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HI was heated in a sealed tube at 400^(@...

`HI` was heated in a sealed tube at `400^(@)C` till the equilibrium was reached. HI was found to be `22%` decomposed. The equilibrium constant for dissociation is

A

0.282

B

0.0796

C

0.0199

D

1.99

Text Solution

Verified by Experts

The correct Answer is:
C

`2HI Leftrightarrow H_(2)+I_(2)`
`{:(,"Initial ",100,0,0),(,"Reacted & formed", 22,11,11),(,"at equilibrium",78,11,11):}`
`K_(c)=([H_(2)] [I_(2)])/([HI]^(2))`
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NARAYNA-CHEMICAL EQUILIBRIUM-Exercise -I (H.W.)
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