Just for the record, this does not need CM. The following is true:
>Let $Z$ be an excellent scheme that admits a dualizing complex. Then $ω_Z$ is torsion-free and $S_2$ on $Z$. If in addition Z is irreducible, then $ω_Z$ is a reflexive $\mathscr O_Z$ -module.
If there is interest in this, I will add a proof.
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As requested, here is a proof:
(The following is Lemma 3.7.5 in [this paper][1].)
>**Lemma** Let $Z$ be an excellent scheme that admits a dualizing complex. Then $\omega_Z$ is torsion-free and $S_2$ on $Z$. If in addition $Z$ is normal, then $\omega_Z$ is a reflexive $\mathscr O_Z$-module.
>*Proof.* The statement is local, so we may assume that $Z$ is a noetherian affine local scheme.
Then, since it admits a dualizing complex, it can be embedded into a finite dimensional Gorenstein affine local scheme W as a closed subscheme by [\[Kaw02, Cor. 1.4\]][2].
Being Gorenstein and local, $W$ must be pure dimensional. Let $r = \mathrm{codim}(Z, W)$. Then
by [\[Mat80, (16.B) Theorem 31(i)\]][3] there exists a length $r$ regular sequence in the ideal of $Z$
in $W$ and let $W'$ be the common zero locus. Then $W'$ is also Gorenstein,
$Z\subseteq W$ is a closed subscheme and $\dim Z = \dim W$ .
> It follows that $\omega_ {W'}^\bullet \simeq
\omega_{W'}[m]$, where $\omega_{W'}$ is a line bundle on ${W'}$. By Grothendieck
duality $\omega^\bullet_ Z\simeq RHom_{W'}(\mathscr O_Z, \omega^\bullet_{W'})$, and since $\dim Z=\dim W'$,
$\omega_Z\simeq Hom_{W'}(\mathscr O_Z, \omega_{W'})$ hence it is indeed torsion-free on
$Z$.
> By [\[Stacks Project, Tag 0AWE\]][4]
$\omega_Z$ is $S_2$. In particular, if $Z$ is normal, then it is reflexive by [\[Stacks Project,Tag 0AVB\]][5].
[1]: https://arxiv.org/abs/1703.02269
[2]: http://www.ams.org/journals/tran/2002-354-01/S0002-9947-01-02817-3/S0002-9947-01-02817-3.pdf
[3]: http://www.cambridge.org/us/academic/subjects/mathematics/algebra/commutative-ring-theory?format=PB#S3kQjWwLamaYB9ch.97
[4]: http://stacks.math.columbia.edu/tag/0AWE
[5]: http://stacks.math.columbia.edu/tag/0AVB