Question

A 20 litre container at 400 K contains CO₂(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO₂ attains its maximum value, will be
(Given That: SrCO₃ (s) ⇋ SrO (s) + CO₂ (g). k_p=1.6 atm)

• A. 5 litre
• B. 10 litre
• C. 4 litre
• D. 2 litre

Answer 1

The Correct Option is A

To solve this problem, we need to understand the equilibrium involved and apply the concepts of chemical equilibrium in a gaseous system.

The equilibrium reaction given is: 

\(\text{SrCO}_3 (s) \rightleftharpoons \text{SrO} (s) + \text{CO}_2 (g)\)

This reaction implies that solid SrCO₃ decomposes to form solid SrO and gaseous CO₂. The equilibrium constant in terms of pressure (\(K_p\)) for the reaction is given as 1.6 atm.

Initially, the container has:

• Volume = 20 litres
• Temperature = 400 K
• Pressure of CO₂, \(P_{CO_2} = 0.4 \text{ atm}\)
• Excess SrO (solid), which means its quantity does not affect the pressure of CO₂

Since SrO is in excess and is a solid, it doesn't appear in the \(K_p\) expression.

The equilibrium constant expression is:

\(K_p = P_{CO_2}\)

As the volume is decreased, the pressure of CO₂ will increase till it reaches the equilibrium pressure. At equilibrium, the pressure of CO₂ should equal \(K_p = 1.6 \text{ atm}\).

Using the ideal gas law for initial and final conditions, and assuming temperature remains constant:

1. Initial moles of CO₂ at 0.4 atm in 20 L can be calculated as follows:
2. Final moles of CO₂ at equilibrium pressure 1.6 atm in volume \(V_f\):
3. Since the chemical equation shows that moles of CO₂ will not change, equating the moles:
4. Solving for \(V_f\) gives:

Therefore, the maximum volume of the container when the pressure of CO₂ attains its maximum value is 5 litre.

This means option "5 litre" is correct.