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

`\Delta S^(0)` for the reaction is -198.762 J/K

Step-by-step 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* S^(0)(NH_(3))_(g)]-[1mole* S^(0)(N_(2))_(g)]-[3* 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* 192.45]-[1* 191.61]-[3* 130.684]J/K=-198.762J/K`