Explanation:
It is known that standard entropies for [tex]N_{2}(g)[/tex] is 191.6 J/mol K, [tex]O_{2}(g)[/tex] = 205 J/mol K, and [tex]NO_{2}(g)[/tex] is 239.7 J/mol K at 298 K.
Therefore, we will calculate the value of [tex]\Delta S^{o}_{reaction}[/tex] from standard absolute entropies as follows.
[tex]\Delta S^{o}_{reaction} = \sum \Delta S^{o}_{products} - \sum \Delta S^{o}_{reactants}[/tex]
= 2 mole of [tex]NO_{2}(g)[/tex] - 1 mole of [tex]N_{2}(g)[/tex] + 2 mole of [tex]O_{2}(g)[/tex]
= [tex]2 \times 239.7 J/mol K - 1 \times 191.6 J/mol K + 2 \times 205 J/mol K[/tex]
= -122.2 J/K
The entropy change for 1.90 moles of [tex]N_{2}(g)[/tex] reacting is as follows.
[tex]\Delta S^{o}_{system}[/tex] = 1.90 moles of [tex]N_{2}(g) \times 122.2 J/K/ 1 \text{mol of} N_{2}(g)[/tex]
= 232.18 J/K
Thus, we can conclude that the entropy change for the given system is 232.18 J/K.
