Gibbs Energy Change and Equilibrium

GIBBS FREE ENERGY

This is another thermodynamic quantity that helps in predicting the spontaneity of a process, is called Gibbs energy (G).

It is defined mathematically by the equation.

G = H - TS

Where H = heat content, S = entropy of the system, T = absolute temperature

Illustration 1. Which of the following will fit into the blank?

When two phases of the same single substance remain in equilibrium with one another at a constant P and T, their molar _________ must be equal.

(A) Internal energy (B) Enthalpy

(C) Entropy (D) Free energy

Solution: (D)

Free energy change

For isothermal process.

$\Delta G=G_2-G_1=\Delta H-T\Delta S$

$\Delta$G = change in Gibbs free energy of the system.

It is that thermodynamic quantity of a system the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.

Illustration 2. Calculate free energy change when one mole of NaCl is dissolved in water at 25°C. Lattice energy = 700 kJ/mol. $\Delta S$at 25°C = 26.5$Jmo{l}^{-1}$, Hydration energy of NaCl = -696 kJ/mol.

(A) -3.9 kJ (B) -8kJ

(C) -12kJ (D) -16kJ

Solution: (A)

RELATIONSHIP BETWEEN FREE ENERGY AND EQUILIBRIUM CONSTANT

The free energy change of the reaction in any state, $\Delta$G (when equilibrium has not been attained) is related to the standard free energy change of the reaction, $\Delta$G⁰ (which is equal to the difference in free energies of formation of the products and reactants both in their standard states) according to the equation.

$\Delta G=\Delta G^0+RT\ln Q$

Where Q is the reaction quotient

When equilibrium is attained, there is no further free energy change i.e. $\Delta$G = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes.

$\Delta G^0=-RT\ln K_{\left(eq.\right)}$

or $\Delta G^0=-2.303RT\log K_{\left(eq.\right)}$

In case of galvanic cells. Gibbs energy change $\Delta$G is related to the electrical work done by the cell.

$\Delta$G = -nFE_(cell) where n = no. of moles of electrons involved

F = the Faraday constant

E = emf of the cell

If reactants and products are in their standard states $\Delta G^0=-nF{E}_{cell}^0$

Illustration 3. Calculate $\Delta$G⁰ for conversion of oxygen to ozone $\frac{3}{2}{O}_2\left(g\right)\to O_3\left(g\right)$at 298 K, if K_p for this conversion is 2.47 x 10^-29.

Solution: $\Delta G^0=-2.303RT\log K_p$

Where R = 8.314 J/K mol, K_p = 2.47 x 10^-29, T = 298K

$\Delta$G⁰ = 16300 J/mol = 163 KJ/mol

THIRD LAW OF THERMODYNAMICS

The entropy of a pure crystalline substance increases with increase of temperature, because molecular motion increases with increase of temperature and vice - versa.

Or the entropy of a perfectly crystalline solid approaches zero as the absolute zero of temperature is approached. This is third law of thermodynamics.