A language $L$ is semidecidable if there is a Turing machine that on input $x$, halts if $x \in L$, and doesn't halt if $x \notin L$. It is decidable if there is a Turing machine that on input $x$, halts at an accepting state if $x \in L$, and halts at a rejecting state if $x \notin L$. It's not hard to check that every decidable language is also semidecidable.
A logical system is semidecidable (decidable) if the language of all theorems in the system is semidecidable (decidable).
Being semidecidable is an upper bound on the complexity of a language. Being NP-hard is a lower bound on the complexity of a language. If all you know about a language is that it is semidecidable, you cannot deduce that it is NP-hard, just as you cannot conclude that $x \geq 0$ from the inequality $x \leq 1$.
The halting problem is $\Sigma_1^0$-complete, which means that it is semidecidable (an upper bound), and moreover, every semidecidable language can be reduced to it using a computable reduction (a lower bound). To be equivalent to the halting problem, you need both an upper bound on your language and a lower bound on your language. The language of theorems in a logical system is guaranteed (by definition) to be semidecidable, but it could also be decidable, in which case it is definitely not equivalent to the halting problem.
Finally, you don't explain what you mean by First Order Logic, but assuming that it can express propositional tautologies, then the corresponding language of theorems is definitely NP-hard, by reduction from SAT. In fact, the language of true quantified Boolean formulas, known as TQBF, is known to be PSPACE-complete.