Answer
The slope of a $ln(k)$ versus $\frac{1}{T}$ plot for a catalyzed reaction would be less negative.
Work Step by Step
The slope of a $ln(k)$ vs $\frac{1}{T}$ plot is equal to $\frac{-E_a}{R}$. A catalyzed reaction has a lower $E_a$ than an uncatalyzed one, and $R$ remains constant, so the slope should be less negative.
