Derive relation between equilibrium constant K, Reaction quotient Q and Gibbs energy change (DeltaG).

Derive relation between equilibrium constant K, Reaction quotient Q an

1.	Derive relation between equilibrium constant K, Reaction quotient Q and Gibbs energy change (DeltaG).
Answer» Solution :According to equilibrium the mathematical THERMODYNAMIC equation is as follow. `DELTAG=DeltaG^(ө)+RT` ln Q ….(Eq.-i) where , `DeltaG^(ө)` = Standard Gibbs force energy change `DeltaG`=Non-standard gibbs energy change R=Gas constant = 8.314 J `"MOL"^(-1) K^(-1)` T = Temperature in Kelvin at equilibrium Q = Reaction quotient Reaction at equilibrium, `DeltaG = 0` and `Q = K_c`. According to equation 7.30 as below, `therefore 0=DeltaG^(ө)+RT` ln K `therefore DeltaG^(ө)`=-RT ln K `therefore` ln K= `-(DeltaG^ө)/(RT)`...(Eq.-ii) In equation taking antilog of both SIDES `K=e^(-DeltaG^(ө//RT))`...(Eq-iii) Reaction of spontaneity and Value of `DeltaG^(ө)` : If `DeltaG lt0`, then `-DeltaG^(ө)//RT` is positive and `e^(-DeltaG^(ө)//RT) gt 1`, making K `gt`1. Which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present PREDOMINANTLY. If `DeltaG^(ө) gt 0` , then `DeltaG^(ө)//RT` is negative and `e^(-DeltaG^(ө)//RT) lt 1`, that is `K lt 1`. Which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.