If S° for $N_{2} {, O}_{2} {, N}_{2} {O, NO,NO}_{2} {and N}_{2} O_{4} are$ are 45.7, 49.0, 48.2, 50.24, 57.24 and 72.77 cal
respectively and $N_{2} + \frac{1}{2} O_{2} ⇌ N_{2} O; ΔH° = 19.49 kcal at 300 K; \frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇌ NO; ΔH° = 21.60 kcal at 300K$
$\frac{1}{2} N_{2} {+ O}_{2} ⇌ {NO}_{2} {; ΔH° = 8.09 kcal at 300 K; N}_{2} {+ 2O}_{2} ⇌ N_{2} O_{4}; ΔH° = 2.19 kcal at 300K$
For which is more +ve:

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$NO$

$N_{2} O$

$N_{2} O_{4}$

${NO}_{2}$

answer is A.

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Detailed Solution

$ΔG° = ΔH° - TΔS°$

${ΔG}_{f_{N_{2} O}}^{°} = 19.49 - [48.2 - 45.7 - \frac{49.0}{2}] {× 300 × 10}^{-3} = 26.09 kcal$

${ΔG}_{f_{NO}}^{°} = 21.60 - [50.24 - \frac{45.7}{2} - \frac{49.0}{2}] {× 300 × 10}^{-3} = 20.73 kcal$

${ΔG}_{f_{{NO}_{2}}}^{°} = 8.09 - [57.24 - \frac{45.7}{2} - \frac{49.0}{2}] {× 300 × 10}^{-3} = 20.73 kcal$

${ΔG}_{{f_{N_{2} O}}_{4}}^{°} = 2.19 - [72.77 - 45.7 - 98.0] {× 300 × 10}^{-3} - 12.47 kcal$

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