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Question

Chemistry Question on Equilibrium

A20A\, 20 litre container at 400K400\, K contains CO2(g)CO _{2}( g ) at pressure 0.4atm0.4\, 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 CO2CO _{2} attains its maximum value, will be (Given that :SrCO3(s)SrO(s)+CO2(g): SrCO _{3}( s ) \rightleftharpoons SrO ( s )+ CO _{2}( g ) Kp=1.6atm)\left. K _{ p }=1.6\, atm \right)

A

10 litre

B

4 litre

C

2 litre

D

5 litre

Answer

5 litre

Explanation

Solution

Max. pressure of CO2=CO _{2}= Pressure of CO2CO _{2} at equilibrium

For reaction,

SrCO3(s)SrO(s)+CO2SrCO _{3}( s ) \rightleftharpoons SrO ( s )+ CO _{2}
Kp=PCO2=1.6atm=K _{ p }= P _{CO _{2}}=1.6\, atm = maximum pressure of CO2CO _{2}

Volume of container at this stage,

V=nRTPV=\frac{ nRT }{ P }...(i)

Since container is sealed and reaction was not earlier at equilibrium

n=\therefore n = constant

n=PVRT=0.4×20RTn =\frac{ PV }{ RT }=\frac{0.4 \times 20}{ RT }...(ii)

Put equation (ii) in equation (i)

V=[0.4×20RT]RT1.6=5LV =\left[\frac{0.4 \times 20}{ RT }\right] \frac{ RT }{1.6}=5\, L