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For the equilibrium in gaseous phase in ...

For the equilibrium in gaseous phase in 2 lit flask we start with 2 moles of `SO_(2)` and 1 mole of `O_(2)` at 3 atm, `2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g))`. When equilibrium is attained, pressure changes to 2.5 atm. Hence, equilibrium constant `K_(c) ` is :

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The correct Answer is:
4
`underset(2-2x)underset(2)(2SO_(2(g)))+underset(1-x)underset(1)(O_(2(g))) harr underset(2x)underset(0)(2SO_(3(g)))`
Moles at equilibrium = 3-x
At equilibrium pressure becomes 2.5 atm since pv= nRT
Hence `p alpha n`
Thus `(3-x)/(3)=(2-5)/(3)`, x=0.5
`[SO_(2)]=(2-2x)/(2)=0.5M`
`[O_(2)]=(1-x)/(2)=0.25M, (SO_(3))=(2x)/(2)=0.5M` `K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=((0.5)^(2))/((0.5)^(2) xx 0.25)=4`
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