Calculate `K_(c)` for the reversible process given below if `K_(p)=167 and T=800^(@)C`
`CaCO_(3)(s)hArr CaO(s)+CO_(2)(g)`

Options

A. 1.95
B. 1.85
C. 1.89
D. 1.6

Answer

To calculate \( K_c \) for the given reversible process, we will use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \cdot R T^{\Delta n_g} \]

### Step 1: Identify the reaction and components
The reaction is:
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]
- Here, \( \Delta n_g \) is the change in the number of moles of gas, which is the difference between the moles of gaseous products and reactants.

### Step 2: Calculate \( \Delta n_g \)
In this reaction:
- Gaseous products: 1 (from \( \text{CO}_2 \))
- Gaseous reactants: 0 (since \( \text{CaCO}_3 \) is a solid)

Thus,
\[ \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - 0 = 1 \]

### Step 3: Convert temperature to Kelvin
The temperature is given as \( 800^\circ C \). To convert to Kelvin:
\[ T(K) = 800 + 273 = 1073 \, K \]

### Step 4: Use the ideal gas constant \( R \)
The value of \( R \) (universal gas constant) is:
\[ R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \]

### Step 5: Substitute values into the equation
We have:
- \( K_p = 167 \)
- \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 1073 \, K \)
- \( \Delta n_g = 1 \)

Substituting these values into the equation:
\[ 167 = K_c \cdot (0.082) \cdot (1073)^{1} \]

### Step 6: Solve for \( K_c \)
Rearranging the equation to solve for \( K_c \):
\[ K_c = \frac{167}{0.082 \cdot 1073} \]

Calculating the denominator:
\[ 0.082 \cdot 1073 \approx 88.046 \]

Now substituting back:
\[ K_c = \frac{167}{88.046} \approx 1.89 \]

### Final Answer
Thus, the value of \( K_c \) is approximately:
\[ K_c \approx 1.89 \]

---