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

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 - [ 1 mol × 191,5 + 2 mol × 205,0 ]=

ΔS Rx = -120, 5 J/K

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

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