--%>

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 : Problem on melting of ice A) It has

    A) It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry out the following calculation to test this hypothesis. Suppose that the width of the skate in contact with the ice has been reduced by sh

  • Q : Describe physical adsorption and its

    When the forces of attraction existing between adsorbate and adsorbent are van der Waal's forces, the adsorption is called physical adsorption. This type of adsorption is also known as physisorption or van der Waal's adsorption. Since the forces existing between adsorbent and adsorbate are very w

  • Q : Problem on molality Select the right

    Select the right answer of the question. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g /ml : (a) 10.43 (b) 20.36 (c) 12.05 (d) 14.05

  • 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 : Basicity order order of decreasing

    order of decreasing basicity of urea and its substituents

  • Q : Describe Transformation Matrices. Each

    Each symmetry operation can be represented by a transformation matrix.You have seen what happens when a molecule is subjected to the symmetry operation that corresponds to any of the symmetry elements of the point group to which the molecule belongs. The m

  • Q : Determining highest normality What is

    What is the correct answer. Which of the given solutions contains highest normality: (i) 8 gm of KOH/litre (ii) N phosphoric acid (iii) 6 gm of NaOH /100 ml (iv) 0.5M H2SO4

  • Q : Reaction of calcium carbonate Give me

    Give me answer of this question. What is the volume of 0.1NHcl required to react completely with 1.0g of pure calcium carbonate : (Ca= 40, C= 12 and o = 16 ) (a)150cm3 (b)250cm3 (c)200cm3 (d)100cm3

    Q : Biodegradable polymers what are the

    what are the examples of biodegradable polymers

  • Q : Schrodinger equation with particle in a

    Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functi