Answer
$\Delta S^{\circ}_{rxn}=119.6\,JK^{-1}$
Since the number of moles of gas stays constant, it is not easy to rationalize the sign of $\Delta S^{\circ}_{rxn}$. Still, we can predict that the change in entropy is small.
Work Step by Step
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[1\,mol\times S^{\circ}(N_{2},g)+ 4\,mol\times S^{\circ}(H_{2}O,g)]-[1\,mol\times S^{\circ}(N_{2}O_{4},g)+4\,mol\times S^{\circ}(H_{2},g)]$
$=[1\,mol (191.6\,JK^{-1}mol^{-1})+4\,mol(188.8\,Jmol^{-1}K^{-1})]-[1\,mol(304.4\,Jmol^{-1}K^{-1})+4\,mol(130.7\,Jmol^{-1}K^{-1})]$
$=119.6\,JK^{-1}$
Since the number of moles of gas stays constant, it is not easy to rationalize the sign of $\Delta S^{\circ}_{rxn}$. Still, we can predict that the change in entropy is small.
