The Drive Toward Equilibrium
In previous subtopics, we calculated \( \Delta G^\circ \) assuming reactions go from pure standard reactants to pure standard products. In reality, reversible reactions in closed systems never reach 100% completion. Instead, they progress toward a state of minimum free energy called equilibrium.
Key Principles
- All chemical systems spontaneously move toward the state of lowest possible free energy.
- When a reaction starts, the driving force (\( \Delta G \)) is at its largest magnitude.
- As the reaction progresses and concentrations shift, the driving force steadily diminishes.
- At equilibrium, \( \Delta G = 0 \). There is no remaining free energy to drive the reaction in either direction.
The Reaction Quotient (Q)
The instantaneous ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. Q has the same mathematical form as K, but can be calculated at any point during the reaction, not just at equilibrium.
By comparing Q to K, we can predict which direction the reaction will shift:
| Condition | Meaning | ΔG | Direction |
|---|---|---|---|
| Q < K | Product ratio is too low | Negative | Forward → |
| Q = K | System is at equilibrium | Zero | No net change |
| Q > K | Product ratio is too high | Positive | ← Reverse |
The Non-Standard Gibbs Equation
To find the instantaneous ΔG at any point during a reaction (not just under standard conditions), we use:
\( \Delta G = \Delta G^\circ + RT\ln Q \)
| Symbol | Meaning | Value / Units |
|---|---|---|
| \( \Delta G \) | Instantaneous Gibbs free energy change | J mol⁻¹ |
| \( \Delta G^\circ \) | Standard Gibbs free energy change | J mol⁻¹ |
| \( R \) | Universal gas constant | 8.314 J K⁻¹ mol⁻¹ |
| \( T \) | Absolute temperature | K |
| \( Q \) | Reaction quotient | Dimensionless |
Calculating K from ΔG°
At equilibrium, two critical substitutions occur: \( \Delta G = 0 \) and \( Q = K \). Substituting these into the non-standard equation gives one of the most powerful equations in the HL syllabus:
\( \Delta G^\circ = -RT\ln K \)
This equation allows you to calculate the equilibrium constant directly from thermodynamic data, without ever performing the experiment.
What ΔG° tells us about K
- ΔG° < 0: \( \ln K > 0 \), so \( K > 1 \). Products are favoured at equilibrium.
- ΔG° > 0: \( \ln K < 0 \), so \( K < 1 \). Reactants are favoured at equilibrium.
- ΔG° = 0: \( K = 1 \). Neither side is favoured.
Worked Example: Calculating K from ΔG°
Given: \( \Delta G^\circ = +4.7 \text{ kJ mol}^{-1} \) at 298 K.
Calculate the equilibrium constant K.
Step 1: Convert ΔG° to J mol⁻¹ (since R is in J):
\( \Delta G^\circ = +4700 \text{ J mol}^{-1} \)
Step 2: Rearrange for ln K:
\( \ln K = \frac{-\Delta G^\circ}{RT} = \frac{-4700}{8.314 \times 298} \)
Step 3: Calculate:
\( \ln K = \frac{-4700}{2477.6} = -1.897 \)
Step 4: Convert to K using \( K = e^{\ln K} \):
\( K = e^{-1.897} = 0.150 \)
Interpretation: Since \( \Delta G^\circ \) is positive, \( K < 1 \), confirming that at 298 K the reactants (N₂O₄) are favoured at equilibrium. The forward decomposition does not proceed very far under standard conditions.
Think About It
If a reaction has a very large negative ΔG° (e.g. −200 kJ mol⁻¹), what does this tell you about K?
A large negative ΔG° gives a very large positive ln K, meaning K is extremely large (e.g. 10³⁵). The reaction essentially goes to completion, with almost no reactants remaining at equilibrium.