Equilibrium constant for the reaction,
`CaCO_(3)(s)hArr CaO(s)+CO_(2)(g)`
at `127^(@)C` in one litre container is `8.21xx10^(-3)` atm. Moles of `CO_(2)` at equilibrium is

Options

A. `5xx10^(-4)`
B. `3.5xx10^(-4)`
C. `2.5xx10^(-4)`
D. `7xx10^(-4)`

Answer

To find the moles of CO₂ at equilibrium for the reaction:

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

we will use the given equilibrium constant \( K_p \) and the ideal gas law.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- The equilibrium constant \( K_p = 8.21 \times 10^{-3} \) atm.
- The reaction occurs in a 1-liter container.
- Temperature \( T = 127^\circ C = 400 \, K \) (after converting to Kelvin).

2. **Write the Expression for \( K_p \):**
Since solids do not appear in the equilibrium expression, we only consider the gaseous product:
\[ K_p = P_{\text{CO}_2} \]
Therefore, at equilibrium, we have:
\[ P_{\text{CO}_2} = 8.21 \times 10^{-3} \, \text{atm} \]

3. **Use the Ideal Gas Law:**
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant \( = 0.0821 \, \text{L atm/(K mol)} \)
- \( T \) = temperature (in Kelvin)

4. **Rearranging the Ideal Gas Law:**
We can rearrange the ideal gas law to solve for \( n \):
\[ n = \frac{PV}{RT} \]

5. **Substituting the Known Values:**
Substitute \( P = 8.21 \times 10^{-3} \, \text{atm} \), \( V = 1 \, \text{L} \), \( R = 0.0821 \, \text{L atm/(K mol)} \), and \( T = 400 \, K \):
\[ n = \frac{(8.21 \times 10^{-3} \, \text{atm}) \times (1 \, \text{L})}{(0.0821 \, \text{L atm/(K mol)}) \times (400 \, K)} \]

6. **Calculating the Number of Moles:**
\[ n = \frac{8.21 \times 10^{-3}}{0.0821 \times 400} \]
\[ n = \frac{8.21 \times 10^{-3}}{32.84} \approx 2.5 \times 10^{-4} \, \text{moles} \]

### Final Answer:
The number of moles of \( \text{CO}_2 \) at equilibrium is approximately:
\[ \boxed{2.5 \times 10^{-4}} \, \text{moles} \]