Calculate the particle pressure of carbon monoxide from the following `CaCO_3(s) overset(Delta)(rarr)CaO(s) + CO_2`
`overset(uarr , K_p = 8 xx 10^(-2))(CO_2)(g) + C(s) to 2CO(g) , K_p = 2`

Options

A. `0.2`
B. `0.4`
C. `1.6`
D. 4

Answer

To calculate the partial pressure of carbon monoxide (CO) from the given reactions, we can follow these steps:

### Step 1: Write down the two equilibrium reactions and their equilibrium constants.

1. The first reaction is:
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]
with \( K_p = 8 \times 10^{-2} \).

2. The second reaction is:
\[ \text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g) \]
with \( K_p = 2 \).

### Step 2: Analyze the first reaction.

For the first reaction, since we are only considering the gaseous products, we can express \( K_p \) as:
\[ K_p = \frac{P_{CO_2}}{1} = P_{CO_2} \]
Thus, we have:
\[ P_{CO_2} = 8 \times 10^{-2} \text{ atm} \]

### Step 3: Analyze the second reaction.

For the second reaction, we can express \( K_p \) as:
\[ K_p = \frac{(P_{CO})^2}{P_{CO_2}} \]
Substituting the known value of \( K_p \) and \( P_{CO_2} \):
\[ 2 = \frac{(P_{CO})^2}{8 \times 10^{-2}} \]

### Step 4: Rearranging to find \( P_{CO} \).

Rearranging the equation gives:
\[ (P_{CO})^2 = 2 \times (8 \times 10^{-2}) \]
\[ (P_{CO})^2 = 1.6 \times 10^{-1} \]

### Step 5: Taking the square root.

Taking the square root of both sides to find \( P_{CO} \):
\[ P_{CO} = \sqrt{1.6 \times 10^{-1}} = \sqrt{0.16} = 0.4 \text{ atm} \]

### Final Answer:
The partial pressure of carbon monoxide (CO) is:
\[ \boxed{0.4 \text{ atm}} \]

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