The most favourable conditions for the formation of ${\mathrm{SO}}_3$ via the equation

${\mathrm{SO}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{SO}}_3$, $\mathrm{\Delta H}=-95 \mathrm{kJ} {\mathrm{mol}}^{-1}$ are

${\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{SO}}_3\left(\mathrm{g}\right)$

$\mathrm{\Delta }S=-ve$

A reaction that has a high concentration of reactants will move forward.
Le Chatelier's principle states that a reaction should benefit from high pressure, however in this case, an acceptable yield is attained at atmospheric pressure.
Low temperature is beneficial to the process because it is exothermic. High pressure encourages reaction since it results in a decrease in the number of moles.