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 follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_p \)

For the reaction given, the equilibrium constant \( K_p \) is defined only in terms of the gaseous products. Since solids do not appear in the expression, we have:

\[ K_p = P_{\text{CO}_2} \]

Where \( P_{\text{CO}_2} \) is the partial pressure of carbon dioxide at equilibrium.

### Step 2: Substitute the given value of \( K_p \)

We are given:

\[ K_p = 8.21 \times 10^{-3} \, \text{atm} \]

Thus, we can write:

\[ P_{\text{CO}_2} = 8.21 \times 10^{-3} \, \text{atm} \]

### Step 3: Use the ideal gas law to find the number of moles

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 L·atm/(K·mol)
- \( T \) = temperature (in Kelvin)

We need to rearrange this equation to solve for \( n \):

\[ n = \frac{PV}{RT} \]

### Step 4: Substitute the known values into the equation

Given:
- \( P = 8.21 \times 10^{-3} \, \text{atm} \)
- \( V = 1 \, \text{L} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 127^\circ C = 400 \, \text{K} \) (since \( 127 + 273 = 400 \))

Now substituting these values into the equation:

\[ n = \frac{(8.21 \times 10^{-3} \, \text{atm}) \times (1 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (400 \, \text{K})} \]

### Step 5: Calculate the number of moles

Calculating the denominator:

\[ 0.0821 \times 400 = 32.84 \, \text{L·atm/(mol)} \]

Now substituting this back into the equation for \( n \):

\[ n = \frac{8.21 \times 10^{-3}}{32.84} \approx 2.50 \times 10^{-4} \, \text{moles} \]

### Final Answer

The number of moles of CO₂ at equilibrium is approximately:

\[ \boxed{2.50 \times 10^{-4}} \, \text{moles} \]

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