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One lit of SO(3) was placed in a two li...

One lit of `SO_(3)` was placed in a two litre vessels of a certain temperature. The following equilibrium was established in the vessel `2SO_(3(g)) hArr 2SO_(2(g)) +O_(2(g))` the equilibrium mixture reacted with 0.2 mole KMnO, in acidic medium. Kc value is `1.25 x10^(-x)` then the value of x is:

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The correct Answer is:
1


Only `SO_(2)` is will oxidised
So, no. of equivalent of `SO_(2)` = equivalent of `KMnO_(4)`
`2x xx 2=0.2 xx 5 implies 2x=0.5`
`Kc=(((0.5)/(2))^(2)((0.25)/(2)))/(((0.5)/(2))^(2))=0.125`
`implies 1.25 xx 10^(-1)=1.25 xx 10^(-x) implies x=1`
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