At a certain temperature , `K_(p)` for dissociation of solid `CaCO_(3)` is `4xx10^(-2)` atm and for the reaction, `C(s)+CO_(2) hArr 2CO` is `2.0` atm, respectively. Calculate the pressure of CO at this temperature when solid `C, CaO, CaCO_(3)` are mixed and allowed to attain equilibrium.

Explanation

To solve the problem, we need to calculate the pressure of CO at equilibrium when solid C, CaO, and CaCO3 are mixed. We will use the given equilibrium constants \( K_p \) for the reactions involved.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants:**
- The dissociation of solid CaCO3 can be represented as:
\[ \text{CaCO}_3 (s) \rightleftharpoons \text{CaO} (s) + \text{CO}_2 (g) \]
Given \( K_{p1} = 4 \times 10^{-2} \) atm.
- The reaction of solid carbon with CO2 to produce CO is:
\[ C (s) + \text{CO}_2 (g) \rightleftharpoons 2 \text{CO} (g) \]
Given \( K_{p2} = 2.0 \) atm.

2. **Write the Expression for \( K_{p1} \):**
- For the first reaction, the equilibrium constant \( K_{p1} \) can be expressed as:
\[ K_{p1} = \frac{P_{\text{CO}_2}}{1} = P_{\text{CO}_2} = 4 \times 10^{-2} \text{ atm} \]

3. **Write the Expression for \( K_{p2} \):**
- For the second reaction, the equilibrium constant \( K_{p2} \) is given by:
\[ K_{p2} = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} = 2.0 \]

4. **Substituting \( P_{\text{CO}_2} \) into the \( K_{p2} \) Expression:**
- From the first reaction, we found \( P_{\text{CO}_2} = 4 \times 10^{-2} \) atm. We can substitute this value into the expression for \( K_{p2} \):
\[ 2.0 = \frac{(P_{\text{CO}})^2}{4 \times 10^{-2}} \]

5. **Solving for \( P_{\text{CO}} \):**
- Rearranging the equation gives:
\[ (P_{\text{CO}})^2 = 2.0 \times (4 \times 10^{-2}) \]
\[ (P_{\text{CO}})^2 = 8 \times 10^{-2} \]
- Taking the square root of both sides:
\[ P_{\text{CO}} = \sqrt{8 \times 10^{-2}} = \sqrt{0.08} \]
\[ P_{\text{CO}} \approx 0.2828 \text{ atm} \]

6. **Final Answer:**
- Rounding to two decimal places, the pressure of CO at equilibrium is:
\[ P_{\text{CO}} \approx 0.28 \text{ atm} \]