What's the validity of the thermodynamic definition of entropy? Thermodynamic Entropy Variation is defined as 
$$\Delta S = \int_i^f \frac{dQ}{T},$$
where $i$ and $f$ are the initial and final states of the process. 
My question is: does this equation apply to quasi-static irreversible processes, or only to reversible processes?
Obviously, it does not apply to processes that go through non-equilibrium states, since Temperature (or any state variable) is not even well defined in these states. But I'm unsure of whether it applies to irreversible processes that are quasi-static (and therefore don't go through non-equilibrium states).
 A: 1) Irreversible processes are the ones, which by definition increase the entropy. And the increase, similarly to non-equilibrium cases, is added to the reversible $\mathrm{d} S$. Hence, for all irreversible processes: $\delta S > \dfrac{\delta Q}{T}$
2) For at least some non-equilibrium cases thermodynamic variables like $T$ actually may well be defined, but for example only locally, or separately for different chemical species etc. Consider a chemical reaction, which goes quasi-statically, is not in equilibrium, hence produces entropy additional entropy and hence is irreversible.
A: The equation is only valid for quasistatic processes, though entropy change is defined for irreversible processes (it's a state function)
I personally have been taught physics from a pure reveraible point of view--where all processes are implied to be reversible. I'm not sure, but I think that this view is carried on throughout physics. In chemistry, on the other hand, I was taught both reversible and irreversible processes, so the chem equation I know is:
$$\Delta S = \int_i^f \frac{\rm dq_{rev}}{T}$$
and sometimes
$$\Delta S = \int_i^f \frac{q_{rev}}{T}$$
The $q$ is lowercase because it's a path function(convention in chem), and it's not really a differential in one of the equations because $q$ and $\rm dq$ are used interchangeably.
Edit: Realized that I missed the 'ir' in quasistatic irreversible :/
No, the equation will not hold--as entropy is a state function, while q is a path function. You must carry out the process $i \to f$ in a reversible manner. Otherwise you could potentially get two values for entropy.
