I want to understand the derivation between gibbs energy and equillibrium constant $$\Delta G=\Delta G^o+RT\ln Q?$$ I have seen a similar post on CSE Derivation of relationship between equilibrium constant and Gibbs free energy change which seems to be incomplete and still confusing so I am again asking this question.
The derivation that was written in the post was as follows:
Using the fundamental equations for the state function (and its natural variables):_ $$dG=-SdT+VdP$$ $$V=(\dfrac {\partial G}{\partial P})_T$$ $$\bar G(T,P_2)=\bar G(T,P_1)+\int_{P_1}^{P_2}\bar V dp$$ Here $\bar x$ represents molar $x$, i.e. $x$ per mole $$\bar V=\frac {RT}{P}$$ $$\bar G(T,P_2)=\bar G(T,P_1)+RT \ln\frac {P_2}{P_1}$$ Defining standard state as $P=1\text{bar}$ and $\bar G=\mu$ $$\mu(T,P)=\mu^o (T)+RT\ln \frac{P}{P_o}$$ consider the general gaseous reaction $aA+bB\rightarrow cC+dD$ $$\Delta G=(c\mu_C+d\mu_D-a\mu_A-b\mu_B)$$ for "unit progress" in reaction. Using $\mu_i=\mu^o_i+RT\ln \frac{P_i}{1bar}$ $$\Delta G=(c\mu^o_C+d\mu^o_D-a\mu^o_A-b\mu^o_B)+RT\ln \frac{P_C^cP_D^d}{P_A^aP_B^b}$$ $$\Delta G=\Delta G^o+RT\ln Q$$
- I know the relation between Gibbs free energy,Enthalpy,Entropy and Temperature as $$G = H - TS$$ how is the relation $dG=-SdT+VdP$ derived from the above formula?
- Why do we take a path where Entropy and volume are constant in the equation $dG=-SdT+VdP$ as in a chemical reaction both volume and entropy can change?