A 20 litre 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) leftharpoons SrO (s) + CO 2(g) , K p = 1.6 atm)

Chemical Equilibrium question

2017 · Q112

Chemical Equilibrium question

2017 · Q112

View written solutionFree

Correct answer: D

Initially, P₁ = 0.4 atm

V₁ = 20 L

T₁ = 400 K

Max. pressure of CO₂ = Pressure of CO₂ at equilibrium

For reaction,

SrCO₃(s) ⇌ SrO(s) + CO₂

K_p = P_CO₂ = 1.6 atm = maximum pressure of CO₂ volume of container at this stage.

V₂ = ${{nRT} \over {{P_2}}}$ ......(i)

Since, reaction was not at equilibrium earlier and also container is sealed.

$\therefore$ n = constant

n = ${{{P_1}{V_1}} \over {RT}}$ .....(ii)

From (i) and (ii)

V₂ = ${{{P_1}{V_1}} \over {RT}} \times {{RT} \over {{P_2}}}$ = ${{0.4 \times 20} \over {1.6}}$ = 5 L

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Chemical Equilibrium question

2017 · Q112

Chemical Equilibrium question

2017 · Q112

View written solutionFree

Correct answer: D

Initially, P₁ = 0.4 atm

V₁ = 20 L

T₁ = 400 K

Max. pressure of CO₂ = Pressure of CO₂ at equilibrium

For reaction,

SrCO₃(s) ⇌ SrO(s) + CO₂

K_p = P_CO₂ = 1.6 atm = maximum pressure of CO₂ volume of container at this stage.

V₂ = ${{nRT} \over {{P_2}}}$ ......(i)

Since, reaction was not at equilibrium earlier and also container is sealed.

$\therefore$ n = constant

n = ${{{P_1}{V_1}} \over {RT}}$ .....(ii)

From (i) and (ii)

V₂ = ${{{P_1}{V_1}} \over {RT}} \times {{RT} \over {{P_2}}}$ = ${{0.4 \times 20} \over {1.6}}$ = 5 L