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.
Gibbs Free Energy Change (\(\Delta G\))
Gibbs energy change (\(\Delta G\)): The thermodynamic quantity that combines enthalpy, entropy, and absolute temperature to determine the maximum non-expansion work obtainable from a closed system at constant pressure. It serves as the ultimate criterion for reaction spontaneity.
The Gibbs Equation
The Gibbs-Helmholtz Equation
\( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)
| Symbol | Meaning | Units |
|---|---|---|
| \( \Delta G^\circ \) | Standard Gibbs 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!) |
Critical Unit Matching Trap (\(\Delta H\) in kJ vs \(\Delta S\) in J)
Critical Unit Trap: \(\Delta H\) is provided in \(\text{kJ mol}^{-1}\), while \(\Delta S\) is in \(\text{J K}^{-1}\text{mol}^{-1}\). You must convert \(\Delta S\) to kiloJoules (divide by 1000) or convert \(\Delta H\) to Joules before substituting into \(\Delta G = \Delta H - T\Delta S\).
Spontaneity Rules
Interpreting the Sign of \(\Delta G\)
- \(\Delta G < 0\) (negative): Reaction is spontaneous in the forward direction.
- \(\Delta G > 0\) (positive): Reaction is non-spontaneous in the forward direction (reverse reaction is spontaneous).
- \(\Delta G = 0\): System is at dynamic equilibrium (\(Q = K\)).
Spontaneous Does Not Mean Fast (Thermodynamics vs Kinetics)
A reaction with \(\Delta G < 0\) is thermodynamically spontaneous, but \(\Delta G\) provides zero information about reaction rate. A spontaneous reaction with a large activation energy (\(E_{\text{a}}\)) will proceed at an imperceptible rate under ambient conditions (e.g. diamond converting to graphite has \(\Delta G < 0\), but is kinetically stable for billions of years).
Worked Example: Calculating ΔG°
Calculating \(\Delta G^\circ\) for Ammonia Synthesis at 298 K
For the Haber process: \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})\)
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 \(298\text{ K}\) and determine if the reaction is spontaneous.
Step 1: Convert \(\Delta S^\circ\) to \(\text{kJ K}^{-1}\text{mol}^{-1}\)
\(\Delta S^\circ = \dfrac{-198.7}{1000} = -0.1987\text{ kJ K}^{-1}\text{mol}^{-1}\)
Step 2: Substitute into Gibbs equation
\(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\)
\(\Delta G^\circ = -92.2 - (298 \times -0.1987) = -92.2 - (-59.2) = \mathbf{-33.0\text{ kJ mol}^{-1}}\)
Conclusion: Because \(\Delta G^\circ\) is negative (\(-33.0\text{ kJ mol}^{-1}\)), the synthesis of ammonia is thermodynamically spontaneous at 298 K.
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 Temperature Boundary for Spontaneity (\(\Delta G = 0\))
At dynamic equilibrium (\(\Delta G = 0\)), the temperature at which a reaction shifts between spontaneous and non-spontaneous is given by setting \(\Delta H^\circ - T\Delta S^\circ = 0\):
\( T_{\text{crossover}} = \dfrac{\Delta H^\circ}{\Delta S^\circ} \)
For the Haber process: \(T = \dfrac{-92.2}{-0.1987} = 464\text{ K} = 191\text{ }^\circ\text{C}\). Above 464 K, the \(-T\Delta S\) penalty exceeds \(\Delta H\), and \(\Delta G\) becomes positive.
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