= Solution
The two remaining <partial pressures> sum to $200-160=40$ bar. Their ratio therefore gives $p_{\mathrm{N_2}}=10$ bar and $p_{\mathrm{NH_3}}=30$ bar. With the ideal-gas activity approximation $a_i=p_i/p^\circ$ and $p^\circ=1$ bar,
$$
K_{800}=\frac{(30/1)^2}{(10/1)(160/1)^3}
=\boxed{2.1973\times10^{-5}}.
$$
The <equilibrium constant> is dimensionless: pressures enter as ratios to the standard <pressure>. Blindly substituting pressures in pascals into an unnormalized expression would give a different, incorrect thermodynamic $K$.
The standard reaction <enthalpy> is $2(-45.90)=-91.80\,\mathrm{kJ\,mol^{-1}}$ per mole of reaction as written, because the elements are in their reference states. Thus
$$
\begin{aligned}
\ln\frac{K_{298}}{K_{800}}
&=-\frac{-91800}{8.3145}\left(\frac1{298}-\frac1{800}\right),\\
K_{298}&=\boxed{2.7466\times10^5}\simeq2.75\times10^5.
\end{aligned}
$$
This calculation uses the supplied constant-enthalpy simplification. At 200 bar, an accurate thermodynamic treatment would use <chemical fugacities> rather than ideal <partial pressures>; no <chemical fugacity> coefficients are supplied. The <iron> catalyst changes the approach to equilibrium, not its thermodynamic value.
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