Entropy residual

When the spectroscopic method is used to evaluate 5m with p set equal to the standard pressure p° = bar, the value is the standard molar entropy, 5, of the substance in the gas phase. This value is useM for thermodynamic calculations even if the substance is not an ideal gas at the standard pressure, as will be discussed in Sec. 7.9. [Pg.155]

Ideally, the molar entropy values obtained by the calorimetric (third-law) method for a gas should agree closely with the values calculated from spectroscopic data. Table 6.1 shows that for some substances this agreement is not present. The table lists values of for ideal gases at 298.15 K evaluated by both the calorimetric and spectroscopic methods. The quantity 5m,o in the last column is the difference between the two 5 values, and is called the molar residual entropy. [Pg.155]

The other substances listed in Table 6.1 have residual entropies that are greater than zero within the uncertainty of the data. What is the meaning of this discrepancy between the calorimetric and spectroscopic results We can assume that the true values of 5 at 298.15 K are the spectroscopic values, because their calculation assumes the solid has only one microstate at 0 K, with an entropy of zero, and takes into account aU of the possible accessible microstates of the ideal gas. The calorimetric values, on the other hand, are based on Eq. 6.2.2 which assumes the solid becomes a perfectly-ordered crystal as the temperature approaches 0 K. [Pg.155]

The conventional explanation of a nonzero residual entropy is the presence of random rotational orientations of molecules in the solid at the lowest temperature at which the heat [Pg.155]

Thermodynamics and Chemistry, secon6 e6il on, version 3 20 by Howard DeVoe. Latest version www.chem.uind.edu/thermobook [Pg.155]

Figure A2.1.10. The impossibility of reaching absolute zero, a) Both states a and p in complete internal equilibrium. Reversible and irreversible paths (dashed) are shown, b) State P not m internal equilibrium and with residual entropy . The true equilibrium situation for p is shown dotted.

The most satisfactory calciilational procedure for thermodynamic properties of gases and vapors requires PVT data and ideal gas heat capacities. The primary equations are based on the concept of the ideal gas state and the definitions of residual enthalpy anci residual entropy ... [Pg.524]

In either case each dimer has two possible orientations, and random disorder between these accounts for the residual entropy of the crystal (6.3JmoH of dimer). More recently ii)... [Pg.446]

Supercooled liquids (glasses111) also have residual entropy at 0 Kelvin. As an example, glycerol supercools badly so that a glass is usually obtained at a low temperature. A crystalline solid can also be obtained if the liquid is cooled in a certain manner to initiate crystalization. Gibson and Giauque9 started with... [Pg.169]

For most substances, the Third Law and statistical calculations of the entropy of the ideal gas are in agreement, but there are exceptions, some of which are summarized in Table 4.2. The difference results from residual entropy, So, left in the solid at 0 Kelvin because of disorder so that St - So calculated from Cp/TdT is less than the St calculated from statistical methods. In carbon monoxide the residual disorder results from a random arrangement of the CO molecules in the solid. Complete order in the solid can be represented schematically (in two-dimensions) by... [Pg.170]

The residual entropy can be calculated. For the para-hydrogen, 7 = 0 at T = 0 K. Since one-fourth of the hydrogen is para, the contribution to the... [Pg.175]

EXAMPLE 7.8 Using the Boltzmann formula to interpret a residual entropy... [Pg.399]

STRATEGY The existence of residual entropy at T = 0 suggests that the molecules are disordered. From the shape of the molecule (which can be obtained by using VSEPR theory), we need to determine how many orientations, W, it is likely to be able to adopt in a crystal then we can use the Boltzmann formula to see whether that number of orientations leads to the observed value of S. [Pg.399]

On the basis of the structures of each of the following molecules, predict which ones would be most likely to have a residual entropy in their crystal forms at T = 0 (a) CO, ... [Pg.424]

Assuming statistical disorder, would you expect a crystal of octahedral c/s-MX2Y4 to have the same, higher, or lower residual entropy than the corresponding trans isomer Explain your conclusion. [Pg.427]

The equation for synchronized resonance with L = 4 and v - 2 gives R In 3/2 for the residual entropy of ice (14). This value differs by only 1.1% from that given by calculations that do not involve the approximations made in our simple treatment. It is likely that the accuracy of Eq. 4 is also reasonably high. [Pg.400]

It should be pointed out that a finite residual entropy calculated for a substance from experimental data obtained at temperatures extending down to a certain temperature, with extrapolation below that point, may arise either from failure of the experimenter to obtain thermodynamic equilibrium in his calorimetric measurements or from error in the extrapolation. Measurements made under ideal conditions and extended to sufficiently... [Pg.799]

The existence of this residual entropy of ice at very low temperatures was discovered by Giauque and Ashley (ref. 7), who preliminarily ascribed it to the persistence of rotation of ortho-water molecules (comprising 3/< of the total) about their electric-moment axes, giving an entropy of 3/ R In 2 = 1.03 E. U. [Pg.799]

Hydrogen bonds between unlike atoms, as in NH4F, would not lead to residual entropy. [Pg.800]

Crystals of cadmium bromide21 and nickel bromide22 prepared in certain ways show a type of randomness which does not lead to any appreciable residual entropy, provided that the crystals are not extremely small. This randomness of structure ( Wechselstruktur, alternating layer structure) consists in a choice between two positions for each layer of a layer structure, leading to an entropy of In 2 for each layer, which remains... [Pg.800]

In a perfect crystal at 0 K all atoms are ordered in a regular uniform way and the translational symmetry is therefore perfect. The entropy is thus zero. In order to become perfectly crystalline at absolute zero, the system in question must be able to explore its entire phase space the system must be in internal thermodynamic equilibrium. Thus the third law of thermodynamics does not apply to substances that are not in internal thermodynamic equilibrium, such as glasses and glassy crystals. Such non-ergodic states do have a finite entropy at the absolute zero, called zero-point entropy or residual entropy at 0 K. [Pg.17]

See also in sourсe #XX -- [ Pg.17 ]

See also in sourсe #XX -- [ Pg.309 ]

See also in sourсe #XX -- [ Pg.40 ]

See also in sourсe #XX -- [ Pg.45 ]

See also in sourсe #XX -- [ Pg.197 , Pg.203 , Pg.206 , Pg.215 , Pg.216 , Pg.217 , Pg.218 , Pg.219 , Pg.220 , Pg.221 ]

See also in sourсe #XX -- [ Pg.200 , Pg.201 ]

See also in sourсe #XX -- [ Pg.56 ]

See also in sourсe #XX -- [ Pg.66 ]

See also in sourсe #XX -- [ Pg.442 ]

See also in sourсe #XX -- [ Pg.155 ]

See also in sourсe #XX -- [ Pg.82 ]

See also in sourсe #XX -- [ Pg.45 ]

See also in sourсe #XX -- [ Pg.303 ]

Big Chemical Encyclopedia

Chemical substances, components, reactions, process design ...