Answer
$\Delta S^{\circ}_{rxn}=133.9\,J/K$
The positive sign of $\Delta S^{\circ}_{rxn}$ is because of the increase in entropy due to the increase in the number of moles of gas.
Work Step by Step
$\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[1\,mol\times S^{\circ}(CO,g)+1\,mol\times S^{\circ}(H_{2},g)]-[1\,mol\times S^{\circ}(C,s)+1\,mol\times S^{\circ}(H_{2}O,g)]$
$=[1\,mol(197.7\,Jmol^{-1}K^{-1})+1\,mol(130.7\,Jmol^{-1}K^{-1})]-[1\,mol(5.7\,Jmol^{-1}K^{-1})+1\,mol(188.8\,Jmol^{-1}K^{-1})]$
$=133.9\,J/K$
The positive sign of $\Delta S^{\circ}_{rxn}$ is because of the increase in entropy due to the increase in the number of moles of gas.
