= Solution
Using the <equilibrium constant> calculated without premature rounding,
$$
\Delta_rG^\circ_{298}=-RT\ln K_{298}
=\boxed{-31.03\,\mathrm{kJ\,mol^{-1}}}.
$$
Then $\Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ$ gives
$$
\Delta_rS^\circ_{298}
=\frac{-91800-(-31029.18)}{298}
=\boxed{-203.93\,\mathrm{J\,K^{-1}\,mol^{-1}}}.
$$
The sign agrees with the reduction in gas stoichiometric amount. The standard reaction <entropy> is not the absolute <entropy> of ammonia: its stoichiometric coefficient is two. Therefore
$$
\begin{aligned}
2S_m^\circ(\mathrm{NH_3})-191.61-3(130.68)&=-203.93,\\
\boxed{S_m^\circ(\mathrm{NH_3},298\,\mathrm K)}&=\boxed{189.86\,\mathrm{J\,K^{-1}\,mol^{-1}}}.
\end{aligned}
$$
This is an absolute <standard molar entropy> inferred within the paper's thermodynamic approximations, not a claim that elemental gases have zero absolute <entropy>. Only their standard enthalpies of formation are zero in this calculation.
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