The change in entropy during any reversible process between 2 states, state 1 and state 2, is given as:
$${\Delta S}_{rev} = \int_1^2 \delta Q / T $$
and if the process between state 1 and 2 is irreversible then the change in entropy is given as:
$${\Delta S}_{irrev} = \int_1^2 \delta Q / T + S_{gen} $$
How can entropy, S, be a state function if it has different values at states 1 and 2 depending on if the process is reversible or irreversible i.e: for the entropy to be a state function shouldn't the change in entropy between the 2 states be the same no matter if the process is reversible or irreversible?
Note:
For reversible process $S_2$ would be:
$$ S_2 = S_1 + \Delta S_{rev} $$
& for irreversible process $S_2$ would be:
$$ S_2 = S_1 + \Delta S_{irrev} $$
So then $S_2$ for the reversible process will be different than $S_2$ for the irreversible process, which shouldn't be the case if entropy is a state function and a property of the system.
Best Answer
I think the way you write the two conditions may be misleading. $$ \begin{align} {\Delta S}_{rev} &= \int_1^2 \delta Q / T \\ {\Delta S}_{irrev} &= \int_1^2 \delta Q / T + S_{gen}. \end{align} $$ The starting point to clarify the issue is the Clausius theorem stating that $$ \oint \frac{\delta Q}{T}\leq 0 $$ in every thermodynamic cycle, where the equality holds for the particular case of a reversible cycle.
Such a property of reversible cycles allows introducing a state function, the entropy, through $$ \Delta S = S(2)-S(1)= \int_1^2 \delta Q_{rev} / T. $$ That means that there is only one difference of entropy, not dependent on the process (since it is a state function) whose definition hinges on the particular case of a reversible transformation. But, this is just the scaffold for defining (and assigning a value of) such a function to every thermodynamic state. Once you have the value of the function at each thermodynamic state, you remove the scaffold, but the function is there.
Therefore, there is nothing like a $\Delta S_{rev}$ different from $\Delta S_{irrev}$. The correct way of rewriting the two equations you wrote is the following: $$ \begin{align} {\Delta S} &= S(2)-S(1)= \int_1^2 \delta Q_{rev} / T = \int_1^2 \delta Q_{irr} / T + S_{gen}. \end{align} $$ where both terms in the right-hand side of the last equality depend on the specific, irreversible process and, as a consequence of Clausius' theorem $S_{gen}>0$.