Additional: Thermodynamic Potentials
Enthalpy ($ H = U + PV $)
In many thermodynamic processes, especially those occurring at constant pressure, the work done is associated with changes in volume ($W = P\Delta V$). It is often convenient to combine the internal energy and this pressure-volume work term into a single state function called enthalpy.
Definition of Enthalpy (H)
Enthalpy ($H$) is a thermodynamic potential and a state function defined as the sum of the internal energy ($U$) of the system and the product of its pressure ($P$) and volume ($V$).
$ H = U + PV $
Enthalpy is an extensive property (depends on the amount of substance), and its units are the same as internal energy, pressure, and work, i.e., Joules (J).
Significance and Usefulness of Enthalpy
The primary usefulness of enthalpy is in analysing processes occurring at constant pressure, which are very common in many applications (e.g., chemical reactions in open containers, heating/cooling at constant atmospheric pressure).
Consider a process at constant pressure $P$. The change in enthalpy ($\Delta H$) is:
$ \Delta H = H_{final} - H_{initial} = (U_{final} + P_{final}V_{final}) - (U_{initial} + P_{initial}V_{initial}) $
Since the pressure is constant, $P_{final} = P_{initial} = P$.
$ \Delta H = (U_{final} - U_{initial}) + (PV_{final} - PV_{initial}) = \Delta U + P(V_{final} - V_{initial}) $
$ \Delta H = \Delta U + P\Delta V $
From the First Law of Thermodynamics, $\Delta U = Q - W$. For a process at constant pressure where the only work done is P-V work ($W = P\Delta V$), the First Law is $\Delta U = Q_P - P\Delta V$, where $Q_P$ is the heat transferred at constant pressure.
Substituting $\Delta U$ into the enthalpy change equation:
$ \Delta H = (Q_P - P\Delta V) + P\Delta V $
$ \Delta H = Q_P $
This is a very important result: For a process occurring at constant pressure with only P-V work, the heat absorbed by the system is equal to the change in its enthalpy.
This is why enthalpy change ($\Delta H$) is often used to quantify the heat involved in chemical reactions or phase changes occurring under constant pressure conditions. Exothermic reactions release heat ($\Delta H < 0$), and endothermic reactions absorb heat ($\Delta H > 0$).
For processes that do not involve volume change ($W=0$, isochoric) or constant volume processes with non-P-V work, $\Delta H$ is not necessarily equal to the heat transferred. However, $\Delta H$ still represents a well-defined change in a state function. The change in internal energy $\Delta U$ is always equal to the heat transferred in a constant volume process ($Q_V$).
Enthalpy is also related to specific heat capacities at constant pressure. For an ideal gas, $dU = nC_V dT$ and $P dV + V dP = nR dT$. At constant pressure, $dP=0$, so $P dV = nR dT$. $dH = dU + d(PV) = dU + P dV + V dP = dU + P dV$. At constant pressure, $dH = dU + P dV = nC_V dT + nR dT = n(C_V+R) dT = nC_P dT$. So, $dH = nC_P dT$, and $\Delta H = nC_P \Delta T$ for an ideal gas at constant pressure.
Helmholtz Free Energy ($ A = U - TS $)
While enthalpy is useful for constant pressure processes, Helmholtz Free Energy ($A$ or $F$) is a thermodynamic potential that is particularly useful for analysing processes occurring at constant volume and constant temperature.
Definition of Helmholtz Free Energy (A)
Helmholtz Free Energy ($A$) is a thermodynamic potential and a state function defined as the difference between the internal energy ($U$) of the system and the product of its absolute temperature ($T$) and entropy ($S$).
$ A = U - TS $
Helmholtz Free Energy is an extensive property, and its units are Joules (J).
Significance and Usefulness of Helmholtz Free Energy
The change in Helmholtz Free Energy ($\Delta A$) provides information about the maximum work that can be obtained from a thermodynamic process occurring at constant temperature and constant volume.
Consider an isothermal process ($\Delta T=0$, $T$ is constant). The change in Helmholtz Free Energy is:
$ \Delta A = A_{final} - A_{initial} = (U_{final} - T_{final}S_{final}) - (U_{initial} - T_{initial}S_{initial}) $
Since $T_{final} = T_{initial} = T$:
$ \Delta A = (U_{final} - U_{initial}) - T(S_{final} - S_{initial}) = \Delta U - T\Delta S $
From the First Law, $\Delta U = Q - W$. Substituting this into the $\Delta A$ equation:
$ \Delta A = (Q - W) - T\Delta S $
Rearranging:
$ W = Q - T\Delta S - \Delta A $
For a process occurring at constant temperature, $T\Delta S$ represents the minimum amount of energy that must be lost to the surroundings as heat due to entropy increase if the process is irreversible (by the Second Law, $Q \le T\Delta S$ for isothermal process, $Q = T\Delta S$ for reversible isothermal process). The maximum work output ($W_{max}$) from an isothermal process is achieved when the process is reversible, where $Q_{rev} = T\Delta S$.
$ W_{max} = Q_{rev} - T\Delta S - \Delta A $
$ W_{max} = T\Delta S - T\Delta S - \Delta A $
$ W_{max} = -\Delta A $
This means that for a reversible process at constant temperature, the work done by the system is equal to the decrease in Helmholtz Free Energy. This is why $A$ is sometimes called the Helmholtz free energy or Helmholtz function or work function – it represents the maximum useful work (total work including P-V work and other forms of work, often called "maximum work") that can be extracted from a system at constant temperature. The decrease in Helmholtz free energy ($\Delta A < 0$) indicates a spontaneous process that can do work at constant T and V.
For processes at constant temperature and constant volume, $W = \int P \, dV = 0$ if only P-V work is considered. In this specific case, the change in Helmholtz Free Energy is: $\Delta A = \Delta U - T\Delta S$. The decrease in $A$ during a process at constant T, V indicates the direction of spontaneous change towards equilibrium. At equilibrium at constant T, V, $A$ is minimum ($dA=0$).
Gibbs Free Energy ($ G = H - TS $)
Gibbs Free Energy ($G$) is arguably the most commonly used thermodynamic potential, especially in chemistry and materials science. It is particularly useful for analysing processes occurring at constant temperature and constant pressure, which are prevalent in everyday life and laboratory conditions.
Definition of Gibbs Free Energy (G)
Gibbs Free Energy ($G$) is a thermodynamic potential and a state function defined as the difference between the enthalpy ($H$) of the system and the product of its absolute temperature ($T$) and entropy ($S$).
$ G = H - TS $
Since $H = U + PV$, Gibbs Free Energy can also be written as:
$ G = U + PV - TS $
Gibbs Free Energy is an extensive property, and its units are Joules (J).
Significance and Usefulness of Gibbs Free Energy
The change in Gibbs Free Energy ($\Delta G$) provides information about the maximum non-P-V work that can be obtained from a thermodynamic process occurring at constant temperature and constant pressure, and more importantly, it is a criterion for spontaneity under these conditions.
Consider a process occurring at constant temperature ($T$) and constant pressure ($P$). The change in Gibbs Free Energy is:
$ \Delta G = G_{final} - G_{initial} = (H_{final} - T_{final}S_{final}) - (H_{initial} - T_{initial}S_{initial}) $
Since $T_{final} = T_{initial} = T$:
$ \Delta G = (H_{final} - H_{initial}) - T(S_{final} - S_{initial}) = \Delta H - T\Delta S $
From the First Law, $\Delta U = Q - W_{total}$, where $W_{total}$ is the total work done by the system (including P-V work and other forms of work, $W_{total} = W_{PV} + W_{nonPV}$). For a process at constant pressure, $W_{PV} = P\Delta V$. So, $W_{total} = P\Delta V + W_{nonPV}$.
$ \Delta U = Q - (P\Delta V + W_{nonPV}) $
$ Q = \Delta U + P\Delta V + W_{nonPV} $
We also know $\Delta H = \Delta U + \Delta(PV)$. At constant pressure, $\Delta H = \Delta U + P\Delta V$. So, $Q_P = \Delta H + W_{nonPV}$.
Substitute this $Q_P$ into the equation for $\Delta G = \Delta H - T\Delta S$ for a constant T, P process: $ Q = \Delta H + W_{nonPV}$.
For a reversible process at constant temperature, $Q_{rev} = T\Delta S$. (This is a general result from the Second Law, not limited to isothermal volume change). Applying this to the process at constant T, P:
$ T\Delta S = \Delta H + W_{nonPV, rev} $
Rearranging for the reversible non-P-V work ($W_{nonPV, rev}$):
$ W_{nonPV, rev} = T\Delta S - \Delta H $
Comparing this with $\Delta G = \Delta H - T\Delta S$, we see that:
$ W_{nonPV, rev} = -\Delta G $
This is a key result: For a reversible process at constant temperature and pressure, the non-P-V work (useful work, i.e., work other than that done by the system expanding against the surroundings) done by the system is equal to the decrease in Gibbs Free Energy. This is why $G$ is called the Gibbs free energy or Gibbs function – it represents the maximum useful work that can be extracted from a system at constant temperature and pressure.
Gibbs Free Energy as a Criterion for Spontaneity
The change in Gibbs Free Energy also provides a criterion for the spontaneity of a process at constant temperature and pressure:
- If $ \Delta G < 0 $: The process is spontaneous (occurs naturally) in the forward direction. The system tends towards a state of lower Gibbs free energy.
- If $ \Delta G > 0 $: The process is non-spontaneous in the forward direction. Work must be done on the system to make the process occur. The reverse process is spontaneous.
- If $ \Delta G = 0 $: The system is at equilibrium. There is no net tendency for the process to occur in either direction.
This is the most important application of Gibbs Free Energy. It predicts the direction of spontaneous change at constant T and P. At equilibrium at constant T, P, $G$ is minimum ($dG=0$).
The relationship $\Delta G = \Delta H - T\Delta S$ also shows that spontaneity depends on the balance between enthalpy change (favouring processes that release heat) and entropy change (favouring processes that increase entropy) at a given temperature.
Example 1. For a certain reaction at 298 K and 1 atm pressure, the enthalpy change is $\Delta H = -50$ kJ and the entropy change is $\Delta S = -100$ J/K. Is the reaction spontaneous under these conditions?
Answer:
Given temperature, $T = 298$ K.
Enthalpy change, $\Delta H = -50$ kJ $= -50 \times 10^3$ J.
Entropy change, $\Delta S = -100$ J/K.
Calculate the change in Gibbs Free Energy using the formula $\Delta G = \Delta H - T\Delta S$ (valid at constant T and P).
$ \Delta G = (-50 \times 10^3 \text{ J}) - (298 \text{ K}) \times (-100 \text{ J/K}) $
$ \Delta G = -50000 \text{ J} - (-29800 \text{ J}) $
$ \Delta G = -50000 \text{ J} + 29800 \text{ J} $
$ \Delta G = -20200 $ J.
$ \Delta G = -20.2 $ kJ.
Since $\Delta G$ is negative ($\Delta G = -20.2 \, \text{kJ} < 0$), the reaction is spontaneous under these conditions (298 K and 1 atm pressure).
 
