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1 mol of XY(g) and 0.2 mol of Y(g) are m...

`1` mol of `XY(g)` and `0.2` mol of `Y(g)` are mixed in `1` L vessel. At equilibrium, `0.6` mol of `Y(g)` is present. The value of `K` for the reaction
`XY(g)hArrX(g)+Y(g)` is

A

`0.04 mol L^(-1)`

B

`0.06 mol L^(-1)`

C

`0.36 mol L^(-1)`

D

`0.40 mol L^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(,XY(g),hArr,X(g),+,Y(g)),("Initial",1,,0,,0.2),("At eq.",1-x,,x,,0.2+x):}`
At eq. `[X]=0.2+x=0.6`
`:. X=0.4 "mol"=0.4/1=0.4 M`
`[XY]=1-0.4 "mol"=0.6 "mol"=0.6/1=0.6 M`
`[X]=x=0.4=0.4/1=0.4 M`
`K=([X][Y])/([XY])=(0.4xx0.6)/0.6=0.4 "mol" L^(-1)`
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