Home | | Chemistry 11th std | Relationship between standard free energy change (ΔG0) and equilibrium constant (Keq)

Chapter: 11th Chemistry : Thermodynamics

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Relationship between standard free energy change (ΔG0) 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 (ΔG0) 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   C + D

The free energy change of the above reaction in any state (ΔG) is related to the standard free energy change of the reaction (ΔG0 ) according to the following equation.

ΔG= ΔG0 + RT ln Q −−−−−−(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. ΔG 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes.

 

ΔG0 = –RT ln Keq

 

This equation is known as Van’t Hoff equation.

 

ΔG0 = –2.303 RT log Keq ----------(7.40)

We also know that

 

ΔG0 = ΔH0 – T ΔS0 = − RT ln Keq

 

Problem: 7.9

 

Calculate ΔG0  for conversion of oxygen to ozone 3/2 O↔ O3(g) at 298 K, if Kp for this conversion is 2.47 u 10−29 in standard pressure units.

 

Solution:

 

ΔG0 = − 2.303 RT log Kp

Where

 

R = 8.314 JK−1mol−1

 

Kp = 2.47 x10−29

 

T = 298K

 

ΔG0=−2.303(8.314)(298)log(2.47u10−29)

 

ΔG0 = 16300 Jmol−1

 

ΔG0 = 16.3 KJ mol−1

 

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