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 reversible process given by the equation:

\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

we start with the known value of \( K_p \) and the temperature. The values provided are:

- \( K_p = 167 \)
- Temperature \( T = 800^\circ C \)

### Step 1: Convert Temperature to Kelvin
To use the ideal gas constant \( R \) in the equation, we need to convert the temperature from Celsius to Kelvin.

\[ T(K) = T(°C) + 273 = 800 + 273 = 1073 \, K \]

### Step 2: Identify the Change in Moles (\( \Delta n \))
Next, we need to determine the change in the number of moles of gas (\( \Delta n \)) in the reaction.

- On the product side, we have 1 mole of \( \text{CO}_2(g) \).
- On the reactant side, we have 0 moles of gas (since \( \text{CaCO}_3 \) and \( \text{CaO} \) are solids).

Thus,

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

### Step 3: Use the Relationship Between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:

\[ K_p = K_c (RT)^{\Delta n} \]

Rearranging this gives us:

\[ K_c = \frac{K_p}{(RT)^{\Delta n}} \]

### Step 4: Substitute Known Values
We know:

- \( K_p = 167 \)
- \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 1073 \, K \)
- \( \Delta n = 1 \)

Now, substituting these values into the equation:

\[ K_c = \frac{167}{(0.082 \times 1073)^{1}} \]

### Step 5: Calculate \( RT \)
First, calculate \( RT \):

\[ RT = 0.082 \times 1073 = 88.0516 \]

### Step 6: Calculate \( K_c \)
Now, substitute \( RT \) back into the equation for \( K_c \):

\[ K_c = \frac{167}{88.0516} \approx 1.89 \]

### Final Answer
Thus, the value of \( K_c \) for the reaction is approximately:

\[ K_c \approx 1.89 \]

---