First order logic cannot express transitive closure: http://math.stackexchange.com/questions/1286141/transitive-closure-and-first-order-logic

The proof works by assuming that there is some formula tc(R,R+) which expresses that R+ is the transitive closure of R. By our assumptions on the nature of transitive closure, we know that an axiom “a R+ b” (under the assumption of tc(R,R+)) must be equivalent to an infinitely large (first-order) axiom:
{a R b} ∨ {∃ x1 : a R x_1 R b} ∨ … ∨ {∃ x_1, x_2, … x_n : a R x_1 R x_2 R … R x_n R b} ∨ …, ∀ n ∈ ℕ.
However, we can form the following (countably infinite) set of first-order axioms:
tc(R,R+)
a R+ b
{¬∃ x_1, x_2, … x_n : a R x_1 R x_2 R …. R x_n R b}, ∀ n ∈ ℕ.

which is equivalent to:
{a R b} ∨ {∃ x_1 : a R x_1 R b} ∨ … ∨ {∃ x_1, x_2, … x_n : a R x_1 R x_2 R … R x_n R b} ∨ …, ∀ n ∈ ℕ.
{¬∃ x_1, x_2, … x_n : a R x_1 R x_2 R … R x_n R b}, ∀ n ∈ ℕ.

But this is of course inconsistent.
If an infinite set of first-order axioms is consistent, then every finite subset must also be consistent. Since the infinite set of axioms is inconsistent, then at least one finite subset must also be inconsistent. However, we can see that any finite subset of these axioms *is* consistent. Any finite subset leaves infinitely many ways to satisfy one of the disjuncts of a R+ b without contradicting with any of the other axioms in the set. Contradiction! We must conclude that tc(R,R+) cannot be expressed as a first-order axiom (the fact that it had to be expressed by an infinitely large axiom was a good clue).
Stronger logics, on the other hand, can express transitive closure.