Aside from the general trend of the boiling points of the hydrides, I noticed how the group 15 hydrides somehow "overtook" the group 17 ones in boiling point, which felt weird.
Notably, H-Br has an electronegativity difference of 0.7, which is enough to be considered polar, which to my understanding means HBr will exist with a permanent dipole. The As-H bond has an even smaller difference, so it can't experience these tiny permanent dipole attractions.
That probably means there's a big difference in London dispersion forces, but both molecules have 36 electrons, so they'd be similarly capable of forming instantaneous dipoles. Their molecule shape might affect their ability to induce dipoles, but I don't know where to start with that.
So if both AsH3 and HBr experience similar London forces, but HBr also experiences weak permanent dipole-dipole interactions, why does AsH3 ($\mathrm{-63^\circ C}$) have a slightly higher boiling point than HBr ($\mathrm{-68^\circ C}$)? (The same could be said for SbH3 and HI, which has a slightly more significant difference in boiling point.)