At 100 k , the pressure of `CO_2` in equilibrium with `CaCO_3 and CaO` is equal to `3.9 xx 10^(-2)` atm. The equilibrium constant for the reaction `C(s) +CO_2 hArr 2CO(g)` , is 1.9 at the same temperature when pressure are in atmospheres, solid carbon CaO and `CaCO_3` are instead allowed to come to equilibrium at 100 K in closed vessel . what is the pressure of CO at equilibrium ?

Answer

To solve the problem, we need to determine the pressure of CO at equilibrium when CaCO3 and CaO are in a closed vessel at 100 K. We will follow these steps:

### Step 1: Write the reactions and given data
1. The first reaction is:
\[ \text{CaCO}_3 (s) \rightleftharpoons \text{CaO} (s) + \text{CO}_2 (g) \]
Given that the pressure of CO2 at equilibrium is \(3.9 \times 10^{-2}\) atm.

2. The second reaction is:
\[ C (s) + \text{CO}_2 (g) \rightleftharpoons 2 \text{CO} (g) \]
The equilibrium constant \(K_c\) for this reaction is given as \(1.9\).

### Step 2: Calculate \(K_p\) for the second reaction
To convert \(K_c\) to \(K_p\), we use the formula:
\[ K_p = K_c \cdot (RT)^{\Delta n} \]
Where:
- \(R = 0.082 \, \text{atm L/(mol K)}\)
- \(T = 100 \, K\)
- \(\Delta n = n_{products} - n_{reactants} = 2 - 1 = 1\)

Calculating \(K_p\):
\[ K_p = 1.9 \cdot (0.082 \cdot 100)^{1} = 1.9 \cdot 8.2 = 15.58 \]

### Step 3: Combine the two reactions
We need to combine the two reactions to find the overall equilibrium constant. The overall reaction can be represented as:
\[ \text{CaCO}_3 (s) \rightleftharpoons \text{CaO} (s) + \text{CO}_2 (g) \quad (1) \]
\[ C (s) + \text{CO}_2 (g) \rightleftharpoons 2 \text{CO} (g) \quad (2) \]
Adding these reactions, we get:
\[ \text{CaCO}_3 (s) + C (s) \rightleftharpoons \text{CaO} (s) + 2 \text{CO} (g) \]

### Step 4: Calculate the overall equilibrium constant \(K_{overall}\)
The overall equilibrium constant \(K_{overall}\) is the product of the individual equilibrium constants:
\[ K_{overall} = K_{1} \cdot K_{2} \]
Where:
- \(K_{1} = \frac{P_{CO_2}}{1} = 3.9 \times 10^{-2}\)
- \(K_{2} = 15.58\)

Calculating \(K_{overall}\):
\[ K_{overall} = (3.9 \times 10^{-2}) \cdot (15.58) = 0.60762 \]

### Step 5: Relate \(K_{overall}\) to the pressures at equilibrium
The equilibrium expression for the overall reaction is:
\[ K_{overall} = \frac{P_{CO}^2}{P_{CO_2}} \]
Substituting the known values:
\[ 0.60762 = \frac{P_{CO}^2}{3.9 \times 10^{-2}} \]

### Step 6: Solve for \(P_{CO}\)
Rearranging the equation to find \(P_{CO}\):
\[ P_{CO}^2 = 0.60762 \cdot (3.9 \times 10^{-2}) \]
Calculating:
\[ P_{CO}^2 = 0.60762 \cdot 0.039 = 0.0237 \]
Taking the square root:
\[ P_{CO} = \sqrt{0.0237} \approx 0.154 \]

### Final Answer
The pressure of CO at equilibrium is approximately **0.154 atm**.

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