Let $C\_1$ and $C\_2$ be two rationally equivalent curves in $\mathbb{P}^3$. Is it true that the dimension of $H^0(\mathcal{N}\_{C\_1|\mathbb{P}^3})$ equal to that of $H^0(\mathcal{N}\_{C\_2|\mathbb{P}^3})$?



Rational equivalence and infinitesimal deformation of curves
No, even if you assume $C\_1$ and $C\_2$ smooth irreducible. A rational equivalence class of curves in $\Bbb{P}^3$ is determined by its degree. Take for $C\_1$ a complete intersection of 2 quadrics, you get 
$\dim H^0(\mathcal{N}\_{C\_1|\mathbb{P}^3})= 16$; and for $C\_2$ a plane quartic curve, you find $\dim H^0(\mathcal{N}\_{C\_2|\mathbb{P}^3})= 17$.


