Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answer using one decimal place. ΔS∘Δ S ∘ = J/K Previous Answer Request Answer Incorrect; Try Again; 3 attempts remaining Part B Complete previous part(s) Provide Feedback

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Answer:

[tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: [tex]\frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)[/tex]

[tex]\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}][/tex]

where [tex]S^{0}[/tex] represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

[tex]\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K[/tex]

So, [tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K

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