--%>

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 : Effect on vapour pressure of dissolving

    Give me answer of this question. When a substance is dissolved in a solvent the vapour pressure of the solvent is decreased. This results in: (a) An increase in the b.p. of the solution (b) A decrease in the b.p. of the solvent (c) The solution having a higher fr

  • Q : Question based on vapour pressure While

    While a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in: (a) An increase in the boiling point of the solution (b) A decrease in the boiling point of solvent (c) The solution having a higher freezing point than

  • Q : Problem on making solution Select the

    Select the right answer of the question. The weight of H2C2O42H2O required to prepare 500ml of 0.2N solution is : (a) 126g (b) 12.6g (c) 63g (d) 6.3g

  • Q : What is Distillation Separation by

    Separation by distillation can be described with a boiling point diagram. The important process of distillation can now be investigated. From the boiling point diagram one can see that if a small amount of vapour were removed from a liquid of composit

  • Q : Biodegradable polymers what are the

    what are the examples of biodegradable polymers

  • Q : Question 6 A student was analyzing an

    A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu

  • Q : The three facts on the evaporation

    Describe briefly the three facts on the evaporation?

  • Q : Lowering of vapour pressure Help me to

    Help me to go through this problem. Lowering of vapour pressure is highest for: (a) urea (b) 0.1 M glucose (c) 0.1M MgSo4 (d) 0.1M BaCl2

  • Q : Crystals of covalent compounds Crystals

    Crystals of the covalent compounds always contain:(i) Atoms as their structural units  (ii) Molecules as structural units  (iii) Ions held altogether by electrostatic forces (iv) High melting pointsAnswer: (i)

  • 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

    Discover Q & A

    Leading Solution Library
    Avail More Than 1455616 Solved problems, classrooms assignments, textbook's solutions, for quick Downloads
    No hassle, Instant Access
    Start Discovering

    18,76,764

    1923079
    Asked

    3,689

    Active Tutors

    1455616

    Questions
    Answered

    Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

    Submit Assignment

    ©TutorsGlobe All rights reserved 2022-2023.