Oxidation of sulfur dioxide to trioxide and its decomposition under same condition at higher temperature
There is an equilibrium between $\ce{SO3}$ and $\ce{SO2 + 1/2 O2}$, at least at high temperatures and with an appropriate catalyst.  We could write the equilibrium as
$$\ce{SO2(g) + 1/2 O2(g) &lt;=&gt; SO3(g)}$$
Just from writing this reaction this way we can glean several things:

The reaction combines 1.5 gas molecules into one gas molecule, a net reduction in the number of molecules.  Since fewer molecules (but the same number of atoms) means there are fewer translational degrees of freedom, the entropy will decrease the more $\ce{SO3}$ &quot;product&quot; is formed.  Higher temperatures will thus disfavor the right-hand $\ce{SO3}$ side of the equilibrium and favor the left-hand $\ce{SO2 + 1/2 O2}$ side.

Total pressure will also affect the equilibrium but in the opposite way.  Higher pressure (crudely speaking) is a force that squeezes things together, so at high pressures the right-hand side will be favored.

The other principle factor that influences the equilibrium would be the partial pressure of $\ce{SO2}$, of $\ce{O2}$, and of $\ce{SO3}$.  So for example  if you had 99.9 mol % oxygen gas and only a trace of $\ce{SO2}$, that would favor the conversion of nearly all sulfur to the LH sulfur trioxide side.

The exact balance will depend on the precise enthalpy change of the reaction, the partial pressure of each species, the temperature, and the total pressure.  You could modify the system to favor one side or the other, and could model this accurately and quantitatively with appropriate numerical data.