--%>

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 : Thermodynamics 1 Lab Report I already

    I already did Materials and Methods section. I uploaded it with the instructions. Also, make sure to see Concept Questions and Thinking Ahead in the instructions that I uploaded. deadline is tomorow at 8 am

  • Q : What are emulsions?Describe its

    Emulsions are colloidal solutions in which disperse phase as well as dispersion medium is both liquids. Emulsions can be broadly classified into two types: (i) Oil in water (O/W type) emulsions: in this type of emulsions, oil acts disperse phase and water acts

  • Q : Molecular Structure type The ionic

    The ionic radii of Rb+ and I- respectively are 1.46 Å and 2.16Å. The very most probable type of structure exhibited by it is: (a) CsCl type  (b) ZnS type  (c) Nacl type  (d) CaF2 type

    Q : Reducible Representations The number of

    The number of times each irreducible representation occurs in a reducible representation can be calculated.Consider the C2v point group as described or Appendix C. you can see that (1) sum of

  • Q : Colligative property problem Which is

    Which is not a colligative property: (a) Refractive index (b) Lowering of vapour pressure (c) Depression of freezing point (d) Elevation of boiling point    

  • Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Reactivity of allyl and benzyl halides

    why allyl halide and haloarenes are more reactive than alkyl halide towards nucleophilic substitution

  • Q : Problem on molarity-normality-molality

    Can someone please help me in getting through this problem. The solution ofAl2(SO4)3 d = 1.253gm/m comprise 22% salt by weight. The molarity, normality and molality of the solution is: (1) 0.805 M, 4.83 N, 0.825 M (2)

  • Q : Calculating Formulae Superphosphate has

    Superphosphate has the formula CaH4(PO4)2 H2O, what is the calculation to get the percentage of Phosphorus, I need to show the calculation. I know it is 30.9737622 u in weight and 2 atoms of the formula, but not sure how to work the calculation backwards.

  • Q : PH of an Alkyl Halide Briefly state the

    Briefly state the pH of an Alkyl Halide?