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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:
1
`H_(2)(g)+l_(2)(g)hArr2Hl(g)`
`K_(c)=([Hl]^(2))/([H_(2)][l_(2)])" "`if`[H_(2)]=[l_(2)]" "rArr" "K_(c)=([Hl]^(2))/([l_(2)]^(2))" "[Hl]^(2)=K_(c)xx[l_(2)]^(2)`
or `([Hl]^(2))/([l_(2)]^(2))=K_(c)" "`or`" "" "([Hl])/([l_(2)]^(2))=sqrt(K_(c))=sqrt(49)=7`
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