The Spontaneity Paradox
The Second Law of Thermodynamics states that the total entropy of the universe must always increase for any process to occur spontaneously. But this creates a problem: how can we measure the entropy change of the entire universe for a reaction in a beaker?
In the 19th century, Josiah Willard Gibbs resolved this by creating a single equation that focuses entirely on the system, wrapping the universe's entropy requirement into one measurable value.
The change in Gibbs free energy relates the enthalpy change (\( \Delta H \)), the entropy change (\( \Delta S \)), and the absolute temperature (\( T \)) to determine whether a reaction is thermodynamically feasible.
The Gibbs Equation
\( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)
| Symbol | Meaning | Units |
|---|---|---|
| \( \Delta G^\circ \) | Standard Gibbs free energy change | kJ mol⁻¹ |
| \( \Delta H^\circ \) | Standard enthalpy change | kJ mol⁻¹ |
| \( T \) | Absolute temperature | K (Kelvin) |
| \( \Delta S^\circ \) | Standard entropy change | J K⁻¹ mol⁻¹ (convert to kJ!) |
Spontaneity Rules
What ΔG tells us
- ΔG < 0 (negative): The reaction is spontaneous (thermodynamically favourable).
- ΔG > 0 (positive): The reaction is non-spontaneous (not thermodynamically favourable).
- ΔG = 0: The system is at equilibrium.
Spontaneous ≠ Fast
A spontaneous reaction is one that is energetically feasible, meaning it can occur without a continuous input of external energy. However, it may still be extremely slow if the activation energy barrier is high. For example, the conversion of diamond to graphite is spontaneous but takes geological timescales. Thermodynamics tells us if a reaction can happen; kinetics tells us how fast.
Worked Example: Calculating ΔG°
Given: \( \Delta H^\circ = -92.2 \text{ kJ mol}^{-1} \) and \( \Delta S^\circ = -198.7 \text{ J K}^{-1}\text{mol}^{-1} \).
Calculate \( \Delta G^\circ \) at 298 K and determine if the reaction is spontaneous.
Step 1: Convert ΔS° to kJ:
\( \Delta S^\circ = \frac{-198.7}{1000} = -0.1987 \text{ kJ K}^{-1}\text{mol}^{-1} \)
Step 2: Substitute into the Gibbs equation:
\( \Delta G^\circ = -92.2 - (298 \times -0.1987) \)
\( \Delta G^\circ = -92.2 - (-59.2) = -92.2 + 59.2 \)
\( \Delta G^\circ = -33.0 \text{ kJ mol}^{-1} \)
Conclusion: Since \( \Delta G^\circ < 0 \), the reaction is spontaneous at 298 K. However, note that both \( \Delta H \) and \( \Delta S \) are negative, so spontaneity depends on temperature (it will become non-spontaneous at high temperatures when TΔS dominates).
The Four Scenarios
The sign of ΔG depends on the interplay between ΔH and TΔS. There are four possible combinations:
| ΔH | ΔS | ΔG | Spontaneous? |
|---|---|---|---|
| − (exo) | + (increase) | Always − | Always ✓ |
| − (exo) | − (decrease) | Depends on T | At low T |
| + (endo) | + (increase) | Depends on T | At high T |
| + (endo) | − (decrease) | Always + | Never ✗ |
Finding the Crossover Temperature
When ΔG = 0 (equilibrium), we can find the temperature at which spontaneity flips. Setting \( \Delta G = 0 \):
\( T = \frac{\Delta H^\circ}{\Delta S^\circ} \)
This only applies to scenarios where ΔH and ΔS have the same sign (both positive or both negative). At this particular temperature, the system is exactly at equilibrium.