Gibbs Free Energy, Equilibrium, and Reaction Quotient in Chemical Thermodynamics

Gibbs Free Energy and Equilibrium

Relationship Between Gibbs Free Energy and Equilibrium Constant

The Gibbs free energy is a thermodynamic quantity that determines the spontaneity of a chemical reaction. Its relationship with the equilibrium constant (Keq) is fundamental in chemical thermodynamics and equilibrium studies.

• Gibbs Free Energy Change (ΔG°): Indicates whether a reaction is spontaneous under standard conditions.

• Equilibrium Constant (Keq): Describes the ratio of product to reactant concentrations at equilibrium.

• The relationship is given by the formula:

$\Delta G^\circ = -RT \ln K_{eq}$

• R: Universal gas constant (8.314 J/mol·K)

• T: Temperature in Kelvin

• Keq: Equilibrium constant

Example

A certain reaction takes place at 25°C and has an equilibrium constant of 2.1 × 104. Determine the Gibbs free energy of the reaction.

• Calculation steps:

• Convert temperature to Kelvin: 25°C + 273.15 = 298.15 K

• Plug values into the formula:

$\Delta G^\circ = - (8.314\, \text{J/mol·K}) \times 298.15\, \text{K} \times \ln(2.1 \times 10^4)$

Result: -24.3 kJ/mol

Practice Problem

For the reaction Ag2CO3 (s) ⇌ 2Ag+ (aq) + CO32− (aq), ΔH° = 79.14 kJ/mol, ΔS° = 102.2 J/mol·K. Determine the equilibrium constant at the temperature 298.15 K.

• First, calculate ΔG° using:

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

• Then, use:

$K_{eq} = e^{-\Delta G^\circ / RT}$

Result: 7.62

Gibbs Free Energy and Reaction Quotient (Q)

Relationship Between Gibbs Free Energy and Reaction Quotient

When a reaction is not at equilibrium, the reaction quotient (Q) is used instead of Keq. The Gibbs free energy change under non-standard conditions is given by:

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

• Q: Reaction quotient, calculated from current concentrations or partial pressures.

• ΔG: Gibbs free energy change under non-standard conditions.

Example

The given reaction has a ΔG° of -374 kJ, and partial pressures of SF6, SF4, S2F2 are 0.43 atm, 0.56 atm, 1.7 atm respectively. Calculate the ΔG for the reaction:

SF6 (g) ⇌ SF4 (g) + S2F2 (g)

• Calculate Q using partial pressures:

$Q = \frac{[SF_4][S_2F_2]}{[SF_6]}$

• Plug values into the formula for ΔG:

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

Result: -370.8 kJ

Practice Problem

Consider a hypothetical reaction at 35°C: X (aq) + 2 Y (aq) ⇌ Z (aq), with ΔG° = -3.2 kJ, and ΔG = -8.4 kJ. Concentrations of reactants and products: [X] = 1.1 M, [Y] = 0.34 M, [Z] = 0.54 M. Calculate Keq of the given reaction.

• First, calculate Q:

$Q = \frac{[Z]}{[X][Y]^2}$

• Then, use the relationship between ΔG, ΔG°, and Q to solve for Keq:

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

Result: 87.4

Summary Table: Gibbs Free Energy Relationships

Condition	Equation	Variables	Application
Standard Conditions (Equilibrium)	$\Delta G^\circ = -RT \ln K_{eq}$	ΔG°, R, T, Keq	Calculating ΔG° from Keq or vice versa
Non-Standard Conditions	$\Delta G = \Delta G^\circ + RT \ln Q$	ΔG, ΔG°, R, T, Q	Calculating ΔG when reaction is not at equilibrium

Additional info: These relationships are essential for understanding chemical equilibrium, spontaneity, and the effect of concentration or pressure changes on reaction direction.