Answer
$\Delta S^{\circ}_{rxn}=-287.5\,J/K$
Because of the decrease in the number of moles of gas, the entropy decreases, and $\Delta S^{\circ}_{rxn}$ is negative.
Work Step by Step
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[2\,mol\times S^{\circ}(HNO_{3},aq)+1\,mol\times S^{\circ}(NO,g)]-[3\,mol\times S^{\circ}(NO_{2},g)+1\,mol\times S^{\circ}(H_{2}O,l)]$
$=[2\,mol(146\,Jmol^{-1}K^{-1})+1\,mol(210.8\,Jmol^{-1}K^{-1})]-[3\,mol(240.1\,Jmol^{-1}K^{-1})+1\,mol(70.0\,Jmol^{-1}K^{-1})]$
$=-287.5\,J/K$
Because of the decrease in the number of moles of gas, the entropy decreases, and $\Delta S^{\circ}_{rxn}$ is negative.
