Is the reaction thermodynamically favored or not? $\ce{CO(g) + 2H2(g) +O2(g) &lt;=&gt; 2CH3OH(g)}$, 

$\Delta H=-128\ \mathrm{kJ/mol}$
$\Delta S=-409.2\ \mathrm{J/(mol \cdot K)}$

Determine if the reaction if thermodynamically favored.

I just wanted someone to see if my logic is right. Since there are 4 moles on the reactants side and and 2 on the products side the reaction would not be thermodynamically favored because the reaction decreases the amount of entities, which decreases the entropy.
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).