Third Law of Thermodynamics

The Third Law of Thermodynamics

Unlike the First and Second Laws, the Third Law doesn't introduce new concepts but restricts the entropy value of crystalline solids. Some question its status as a 'law,' but its experimentally verified conclusions align with scientific standards.

Nernst Heat Theorem

• A precursor to the Third Law, the Nernst Heat Theorem is still relevant.
• From the Gibbs-Helmholtz equation: $\Delta G - \Delta H = T \frac{\partial(\Delta G)}{\partial T}_p$ where $\,\Delta G$ is the change in free energy and $\,\Delta H$ is the change in enthalpy.
• At absolute zero (T=0), $\,\Delta G = \Delta H$.
• Richards discovered that $\frac{\partial(\Delta G)}{\partial T}$ decreases with decreasing temperature, implying $\,\Delta G$ and $\,\Delta H$ converge as temperature drops.
• Nernst proposed that $\frac{\partial(\Delta G)}{\partial T}$ approaches zero as temperature nears absolute zero – the Nernst heat theorem.
• $\,\Delta G$ and $\,\Delta H$ become equal at absolute zero and approach each other asymptotically at near-absolute-zero temperatures.
• Mathematically: $Lt<em>{T \to 0} {\frac{\partial(\Delta G)}{\partial T}}</em>p = Lt<em>{T \to 0} {\frac{\partial(\Delta H)}{\partial T}}</em>p = 0$
• From the Second Law of Thermodynamics:
  • ${\frac{\partial(\Delta G)}{\partial T}}_p = -\Delta S$
  • ${\frac{\partial(\Delta H)}{\partial T}}<em>p = \Delta C</em>p$ (Kirchhoff equation)
• Where $\,\Delta S$ is the entropy change and $\,\Delta C_p$ is the difference in heat capacities.

Implications of the Nernst Theorem

• From the equations above:
  • $Lt_{T \to 0} \Delta S = 0$
  • $Lt<em>{T \to 0} \Delta C</em>p = 0$
• The entropy change of a reaction and the difference in heat capacities both tend to zero as temperature approaches absolute zero.
• The Nernst theorem applies only to pure solids.

Third Law of Thermodynamics

• According to Eq. 6, $\,\Delta C_p$ tends to approach zero at 0 K, implying that at absolute zero, products and reactants in solid state have identical heat capacities.
• At absolute zero, all substances share the same heat capacity.
• Quantum theory indicates that heat capacities of solids tend to become zero at 0 K.
• The Nernst heat theorem can be expressed as: $Lt<em>{T \to 0} C</em>p = 0$
• According to Eq. 5, $\,\Delta S$ becomes zero at absolute zero; the absolute entropies of solid products and reactants are identical.
• Planck suggested that entropies of all pure solids approach zero at 0 K: $Lt_{T \to 0} S = 0$
• Statement of the Third Law: At absolute zero, the entropy of every substance may become zero, and it does so in the case of a perfectly crystalline solid.
• A perfect crystal at absolute zero exhibits perfect order (zero disorder) and, consequently, zero entropy.
• Walther Nernst (1864-1941) received the 1920 Chemistry Nobel Prize for his work in thermochemistry and made significant contributions to electrochemistry.

Determination of Absolute Entropies

• For an infinitesimally small change of state: $dS = \frac{dq}{T}$
• At constant pressure: $({\frac{\partial S}{\partial T}})<em>p = {(\frac{\partial q}{\partial T})</em>p} \times \frac{1}{T}$
• By definition: $({\frac{\partial q}{\partial T}})<em>p = C</em>p$
• Therefore: $({\frac{\partial S}{\partial T}})<em>p = \frac{C</em>p}{T}$
• At constant pressure: $dS = {\frac{C_p}{T}} dT$
• For a perfectly crystalline substance, the absolute entropy S=0 at T=0. Therefore, we may write
• $S = S_T$
• $\int<em>{T=0} dS = \int</em>{T=0}^{T=T} (\frac{C_p}{T})dT$
• $S<em>T = \int</em>{0}^{T} C_p d(lnT)$
• Where $S_T$ is the absolute entropy at temperature T.

Calculating Absolute Entropy

1. Measure $C_p$ at various temperatures between T=0 and the desired temperature T.
2. Plot $C<em>p$ against $\,ln T$ and find the area under the curve between T=0 and T. This area equals $S</em>T$.
3. Since obtaining $C_p$ at absolute zero is impossible, measure heat capacities as low as possible (usually up to 15 K) and extrapolate to absolute zero.
4. Determine heat capacities from approximately 15 K to the required temperature T.
5. Extrapolate a graph of $C_p$ vs $\,ln T$ to absolute zero.
6. The area under the graph represents the absolute entropy of the substance at temperature T.

Debye Theory

• $S<em>T = \int</em>{0}^{T^<em>} \frac{C<em>p}{T} dT + \int</em>{T^</em>}^{T} \frac{C_p}{T} dT$
• Where 0 < T^* < 15 K.
• At very low temperatures (0 < T < 15 K), according to the Debye theory:
  • $\frac{C_p}{T} = aT^3$
  • Where a is an empirical constant (Debye $T^3$ law).
• Therefore:
  • $S<em>T = \int</em>{0}^{T^<em>} Ta^3 T + \int<em>{T^</em>}^{T} \frac{C</em>p}{T} dT$
  • $S<em>T = \int</em>{0}^{T^<em>} \frac{C<em>p}{T} dT + \int</em>{T^</em>}^{T} \frac{C_p}{T} dT$
• The second integral is evaluated from experimental heat capacity measurements.

Determining Absolute Entropy Combining Heat Capacity and Enthalpy Data

Combine heat capacity data with enthalpy data on phase transformations to determine the absolute entropy of a substance (solid, liquid, or gas) at temperature T.
Start with the substance in the crystalline solid state at absolute zero, where its absolute entropy is zero. The total absolute entropy is the sum of all entropy changes the substance undergoes to reach the given state at the given temperature.

Calculating Absolute Entropy of a Gas at 25°C

1. Heating the crystalline solid from absolute zero to $T^<em>$ (0 < $T^</em>$ < 15 K) using Debye's theory:
   • $\Delta S<em>1 = \int</em>{0}^{T^<em>} aT^3 dT = \frac{a(T^</em>)^3}{3}$
2. Heating the crystalline solid from $T^*$ to $T_{tr}$, the transition temperature:
   • $\Delta S<em>2 = \int</em>{T^*}^{T<em>{tr}} \frac{C</em>{ps}(\alpha)}{T} dT$
   • Where $C<em>{ps}(\alpha)$ is the heat capacity of the solid in allotropic form α. Evaluate $\Delta S</em>2$ by graphical integration of heat capacity data.
3. Transition of the solid from allotropic form α to β at $T_{tr}$:
   • $\Delta S<em>3 = \frac{\Delta H}{T</em>{tr}}$
   • Where $\,\Delta H$ is the molar enthalpy of transition.
4. Heating the solid in allotropic form β up to its fusion point, $T_{fus}$:
   • $\Delta S<em>4 = \int</em>{T<em>{tr}}^{T</em>{fus}} \frac{C_{ps}(\beta)}{T} dT$
   • Where $C_{ps}(\beta)$ is the heat capacity of the solid in allotropic form β.
5. Changing the solid in allotropic form β into the liquid state at the fusion temperature $T_{fus}$:
   • $\Delta S<em>5 = \frac{\Delta H</em>{fus}}{T_{fus}}$
   • Where $\,\Delta H_{fus}$ is the molar enthalpy of fusion.
6. Heating the liquid from its freezing point ($T<em>{fus}$) to its boiling point ($T</em>b$):
   • $\Delta S<em>6 = \int</em>{T<em>{fus}}^{T</em>b} C_{p,l} d \ln T$
   • Where $C<em>{p,l}$ is the heat capacity of the substance in the liquid state. Evaluate by plotting $C</em>{p,l}$ vs $\,ln T$ and finding the area under the graph.
7. Changing the liquid into the gaseous state at the temperature $T_b$:
   • $\Delta S<em>7 = \frac{\Delta H</em>{vap}}{T_b}$
   • Where $\,\Delta H_{vap}$ is the enthalpy of vaporisation per mole.
8. Heating the gas from $T_b$ to the required temperature, i.e., 25°C (298.15 K):
   • $\Delta S<em>8 = \int</em>{T<em>b}^{298.15} C</em>{p,g} d \ln T$
   • Where $C<em>{p,g}$ is the heat capacity of the substance in the gaseous state. Evaluate by plotting $C</em>{p,g}$ vs $\,ln T$ and finding the area under the curve.

The absolute entropy of the gas at 298.15 K (25°C), $S_T$, is the sum of all entropy changes:

$S<em>T = \Delta S</em>1 + \Delta S<em>2 + \Delta S</em>3 + \Delta S<em>4 + \Delta S</em>5 + \Delta S<em>6 + \Delta S</em>7 + \Delta S_8$

Experimental Verification of the Third Law

Heat capacity and enthalpy data on substances existing in different crystalline forms can verify the Third Law. For a reversible isothermal transition, α → β:

$\Delta S<em>{tr} = S</em>\beta - S<em>\alpha = \frac{\Delta H}{T</em>{tr}}$

Where $\,\Delta H$ and $T_{tr}$ are the experimentally determined enthalpy of transition and the temperature of transition, respectively.

$\Delta S = S<em>T(\beta) - S</em>T(\alpha) = \int<em>{0}^{T</em>{tr}} \frac{C<em>{ps}(\beta)}{T} dT - \int</em>{0}^{T<em>{tr}} \frac{C</em>{ps}(\alpha)}{T} dT = \frac{\Delta H}{T_{tr}}$

If:

$\int<em>{0}^{T</em>{tr}} \frac{C<em>{ps}(\beta)}{T} dT - \int</em>{0}^{T<em>{tr}} \frac{C</em>{ps}(\alpha)}{T} dT = \frac{\Delta H}{T<em>{tr}}$, then $S</em>0(\beta) = S_0(\alpha)$

This indicates that both crystalline modifications have equal entropies at 0 K, consistent with the Third Law.

Experimental Validation

Experiments on systems like sulphur, tin, and phosphine have validated the Third Law.

For phosphine:

$\Delta S<em>{tr} = \frac{\Delta H}{T</em>{tr}} = \frac{185.7 Jmol^{-1}}{49.43 K} = 15.73 JK^{-1}mol^{-1}$

The difference of the two integrals in Eq. 30 is experimentally found to be 15.69 J K-1 mol-1:

$\int<em>{0}^{T</em>{tr}} \frac{C<em>{ps}(\beta)}{T} dT - \int</em>{0}^{T<em>{tr}} \frac{C</em>{ps}(\alpha)}{T} dT = 15.69 JK^{-1}mol^{-1}$

Comparison shows that the Third Law is valid for phosphine within experimental error limits.

Entropies of Real Gases

From Maxwell relations:

$({\frac{\partial S}{\partial P}})<em>T = -({\frac{\partial V}{\partial T}})</em>P$

$dS = -({\frac{\partial V}{\partial T}})_P dP$

Integrating between pressures $P<em>1$ and $P</em>2$ at constant T:

$\int<em>{S</em>1}^{S<em>2} dS = - \int</em>{P<em>1}^{P</em>2} ({\frac{\partial V}{\partial T}})_P dP$

For a real gas behaving ideally at low pressures, let $P^*$ be this pressure. Let $(S<em>T)</em>1$ be the entropy of the real gas at 1 atm pressure and $(S<em>T)</em>P$ be the entropy at pressure P. Then, at constant temperature:

$(S<em>T)</em>1 - (S<em>T)</em>P = - \int<em>{P}^{1} ({\frac{\partial V}{\partial T}})</em>P dP$

Entropy of Real Gases at Low Pressures

*For an ideal gas $({\frac{\partial V}{\partial T}})_P = (\frac{R}{P})$, and the entropy of an ideal gas at 1 atm and P are, respectively:

$(S<em>T)</em>1^{ideal} - (S<em>T)</em>P^{ideal} = \int_{P}^{1} (\frac{R}{P}) dP$

$(S<em>T)</em>P^{ideal} = (S<em>T)</em>1^{ideal} + \int_{P}^{1} (\frac{R}{P}) dP$

We can equate $(S<em>T)</em>P$ with $(S<em>T)</em>P^{ideal}$ (because at the low pressure P, the real gas behaves ideally). Adding Eqs 38 and 36, we get

$S^\circ = (S<em>T)</em>1 = S + \int<em>{P}^{1} [({\frac{\partial V}{\partial T}})</em>P - \frac{R}{P}] dP$

Where $S^\circ$ is the standard entropy and S is the entropy of real gas both determined at 1 atm. We need to determine S. Here $S^\circ$ can be given by

$S^\circ = (S<em>T)</em>P^{ideal} + \int<em>{P}^{1} [({\frac{\partial V}{\partial T}})</em>P - \frac{R}{P}] dP$

Using Berthelot's Equation of State

Though the integrals can be evaluated graphically, using Berthelot's equation of state is more convenient than van der Waals equation:

$PV = RT + P \frac{ab}{TV^2}$

Multiplying and rearranging, we get

$PV = RT + P \frac{b - a}{TV} - \frac{ab}{TV}$

The term $\frac{ab}{TV}$ in the above equation is negligible compared to other terms since the Berthelot constants a and b are small.

$PV = RT + P \frac{b - a}{TV} = RT + P [b - \frac{a}{RT^2}]$

$(V = \frac{RT}{P})$

$PV = RT + P [b - \frac{a}{RT^2}] = RT + P [\frac{b}{RT} - \frac{a}{R^2T^2}]$

For Berthelot's equation of state, the constants a, b, and R can be written in terms of the critical constants. Accordingly,

$a = (\frac{16}{3}) P<em>cV</em>cT_c$

$b = (\frac{V_c}{4})$

$R = (\frac{32}{9}) \frac{P<em>cV</em>c}{T_c}$

Hence,

$PV = RT + P [\frac{b}{RT} - \frac{a}{R^2T^2}] = RT + P [(\frac{V<em>c}{4}) - \frac{(\frac{16}{3}) P</em>cV<em>cT</em>c}{((\frac{32}{9}) P<em>cV</em>c/T_c)^2T^2}]$

PV = RT + P (\frac{Vc}{4} - \frac{(\frac{16}{3}) PcVcTc}{((32/9)^2}){\frac{Pc^2Vc^2}{T_c^2}}T^2})

$PV = RT + P (\frac{V<em>c}{4} - \frac{(16/3)}{(32/9)^2} \frac{T</em>c^3}{P<em>cV</em>cT^2}) = RT + P (\frac{V<em>c}{4} - \frac{(16 ∗ 9 ∗ 9)}{(3 ∗ 32 ∗ 32)} \frac{T</em>c^3}{P<em>cV</em>cT^2})$

Dividing by P, we have

$V = \frac{RT}{P} + (\frac{V<em>c}{4} - \frac{27}{64} \frac{RT</em>c^3}{P_cT^2}))$

$\frac{\partial V}{\partial T} = \frac{R}{P} + (\frac{27RT<em>c^3}{32P</em>cT^3})$

Substituting for $(\frac{\partial V}{\partial T})_P$ in Eq. 37, we get

$S^\circ = S + \int<em>{0}^{1} (\frac{27RT</em>c^3}{32P_cT^3}) dP$

Let's assume that it goes from $P = 0$ to $P = 1$ to be at standard conditions.

$S^\circ = S + (\frac{27RT<em>c^3}{32P</em>cT^3}) \int<em>{0}^{1} dP = S + (\frac{27RT</em>c^3}{32P_cT^3}) (1 - 0)$

$S^\circ = S + (\frac{27RT<em>c^3}{32P</em>cT^3})$

$S = S^\circ - (\frac{27RT<em>c^3}{32P</em>cT^3})$

As mentioned earlier, $S^\circ$ is given by Eq. 38. In Eq. 46, the second term is the correction term that should be added to $S$ to give $S^\circ$, the standard entropy for a real gas.

Entropy Changes in Chemical Reactions

We can calculate $\,\Delta S$ for a chemical reaction from the tabulated standard entropy values for the reactants and products at 298 K. This is one of the most important applications of the Third Law.

For a reaction occurring in the standard state:

aA + bB + … → lL + mM + …

The standard entropy change, $\,\Delta S^\circ$, is given by:

$\Delta S^\circ = [lS<em>L + mS</em>M + …] - [aS<em>A + bS</em>B + …]$

$\Delta S^\circ = \sum S<em>{products} - \sum S</em>{reactants}$

Where $S^\circ$s are the molar standard entropies of the species involved, and a, b, l, m, etc., are the stoichiometric coefficients.

$\Delta S^\circ$ at any other temperature:

$\frac{d(\Delta S^\circ)}{dT} = \sum (\frac{\partial S}{\partial T})<em>{products} - \sum (\frac{\partial S}{\partial T})</em>{reactants}$

Since $\,(\frac{\partial S}{\partial T})<em>P = \frac{C</em>P}{T}$, we have

$[\frac{d(\Delta S^\circ)}{dT}]<em>P = \sum (\frac{C</em>P}{T})<em>{products} - \sum (\frac{C</em>P}{T})_{reactants}$

Rearranging and integrating between 298 K and T K, we have

$(\Delta S)<em>T = \int</em>{298}^{T} (\frac{\Delta C_P}{T}) dT$

$(\Delta S)<em>T = \int</em>{298}^{T} (\frac{\Delta C_P}{T}) dT$

$(\Delta S)<em>T = (\Delta S)</em>{298} + \int<em>{298}^{T} (\frac{\Delta C</em>P}{T}) dT = (\Delta S)<em>{298} + \int</em>{298}^{T} {\Delta C_P} d \ln T$

Eq. 50 is applicable to chemical reactions involving solids, liquids, or gases. Action