From R1.4.2, we know that \( \Delta G = \Delta H - T\Delta S \). Two of the four ΔH/ΔS combinations always give a clear answer. But when ΔH and ΔS share the same sign, spontaneity becomes temperature-dependent.
The Four-Quadrant Spontaneity Matrix
This is one of the most important tables in the entire HL syllabus. You should memorise it:
| ΔH | ΔS | Spontaneous? | Explanation |
|---|---|---|---|
| − (exo) | + (increase) | Always | Both terms favour spontaneity. ΔG is always negative regardless of temperature. |
| − (exo) | − (decrease) | At low T only | Favourable enthalpy vs unfavourable entropy. At low T the small TΔS term cannot overcome ΔH. At high T the TΔS term grows and makes ΔG positive. |
| + (endo) | + (increase) | At high T only | Unfavourable enthalpy vs favourable entropy. At high T the large TΔS term outweighs the positive ΔH, making ΔG negative. |
| + (endo) | − (decrease) | Never | Both terms oppose spontaneity. ΔG is always positive regardless of temperature. |
Calculating the Transition Temperature
For the two temperature-dependent cases (same-sign ΔH and ΔS), we can find the exact temperature at which spontaneity flips. This occurs when \( \Delta G = 0 \) (equilibrium):
The Crossover Temperature Equation
\( T = \dfrac{\Delta H^\circ}{\Delta S^\circ} \)
Derived by setting \(\Delta G^\circ = 0\) at the thermodynamic boundary.
Unit Matching in Temperature Calculations (\(\Delta H\) vs \(\Delta S\))
Unit Conversion Trap: Since \(\Delta H\) is given in \(\text{kJ mol}^{-1}\) and \(\Delta S\) in \(\text{J K}^{-1}\text{mol}^{-1}\), you must convert \(\Delta S\) to kiloJoules (divide by 1000) before dividing. Forgetting this creates a 1000-fold temperature calculation error.
Worked Example: Finding the Transition Temperature
Finding Crossover Temperature for an Endothermic Reaction
Steam reforming of methane: \(\text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{CO}(\text{g}) + 3\text{H}_2(\text{g})\)
Given: \(\Delta H^\circ = +206\text{ kJ mol}^{-1}\) and \(\Delta S^\circ = +215\text{ J K}^{-1}\text{mol}^{-1}\).
Calculate the temperature above which this reaction becomes spontaneous.
Step 1: Convert \(\Delta S^\circ\) to \(\text{kJ K}^{-1}\text{mol}^{-1}\)
\(\Delta S^\circ = \dfrac{+215}{1000} = +0.215\text{ kJ K}^{-1}\text{mol}^{-1}\)
Step 2: Apply the transition temperature formula
\(T = \dfrac{\Delta H^\circ}{\Delta S^\circ} = \dfrac{+206}{+0.215} \)
Step 3: Calculate
\(T = 958\text{ K} \quad (\mathbf{685\ ^\circ\text{C}})\)
Interpretation: Both \(\Delta H\) and \(\Delta S\) are positive. At low temperatures, the endothermic enthalpy dominates (\(\Delta G > 0\)). Above \(958\text{ K}\), the favourable \(T\Delta S\) term overcomes \(\Delta H\), making the process spontaneous.
Worked Example: Low-Temperature Spontaneity
Finding Spontaneity Limit for an Exothermic Dimerisation
For the reaction: \(2\text{NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\)
Given: \(\Delta H^\circ = -57.2\text{ kJ mol}^{-1}\) and \(\Delta S^\circ = -175.8\text{ J K}^{-1}\text{mol}^{-1}\).
Below what temperature is this reaction spontaneous?
Step 1: Convert \(\Delta S^\circ\) to \(\text{kJ K}^{-1}\text{mol}^{-1}\)
\(\Delta S^\circ = \dfrac{-175.8}{1000} = -0.1758\text{ kJ K}^{-1}\text{mol}^{-1}\)
Step 2: Apply the transition formula
\(T = \dfrac{\Delta H^\circ}{\Delta S^\circ} = \dfrac{-57.2}{-0.1758} = 325\text{ K} \quad (\mathbf{52\ ^\circ\text{C}})\)
Interpretation: Both \(\Delta H\) and \(\Delta S\) are negative. Below \(325\text{ K}\), the favourable exothermic enthalpy dominates (\(\Delta G < 0\)). Above \(325\text{ K}\), the unfavourable entropy penalty makes \(\Delta G\) positive.
Rigorous Structuring for Spontaneity Justifications
When justifying reaction spontaneity in an examination, always provide the complete 3-part logical progression:
- State the mathematical equation: \(\Delta G = \Delta H - T\Delta S\).
- Evaluate the sign and magnitude of the \(T\Delta S\) term relative to \(\Delta H\).
- Explicitly state: "Because \(\Delta G < 0\), the reaction is spontaneous at this temperature."
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