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Question

A reaction system in equilibrium according to reaction : 2SO2(g)+O2(g)2SO3(g) in one litre vessel at a given temperature was found to be 0.12 mole each of SO2 and SO3 and 5 mole of O2. In another vessel of one litre contains 32g of SO2 at the same temperature. What mass of O2 must be added to this vessel in order that at equilibrium 20% of SO2 is oxidised to SO3:

A
0.4125 g
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B
13.2 g
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C
1.6 g
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D
11.6 g
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Solution

The correct option is D 11.6 g
Kc=[SO3]2[SO2]2[O2]=0.1220.122×5=0.2

At constant temperature, Kc will remain constant.
Note : 32g of SO2 = 0.5 mol
Assume x moles of O2 is added

2SO20.5 +O2x 2SO30

2SO20.52y +O2xy 2SO32y

According to question,
2y=0.2×0.5
y=0.05 moles

Now,
Kc=[SO3]2[SO2]2[O2]=(2y)2(0.52y)2×(xy)=0.2

Solving the equation by putting y=0.05, we get

x=0.3625 moles=0.3625×32 g=11.6 g

Hence, option D is correct

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