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Q.

If S° for  N2, O2, N2O, NO,NO2 and N2O4 are are 45.7, 49.0, 48.2, 50.24, 57.24 and 72.77 cal

respectively and N2 + 12 O2 N2O; ΔH° = 19.49 kcal at 300 K; 12N2 + 12 O2 NO; ΔH° = 21.60 kcal at 300K

12 N2 + O2 NO2; ΔH° = 8.09 kcal at 300 K; N2 + 2O2 N2O4; ΔH° = 2.19 kcal at 300K 

For which is more +ve:

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a

N2O

b

NO

c

NO2

d

N2O4

answer is A.

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Detailed Solution

ΔG° = ΔH°  TΔS°

ΔGfN2O° = 19.49 [48.2 45.7 49.02] × 300 × 10-3 = 26.09 kcal

ΔGfNO° = 21.60 [50.24 - 45.7249.02] × 300 × 10-3 = 20.73 kcal

ΔGfNO2°= 8.09 [57.24 45.7249.02] × 300 × 10-3 = 20.73 kcal

ΔGfN2O4° = 2.19 [72.77  45.7  98.0] × 300 × 10-3 12.47 kcal

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