Answer
$\Delta S^{\circ}=-89.3\,J/K$
The change in entropy is negative because of the decrease in the moles of gas.
Work Step by Step
The equation for the formation of $CH_{2}Cl_{2}$ is
$C(s)+H_{2}(g)+Cl_{2}(g)\rightarrow CH_{2}Cl_{2}(g)$
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[1\,mol\times S^{\circ}(CH_{2}Cl_{2},g)]-[1\,mol\times S^{\circ}(C,s)+1\,mol\times S^{\circ}(H_{2},g)+1\,mol\times S^{\circ}(Cl_{2},g)]$
$=[1\,mol(270.2\,Jmol^{-1}K^{-1})]-[1\,mol(5.7\,Jmol^{-1}K^{-1})+1\,mol(130.7\,Jmol^{-1}K^{-1})+1\,mol(223.1\,Jmol^{-1}K^{-1})]$
$=-89.3\,J/K$
The change in entropy is negative because of the decrease in the moles of gas.
