Question

Calculate the change in the standard entropy of the system, delta s degree for the synthesis of ammonia from N_2(g) and H_2(g) at 298 K: N_2(g) + 3H_2(g) rightarrow 2 NH_3(g)

Answer 1

Answer:

$\Delta S^{0}$ for the reaction is -198.762 J/K

Explanation:

$N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)$

Standard change in entropy for the system ($\Delta S^{0}$) is given by-

$\Delta S^{0}=[2moles\times S^{0}(NH_{3})_{g}]-[1mole\times S^{0}(N_{2})_{g}]-[3\times S^{0}(H_{2})_{g}]$

where $S^{0}$ represents standard entropy.

Here $S^{0}(NH_{3})_{g}=192.45J/(K.mol)$, $S^{0}(N_{2})_{g}=191.61J/(K.mol)$ and $S^{0}(H_{2})_{g}=130.684J/(K.mol)$

So, $\Delta S^{0}=[2\times 192.45]-[1\times 191.61]-[3\times 130.684]J/K=-198.762J/K$