education began at the Aachen Institute of Technology (Technische Hochschule) where he earned a degree in electrical technology in 1905. In 1906 Debye accepted a position in Theoretical Physics at Munich University, where he qualified as a University lecturer in 1910 (having obtained this University's Ph.D obtained this University's Ph.D. in Physics in 1908).

In 1911 Debye became Professor of Theoretical Physics at Zurich University, where he remained for two years. He returned to The Netherlands in 1912 when he was appointed Professor of Theoretical Physics at Utrecht University, and in 1914 he moved to the University of Göttingen, to take charge of the Theoretical Department of the Physical Institute. Later, he became Director of the entire Institute and lectured on experimental physics until 1920. Debye returned to Zurich in 1920, as Professor of Physics, and Principal of the Eidgenössische Technische Hochschule. In 1927 he held the same post at Leipzig and from 1934 to 1939 he was Director of the Max Planck Institute of the Kaiser Wilhelm Institute for Physics in Berlin-Dahlem and Professor of Physics at the University of Berlin. In 1915 Professor Debye became Editor of Physikalische Zeitschrift and continued to act in this capacity until 1940. In 1940 he became Professor of Chemistry and Principal of the Chemistry Department of Cornell University, Ithaca, New York, taking American citizenship in 1946.

27.5

He holds the Rumford Medal of the Royal Society, London, and the Franklin and Faraday Medals, the Lorentz Medal of the Royal Netherlands Academy, the Max Planck Medal, the Willard Gibbs Medal, the Nichols Medal, the Kendall Award, and the Priestley Medal of the American Chemical Society.

Peter Debye contributed (along with Huckel) to the understanding of long range ionic interactions in aqueous solutions of electrolytes. The widely used Debye-Huckel equation allows activity coefficients to be calculated for aqueous ions. He also successfully predicted the low temperature dependence of heat capacity of metals and ionic solids. In 1936 Peter Debye was awarded the Nobel Prize in Chemistry for his contributions to our knowledge of molecular structure through his investigations on dipole moments (1Debye = 4.803 x 10-10 esu cm is a unit for measuring dipole moments) and on the diffraction of X-rays and electrons in gases (the Debye Scherrer method of X-ray analysis is named after him and Paul Scherrer).

Professor Debye was married to Mathilde Alberer and had a son Peter Paul Rupprecht (b. 1916) and a daughter Mathilde Maria (b.1921). Peter Debye died in 1966.

Much of this biograpical material was taken from the WEB site, http://www.nobel.se/chemistry/laureates/1936/. The picture of Peter Debye is from a ChemTeam WEB page maintained by John L. Park at http://dbhs.wvusd.k12.ca.us/Gallery/GalleryMenu.html .

27.6

The contributions to the entropy that result from warming a particular phase over the temperature range over which the phase is stable are often evaluated by measuring the area under the curve in a plot of Cp / T versus T. Such a plot and the contributions to the entropy for the three solid phases, the liquid phase, and the vapor phase of O2 is shown below:

Third Law Entropy of O2

0.0000

0.0500

0.1000

0.1500

0.2000

0.2500

0.3000

0.3500

0 50 100 150 200 250 300

Temperature (K)

Cp

/ T

(ca

l /

K^

2)

area under curve is the entropy increase for warming alpha solid phase and equals 1.500 cal / K

beta solid phase area = 4.661 cal / K gamma solid phase

area = 2.397 cal / K

liquid phase area = 6.462 cal / K gas phase area

= 8.25 cal / K

27.7

The largest contributions to the 3rd law entropies come from the entropy increases associated with phase changes. For example, when solid alpha oxygen converts to solid beta oxygen:

23.66 K

O2 () -----------> O2 () Ho23.66 K = 22.42

cal

the entropy increase is:

So = Ho

23.66 K / T = 22.42 cal / 23.66 K = + 0.9476 cal / K

The other entropy increases associated with phase changes are calculated similarly and we will just include them below in the overall calculation of the standard absolute 3rd law entropy of molecular oxygen at 298.15 K:

So298.15 K, O2

= 0 + 0.3 cal / K + 1.500 cal / K + 0.9476 cal / K

+ 4.661 cal / K + (Ho = 177.6 cal) / 43.76 K

+ 2.397 cal / K + (Homelt = 106.3 cal) / 54.39 K

+ 6.462 cal / K + (Hovap = 1628.8 cal) / 90.13 K

+ 8.25 cal / K

= + 48.6 cal / K

Could you associate each of the terms in the above calculation with the physical process that was occuring as molecular oxygen is warmed from 0 K to 298.15 K?

27.8

The utility of tabluated standard 3rd law entropies is that they can be used to calculate standard entropy changes for reactions. Consider the reaction of hydrogen and oxygen to form water vapor under standard conditions:

298.2 K H2 (g) + 1/2 O2 (g) --------> H2O (g)

1.00 bar

So298.2 K = 1 So

298.2 K, H2O (g) - 1 So298.2 K, H2

(g) - 1/2 So298.2 K, O2

(g)

= + (1 mole) (+ 45.13 cal / mole K)

- (1 mole) (+ 31.23 cal / mole K)

- (1/2 mole) (+ 48.6 cal / mole K)

= - 10.4 cal / K

Why do you think the standard entropy change for this reaction is negative?