Consider the synthesis of ammonia:
`{:(N_(2) (g) + 3 H_(2) (g) rarr 2NH_(3) (g),,,Delta_(r )H^(@)=-92.6 kJ):}`
If absolute entropies of `N_(2) (g), H_(2) (g)`, and `NH_(3) (g)` are `192 j K^(-1) mol^(-1), 131 jK^(-1) mol^(-1)`, and `193 j k^(-1)`, respectively, at `25^(@)C`, predict whether the reaction is spontaneous or not.
Strategy : Calculate `Delta S_("univ")` using `Delta S_("says")` and `Delta S_("surr")`. For calculating `Delta S_("rays")` use the absolute entropies of reactants and product. For calculating `Delta S_("surr")`, use `Delta_(r) H^(@)` and `T`.

Explanation

We can calculate `Delta_(r) S^(@) (= Delta S_("says"))` by using Eq.
`Delta_(r )S^(@)=[2S^(@)._(m)(NH_(3))]-[S^(@)._(m)(N_(2))+3S^(@)._(m)(N_(2))]`
`= (2 mol) (193 J K^(-1) mol^(-1)) - [(1 mol) (192 J K^(-1) mol^(-1)) + (3 mol) (131 J K^(-1) mol^(-1))] = - 199 J K^(-1)`
This result shows that when 1 mol of `N_(2) (g)` reacts with 3 mol of gaseous `H_(2)` form 2 mol of `NH_(3) (g)`, there is a decrease in the entropy of system equal to `- 199 J K^(-1)`.
Substituting `Delta_("says") (= Delta_(r) H^(@))` and `T` in Eq. we obtain
`Delta S_("surr") = (-(- 92.6 xx 10^(3) J))/((298 K)) = 311 J K^(-1)`
Thus, the change in the entropy of the universe is
`Delta S_("univ")=Delta S_("says")+Delta S_("surr")`
`=(-199 J K^(-1))+(311 JK^(-))`
`= 112 J K^(-1)`
`DeltaS_("univ")` is positive, we priedict that the reaction is spontaneous at `25^(@)C`.