Is the reaction thermodynamically favored or not?
Your reasoning is correct but there is more to it.

Look at it from the point of view of Gibbs Free Energy,

$\Delta G = \Delta H - T \Delta S$

For a given reaction if:

$\Delta G &gt;0$, the reaction is nonspontaneous in the forward direction, not thermodynamically favourable

$\Delta G &lt; 0 $, the reaction is spontaneous in the forward direction, thermodynamically favourable

If we substitute the values for $\Delta H$ and $\Delta S$:

$\Delta G=(-128\ \mathrm{kJ/mol}) - T\left(-0.409\ \mathrm{kJ/(mol \cdot K)}\right)$

Mathematically, $\Delta G$ will become positive only when $T$ is greater than 313K. Because of this,  the reaction is thermodynamically favorable at any temperature less than 313K (since $\Delta G$ will be negative), but is not thermodynamically favorable at any temperature greater than 313K (since $\Delta G$ will be positive).