In which situation the potential energy of a system is equal to its Gibbs free energy?

Question

The other day I was giving a presentation about Transition State Theory, and I was showing both pictures of some potential energy surfaces (PES) and some Gibbs Free energy vs. reaction coordinate plots. At some point the professor asked me a question that I didn't fully understand to be honest, as I indeed didn't answer. She asked me when would the potential energy equal the Gibbs free energy. Can anyone explain when this would happen?

Answer 1

I can imagine situations that PES is equal to G, when we have two identities:

1. PES = U (ideal conditions)
2. U = G (Equilibrium or "no action" = zero Lagrangian)

The potential energy surface (PES) can be considered equal to the internal energy ( $ U $) of a system under certain idealized conditions. This equivalence can make sense and be particularly useful in specific contexts.

$ U - G $ can be considered analogous to a Lagrangian in the context of thermodynamics. To see this clearly, let's explore the expression $ U - G $ and how it relates to thermodynamic potentials and their differentials.

Relationship Between $ U $ and $ G $

The Gibbs free energy $ G $ is defined as: $$ G = H - TS = U + PV - TS $$

Rewriting this, we get: $$ G = U + PV - TS $$

Subtracting $ G $ from $ U $: $$ U - G = U - (U + PV - TS) $$ $$ U - G = - PV + TS $$

So, the expression $ U - G $ simplifies to: $$ U - G = TS - PV $$

Thermodynamic Interpretation

In thermodynamics, we often look for quantities that are minimized or maximized under certain conditions to describe equilibrium. The expression $ TS - PV $ combines the effects of entropy and volume changes, weighted by temperature and pressure respectively. Let's consider how $ U - G $ (or equivalently, $ TS - PV $) could serve as a Lagrangian-like function.

Potential Role of $ U - G $ as a Lagrangian

To explore this concept, we can draw an analogy with classical mechanics where the Lagrangian $ L $ is defined as the difference between kinetic energy ( $ T $) and potential energy ( $ V $): $$ L = T - V $$

In thermodynamics, $ U - G = TS - PV $ combines terms related to entropy and pressure-volume work. If we consider processes at constant temperature and pressure, minimizing $ G $ leads to equilibrium, much like how minimizing the action (integral of the Lagrangian) leads to the equations of motion in mechanics.

Differential Forms

• Internal Energy: $ dU = TdS - PdV $
• Gibbs Free Energy: $ dG = VdP - SdT $

From these differentials, subtracting $ dG $ from $ dU $ gives: $$ d(U - G) = dU - dG = (TdS - PdV) - (VdP - SdT) $$ $$ d(U - G) = TdS - SdT - PdV - VdP $$

If we interpret $ TS - PV $ as a Lagrangian-like function, we can consider the path of a thermodynamic process that extremizes this quantity.

Conceptual Framework

In this framework:

• $ S $ and $ V $ can be seen as analogs of generalized coordinates in mechanics.
• $ T $ and $ P $ can be seen as conjugate variables (analogous to generalized momenta).

The expression $ TS - PV $ then captures the "cost" of entropy and volume changes under constant temperature and pressure conditions.

Action Integral

Analogous to classical mechanics, where the action $ \mathcal{A} $ is the integral of the Lagrangian over time, we could define a thermodynamic action $ \mathcal{A} $ as: $$ \mathcal{A} = \int (TS - PV) \, dt $$

This integral would be evaluated over the path of a thermodynamic process.

Yes, we can consider specific examples where the Lagrangian-like function $ U - G $ (or equivalently $ TS - PV $) could be zero. Let's examine a few scenarios where this might occur.

Example: A Crystal at Zero Kelvin

At absolute zero temperature (0 K), the entropy $ S $ of a perfect crystal is zero according to the third law of thermodynamics. Therefore, the term $ TS $ becomes zero. Additionally, if we consider a system where the volume is held constant (e.g., a crystal lattice with negligible volume change), the $ PV $ term can also be zero if the pressure is zero or if the system is incompressible at low temperatures.

For a perfect crystal at zero Kelvin:

• $ T = 0 $
• $ S = 0 $
• $ P \approx 0 $ (if we assume the system is at atmospheric pressure and the volume is constant or negligible).

Under these conditions: $$ TS - PV = 0 \cdot 0 - 0 \cdot V = 0 $$

Thus, $ U - G = TS - PV = 0 $.

Example: Ideal Gas at Zero Pressure

Consider an ideal gas at zero pressure (which is a theoretical limit but useful for understanding the concept). In this case, the pressure $ P $ is zero, so the $ PV $ term is zero. If we also consider the temperature to be zero (again, a theoretical limit):

For an ideal gas at zero pressure and zero temperature:

• $ T = 0 $
• $ P = 0 $

Under these conditions: $$ TS - PV = 0 \cdot S - 0 \cdot V = 0 $$

Thus, $ U - G = TS - PV = 0 $.

General Conditions for Zero Lagrangian-like Function

The Lagrangian-like function $ U - G = TS - PV $ can be zero under the following general conditions:

1. Absolute Zero Temperature: When $ T = 0 $, the $ TS $ term is zero regardless of the entropy $ S $.
2. Zero Pressure: When $ P = 0 $, the $ PV $ term is zero regardless of the volume $ V $.
3. Zero Entropy: When $ S = 0 $, typically in perfect crystals at 0 K, making the $ TS $ term zero.
4. Zero Volume: In theoretical constructs where the volume is negligible, making the $ PV $ term zero.

Example: Phase Transition at Equilibrium

During a phase transition at equilibrium (e.g., liquid-gas equilibrium), the Gibbs free energy is minimized and the system is in a state where the temperature and pressure are constant. If the entropy and volume changes are balanced such that their contributions to the Lagrangian-like function are equal and opposite, $ TS - PV $ could be zero.

For example, at equilibrium during phase transition:

• $ T \Delta S = \Delta H $ (latent heat of phase transition)
• $ P \Delta V = \Delta G $

If $ \Delta G = 0 $ at equilibrium (common in reversible phase transitions), then: $$ \Delta (U - G) = T \Delta S - P \Delta V = 0 $$

Circumstances Where PES is Equal to $ U $

1. Isolated Systems:

   • For isolated systems with no interaction with the surroundings (no exchange of heat or work), the total energy is conserved. In such cases, the internal energy $ U $ is the primary energy of interest, and the PES represents the variations of $ U $ with respect to different configurations of the system.

2. Constant Volume and No Heat Exchange (Adiabatic Processes):

   • In processes where the volume remains constant ( $ \Delta V = 0 $) and there is no heat exchange ( $ \Delta Q = 0 $), the change in internal energy is equal to the work done. Since there is no heat transfer, $ U $ reflects the total energy changes purely due to the internal interactions. Therefore, the PES can be seen as representing $ U $.

3. Quantum Mechanical Systems:

   • In quantum chemistry and molecular physics, the PES often represents the potential energy of a system of nuclei and electrons, assuming the Born-Oppenheimer approximation where kinetic energy contributions are separated out. In this context, the PES corresponds to the electronic potential energy, which closely aligns with the internal energy $ U $ of the electrons for fixed nuclei positions.

4. No External Work Interaction:

   • When there is no external work being done on or by the system (i.e., $ P \Delta V = 0 $), the internal energy $ U $ becomes the relevant energy term to describe the state of the system. Under these conditions, the PES can be used to represent $ U $.

Thermodynamic Context

In thermodynamics, the internal energy ( $ U $) is a state function that includes all the microscopic kinetic and potential energies of the particles within the system. The potential energy surface (PES) describes how this energy varies with the positions of the particles (e.g., atomic configurations).

Example Scenarios

1. Ideal Gas in a Closed Container:

   • Consider an ideal gas in a perfectly insulated, rigid container. Since the container is rigid, the volume does not change ( $ \Delta V = 0 $), and since it is insulated, there is no heat exchange with the surroundings ( $ \Delta Q = 0 $). The internal energy $ U $ of the gas changes only due to internal interactions. Thus, the PES can be considered equal to $ U $.

2. Chemical Reactions in a Closed System:

   • For a chemical reaction occurring in a closed, insulated container where no work is done on or by the system (other than changes in internal energy due to bond breaking/forming), the PES describing the reaction pathway can be seen as representing the internal energy $ U $ of the system at various stages of the reaction.

3. Molecular Systems at Low Temperatures:

   • At very low temperatures, kinetic energy contributions can be minimal, and the internal energy $ U $ is dominated by potential energy interactions. Thus, the PES, which maps out these potential energy interactions, effectively represents $ U $.

Why It Makes Sense

• Simplified Analysis: Equating the PES with $ U $ allows for simplified analysis of systems where kinetic energy changes and work interactions are negligible or constant.
• Energy Landscape Understanding: In many chemical and physical processes, understanding the potential energy landscape (PES) provides insights into the stability, transition states, and reaction mechanisms of the system.
• Quantum Mechanics: In quantum mechanical treatments, the PES often directly represents the internal energy of the system, particularly within the Born-Oppenheimer approximation.