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Calculate AS° for the reaction below: N2(g)+202(g) 2NO2(g) where ASo for N2(g), O2(g), & NO2(g), respectively, is 191.5, 205.0, & 240.5 J/mol-K -156.0 J/K 156.0 J/K 120.5 J/K -120.5 J/K O OOO

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Answer:

ΔS Rx = -120, 5 J/K

Explanation:

The ΔS in a reaction is defined thus:

ΔS Rx = ∑ n S°products - ∑ m S°reactants

For the reaction:

N₂(g) + 2 O₂(g) → 2NO₂(g)

ΔS Rx = 2 mol × 240,5 [tex]\frac{J}{mol.K}[/tex] - [ 1 mol × 191,5 [tex]\frac{J}{mol.K}[/tex] + 2 mol × 205,0 [tex]\frac{J}{mol.K}[/tex]]=

ΔS Rx = -120, 5 J/K

A negative value in ΔS means a negative entropy of the process. Doing this process entropycally unfavorable.

I hope it helps!

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