Derivation of Gibbs free energy formula Okay, so I'm trying to derive an important formula which states that -
                             ∆G(r) = ∆G° + RTln( RQ )
where ∆G(r) is the instantaneous rate of change of Gibbs energy with respect to moles of reagent consumed (am I right on this?) I'm hight confused because I have 'almost' derived it but due to some conceptual difficulty I'm not able to get the exact formula!

NOTE: 
(1) Please ignore any non expansion work done on the system and tell my mistake as simply as possible because I have just started thermodynamics.

(2) The line above G represent molar Gibbs energy of substance. The small letters on the side of the symbol of reactent are the stochiometric coefficients. 

(3) Many times I see variations of this formula, instead of ∆G(r) they simply use ∆G, there is a difference between two. Is this formula valid if ∆G is used instead of ∆G(r) in left hand side of equation? Also don't see any 'sense' in using ∆G because if ∆G is there then we need the initial and final quantities of the substances in the reaction but in the reaction quotient we plug the instantaneous activities of reactents and products.



I'm getting the above equation but a stochiometric coefficient is also included in this ! Obviously it is wrong. Kindly tell me my mistake.
I think you are using the first approach that I described.  So, here is the analysis based on the first approach.  Suppose you have the following numbers of moles of A, B, C, and D in the reaction mixture at some time: $n_A$, $n_B$, $n_C$, and $n_D$. 

And the total pressure of the mixture (in bars) is held constant at P.  So the partial pressure of species A is $$p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$$And the partial molar free energy of species A is $$\bar{G}_A=G^0_A+RT\ln{\left(\frac{n_A}{n_A+n_B+n_C+n_D}P\right)}$$where $G^0_A$ is the free energy of pure species A at the reference pressure of 1 bar.  Similar equations can be written for the other species.  The total free energy of the mixture at any point in time during the reaction is:  $$G=n_A\bar{G}_A+n_B\bar{G}_B+n_C\bar{G}_C+n_D\bar{G}_D$$
Now, let the concentrations of the chemical species in the reaction mixture be expressed as follows:
$$n_A=n_{A0}-a\xi$$
$$n_B=n_{B0}-b\xi$$
$$n_C=n_{C0}+c\xi$$
$$n_A=n_{D0}+d\xi$$
where the subscript 0 refers to the initial state and $\xi$ is the conversion.  To find the equilibrium condition, you want to find the value of $ \xi $ that minimizes the free energy of the solution mixture.  To do that, you have a calculus problem to solve.