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If 0.5 mole H(2) is reacted with 0.5 mol...

If `0.5 `mole `H_(2)` is reacted with 0.5 mole `I_(2)` in a ten `-` litre container at `444^(@)C` and at same temperature value of equilibrium constant `K_(C)` is 49, the ratio of `[Hl]` and `[l_(2)]` will be `:`

A

`7`

B

`(1)/(7)`

C

`sqrt((1)/(7))`

D

`49`

Text Solution

Verified by Experts

The correct Answer is:
A
`H_(2)(g)+l_(2)(g)hArr2HI(g)`
`K_(c)=([HI]^(2))/([H_(2)][l_(2)]) "if" [H_(2)]=[l_(2)]`
`K_(c)=([HI]^(2))/([l_(2)]^(2)) [HI]^(2)=K_(c)xx[l_(2)]`
or `([HI]^(2))/([l_(2)]^(2))=K_(c) "or" ([HI])/([l_(2]])=sqrtK_(c)=sqrt49=7`
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