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A sample of HI was found to be 22% disso...

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.

A

`0.065`

B

`0.085`

C

`0.037`

D

`0.052`

Text Solution

Verified by Experts

The correct Answer is:
B
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 _(x) = ([ 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.
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