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At certain temperature 50% of HI is diss...

At certain temperature `50%` of `HI` is dissociated into `H_(2)` and `I_(2)` the equilibrium constant is

A

`1.0`

B

`3.0`

C

`0.5`

D

`0.25`

Text Solution

Verified by Experts

The correct Answer is:
A
`HIhArr(1)/(2)H_(2)+(1)/(2)I_(2)`
at `t=0" "0" "0`
at eq. `(1-.05)" "(0.5)" "(0.5)" "("":.alpha=(50)/(100)=0.5)`
Now `K_(C)=((0.5)^(1//2)(0.5)^(1//2))/(0.5)=1`
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DISHA PUBLICATION-EQUILIBRIUM-Exercise-2 : Concept Applicator
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