Question

Consider the following reaction at constant P. Determine the value of ΔSsurroundings at 298 K? Predict whether or not this reaction will be spontaneous at this temperature. N₂(g) + 2 O₂(g) → 2 NO₂(g); ΔH = +66.4 kJ

(Please show work and thank you for taking the time to answer.)

Answer 1

N2(g) + 2 O2(g) → 2 NO2(g); ΔH = + 66.4 KJ
at temperature =298 K
ΔSsurroundings = - [ΔH ]/T= - 66.4 x10^3 J/ 298 K
= -222.8 j/k

rounding to 3 significant figures

ΔSsurroundings = -223. J/ K
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N2(g) + 2 O2(g) → 2 NO2(g)

this shows 3 mols gaseous reactants produce 2 mols of product. Since there
is a decrease in number of mols of product , entropy of system has been
reduced. Because entropy is a measure of disorder. More Gaseous particle
higher the entropy of system.
Thus entropy of this system [reaction] is reduced , that means

ΔSsystem is negative.
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According to 2nd law , ΔSuniverse = ΔSsystem + ΔSsurrounding

when ΔSuniverse > 0 [ or positive value] the reaction will be spontaneous

but here ΔSsystem and ΔSsurrounding are negative . SO
ΔSuniverse will be negative or ΔSuniverse <0

thus the reaction is non spontaneous at this temperature
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