For the reaction, $I_2\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons 2ICl\left(g\right);\Delta H=16kJmo{l}^{-1}$ .
If $K_c=80$ at ${27}^{\circ }C,$ the value of $K_p$ at ${127}^{\circ }C$ is_______
[Given $\log 2=0.3,\ln x=2.3\log x,R=8.3J{K}^{-1}mo{l}^{-1}$ ]

As  $\Delta n_g=0,K_p=K_c$
 $\log \frac{K_{p2}}{K_{p1}}=\frac{\Delta H}{2.3R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]\Rightarrow \log K_{p2}-\log 80=\frac{16000}{2.3\times 8.3}\left[\frac{400-300}{400\times 300}\right]$
 $\Rightarrow \log K_{p2}-1.9=\frac{16000}{2.3\times 8.31}\left[\frac{400-300}{400\times 300}\right]\Rightarrow \log K_{p2}=2.598\approx 2.6\Rightarrow K_{p2}=400$