`C(s)+CO_(2)(g)hArr2CO(g) K_(p)=1atm`
`CaCO_(3)(s)hArrCaO(s)+CO_(2)(s) K_(p)=4xx10^(-2)`
Solid `C,CaO "and" CaCO_(3) "are mixed and allowed to attain equilibrium. Calculate final pressure of" CO`.

Options

A. `0.4`atm
B. `0.2` atm
C. `8` atm
D. `0.01` atm

Answer

To solve the problem step by step, we will analyze the two reactions provided, write their equilibrium expressions, and then find the final pressure of CO.

### Step 1: Write the Equations and Equilibrium Expressions

1. **First Reaction**:
\[ C(s) + CO_2(g) \rightleftharpoons 2CO(g) \]
The equilibrium constant \( K_p \) for this reaction is given by:
\[ K_{p1} = \frac{(P_{CO})^2}{(P_{CO_2})} \]
Given \( K_{p1} = 1 \, \text{atm} \).

2. **Second Reaction**:
\[ CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \]
The equilibrium constant \( K_p \) for this reaction is given by:
\[ K_{p2} = P_{CO_2} \]
Given \( K_{p2} = 4 \times 10^{-2} \, \text{atm} \).

### Step 2: Determine \( P_{CO_2} \)

From the second reaction, we know:
\[ P_{CO_2} = K_{p2} = 4 \times 10^{-2} \, \text{atm} \]

### Step 3: Substitute \( P_{CO_2} \) into the First Reaction's Equilibrium Expression

Using the value of \( P_{CO_2} \) in the first reaction's equilibrium expression:
\[ K_{p1} = \frac{(P_{CO})^2}{P_{CO_2}} \]
Substituting the known values:
\[ 1 = \frac{(P_{CO})^2}{4 \times 10^{-2}} \]

### Step 4: Solve for \( P_{CO} \)

Rearranging the equation to solve for \( P_{CO} \):
\[ (P_{CO})^2 = 1 \times (4 \times 10^{-2}) = 4 \times 10^{-2} \]
Taking the square root of both sides:
\[ P_{CO} = \sqrt{4 \times 10^{-2}} = 2 \times 10^{-1} = 0.2 \, \text{atm} \]

### Final Answer

The final pressure of CO is:
\[ \boxed{0.2 \, \text{atm}} \]