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If 1.0 mole of I(2) is introduced into 1...

If 1.0 mole of `I_(2)` is introduced into 1.0 litre flask at 1000 K, at quilibrium `(K_(e )=10^(-6))`, which one is correct ?

A

`[I_(2)(g)] gt [1^(-1)(g)]`

B

`[I_(2)(g)] lt [1^(-)(g)]`

C

`[I_(2)(g)]=[I^(-)(g)]`

D

`[I_(2)(g)]=(1)/(2)[I^(-)(g)]`

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(I-x)(I_(2)) hArr underset(2x)(2I^(-))`
`K_(e )=((2x)^(2))/((1-x))=10^(-6)`
So In. shows that `(1-x)gt 2x therefore [I_(2)(g)] gt [I^(-)(g)]`
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Knowledge Check

  • For the reaction I_(2)(g) hArr 2I(g) , K_(c) 37.6 xx 10^(-6) at 1000K . If 1.0 mole of I_(2) is introduced into a 1.0 litre flask at 1000K, at equilibrium

    A
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    B
    `[I_(2)] = [I]`
    C
    `[I_(2)] lt [I]`
    D
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    A
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  • If 10^(-4) dm^(3) of water is introduced into a 1.0 dm^(3) flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established ? (Given : Vapour pressure of H_(2)O at 300 K is 3170 Pa , R=8.314 JK^(-1)mol^(-1) )

    A
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    B
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    C
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    D
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