The third law of thermodynamics

The first statement of the third law is attributed to Nernst who observed that the entropy change for reactions approaches zero as temperature approaches zero. Nernsts postulate can be expressed in symbolic form:

DrS à 0 as T à 0

This implies that all substances have the same entropy at absolute zero.

A more exact statement is that all pure crystalline substances have the same entropy at absolute zero. Planck postulated further that the entropy of all pure crystalline substances is zero at absolute zero.

S à 0 as T à 0.

The third law of thermodynamics states that every substance has a positive entropy, but at zero Kelvin the entropy is zero for a perfectly crystalline substance.

Entropy at zero Kelvin

The third law introduces a numerical scale for the entropy. One of the earliest formulations was that of Boltzmann. If we recall the function W that represents the number of ways we can distribute N particles into a number of states the entropy can be expressed as

S = kB ln W

At zero Kelvin the system is in its lowest energy state. For a perfect crystal there is only one way to distribute the energy and W = 1, therefore S = 0.

We can also begin with the statistical view of entropy

At T = 0 K, P0 = 1 because only the ground state will be populated. Since ln(1) = 0, S = 0 as well.

Absolute Entropy

We have seen calculations for the entropy change for processes. However, it is also possible to calculate the absolute entropy. We can begin with the definition dS = dqrev/T. The heat transferred during a process at constant volume is dqv,rev = CvdT. Substituting this value into the expression for the entropy and assuming reversibility leads to

If we wish to find the entropy at a given temperature T we can integrate this expression

We have kept the equation general by showing Cv(T) as a function of temperature. This calculation of the entropy is valid only at constant V. At constant P we find an analogous expression. Starting with the heat transferred, dqp,rev = CpdT. We have

which leads to the integrated expression for the absolute entropy at constant pressure

These equations tell us that if we know the entropy at T = 0 K we can determine the absolute entropy at the temperature of interest by integration of the temperature dependent heat capacity.

Phase transitions

If there is a phase transition between 0 and T we can also calculate the contribution to the entropy from the transition.

A phase transition is a reversible process. Moreover, since phase transitions that normally consider occur at constant pressure we have qp,rev = DHtrs.

To calculate the absolute entropy we integrate Cp(T)/T to the first phase transition, add a term for the phase transition and continue the integration. The absolute entropy is then calculated as

The third law of thermodynamics implies that Cp ® 0 as T ® 0 K as well. It is necessary to leave Cp(T) as a function of temperature for this reason alone. The Einstein and Debye theories of heat capacity can be used to determine the functional form of the heat capacity of at these low temperatures (see section 17-4 and Mathchapter I). For practical calculation of the entropy, however, the experimental values can be used and the integrals above are evaluated numerically (Section 21-5).

So is the entropy not equal to zero at T = 0 K if the substance is not a perfect crystal. The answer is yes. Let us consider the example of CO. CO has a very small dipole moment and there is a finite chance that CO will crystallize as CO:OC:CO instead of CO:CO:CO. For each CO molecule there are two possible orientations of the molecule, therefore there are two ways each CO can exist in the lattice. w = 2 for each CO. If we have N CO molecules there are wN ways or 2N ways that all of the CO can be distributed. Therefore, the entropy at zero Kelvin is

S = k ln W = k ln(wN) = Nk ln w = nR ln 2.

The entropy at zero Kelvin is known as residual entropy. The CO crystal is considered imperfect since the sites do not all have CO aligned with the same orientation. There is significant residual entropy in a number of common substances. For example, water has residual entropy at zero Kelvin due to the fact that orientation of the hydrogens and lone pairs does not have same orientation in all of the molecules. The residual entropy is responsible for the deviation of calculated from experimental values of the entropy discussed in section 21-8.

Calculation of entropy of gases from the partition function

We have seen that

S = U/T + k lnQ

and that U itself can be calculated from the partition function. This leads to a calculation of the entropy in terms of the partition function.

The entropy can be also expressed in terms of the molecular partition function. Recall that the molecules in a gas are not distinguishable. Therefore, Q = qN/N!. Substituting this into the expression for the entropy we have

Using Stirling’s approximation: ln N! = NlnN-N, we find

Recall that the molecular partition function is the product of the partition functions for various molecular energy levels, translation, rotation, vibration, and electronic, q = qtransqrotqvibqelec. For our discussion we consider only the translational partition function. This one is by far the largest and this is the only one that is relevant for an ideal gas. After all an ideal gas can translate, but has not rotation, vibrational or electronic states. Application of the equation to real gases as shown on page 862 of McQuarrie and Simon is complicated in appearance, but is really no more difficult than application to qtrans. To calculate the entropy of an ideal monatomic gas we substitute qtrans into (II). We have already shown that

To substitute this expression into (II) we need to take the derivative

The first an last terms are zero since they have no temperature dpendence. The middle term evaluates to 3/2T. Substituting this into (II) we have

Recalling that Nk = nR where N is the number of molecules and n is the number of moles we can express the molar entropy (S = S/n) as

The dependence of entropy on molecular structure

The entropies of gases are greater than liquids and liquids are greater than solids. This follows directly from the calculation using equation (I). Equation (II) shown for an ideal gas above indicates that the entropy of a gas will increase as its mass increases. Thus, the trend in the noble gases is for the entropy to increase as we descend the periodic table (S(Xe) > S(Kr) > S (Ar) > S(Ne)). Although we have not considered rotational and vibrational contributions mathematically, it is intuitively understandable that the greater the number of accessible levels, the greater the entropy. Therefore, molecules with a large number of atoms (and heavier atoms) will tend to have a larger entropy. This is primarily because of rotational levels. A larger number of these can be populated compared to vibrational levels.

Absolute entropies can be used to calculate reaction entropies

Entropies are tabulated in order to facilitate the calculation of the entropy change of chemical reactions. For the general reaction

aA + bB ® yY + zZ

the standard entropy change is given by

DrSo = ySo[Y] + zSo[Z] - aSo[A] - bSo[B]

where the absolute entropies So are molar quantities.