Consider the following reaction at constant P. Determine the value of ΔSsurroundings at 298 K? Predict...

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.)

Expert Solution

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
*******************************
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.
***
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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queen_honey_blossom answered 1 year ago

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