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.

I hope it helps!

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )