--%>

Molecular basis of third law.

The molecular, or statistical, basis of the third law can be seen by investigating S = k in W.

The molecular deductions of the preceding sections have led to the same conclusions as that stated in the third law of thermodynamics, namely, that a value can be assigned to the entropy of any substance. When the entropy values calculated from the details of the molecular energies are compared with those obtained calorimetric third law measurements, arrangements with in experimental error in usually found, but there are some exceptions. It is the perfectly ordered state of the crystal, with all the molecules in the same lowest energy level that is the molecular basis of the third law that the entropy is zero at absolute zero.

The third law value obtained for the entropy of CO at 1 bar and 298.15 K is 193.3 JK -1 mol-1. This value is obtained lower than the statistical result of 197.6 J K mol-1 obtained by the methods of the preceding section. Similar descriptions are found for NO and N2O. The third law result forH2O vapour is lower than the statistically calculated value by 3.3 J K-1 mol-1  there discrepancies can now be attributed to the failure of these materials to form the perfect crystalline state required at absolute zero for the third law to be applied. It is the perfectly ordered state of the crystal, with all the molecules in the same lowest energy level that is the molecular basis of the third law that the entropy is zero at absolute zero. (The positive value at the entropies of all compounds at temperature above absolute zero result from the fact as the temperature is raised, more and more energy levels become accessible to the molecules. The entropy at such temperature is, of course very characteristic of the individual molecule, since each molecule has its own particular energy level pattern.)

The discrepancies between calculated and third law entropies can now be attributed to a nonzero value of the entropy at absolute zero. Thus we must explain absolute zero entropy of, for example, about 4.3 J K-1 mol-1 for CO.

A disorder to be expected for such a material is that in which the molecular alignment in the crystal is not CO CO CO CO .... But rather a disorder pattern in the crystal like CO CO OC CO.... a crystal formed initially in this way could have the disorder "frozen" in as the temperature is lowered, there being too little thermal energy for the molecules to rearrange to the ordered structure. Thus, instead of each molecule having a single state to occupy, the randomness makes two states available to each molecule. The entropy of such a crystal can then be expected to be greater by k In 2N = R In 2 = 5.8 JK-1 mol-1 than it would be for a perfect crystal. This is, in fact, the approximate discrepancy found for CO.

Other types of disorder can now be expected to persist at absolute zero and to lead to apparent discrepancies in the third law. For example, a glassy material at entropy of zero will not have the necessary molecular order to guarantee as entropy of zero at absolute zero. In view of such difficulties, the third law statement must include the restriction that only perfectly ordered crystalline materials have zero entropy at absolute zero.

   Related Questions in Chemistry

  • Q : Help 1) Chromium(III) hydroxide is

    1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer. 2) Explain how dissolving t

  • Q : Determining Mole fraction of water Can

    Can someone please help me in getting through this problem. The mole fraction of water in 20% aqueous solution of H2O2 is: (a) 77/68 (b) 68/77 (c) 20/80  (d) 80/20

  • Q : Calculating density of water using

    What is the percent error in calculating the density of water using the ideal gas law for the following conditions:  a. 110 oC, 1 bar   b. 210 oC 10 bar  c. 374 o

  • Q : Preparation of normal solution Give me

    Give me answer of this question. What weight of ferrous ammonium sulphate is requiored to prepare 100 ml of 0.1 normal solution (mol. wt. 392): (a) 39.2 gm (b) 3.92 gm (c)1.96 gm (d)19.6 gm

  • Q : Chem Explain how dissolving the Group

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.

  • Q : Describe the properties of the

    Briefly describe the properties of the carbohydrates?

  • Q : Molecular Symmetry Types The number of

    The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop

  • Q : Group IV Cations Chromium(III)

    Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer.

  • Q : Number of moles of potassium chloride

    Choose the right answer from following. The number of moles of KCL in 1000ml of 3 molar solution is: (a)1 (b)2 (c)3 (d)1.5

  • Q : Vitamines 7 enzyme cofactor what is the

    what is the relationship between vitamins and enzyme cofactors