Let $\mathcal Q$ be some qualification on formulas in the first order language of set theory (FOL($\in$)), *that is met by at least one formula*; Let $T$ be the first order set theory whose extra-logical axioms are the following *sole* axiom schema:

$\mathcal Q$-Comprehension schema: if $\phi(y)$ is a formula that meets qualification $\mathcal Q$, in which $x$ doesn't occur, and in which the symbol $y$ occurs free, and only free; then all closures of: $$\exists x \forall y (y \in x \leftrightarrow \phi(y))$$, are axioms.

Now suppose that $T$ is consistent, and that $\psi(x)$ is some formula in one free variable $x$, and $x$ only occur free in it, and suppose that theory $T$ proves that per the same conditions written above for $\mathcal Q$-Comprehension, all closures of following: $$\forall x [\forall y (y \in x \leftrightarrow \phi(y)) \to \psi(x)]$$, are theorems.

Would it always follow that: $T + \forall x (\psi(x))$ is consistent?

The other question is:

If not, then: are there known conditions that if qualifcation $\mathcal Q$ meets then $T + \forall x (\psi(x))$ would be consistent?