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I have studied thermodynamics, in which the infamous Gibbs equation came.
$$ \mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S $$
Then my book said the criteria of spontaneity based on gibbs energy is at constant pressure and temperature.
Since $T\,\mathrm{d}S = q = \mathrm{d}H$ (since at constant pressure $\mathrm{d}H = q$), shouldn't $\mathrm{d}G$ all always be equal to $0$?
That is not the Gibbs Free Energy at constant pressure and temperature.
Alphabetic soup definitions:
Definition of $G$ is $$G = H - TS$$
Definition of $H$ is $$H = U + PV$$
The total differential of $G$ is
$$dG = dU + PdV + VdP - TdS - SdT$$
The definition for the change in $U$ is $$dU = dq + dw$$
There are two kinds of work: pressure-volume work (pv) and non-pressure-volume work (non-pv)
$$dw = dw_{\mathrm{pv}} + dw_\mathrm{non-pv}$$
Thus, substituting we get
$$dG = dq - PdV + dw_\mathrm{non-pv} + PdV + VdP - TdS - SdT$$
If the process is (approximately) reversible: $$T =\dfrac{dq}{dS}$$
Substituting $q = TdS$ and simplifying, we get:
$$dG = VdP - SdT + dw_\mathrm{non-pv}$$
At constant pressure, $dP=0$, and at constant temperature $dT = 0$.
Therefore: $$dG = dw_\mathrm{non-pv}$$
Hence, Gibbs free energy is sort of a measure of useful work we can get out of a system.
$$w_\mathrm{non-pv\ by\ system} = - w_\mathrm{non-pv\ on\ system}$$
Although the answer @ELiT gave is the perfect answer, I would like to answer my own question. What a silly mistake I made!
$\mathrm{d}S = \frac{q}{T}$ only for reversible processes. In reversible processes, what I said is true, as $\mathrm{d}S_\text{total} = 0$, thus $\mathrm{d}G = 0$. But $\mathrm{d}S \neq \frac{q}{T}$ for non-reversible processes. Hence I cannot substitute that value here.
