Answer
$-390.8\,J/K$
The decrease in the number of moles of gas results in the decrease of entropy and therefore $\Delta S^{\circ}_{rxn}$ is negative.
Work Step by Step
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[2\,mol\times S^{\circ}(H_{2}O,l)+2\,mol\times S^{\circ}(SO_{2},g)]-[2\,mol\times S^{\circ}(H_{2}S,g)+3\,mol\times S^{\circ}(O_{2},g)]$
$=[2\,mol(70.0\,Jmol^{-1}K^{-1})+2\,mol(248.2\,Jmol^{-1}K^{-1})]-[2\,mol(205.8\,Jmol^{-1}K^{-1})+3\,mol(205.2\,Jmol^{-1}K^{-1})]$
$=-390.8\,J/K$
As expected because of the decrease in the number of moles of gas, $\Delta S^{\circ}_{rxn}$ is negative.
