\(P_{\text {initial}} 0.5 \;a t m\) \(0\)
\(P_{\text {final}}(0.5-x)\;\,atm\) \( 2 x\; \,atm\)
Total \(P\) at equilibrium
\(=0.5-x+2 x=0.5+x\) atm
\(0.8=0.5+x\)
\(\therefore x=0.8-0.5=0.3 \,a t m\)
Now \(k_{p}=\left(P_{C O}\right)^{2} / P_{C O_{2}}\)
\(=\frac{(2 \times 0.3)^{2}}{(0.5-0.3)}=\frac{(0.6)^{2}}{(0.2)}\)
\(=1.8 \mathrm{\,atm}\)