Answer
$\Delta S^{\circ}_{rxn}=-94.0\,J/K$
There is a decrease in the moles of gas and this decreases entropy. So, $\Delta S^{\circ}_{rxn}$ is negative.
Work Step by Step
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[1\,mol\times S^{\circ}(SO_{3},g)]-[1\,mol\times S^{\circ}(SO_{2},g)+\frac{1}{2}\,mol\times S^{\circ}(O_{2},g)]$
$=[1\,mol(256.8\,Jmol^{-1}K^{-1})]-[1\,mol(248.2\,Jmol^{-1}K^{-1})+\frac{1}{2}\,mol(205.2\,Jmol^{-1}K^{-1})]$
$=-94.0\,J/K$
There is a decrease in the moles of gas and this decreases entropy. So, $\Delta S^{\circ}_{rxn}$ is negative.
