A sample of HI was found to be 22% dissociated when equilibrium was reached. What will be the degree of dissociation if hydrogen is added in the proportion of mol for very mole of HI present originally ? Assume temperature and pressure to be constant.

Options

A. 0.065
B. 0.085
C. 0.027
D. 0.052

Answer

C

Explanation

Given that the percentage dissociation = 22, so degree of dissociation = 0.22. The reaction involved is
`{:(,2HI,hArr,H_(2),+,I_(2)),("Initial moles",1,,0,,0),("Moles at equilibrium",1-alpha = 0.78,,alpha//2 = 0.11,,alpha//2 = 0.11):}`
Therefore, `K_(c ) = ([H_(2)][I_(2)])/([HI]^(2)) = (0.11 xx 0.11)/((0.78)^(2)) = 0.0199`
When 1 mol of hydrogen is added starting with 1 mol of HI, we get
`K_(c )= ([H_(2)][I_(2)])/([HI]^(2)) = ((alpha//2) xx (alpha//2+1))/((1-alpha)^(2)) = 0.0199`
Solving, we get `alpha = 0.037`. So, the addition of 1 mol of `H_(2)` suppresses the dissociation of HI.