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1 mol of of XY(g) and 0.2 mol of Y(g) a...

1 mol of of `XY_(g)` and 0.2 mol of `Y_(g)` are mixed in 1L vessel. At equilibrium 0.6 mol of `Y _(g)` is present .The value of K for the reaction :-
` XY(g) hArr X(g)+Y(g)` is

A

0.04

B

0.06

C

0.36

D

0.4

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(,XY_((g)),hArr,X_((g))+Y_((g))),(t=0,1 "mole",,0.2 "mole"):}`
at equilibrium 1-x x 0.2+x
`:. 0.2+x=0.6`
x=0.4mole
`K=(0.6xx0.4)/(0.6)=0.4`
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