# Confusion between various formulas of Gibbs free energy [closed]

According to the assumptions of Constant P and T and no non mechanical work, we get

ΔG=ΔH-TΔS

But there is another form of the Gibbs free energy equation which takes Reversible process and no non mechanical work as assumption

dG=Vdp-SdT

If I Use the assumptions constant Pressure and Temperature, according to the above equation we get dG=0,which contradicts the first equation.

What is the mistake I am making? Also, when ΔGsystem=-TΔStotal

• For T const, p const, zero nonmechanical work and for reversible system, Delta H = Q = T Delta S, so delta G = 0. Commented Feb 7, 2021 at 17:31
• You omitted the effect on G of changes in the number of moles of the reacting chemical species. Commented Feb 7, 2021 at 19:10
• @Chet Miller, Can u elaborate a bit Commented Feb 8, 2021 at 1:50
• See my answer at chemistry.stackexchange.com/questions/138563/… for the restrictions on $dG=Vdp-SdT$. Commented Feb 8, 2021 at 5:34
• The differential you took as formula for Gibbs free energy is merely that for a compression/expansion, there is no reaction involved. A compression at constant T and P is a non-process. Commented Feb 8, 2021 at 8:48

The Gibbs free energy of a multicomponent mixture is a function not only of temperature and pressure, but also of the number of moles of the various species in the mixture: $$G=G(T, P, n_1, n_2. m_3. ...)$$So, $$dG=\frac{\partial G}{\partial T}dT+\frac{\partial G}{\partial P}dP+\frac{\partial G}{\partial n_1}dn_1+\frac{\partial G}{\partial n_2}dn_2+\frac{\partial G}{\partial n_3}dn_3+...$$or $$dG=-SdT+VdP+\mu_1dn_1+\mu_2dn_2+\mu_3dn_3+...$$ where the $$\mu's$$ are the partial derivatives of the Gibbs free energy with respect to the number of moles of each of the species, and are referred to as the so-called Chemical Potentials of the various species in the solution. The numbers of moles of the species in the mixture change as the reaction proceeds, and that causes the Gibbs free energy to change.