Answer
-294 kJ/mol
Work Step by Step
The reactions with known $\Delta H$'s:
$S+\frac{3}{2}O_{2}\rightarrow SO_{3},\,\Delta H=-204\,kJ/mol\qquad (1)$
$SO_{2}+\frac{1}{2}O_{2}\rightarrow SO_{3},\,\Delta H=89.5\,kJ/mol\qquad(2)$
The reaction of interest is:
$S+O_{2}\rightarrow SO_{2}$
From reaction (2), we can write
$SO_{3}\rightarrow SO_{2}+\frac{1}{2}O_{2},\,\Delta H=-89.5\,kJ/mol\qquad(3)$
Adding reaction (1) and reaction (3), we get the reaction of interest:
$S+\frac{3}{2}O_{2}+SO_{3}\rightarrow SO_{3}+SO_{2}+\frac{1}{2}O_{2}$
That is, $S+O_{2}\rightarrow SO_{2}$
According to Hess's law, $\Delta H$ for this reaction is just the sum of $\Delta H$'s for reaction (1) and reaction (3). That is
$\Delta H= -204\,kJ/mol-89.5\,kJ/mol=-294\,kJ/mol$
