1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K.If Kp for CaCO3=CaO+CO2 at this temperature is 2 atm , equilibrium concentration of CO2 (in mol/lit) is?

To solve this question, we will proceed step by step.

Given data:
The reaction:
CaCO₃ (s) ⇌ CaO (s) + CO₂ (g)
Kp = 2 atm at 819 K
Volume of the vessel = 11.2 L
We need to find the equilibrium concentration of CO₂ in mol/L.
Step 1: Understanding the Equilibrium Condition
Since CaCO₃ and CaO are solids, they do not appear in the equilibrium expression. The equilibrium constant in terms of pressure is:

Kp = P(CO₂)

This means that at equilibrium, the partial pressure of CO₂ is 2 atm.

Step 2: Using the Ideal Gas Law
The concentration of CO₂ can be determined using the ideal gas equation:

PV = nRT

Rearrange to find n/V:

n/V = P / RT

Substituting the given values:

P = 2 atm
R = 0.0821 L·atm·mol⁻¹·K⁻¹
T = 819 K
n/V = (2) / (0.0821 × 819)
n/V = 2 / 67.24
n/V ≈ 0.0297 mol/L

Final Answer:
The equilibrium concentration of CO₂ is 0.0297 mol/L.