# Why is Gibbs Free Energy not equal to 0? Its definition itself makes it 0

Gibbs Free Energy is defined as

$$G = H - TS$$ and the change in $G,$

$$\Delta G = \Delta H - T\Delta S,$$ (isothermal and isobaric)

Now at constant pressure, $q = \Delta H$

and Entropy is defined as , $\Delta S = \dfrac{q}{T}$ which means that $T\Delta S = q.$

Thus shouldn't Gibbs Free Energy automatically be zero. Or is there something I'm missing, like a specific condition?

For a quasi-static, isobaric process, $$\Delta H = \Delta (U + pV) = \Delta U + p\Delta V = Q + W + p\Delta V = Q -p\Delta V + p\Delta V = Q,$$ and so the change in enthalpy during the process is exactly equal to the heating of the system. For a quasi-static, isothermal process, $$Q = T\Delta S.$$ Therefore, for a process that is both isothermal and isobaric, $$\Delta G = \Delta (H - TS) = \Delta H - T\Delta S = Q - Q = 0.$$ In other words, yes! - the Gibbs free energy of the whole system is constant during such a process, but note that this is a very special process. It is rare that a system can undergo both a process in which both the temperature and pressure can remain constant at the same time, but such a process$-$*if quasi-static*$-$is a process of constant Gibbs free energy.
So: as long as there is a system containing a single substance in two different phases, and the phases are in thermal, mechanical, and phase equilibrium, then any process occurring at constant pressure also occurs at constant temperature, and so $\Delta G = 0$.
The differential of the Gibbs free energy of $n$ species at temperature $T$ and pressure $P$ is $$\text{d}G = -S \text{d} T + V \text{d}P + \sum_{i=1}^n \mu_i \text{d}N_i.$$ At fixed temperature and pressure, the only change in the Gibbs free energy will come from chemical reactions which change the particle number for some species $N_i$. This is presumably why chemists are so interested in $G$, since any equilibrium process occurring in open air will be at fixed pressure and temperature.