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Equimolar concentrations of H(2) and I(2...

Equimolar concentrations of `H_(2)` and `I_(2)` are heated to equilibrium in a 2 L flask. At equilibrium, the forward and backward rate constants are found to be equal. What percentage of initial concentration of `H_(2)` has reached at equilibrium ?

A

`33%`

B

`66%`

C

`50%`

D

`40%`

Text Solution

Verified by Experts

The correct Answer is:
A
We know that
`K = (k_f)/(k_b) " "` here K =1
`H_(2) + I_(2) hArr 2HI`
`K=([HI]^2)/([HI]_2[I_2])" as" [H_2]=[I_2]`
`I= ([HI])/([H_2])`
`[H_2]=[HI]`
Initially concentration of `H_2` is 100% and let y% get reacted so 2y% of HI will be formed as stoichiometry as
`[HI] = [H_2]`
`(100 - y) = 2y`
100 = 3y
y = 33.33%
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