Gibbs Energy and Equilibrium

Gibbs Energy Change And Equilibrium

---

Gibbs Free Energy ($G$): As previously discussed, Gibbs free energy is a thermodynamic potential that combines enthalpy ($H$) and entropy ($S$) to determine the spontaneity of a process at constant temperature ($T$) and pressure ($P$). It is defined as $G = H - TS$. The change in Gibbs free energy ($\Delta G$) for a process at constant $T$ and $P$ is given by:

$$\Delta G = \Delta H - T\Delta S$$

Spontaneity Criterion:

• $\Delta G < 0$: Process is spontaneous (exergonic).
• $\Delta G > 0$: Process is non-spontaneous (endergonic); the reverse process is spontaneous.
• $\Delta G = 0$: The system is at equilibrium.

Gibbs Energy and Equilibrium Constant ($K$):

For a reversible reaction at equilibrium, the Gibbs free energy change is zero ($\Delta G = 0$). We can relate the standard Gibbs free energy change ($\Delta G^\circ$) to the equilibrium constant ($K$) of the reaction.

Derivation of the Relationship:

Consider a reversible reaction in the gaseous phase:

$$aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$$

The Gibbs free energy change ($\Delta G$) for this reaction under non-standard conditions (where the partial pressures are not standard pressures) can be related to the standard Gibbs free energy change ($\Delta G^\circ$) and the reaction quotient ($Q$):

$$\Delta G = \Delta G^\circ + RT \ln Q$$

Where:

• $R$ is the ideal gas constant.
• $T$ is the absolute temperature.
• $Q$ is the reaction quotient, defined as $Q = \frac{(P_C/P^\circ)^c (P_D/P^\circ)^d}{(P_A/P^\circ)^a (P_B/P^\circ)^b}$, where $P^\circ$ is the standard pressure (1 bar). For simplicity, often written as $Q = \frac{P_C^c P_D^d}{P_A^a P_B^b}$ if standard pressure is implicitly handled in $P_i$.

At Equilibrium:

When the reaction reaches equilibrium:

• $\Delta G = 0$ (no net driving force for reaction).
• The reaction quotient ($Q$) becomes equal to the equilibrium constant ($K$).

Substituting these into the equation $\Delta G = \Delta G^\circ + RT \ln Q$:

$$0 = \Delta G^\circ + RT \ln K$$

Rearranging this equation gives the fundamental relationship between standard Gibbs free energy change and the equilibrium constant:

$$\Delta G^\circ = -RT \ln K$$

This equation is incredibly powerful. It connects a thermodynamic quantity ($\Delta G^\circ$) that relates to spontaneity and energy changes to a kinetic/equilibrium quantity ($K$) that describes the extent of a reaction.

Interpretation of the Relationship:

• If $\Delta G^\circ < 0$ (Spontaneous under standard conditions):
• $-RT \ln K < 0$
• $\ln K > 0$
• $K > 1$

This means the reaction favors the formation of products at equilibrium, and the equilibrium lies to the right.

• If $\Delta G^\circ > 0$ (Non-spontaneous under standard conditions):
• $-RT \ln K > 0$
• $\ln K < 0$
• $K < 1$

This means the reaction favors reactants at equilibrium, and the equilibrium lies to the left. The reverse reaction is spontaneous.

• If $\Delta G^\circ = 0$ (At equilibrium under standard conditions):
• $-RT \ln K = 0$
• $\ln K = 0$
• $K = 1$

This means the concentrations of reactants and products are comparable at equilibrium, and the reaction is neither strongly product-favored nor reactant-favored under standard conditions.

Calculating $\Delta G^\circ$ from $\Delta H^\circ$ and $\Delta S^\circ$:

We can also calculate $\Delta G^\circ$ using standard enthalpies of formation ($\Delta H^\circ_f$) and standard absolute entropies ($S^\circ$):

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

This allows us to predict the equilibrium constant ($K$) for a reaction even without direct experimental equilibrium measurements, provided enthalpy and entropy data are available.

Example: Calculate the equilibrium constant ($K_p$) for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ at 298 K, given $\Delta H^\circ = -92.4 \text{ kJ/mol}$ and $\Delta S^\circ = -198.7 \text{ J/mol K}$.

Answer: