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):
\( T = \frac{\Delta H^\circ}{\Delta S^\circ} \)
(Both must be in the same energy unit, either both kJ or both J)
Worked Example: Finding the Transition Temperature
CH₄(g) + H₂O(g) → CO(g) + 3H₂(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 ΔS° to kJ:
\( \Delta S^\circ = \frac{+215}{1000} = +0.215 \text{ kJ K}^{-1}\text{mol}^{-1} \)
Step 2: Apply the transition temperature formula:
\( T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{+206}{+0.215} \)
Step 3: Calculate:
\( T = 958 \text{ K} \quad (685 \text{ °C}) \)
Interpretation: Because both ΔH and ΔS are positive, this reaction is only spontaneous at high temperatures. Below 958 K, the endothermic enthalpy demand dominates and ΔG is positive (non-spontaneous). Above 958 K, the favourable entropy term TΔS finally outweighs the enthalpy cost, making ΔG negative and the forward reaction spontaneous.
Worked Example: Low-Temperature Spontaneity
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 ΔS°:
\( \Delta S^\circ = -0.1758 \text{ kJ K}^{-1}\text{mol}^{-1} \)
Step 2: Apply the formula:
\( T = \frac{-57.2}{-0.1758} = 325 \text{ K} \quad (52 \text{ °C}) \)
Interpretation: Both ΔH and ΔS are negative. Below 325 K, the favourable exothermic enthalpy dominates and the reaction is spontaneous. Above 325 K, the unfavourable entropy term grows large enough to make ΔG positive, stopping the forward reaction.
Exam Strategy
When justifying a spontaneity prediction in an exam, always write the complete logic: "Because the calculated ΔG is negative, the reaction is spontaneous at this temperature." Simply calculating a number is often not enough for full marks.