The standard free energy change `(DeltaG^0)` is related to equilibrium constant (K ) as ____

Explanation

(i)In a reversible process, system is at all times in perfect equilibrium with its surroundings .
(ii) A reversible chemical reaction can proceed in either direction simultaneously, dynamic equilibrium is set up.
(iii)This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible.
(iv)It is possible only if at equilibrium, the free energy of a system minimum
(v) Lets consider a general equilibrium reaction,
A+B `hArr` C+D
The free energy change of the above reaction in any state (`DeltaG`) is related to the standard free energy change of the reaction (`DeltaG^0`) according to the following equation.
`DeltaG=DeltaG^0+RT ln Q`....(1)
where Q is reaction quotient and is defined as the ratio of concentration of the products to the concentration of the reactants under non-equilibrium condition.
(vi)When equilibrium is attained, there is no further free energy change i.e. `DeltaG=0` and Q becomes equal to equilibrium constant. Hence the above equation becomes ,
`DeltaG^0=-RT ln K_(eq)` ...(2 )
This equation is known as Van.t Hoff equation.
`DeltaG^0`=-2.303 RT log `K_(eq)`...(3)
We also know that ,
`DeltaG^0=DeltaH^0 -TDeltaS^0=-RT ln K_(eq)` ....(4)