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Kelvi

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A 20 L container at 400 K contains CO $_2$ (g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO $_2$ attains its maximum value, will be: $\\$ (Given that: SrCO $_3$ (s) $\rightleftharpoons$ SrO(s) + CO $_2$ (g), $K_p = 1.6$ atm)

hard

Equilibrium

2017

chemistry

$10$ L

$4$ L

$2$ L

$5$ L

Explanation

To solve this problem, we need to understand the equilibrium between SrCO $_3$ (s), SrO(s), and CO $_2$ (g). $\\$ The reaction is: $\\$ SrCO $_3$

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(s)

\rightleftharpoons

SrO(s) + CO

_2

(g)

\\

Given:

\\

• Initial pressure of CO

_2 = 0.4

atm

\\

• Volume of the container

V_1 = 20

L

\\

• Temperature

T = 400

K

\\

• Equilibrium constant

K_p = 1.6

atm

\\

At equilibrium, the pressure of CO

_2

is

P_{CO_2}. \\

Since SrO is in excess, it will not affect the equilibrium.

\\

The equilibrium condition is given by:

\\ P_{CO_2} = K_p = 1.6

atm

\\

Initially, the pressure of CO

_2

is 0.4 atm. As the volume decreases,

\\

the pressure will increase until it reaches equilibrium at 1.6 atm.

\\

Using the ideal gas law,

PV = nRT,

we can find the initial moles of CO

_2: \\ n_{initial} = \frac{P_{initial} \cdot V_1}{RT} = \frac{0.4 \cdot 20}{R \cdot 400} \\

At equilibrium, the moles of CO

_2

will be:

\\ n_{equilibrium} = \frac{P_{equilibrium} \cdot V_{equilibrium}}{RT} = \frac{1.6 \cdot V_{equilibrium}}{R \cdot 400} \\

Since the moles of CO

_2

are conserved, we equate the two expressions:

\\ \frac{0.4 \cdot 20}{R \cdot 400} = \frac{1.6 \cdot V_{equilibrium}}{R \cdot 400} \\

Simplifying, we find:

\\ 0.4 \cdot 20 = 1.6 \cdot V_{equilibrium} \\ 8 = 1.6 \cdot V_{equilibrium} \\ V_{equilibrium} = \frac{8}{1.6} \\ V_{equilibrium} = 5

L

\\

Therefore, the maximum volume of the container when the pressure of CO

_2 \\

attains its maximum value is 5 L. This corresponds to Option 4.