Entropy and spontaneity: R1.4.3 Spontaneity and Gibbs free energy IB DP Chemistry Study Notes - New Syllabus 2025
Entropy and spontaneity – IB DP Chemistry- Study Notes
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Reactivity 1.4.3 — Spontaneity and Gibbs Free Energy at Constant Pressure
Reactivity 1.4.3 — Spontaneity and Gibbs Free Energy at Constant Pressure
What is Spontaneity?
A chemical or physical process is said to be spontaneous if it occurs naturally under a given set of conditions, without needing continuous input of external energy (beyond initial activation energy).
Note that spontaneity does not imply speed. A spontaneous reaction may occur slowly if the activation energy is high (e.g., rusting of iron).
Gibbs Free Energy and Spontaneity
The Gibbs free energy change, \( \Delta G \), is a thermodynamic quantity that combines enthalpy and entropy to predict the spontaneity of a reaction at constant temperature and pressure.
$ \Delta G = \Delta H – T\Delta S $
- \( \Delta G \): Gibbs free energy change (kJ mol⁻¹)
- \( \Delta H \): Enthalpy change (kJ mol⁻¹)
- \( \Delta S \): Entropy change (J K⁻¹ mol⁻¹) — must be converted to kJ by dividing by 1000
- \( T \): Absolute temperature in Kelvin (K)
Criteria for Spontaneity
| \( \Delta G \) | Spontaneity | Description |
|---|---|---|
| \( \Delta G < 0 \) | Spontaneous | Reaction occurs naturally under the given conditions.(feasible) |
| \( \Delta G > 0 \) | Non-spontaneous | Reaction requires external energy to proceed.(not feasible) |
| \( \Delta G = 0 \) | At equilibrium | No net change in the system; forward and reverse rates are equal. |
Factors Affecting Gibbs Free Energy Change (\( \Delta G \))
1. Enthalpy Change (\( \Delta H \))
- Represents the heat released or absorbed at constant pressure.
- A negative \( \Delta H \) (exothermic) favors spontaneity and makes \( \Delta G \) more negative.
- A positive \( \Delta H \) (endothermic) increases \( \Delta G \), possibly making it non-spontaneous unless offset by entropy.
2. Entropy Change (\( \Delta S \))
- Indicates the degree of disorder or dispersal of energy in a system.
- A positive \( \Delta S \) increases the \( -T\Delta S \) term, favoring spontaneity.
- A negative \( \Delta S \) can make \( \Delta G \) more positive, reducing spontaneity.
3. Temperature (T, in Kelvin)
- Plays a crucial role in the balance between \( \Delta H \) and \( \Delta S \).
- At high T, the \( -T\Delta S \) term dominates, making entropy-driven reactions more spontaneous.
- At low T, enthalpy has a greater effect, favoring exothermic reactions.
4. Physical State of Reactants and Products
- Reactions that produce gases generally have higher entropy and may result in more negative \( \Delta G \).
- Reactions going from solid → liquid → gas increase \( \Delta S \) and favor spontaneity.
5. Concentration and Pressure (for non-standard conditions)
- In real systems (not standard state), changes in pressure or concentration affect reaction quotient \( Q \), modifying actual \( \Delta G \):
$ \Delta G = \Delta G^\circ + RT \ln Q $ - Where \( R \) is the gas constant and \( Q \) is the reaction quotient.
Entropy Changes in System vs Surroundings
Gibbs energy includes both:
- System entropy change: \( \Delta S_{\text{system}} \)
- Entropy change of surroundings: accounted for via \( \Delta H/T \)
This allows \( \Delta G \) to predict the net entropy change of the universe and determine spontaneity.
Example
Determine whether the following reaction is spontaneous at 298 K:
$ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) $
Given:
- \( \Delta H^\circ = -92.4 \ \text{kJ mol}^{-1} \)
- \( \Delta S^\circ = -198.3 \ \text{J K}^{-1} \text{mol}^{-1} \)
▶️Answer/Explanation
Step 1: Convert entropy to kJ:
$ \Delta S^\circ = \frac{-198.3}{1000} = -0.1983 \ \text{kJ K}^{-1} \text{mol}^{-1} $
Step 2: Apply the Gibbs free energy equation:
$ \Delta G^\circ = \Delta H^\circ – T\Delta S^\circ $
$ \Delta G^\circ = -92.4 – (298)(-0.1983) = -92.4 + 59.1 = -33.3 \ \text{kJ mol}^{-1} $
Conclusion: Since \( \Delta G^\circ \) is negative, the reaction is spontaneous under standard conditions at 298 K.
Temperature Dependence of Spontaneity
The spontaneity of a reaction depends on the signs of \( \Delta H \) and \( \Delta S \):
| \( \Delta H \) | \( \Delta S \) | \( \Delta G \) Behavior | Spontaneity |
|---|---|---|---|
| – | + | Always negative | Spontaneous at all temperatures |
| + | – | Always positive | Never spontaneous |
| – | – | Negative at low T | Spontaneous at low temperatures |
| + | + | Negative at high T | Spontaneous at high temperatures |
Calculating the Temperature at Which a Reaction Becomes Spontaneous
To find the temperature where \( \Delta G = 0 \), set the Gibbs equation to zero:
$ 0 = \Delta H – T\Delta S \Rightarrow T = \frac{\Delta H}{\Delta S} $
Important: Ensure unit consistency. If \( \Delta H \) is in kJ mol⁻¹ and \( \Delta S \) is in J K⁻¹ mol⁻¹, convert entropy to kJ:
$ \Delta S \ (\text{kJ K}^{-1} \text{mol}^{-1}) = \frac{\Delta S \ (\text{J K}^{-1} \text{mol}^{-1})}{1000} $
Example
Given: \( \Delta H = +40.0 \ \text{kJ mol}^{-1} \), \( \Delta S = +100 \ \text{J K}^{-1} \text{mol}^{-1} \). At what temperature does the reaction become spontaneous?
▶️Answer/Explanation
Step 1: Convert entropy to kJ mol⁻¹ K⁻¹:
$ \Delta S = \frac{100}{1000} = 0.100 \ \text{kJ K}^{-1} \text{mol}^{-1} $
Step 2: Set \( \Delta G = 0 \) and solve for T:
$ T = \frac{\Delta H}{\Delta S} = \frac{40.0}{0.100} = 400 \ \text{K} $
Conclusion: The reaction becomes spontaneous above 400 K.
Example:
Predict how the spontaneity of the following reaction changes with temperature:
\( \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \)
- \( \Delta H^\circ = -92 \, \text{kJ mol}^{-1} \)
- \( \Delta S^\circ = -198 \, \text{J K}^{-1} \text{mol}^{-1} \)
▶️Answer/Explanation
Step 1: Identify Signs of ΔH and ΔS
- \( \Delta H^\circ < 0 \): Exothermic reaction (favorable for spontaneity).
- \( \Delta S^\circ < 0 \): Entropy decreases (unfavorable).
Step 2: Analyze Using the Gibbs Equation
\( \Delta G^\circ = \Delta H^\circ – T\Delta S^\circ \)
Since both ΔH and ΔS are negative:
- At low temperatures: The \( T\Delta S \) term is small → \( \Delta G \) likely negative → reaction spontaneous.
- At high temperatures: The \( T\Delta S \) term becomes more significant → \( \Delta G \) could become positive → reaction becomes non-spontaneous.
Conclusion:
The reaction is spontaneous only at low temperatures. Increasing the temperature reduces the favorability due to the unfavorable entropy change.
This explains why industrial ammonia production (Haber process) is carried out at moderate temperatures to balance rate and yield.