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.
Thermodynamic Equilibrium and Minimum Free Energy
- All chemical systems spontaneously evolve toward the state of lowest possible Gibbs free energy (\(G\)).
- Dynamic equilibrium is established at the precise minimum of the free energy curve, where \(\Delta G = 0\).
- At equilibrium, the rate of forward and reverse reactions are equal, and macroscopic composition remains constant.
The Reaction Quotient (Q)
Reaction Quotient (Q) vs Equilibrium Constant (K)
Reaction Quotient (\(Q\)): The instantaneous ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients at any given point during a reaction. When \(Q < K\), the forward reaction is spontaneous (\(\Delta G < 0\)); at equilibrium, \(Q = K\) and \(\Delta G = 0\).
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:
Gibbs Free Energy Under Non-Standard Conditions
\( \Delta G = \Delta G^\circ + RT\ln Q \)
| Symbol | Meaning | Value / Units |
|---|---|---|
| \( \Delta G \) | Instantaneous Gibbs energy change | J mol⁻¹ |
| \( \Delta G^\circ \) | Standard Gibbs 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:
Connecting Standard Gibbs Energy to Equilibrium Constant
\( \Delta G^\circ = -RT\ln K \)
This equation allows you to calculate the equilibrium constant directly from thermodynamic data, without ever performing the experiment.
Relationship Between \(\Delta G^\circ\) and \(K\)
- \(\Delta G^\circ < 0\): \(\ln K > 0 \Rightarrow K > 1\). Products predominate at equilibrium.
- \(\Delta G^\circ = 0\): \(\ln K = 0 \Rightarrow K = 1\). Products and reactants are in equal thermodynamic balance.
- \(\Delta G^\circ > 0\): \(\ln K < 0 \Rightarrow K < 1\). Reactants predominate at equilibrium.
Worked Example: Calculating K from ΔG°
Calculating Equilibrium Constant \(K\) from \(\Delta G^\circ\)
For the reaction: \(\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})\)
Given: \(\Delta G^\circ = +4.7\text{ kJ mol}^{-1}\) at \(298\text{ K}\).
Calculate the equilibrium constant \(K\).
Step 1: Convert \(\Delta G^\circ\) to \(\text{J mol}^{-1}\)
\(\Delta G^\circ = +4700\text{ J mol}^{-1}\) (matching gas constant \(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\))
Step 2: Rearrange for \(\ln K\)
\(\ln K = \dfrac{-\Delta G^\circ}{RT} = \dfrac{-4700}{8.314 \times 298} = \dfrac{-4700}{2477.6} = -1.897\)
Step 3: Calculate \(K\)
\(K = e^{-1.897} = \mathbf{0.150}\)
Interpretation: Since \(\Delta G^\circ > 0\), \(K < 1\) (\(0.150\)), confirming that at 298 K reactants (\(\text{N}_2\text{O}_4\)) predominate at equilibrium.
Gas Constant R Unit Trap (\(\Delta G^\circ = -RT\ln K\))
Gas Constant Unit Trap: When calculating \(K\) using \(\Delta G^\circ = -RT\ln K\), ensure \(\Delta G^\circ\) units match \(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\). You must convert \(\Delta G^\circ\) from \(\text{kJ mol}^{-1}\) to \(\text{J mol}^{-1}\) (multiply by 1000).
Physical Meaning of Extreme \(\Delta G^\circ\) and \(K\) Values
If a reaction has a very large negative \(\Delta G^\circ\) (e.g. \(-200\text{ kJ mol}^{-1}\)), \(\ln K\) becomes a large positive value, yielding an astronomical \(K\) (e.g. \(10^{35}\)). The reaction essentially proceeds to 100% complete conversion with no detectable reactants at dynamic equilibrium.
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