Solid carbon, CaO and CaCO$$_3$$ are mixed and allowed to attain equilibrium at T K.
$$\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)} \quad K_{p_1} = 0.08$$ atm
$$\text{C(s)} + \text{CO}_2\text{(g)} \rightleftharpoons 2\text{CO(g)} \quad K_{p_2} = 2$$ atm
The partial pressure of CO is __________ $$\times 10^{-1}$$ atm

For a heterogeneous equilibrium, the equilibrium constant expression ($$K_p$$) only includes the partial pressures of gaseous components, as the activities of pure solids are taken as $$1$$.

From Reaction 1:

$$K_{p1} = P_{\text{CO}_2}$$

Given that $$K_{p1} = 0.08\text{ atm}$$, we can directly state:

$$P_{\text{CO}_2} = 0.08\text{ atm}$$

Since both reactions occur simultaneously in the same container, the $$\text{CO}_2$$ gas produced in the first reaction is utilized in the second reaction. The system will adjust until both equilibria are satisfied at the same time.

From Reaction 2:

$$K_{p2} = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}}$$

Substitute the values of $$K_{p2}$$ and $$P_{\text{CO}_2}$$ into this expression:

$$2 = \frac{(P_{\text{CO}})^2}{0.08}$$

$$(P_{\text{CO}})^2 = 2 \times 0.08$$

$$(P_{\text{CO}})^2 = 0.16$$

Taking the square root on both sides:

$$P_{\text{CO}} = \sqrt{0.16} = 0.4\text{ atm}$$

Converting $$0.4\text{ atm}$$ into scientific notation:

$$0.4\text{ atm} = 4 \times 10^{-1}\text{ atm}$$