Answer
$\Delta S^{\circ}= -278.4\,J/K$
$\Delta S^{\circ}$ is negative because of the decrease in the moles of gas.
Work Step by Step
The reaction can be written as
$N_{2}(g)+3F_{2}(g)\rightarrow 2NF_{3}(g)$
Recall: $\Delta S^{\circ}_{rxn}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
Then,
$\Delta S^{\circ}=[2\,mol\times S^{\circ}(NF_{3},g)]-[1\,mol\times S^{\circ}(N_{2},g)+3\,mol\times S^{\circ}(F_{2},g)]$
$=[2\,mol(260.8\,Jmol^{-1}K^{-1})]-[1\,mol(191.6\,Jmol^{-1}K^{-1})+3\,mol(202.79\,Jmol^{-1}K^{-1})]$
$=-278.4\,J/K$
The change in entropy is negative as the entropy decreases due to the decrease in the moles of gas.
