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# Relationship between standard free energy change (╬öG0) and equilibrium constant (Keq)

In a reversible process, the system is in perfect equilibrium with its surroundings at all times.

Relationship between standard free energy change (╬öG0) and equilibrium constant (Keq):

In a reversible process, the system is in perfect equilibrium with its surroundings at all times. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible. It is possible only if at equilibrium, the free energy of a system is minimum. Lets consider a general equilibrium reaction

A+B Ōćī  C + D

The free energy change of the above reaction in any state (╬öG) is related to the standard free energy change of the reaction (╬öG0 ) according to the following equation.

╬öG= ╬öG0 + RT ln Q ŌłÆŌłÆŌłÆŌłÆŌłÆŌłÆ(7.39)

where Q is reaction quotient and is defined as the ratio of concentration of the products to the concentrations of the reactants under non equilibrium condition.

When equilibrium is attained, there is no further free energy change i.e. ╬öG 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes.

╬öG0 = ŌĆōRT ln Keq

This equation is known as VanŌĆÖt Hoff equation.

╬öG0 = ŌĆō2.303 RT log Keq ----------(7.40)

We also know that

╬öG0 = ╬öH0 ŌĆō T ╬öS0 = ŌłÆ RT ln Keq

### Problem: 7.9

Calculate ╬öG0  for conversion of oxygen to ozone 3/2 OŌåö O3(g) at 298 K, if Kp for this conversion is 2.47 u 10ŌłÆ29 in standard pressure units.

### Solution:

╬öG0 = ŌłÆ 2.303 RT log Kp

Where

R = 8.314 JKŌłÆ1molŌłÆ1

Kp = 2.47 x10ŌłÆ29

T = 298K

╬öG0=ŌłÆ2.303(8.314)(298)log(2.47u10ŌłÆ29)

╬öG0 = 16300 JmolŌłÆ1

╬öG0 = 16.3 KJ molŌłÆ1

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11th Chemistry : UNIT 7 : Thermodynamics : Relationship between standard free energy change (╬öG0) and equilibrium constant (Keq) |