Consider the following heterogeneous equilibrium, `CaCO_(3)(s)CaO(s)+CO_(2)(g)`. At `800^(@)C`, the pressure of `CO_(2)` is 0.236 atm. The value of `K_(c)` for the abvoe given equilibrium reaction at this temperature is

Options

A. `4xx10^(-5)`
B. `2.68xx10^(-3)`
C. `2.64xx10^(-5)`
D. `1.5xx10^(-3)`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Write the equilibrium expression
For the reaction:
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

The equilibrium constant \( K_c \) is defined only for the gaseous and aqueous species. Since both CaCO3 and CaO are solids, they do not appear in the equilibrium expression. Thus, the equilibrium expression is:
\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} = P_{\text{CO}_2} \]

### Step 2: Identify the given values
From the problem, we know that the pressure of \( \text{CO}_2 \) at 800°C is:
\[ P_{\text{CO}_2} = 0.236 \, \text{atm} \]

### Step 3: Calculate \( K_p \)
Since \( K_c \) is related to \( K_p \) through the equation:
\[ K_p = K_c (RT)^{\Delta n} \]
where \( \Delta n \) is the change in the number of moles of gas.

In this case:
- There is 1 mole of \( \text{CO}_2 \) produced (product side).
- There are no gaseous reactants.

Thus, \( \Delta n = 1 - 0 = 1 \).

Therefore, we can say:
\[ K_p = P_{\text{CO}_2} = 0.236 \]

### Step 4: Calculate \( K_c \)
Using the relationship between \( K_p \) and \( K_c \):
\[ K_p = K_c (RT)^{\Delta n} \]

Substituting \( \Delta n = 1 \):
\[ K_p = K_c (RT) \]

Rearranging gives:
\[ K_c = \frac{K_p}{RT} \]

### Step 5: Substitute the values
We need to convert the temperature from Celsius to Kelvin:
\[ T = 800 + 273 = 1073 \, \text{K} \]

The ideal gas constant \( R \) is:
\[ R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \]

Now substituting the values:
\[ K_c = \frac{0.236}{(0.0821)(1073)} \]

### Step 6: Calculate \( K_c \)
Calculating the denominator:
\[ (0.0821)(1073) \approx 88.1 \]

Now substituting back:
\[ K_c = \frac{0.236}{88.1} \approx 0.00268 \]

Expressing this in scientific notation:
\[ K_c \approx 2.68 \times 10^{-3} \]

### Final Answer
Thus, the value of \( K_c \) for the given equilibrium reaction at 800°C is:
\[ K_c \approx 2.68 \times 10^{-3} \]

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