Answer:
[tex]\Delta S^{0}[/tex] for the reaction is -198.762 J/K
Explanation:
[tex]N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)[/tex]
Standard change in entropy for the system ([tex]\Delta S^{0}[/tex]) is given by-
[tex]\Delta S^{0}=[2moles\times S^{0}(NH_{3})_{g}]-[1mole\times S^{0}(N_{2})_{g}]-[3\times S^{0}(H_{2})_{g}][/tex]
where [tex]S^{0}[/tex] represents standard entropy.
Here [tex]S^{0}(NH_{3})_{g}=192.45J/(K.mol)[/tex], [tex]S^{0}(N_{2})_{g}=191.61J/(K.mol)[/tex] and [tex]S^{0}(H_{2})_{g}=130.684J/(K.mol)[/tex]
So, [tex]\Delta S^{0}=[2\times 192.45]-[1\times 191.61]-[3\times 130.684]J/K=-198.762J/K[/tex]
