Interactive Lecture
Enthalpy, Entropy, Gibbs and Helmholtz Free Energy
by Dr Arnaldo Delli Carri
Thermodynamic state variables (or state functions) are properties that depend only on the current state of a system, not on how the system reached that state. They are what let you characterise a thermodynamic system and predict how it will behave.
State variables are path-independent. The change in a state variable between two equilibrium states is the same regardless of the process path taken — unlike heat and work, which are path functions and depend on how you get there.
The basic state variables include:
From these fundamental variables, we derive composite state functions that provide deeper insights into system behaviour.
Enthalpy is a state function particularly useful for processes occurring at constant pressure, which are common in engineering applications.
Definition:
\[H = U + PV\]
Where:
Differential form:
\[dH = dU + PdV + VdP\]
For constant pressure processes:
\[\Delta H = Q_P\]
The enthalpy change equals the heat transferred at constant pressure.
Enthalpy is the total heat content of a system at constant pressure. It accounts for both the internal energy and the energy required to "make room" for the system (the \(PV\) term is flow work).
Entropy is a measure of disorder or energy dispersal in a system. It is central to the Second Law of Thermodynamics.
Thermodynamic Definition:
\[dS = \frac{\delta Q_{rev}}{T}\]
For a reversible process between states 1 and 2:
\[\Delta S = \int_1^2 \frac{\delta Q_{rev}}{T}\]
Statistical Definition (Boltzmann):
\[S = k_B \ln(\Omega)\]
Where:
Second Law of Thermodynamics:
\[\Delta S_{universe} \geq 0\]
Entropy quantifies the degree of disorder or randomness. Systems naturally evolve towards states of higher entropy. Entropy also measures the unavailability of energy to do useful work.
The Gibbs free energy determines the spontaneity of processes at constant temperature and pressure - the most common conditions in chemistry and biology.
Definition:
\[G = H - TS = U + PV - TS\]
Differential form:
\[dG = dH - TdS - SdT\]
Natural variables \(P\) and \(T\):
\[dG = VdP - SdT\]
At constant temperature:
\[\Delta G = \Delta H - T\Delta S\]
Criterion for Spontaneity:
Gibbs free energy is the maximum useful work obtainable at constant temperature and pressure. It balances the enthalpy drive (energy release) against the entropy drive (disorder increase).
The Helmholtz free energy is useful for processes at constant temperature and volume, common in theoretical studies and closed rigid systems.
Definition:
\[A = U - TS\]
Differential form:
\[dA = dU - TdS - SdT\]
Natural variables \(V\) and \(T\):
\[dA = -PdV - SdT\]
At constant temperature:
\[\Delta A = \Delta U - T\Delta S\]
Criterion for Spontaneity (constant T, V):
Helmholtz free energy is the maximum useful work obtainable at constant temperature and volume, excluding expansion work. It's the "free" energy available to do non-expansion work.
The four thermodynamic potentials lead to important mathematical relationships called Maxwell relations:
From Internal Energy \(dU = TdS - PdV\):
\[\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V\]
From Enthalpy \(dH = TdS + VdP\):
\[\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P\]
From Helmholtz Free Energy \(dA = -SdT - PdV\):
\[\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V\]
From Gibbs Free Energy \(dG = -SdT + VdP\):
\[\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P\]
These state functions are intimately related through thermodynamic identities:
\[G = H - TS = U + PV - TS = A + PV\]
\[A = U - TS = G - PV\]
\[H = U + PV = G + TS\]
\[U = H - PV = A + TS = G - PV + TS\]
Explore how thermodynamic state variables change during different processes. Use Explore mode to study individual processes, Compare Paths to see why state functions are path-independent, or Build Cycle to construct thermodynamic cycles and watch state functions return to zero.
| State Variable | Definition | Differential Form | Physical Meaning |
|---|---|---|---|
| Internal Energy (U) | Fundamental property | \(dU = TdS - PdV\) | Total microscopic energy |
| Enthalpy (H) | \(H = U + PV\) | \(dH = TdS + VdP\) | Heat content at constant P |
| Entropy (S) | \(dS = \frac{\delta Q_{rev}}{T}\) | \(dS \geq \frac{\delta Q}{T}\) | Measure of disorder |
| Gibbs Free Energy (G) | \(G = H - TS\) | \(dG = VdP - SdT\) | Available work at const. T, P |
| Helmholtz Free Energy (A) | \(A = U - TS\) | \(dA = -PdV - SdT\) | Available work at const. T, V |
Two moles of nitrogen gas (diatomic, \(C_P = 29.1\) J/(mol·K)) are heated from 300 K to 500 K at constant pressure of 200 kPa. Calculate:
(a) The change in internal energy
(b) The change in enthalpy
(c) The work done by the gas
(d) The heat transferred
Show Solution
Given:
(a) Change in Internal Energy:
\[\Delta U = nC_V\Delta T = 2 \times 20.786 \times (500 - 300)\]
\[\Delta U = 2 \times 20.786 \times 200 = 8{,}314 \text{ J}\]
(b) Change in Enthalpy:
\[\Delta H = nC_P\Delta T = 2 \times 29.1 \times 200\]
\[\Delta H = 11{,}640 \text{ J}\]
(c) Work Done by Gas:
For isobaric process: \(W = P\Delta V = nR\Delta T\)
\[W = 2 \times 8.314 \times 200 = 3{,}326 \text{ J}\]
(d) Heat Transferred:
At constant pressure: \(Q = \Delta H\)
\[Q = 11{,}640 \text{ J}\]
Verify with First Law: \(Q = \Delta U + W = 8{,}314 + 3{,}326 = 11{,}640\) J ✓
Answer: (a) \(\Delta U = 8{,}314\,\text{J}\); (b) \(\Delta H = 11{,}640\,\text{J}\); (c) \(W = 3{,}326\,\text{J}\); (d) \(Q = 11{,}640\,\text{J}\)
One mole of an ideal gas expands isothermally at 400 K from volume 10 L to 30 L. Calculate:
(a) The entropy change of the gas
(b) The heat transferred
(c) Is this process reversible or irreversible if it occurs spontaneously?
Show Solution
Given:
(a) Entropy Change:
For isothermal expansion of ideal gas:
\[\Delta S = nR\ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times \ln\left(\frac{30}{10}\right)\]
\[\Delta S = 8.314 \times \ln(3) = 8.314 \times 1.099 = 9.13 \text{ J/K}\]
(b) Heat Transferred:
For isothermal process: \(\Delta U = 0\), so \(Q = W\)
\[Q = nRT\ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times 400 \times 1.099\]
\[Q = 3{,}653 \text{ J}\]
Alternative: \(Q = T\Delta S = 400 \times 9.13 = 3{,}652\) J ✓
(c) Reversibility:
The gas entropy increases by 9.13 J/K. If the expansion is into a vacuum (free expansion), no heat is transferred from surroundings, so:
\[\Delta S_{total} = \Delta S_{gas} + \Delta S_{surroundings} = 9.13 + 0 = 9.13 \text{ J/K} > 0\]
Since \(\Delta S_{total} > 0\), the process is irreversible.
For a reversible isothermal expansion, heat would be transferred from a reservoir at 400 K, giving \(\Delta S_{surroundings} = -Q/T = -9.13\) J/K, making \(\Delta S_{total} = 0\).
Answer: (a) \(\Delta S = 9.13\,\text{J/K}\); (b) \(Q = 3{,}653\,\text{J}\); (c) irreversible if it occurs spontaneously (\(\Delta S_{total} > 0\))
A chemical reaction at 298 K and 1 atm has \(\Delta H = -92.4\) kJ and \(\Delta S = -198.7\) J/K. Calculate:
(a) The Gibbs free energy change at 298 K
(b) Is the reaction spontaneous at 298 K?
(c) At what temperature does the reaction become non-spontaneous?
Show Solution
Given:
(a) Gibbs Free Energy Change:
\[\Delta G = \Delta H - T\Delta S\]
\[\Delta G = -92{,}400 - 298 \times (-198.7)\]
\[\Delta G = -92{,}400 + 59{,}213 = -33{,}187 \text{ J} = -33.2 \text{ kJ}\]
(b) Spontaneity at 298 K:
Since \(\Delta G = -33.2\) kJ \(< 0\), the reaction is spontaneous at 298 K.
(c) Temperature at Which Reaction Becomes Non-spontaneous:
The reaction becomes non-spontaneous when \(\Delta G > 0\).
At the transition point: \(\Delta G = 0\)
\[0 = \Delta H - T\Delta S\]
\[T = \frac{\Delta H}{\Delta S} = \frac{-92{,}400}{-198.7} = 465 \text{ K}\]
The reaction is spontaneous below 465 K and becomes non-spontaneous above 465 K.
Interpretation: This is an exothermic reaction (\(\Delta H < 0\)) with decreasing entropy (\(\Delta S < 0\)). At low temperatures, the enthalpy term dominates, making it spontaneous. At high temperatures, the entropy term dominates, making it non-spontaneous.
Answer: (a) \(\Delta G = -33.2\,\text{kJ}\); (b) spontaneous at 298 K (\(\Delta G < 0\)); (c) non-spontaneous above \(T = 465\,\text{K}\)
An ideal gas undergoes a cycle consisting of three processes:
For 1 mole of gas with \(C_V = 12.5\) J/(mol·K), calculate the total change in:
(a) Internal energy for the complete cycle
(b) Enthalpy for the complete cycle
(c) Entropy for the complete cycle
Show Solution
Given:
\(U\), \(H\), and \(S\) are state functions, so their changes over a complete cycle must be zero.
(a) Total Change in Internal Energy:
\[\Delta U_{cycle} = U_1 - U_1 = 0\]
For any thermodynamic cycle, \(\Delta U = 0\) because the system returns to its initial state.
Verification by calculating each process:
Process 1→2 (isothermal): \(\Delta U_{12} = 0\) (constant T)
Process 2→3 (isochoric): \(\Delta U_{23} = nC_V\Delta T = 1 \times 12.5 \times (300-400) = -1{,}250\) J
Process 3→1 (isobaric): \(\Delta U_{31} = nC_V\Delta T = 1 \times 12.5 \times (400-300) = 1{,}250\) J
Total: \(\Delta U_{cycle} = 0 - 1{,}250 + 1{,}250 = 0\) ✓
(b) Total Change in Enthalpy:
\[\Delta H_{cycle} = H_1 - H_1 = 0\]
Same reasoning: state function over a cycle equals zero.
(c) Total Change in Entropy:
\[\Delta S_{cycle} = S_1 - S_1 = 0\]
Again, state function over a cycle equals zero.
Important Conclusion: This problem illustrates the fundamental property of state functions: their change depends only on the initial and final states, not the path. Over any complete cycle, all state functions return to their original values, giving zero net change.
Answer: (a) \(\Delta U_{cycle} = 0\); (b) \(\Delta H_{cycle} = 0\); (c) \(\Delta S_{cycle} = 0\)