Equilibrium - Result Question 2
####2. 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: $SrCO_3(s) \rightarrow SrO(s)+CO_2(g), Kp$ $=1.6 atm$ )
(a) 10 litre
(b) 4 litre
(c) 2 litre
(d) 5 litre
[2017]
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Solution:
- (d) Max. pressure of $CO_2=$ Pressure of $CO_2$ at equilibrium
For reaction,
$SrCO_3(s) \rightarrow SrO(s)+CO_2(g)$ $K_p=P _{CO_2}=1.6 atm=$ maximum pressure of $CO_2$ volume of container at this stage.
$V=\frac{n R T}{P}$
Since container is sealed and reaction was not earlier at equilibrium.
$\therefore \quad n=$ constant.
$n=\frac{P V}{R T}=\frac{0.4 \times 20}{R T}$
Put equation (ii) in equation (i)
$V=[\frac{0.4 \times 20}{R T}] \frac{R T}{1.6}=5 L$
