An empty steel container is filled with $5\mathrm{.}90$atm of $H_2$

and $5\mathrm{.}90$atm of $F_2.$ The system is allowed to reach

equilibrium. If $Kp=0\mathrm{.}450$ for the reaction below,

what is the equilibrium partial pressure of HF$?$

$H_2(g)+F_2(g)2HF(g)$