= Solution
For a reversible <temperature> change at constant <pressure> and fixed composition, $\mathrm dS_m=\delta q_{\rm rev,m}/T$ and $\delta q_{\rm rev,m}=C_{p,m}\,\mathrm dT$. Hence
$$
\mathrm dS_m^\circ=\frac{C_{p,m}^\circ(T)}{T}\,\mathrm dT.
$$
With $T_0=298\,\mathrm K$, termwise <integration> yields
$$
\boxed{\begin{aligned}
S_m^\circ(T)-S_m^\circ(T_0)
={}&a\ln\frac{T}{T_0}+b(T-T_0)+\frac c2(T^2-T_0^2)\\
&+\frac d3(T^3-T_0^3)-\frac e2(T^{-2}-T_0^{-2}).
\end{aligned}}
$$
The last sign follows from integrating $eT^{-3}$, not $eT^{-2}$. Every term has units $\mathrm{J\,K^{-1}\,mol^{-1}}$; the logarithm has a dimensionless argument. Inserting 800 K gives an <entropy> increase of $41.733\,\mathrm{J\,K^{-1}\,mol^{-1}}$, and therefore
$$
\boxed{S_m^\circ(\mathrm{NH_3},800\,\mathrm K)=231.59\,\mathrm{J\,K^{-1}\,mol^{-1}}.}
$$
The standard <pressure> remains fixed during this <integration>; the 200-bar equilibrium mixture is not the <standard state>. Differentiating the integrated expression recovers $C_p/T$, providing a check on all five terms. Using the ammonia heat-capacity polynomial here does not retroactively replace the explicitly requested constant-reaction-enthalpy approximation in the preceding equilibrium calculation.
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