Equilibrium constant Kp for the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) is 0.82 atm at7270C . If 1 mole of CaCO₃ is placed in a closed container of 20L and heated to this temperature, what amount of CaCO₃ would dissociate at equilibrium ?

To determine the amount of calcium carbonate (CaCO₃) that dissociates at equilibrium, we start with the decomposition reaction:

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Given:

• Equilibrium constant (Kₚ) at 727°C (1000 K): 0.82 atm
• Volume of the container: 20 L
• Initial moles of CaCO₃: 1 mole

Since CaCO₃ is a solid, its concentration remains constant and does not appear in the expression for Kₚ. Therefore, Kₚ is defined solely by the partial pressure of CO₂:

Kₚ = PCO₂

At equilibrium, the partial pressure of CO₂ (PCO₂) equals Kₚ:

PCO₂ = 0.82 atm

Using the ideal gas law, we can calculate the number of moles of CO₂ at equilibrium:

PCO₂ × V = nCO₂ × R × T

Where:

• PCO₂ = 0.82 atm
• V = 20 L
• R = 0.0821 L·atm/(K·mol) (ideal gas constant)
• T = 1000 K

Rearranging to solve for nCO₂:

nCO₂ = (PCO₂ × V) / (R × T)

Substituting the known values:

nCO₂ = (0.82 atm × 20 L) / (0.0821 L·atm/(K·mol) × 1000 K)

Calculating this:

0.199756 moles of CO2

0.82 * 20 / 0.0821 / 1000=0.199756This means that approximately 0.1998 moles of CaCO₃ have dissociated, as each mole of CaCO₃ produces one mole of CO₂.To find the mass of dissociated CaCO₃:

Mass = moles × molar mass

The molar mass of CaCO₃ is:

• Ca: 40.08 g/mol
• C: 12.01 g/mol
• O₃: 3 × 16.00 g/mol = 48.00 g/mol

Total molar mass = 40.08 + 12.01 + 48.00 = 100.09 g/mol

Therefore:

Mass = 0.1998 moles × 100.09 g/mol

Calculating this:

Mass ≈ 20.0 grams

Thus, at equilibrium, approximately 20 grams of CaCO₃ would dissociate.