1.

HI was heated in a closed tube at 440^(@) C till equilibrium is obtained. At this temperaturee 22% of HI was dossociated. The equilbrium constant for this dissociation wil be

Answer»

`0.282`
`0.0769`
`0.0199`
`1.99`

Solution :`2HIhArrH_(2)+I_(2)`
`{:("Initial CONC.", "2 MOLES",0,0),("At EQULIBRIUM",22/100xx2,0.22,0.22):}`
`=2-0.44=1.56`
`K=([H_(2)][I_(2)])/([HI]^(2))=(0.22xx0.22)/([1.56]^(2))=0.0199.`


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