Entropy and spontaneity: R1.4.2 Gibbs free energy (ΔG) IB DP Chemistry Study Notes - New Syllabus 2025
Entropy and spontaneity – IB DP Chemistry- Study Notes
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Reactivity 1.4.2 — Gibbs Free Energy
Reactivity 1.4.2 — Gibbs Free Energy
Gibbs Free Energy (ΔG) is a thermodynamic quantity that determines the spontaneity of a chemical or physical process at constant temperature and pressure.
It combines two important factors:
- Enthalpy change (ΔH): Heat absorbed or released at constant pressure.
- Entropy change (ΔS): Measure of disorder or dispersal of energy in the system.
The relationship is given by the Gibbs equation.
Calculating ΔG°, ΔH°, and ΔS° using Gibbs Free Energy Equation
Gibbs Free Energy Equation
$ \Delta G^\circ = \Delta H^\circ – T\Delta S^\circ $
- \( \Delta G^\circ \): Standard Gibbs free energy change (kJ mol⁻¹)
- \( \Delta H^\circ \): Standard enthalpy change (kJ mol⁻¹)
- \( T \): Temperature in kelvin (K)
- \( \Delta S^\circ \): Standard entropy change (J K⁻¹ mol⁻¹)
Important: Ensure unit consistency. Since ΔH° and ΔG° are in kJ mol⁻¹ and ΔS° is in J K⁻¹ mol⁻¹, convert ΔS° to kJ by dividing by 1000:
$ \Delta G = \Delta H – T \left( \frac{\Delta S}{1000} \right) $
Interpreting the Sign of ΔG°:
- ΔG° < 0: Spontaneous under standard conditions.
- ΔG° > 0: Non-spontaneous under standard conditions.
- ΔG° = 0: Reaction is at equilibrium.
Effect of Temperature on Spontaneity:
The temperature plays a crucial role in determining spontaneity. Consider the signs of ΔH and ΔS:
| ΔH | ΔS | Spontaneity | Effect of Temperature |
|---|---|---|---|
| − | + | Always spontaneous (ΔG < 0) | Spontaneous at all T |
| − | − | Spontaneous at low T | Becomes non-spontaneous at high T |
| + | + | Spontaneous at high T | Non-spontaneous at low T |
| + | − | Never spontaneous (ΔG > 0) | Non-spontaneous at all T |
Note:
The symbol \( \Delta G^\circ \) refers to the standard Gibbs free energy change, which is measured under standard conditions (298 K, 1 atm pressure, 1 mol dm−3 concentration). The symbol \( \Delta G \), without the superscript, refers to non-standard conditions, where temperature, pressure, and concentration may vary. Use \( \Delta G^\circ \) when using standard thermodynamic data values from the IB Chemistry data booklet.
How to Apply the Equation
- Convert \( \Delta S^\circ \) from J to kJ (divide by 1000).
- Substitute values into the formula.
- Solve for the unknown quantity.
Example
Given: \( \Delta H^\circ = -92.0\ \text{kJ mol}^{-1} \), \( \Delta S^\circ = -198\ \text{J K}^{-1} \text{mol}^{-1} \), T = 298 K
Calculate \( \Delta G^\circ \).
▶️Answer/Explanation
\( \Delta S^\circ = -198\ \text{J K}^{-1} \text{mol}^{-1} = -0.198\ \text{kJ K}^{-1} \text{mol}^{-1} \)
$ \Delta G^\circ = -92.0\ \text{kJ mol}^{-1} – (298\ \text{K} \times -0.198\ \text{kJ K}^{-1} \text{mol}^{-1}) \\ \Delta G^\circ = -92.0 + 59.0 = -33.0\ \text{kJ mol}^{-1} $
So, the reaction is spontaneous at 298 K since \( \Delta G^\circ \lt 0 \).
Example
Given: \( \Delta H^\circ = -25.0\ \text{kJ mol}^{-1} \), \( \Delta S^\circ = -75\ \text{J K}^{-1} \text{mol}^{-1} \), T = 500 K
Calculate \( \Delta G^\circ \) and assess spontaneity.
▶️Answer/Explanation
\( \Delta S^\circ = -75\ \text{J K}^{-1} \text{mol}^{-1} = -0.075\ \text{kJ K}^{-1} \text{mol}^{-1} \)
$ \Delta G^\circ = -25.0 – (500 \times -0.075) = -25.0 + 37.5 = +12.5\ \text{kJ mol}^{-1} $
Conclusion: Since \( \Delta G^\circ \gt 0 \), the reaction is non-spontaneous at 500 K.
Because both ΔH and ΔS are negative, the reaction is only spontaneous at lower temperatures.
Example
The reaction:
\( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \)
has the following thermodynamic values:
- \( \Delta H^\circ = +58.0\ \text{kJ mol}^{-1} \)
- \( \Delta S^\circ = +176\ \text{J K}^{-1} \text{mol}^{-1} \)
Determine the temperature above which the reaction becomes spontaneous (i.e., when \( \Delta G^\circ < 0 \)).
▶️Answer/Explanation
Step 1: Convert entropy to kJ:
\( \Delta S^\circ = \frac{176}{1000} = 0.176\ \text{kJ K}^{-1} \text{mol}^{-1} \)
Step 2: Set \( \Delta G^\circ = 0 \) to find the transition temperature:
$ \Delta G^\circ = \Delta H^\circ – T\Delta S^\circ = 0 $ $ T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{58.0}{0.176} = 329.5\ \text{K} $
Conclusion: The reaction is:
- Non-spontaneous below 329.5 K
- Spontaneous above 329.5 K
Reasoning: Endothermic reaction (\( \Delta H^\circ > 0 \)) with increased disorder (\( \Delta S^\circ > 0 \)) becomes favorable only at higher temperatures where the entropy term outweighs the enthalpy input.