Partial Pressure of Carbon Monoxide at Equilibrium

To find the partial pressure of $\text{CO}$ at equilibrium, we first determine the partial pressure of $\text{CO}_2$ from the equilibrium constant of the first reaction, $K_{p1} = P_{\text{CO}_2} = 0.08\text{ atm}$. Using the equilibrium expression for the second reaction, $K_{p2} = \frac{P_{\text{CO}}^2}{P_{\text{CO}_2}}$, we substitute $K_{p2} = 2\text{ atm}$ and $P_{\text{CO}_2} = 0.08\text{ atm}$ to get $P_{\text{CO}}^2 = 2 \times 0.08 = 0.16\text{ atm}^2$. Taking the square root gives $P_{\text{CO}} = 0.4\text{ atm} = 4 \times 10^{-1}\text{ atm}$. Thus, the required value is $4$.