Answer
$312.4\,J mol^{-1}K^{-1}$
Work Step by Step
The decomposition reaction is
$N_{2}O_{3}(g)\rightarrow NO(g)+NO_{2}(g)$
$\Delta S^{\circ}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[1\,mol\times S^{\circ}(NO,g)+1\,mol\times S^{\circ}(NO_{2},g)]-[1\,mol\times S^{\circ}(N_{2}O_{3},g)]$
$=(210.8\,J/K+240.1\,J/K)-(1\,mol\times S^{\circ}(N_{2}O_{3},g)=138.5\,J/K$
$\implies S^{\circ}(N_{2}O_{3},g)=\frac{(210.8+240.1-138.5)\,J/K}{1\,mol}$
$=312.4\,J mol^{-1}K^{-1}$
