Relationship: $K, Q$ and $\Delta G$
Linking Thermodynamics and Equilibrium | Class 11
1. Reaction Quotient ($Q$) vs Equilibrium Constant ($K$)
Reaction Quotient ($Q$): Has the same mathematical expression as $K_c$ or $K_p$, but the concentrations/pressures are taken at any point in time (not necessarily at equilibrium).
For $aA + bB \rightleftharpoons cC + dD$:
$$ Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} $$Predicting Direction of Reaction
| Relation | Meaning | Direction |
|---|---|---|
| $Q < K$ | Product ratio is too low | Forward (Reactants $\to$ Products) |
| $Q > K$ | Product ratio is too high | Backward (Products $\to$ Reactants) |
| $Q = K$ | System is stable | Equilibrium (No net change) |
2. Relation between Gibbs Energy and $Q$
The fundamental thermodynamic equation relating non-standard free energy change ($\Delta G$) to standard free energy change ($\Delta G^\circ$) and reaction quotient ($Q$) is:
$$ \Delta G = \Delta G^\circ + RT \ln Q $$
$$ \Delta G = \Delta G^\circ + 2.303 RT \log Q $$
Where:
- $\Delta G$ = Free energy change at any condition (Determines Spontaneity).
- $\Delta G^\circ$ = Standard Free energy change (Constant for a reaction at T).
- $R$ = Gas constant ($8.314 J \cdot K^{-1} \cdot mol^{-1}$).
3. At Equilibrium ($Q = K$)
At equilibrium, the system does no net work, so the change in Gibbs energy is zero.
Condition: $\Delta G = 0$ and $Q = K_{eq}$
Substituting this into the master equation:
$$ 0 = \Delta G^\circ + RT \ln K $$
$$ \Delta G^\circ = -RT \ln K $$
$$ \Delta G^\circ = -2.303 RT \log K $$
Rearranging for $K$: $K = e^{-\Delta G^\circ / RT}$
4. Interpreting $\Delta G^\circ$ and $K$
| Value of $\Delta G^\circ$ | Value of $K$ | Interpretation |
|---|---|---|
| Negative ($\Delta G^\circ < 0$) | $K > 1$ | Reaction is Product-Favored (Spontaneous in forward direction under standard conditions). |
| Positive ($\Delta G^\circ > 0$) | $K < 1$ | Reaction is Reactant-Favored (Non-spontaneous forward, spontaneous backward). |
| Zero ($\Delta G^\circ = 0$) | $K = 1$ | Reactants and Products are equally favored at standard state. |
Practice Quiz
Test your ability to link Thermodynamics and Equilibrium.
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