Using the Standard Entropy Values:
NO(g), S° = +210 $\rarr$ NO₂(g)
A)-72

Using the standard entropy values:
NO(g) , S° = +210.6 J mol^-1 K^-1
O₂(g) , S° = +205.0 J mol^-1 K^-1
NO₂(g) , S° = +240.5 J mol^-1 K^-1
Calculate the standard entropy change, S°, for the reaction:NO(g) + ½ O₂(g) $\rarr$ NO₂(g)

A) -72.6 J mol^-1 K^-1
B) -175.1 J mol^-1 K^-1
C) +72.6 J mol^-1 K^-1
D) +175.1 J mol^-1 K^-1
E) -656.1 J mol^-1 K^-1