 Solution

For a reversible temperature change at constant pressure and fixed composition, $\mathrm{d}{S}_m=\delta q_{\mathrm{rev},\mathrm{m}}/T$ and $\delta q_{\mathrm{rev},\mathrm{m}}=C_{p,m}\mathrm{d}T$. HenceWith $T_0=298\mathrm{K}$, termwise integration yieldsThe last sign follows from integrating $e{T}^{-3}$, not $e{T}^{-2}$. Every term has units $\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$; the logarithm has a dimensionless argument. Inserting 800 K gives an entropy increase of $41.733\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$, and thereforeThe 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.