Entropy and spontaneity: R1.4.4 Gibbs free energy and equilibrium IB DP Chemistry Study Notes - New Syllabus 2025
Entropy and spontaneity – IB DP Chemistry- Study Notes
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Reactivity 1.4.4 : Change in Gibbs Free Energy as a Reaction Approaches Equilibrium
Reactivity 1.4.4 : Change in Gibbs Free Energy as a Reaction Approaches Equilibrium
In thermodynamics, the spontaneity and direction of a chemical reaction are governed by the change in Gibbs free energy, denoted as \( \Delta G \).
Gibbs Free Energy and Equilibrium
- \( \Delta G \) indicates whether a reaction is spontaneous at constant temperature and pressure.
- As a reaction proceeds toward equilibrium, the composition of the system changes and so does the value of \( \Delta G \).
Key Idea: As a chemical reaction proceeds toward equilibrium, the magnitude of \( \Delta G \) decreases, eventually reaching zero at equilibrium.
Behaviour of \( \Delta G \) Over the Course of a Reaction
- At the beginning of a spontaneous reaction, \( \Delta G < 0 \) (negative), meaning the reaction is energetically favorable.
- As the reaction progresses and the concentrations of reactants and products change, the system moves closer to equilibrium.
- This leads to a reduction in the magnitude of \( \Delta G \) — it becomes less negative over time.
- At equilibrium, the system has no driving force to change further, and therefore:
- \( \Delta G = 0 \)
Thermodynamic Perspective
From a thermodynamic standpoint, the system seeks to minimize its Gibbs free energy. The equilibrium position represents the point of minimum Gibbs free energy under the given conditions.
Graphical Representation
In a plot of Gibbs free energy (\( G \)) versus reaction progress:
- The graph typically shows a downward curve as the reaction proceeds spontaneously.
- The minimum point on the curve corresponds to equilibrium.
- To the left of equilibrium, \( \Delta G < 0 \); to the right, \( \Delta G > 0 \).
Microscopic Interpretation
At the molecular level, the system constantly adjusts the rates of the forward and reverse reactions. Equilibrium is reached when these rates are equal, and the overall free energy change becomes zero.
Summary Table
| Stage | \( \Delta G \) | System Behavior |
|---|---|---|
| Initial (far from equilibrium) | Very negative | Reaction proceeds spontaneously |
| Midway | Less negative | Reaction still proceeds, but less vigorously |
| At equilibrium | Zero | No net change; dynamic balance |
Example
Consider the reaction: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
Initially, only nitrogen and hydrogen are present. Predict how \( \Delta G \) changes as the reaction proceeds toward equilibrium.
▶️Answer/Explanation
At the beginning, the concentration of ammonia is zero, and the forward reaction is strongly favored, so \( \Delta G \) is highly negative. As ammonia forms and its concentration increases, the reaction mixture approaches equilibrium. The value of \( \Delta G \) becomes less negative because the driving force to form more ammonia decreases. Eventually, equilibrium is reached, and at that point, \( \Delta G = 0 \). This means the forward and reverse reactions occur at the same rate and the system’s free energy is at a minimum.
Calculations Using Gibbs Free Energy and Equilibrium Relationships
Gibbs Free Energy Under Non-Standard Conditions
To determine how far a system is from equilibrium or to predict the direction of spontaneous change under non-standard conditions, the equation used is:
\( \Delta G = \Delta G^\circ + RT \ln Q \)
Here,
- \( \Delta G \) is the Gibbs free energy change under actual conditions,
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( R \) is the gas constant (8.314 J mol⁻¹ K⁻¹),
- \( T \) is temperature in Kelvin, and
- \( Q \) is the reaction quotient.
Reaction Quotient \( Q \)
The reaction quotient \( Q \) reflects the ratio of the concentrations or partial pressures of products and reactants at a given moment.
Equilibrium Constant \( K \)
The equilibrium constant \( K \) expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients. It is calculated using the same formula as the reaction quotient \( Q \), but specifically under equilibrium conditions.
- If \( Q < K \), the reaction proceeds in the forward direction and \( \Delta G < 0 \);
- if \( Q > K \), the reaction proceeds in the reverse direction and \( \Delta G > 0 \).
- When \( Q = K \), the system is at equilibrium and \( \Delta G = 0 \).
Relationship Between Gibbs Free Energy and Equilibrium Constant
At equilibrium, since \( \Delta G = 0 \), the above equation simplifies to:
\( \Delta G^\circ = -RT \ln K \), linking thermodynamic favorability with equilibrium position.
- A large, negative \( \Delta G^\circ \) corresponds to a large \( K \), indicating a product-favored reaction at equilibrium.
- A positive \( \Delta G^\circ \) yields a small \( K \), favoring the reactants at equilibrium.
This relationship helps chemists determine how favorable a reaction is, estimate equilibrium constants from thermodynamic data, or vice versa, using appropriate temperature and standard Gibbs energy values provided in data booklets.
Example
The standard Gibbs energy change for the reaction \[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \] is \( \Delta G^\circ = -33.0 \, \text{kJ mol}^{-1} \) at 298 K. Calculate the equilibrium constant \( K \) at this temperature.
▶️Answer/Explanation
Step 1: Use the equation \( \Delta G^\circ = -RT \ln K \)
Step 2: Rearranged: \( \ln K = -\frac{\Delta G^\circ}{RT} \)
Convert \( \Delta G^\circ \) to J mol⁻¹: \( -33.0 \times 10^3 = -33000 \, \text{J mol}^{-1} \)
Substitute: \[ \ln K = -\frac{-33000}{8.314 \times 298} = \frac{33000}{2477.572} \approx 13.32 \]
Step 3: Exponentiate to solve for \( K \): \[ K = e^{13.32} \approx 6.1 \times 10^5 \]
Since \( K \gg 1 \), this means the reaction strongly favors the products (ammonia) at equilibrium.
Example:
Consider the reaction: \( \text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g) \) At 298 K, the following data are given:
- Standard Gibbs free energy change: \( \Delta G^\circ = -28.6 \, \text{kJ mol}^{-1} \)
- Gas constant: \( R = 8.31 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Reaction quotient: \( Q = \dfrac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = 0.20 \)
Calculate the actual Gibbs free energy change \( \Delta G \) under these non-standard conditions, and state whether the reaction will proceed spontaneously in the forward or reverse direction.
▶️Answer/Explanation
Step 1: Convert \( \Delta G^\circ \) to J/mol: \( -28.6 \, \text{kJ mol}^{-1} = -28600 \, \text{J mol}^{-1} \)
Step 2: Apply the equation:
\[ \Delta G = -28600 + (8.31)(298) \ln(0.20) \] \[ \ln(0.20) \approx -1.609 \Rightarrow \Delta G = -28600 + (8.31 \times 298 \times -1.609) \] \[ \Delta G \approx -28600 – 3983 \approx -32583 \, \text{J mol}^{-1} \]
Conclusion: Since \( \Delta G < 0 \), the reaction is spontaneous under the given non-standard conditions. A small value of \( Q \) suggests a shift toward products to reach equilibrium.
Applications of the Gibbs Free Energy Equilibrium Relationships
1. Predicting Equilibrium Position:
Using the equation \( \Delta G^\circ = -RT \ln K \), we can predict whether a reaction is product-favored or reactant-favored at equilibrium.
2. Estimating Feasibility of Industrial Processes:
Chemical engineers use \( \Delta G^\circ \) and \( K \) to determine optimal reaction conditions (like temperature and pressure) for maximizing yield in equilibrium-limited processes such as ammonia synthesis (Haber process) or sulfuric acid production (Contact process).
3. Coupling Reactions in Biochemistry:
Biological systems often drive thermodynamically unfavorable reactions (positive \( \Delta G^\circ \)) by coupling them with highly favorable ones (like ATP hydrolysis with very negative \( \Delta G^\circ \)). This enables cellular work such as muscle contraction, nerve impulses, and biosynthesis.
4. Electrochemical and Fuel Cell Calculations:
In electrochemistry, Gibbs free energy is used to relate the standard cell potential to the equilibrium constant via \( \Delta G^\circ = -nFE^\circ \) and subsequently to \( K \) using \( \Delta G^\circ = -RT \ln K \). This is key to designing batteries and fuel cells with efficient performance.
5. Environmental Chemistry and Solubility:
Thermodynamic data is applied to assess solubility equilibria and the stability of pollutants in environmental systems. Knowing the value of \( \Delta G \) helps determine whether precipitation, dissolution, or adsorption processes are spontaneous under natural conditions.