Calculate the degree of dissociation of HI at `450^(@)C` if the equilibrium constant for the dissociation reaction is `0.263`.

Answer

B

Explanation

`{:(,HI,hArr,1/2H_(2),+,1/2I_(2)),("Initial",1,,0,,0),("At equilibrium",1-x,,x/2,,x/2),(,(1-x)/v,,x/(2v),,x/(2v)):}`
`K_(c)=((x/(2v))^(1//2)(x/(2v))^(1//2))/((1-x)/v)`
`0.263=(x xx v)/(2vxx(1-x))`
`0.263=x/2(1-x~~1)` (x is very small)
`x=0.526`