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2 mol of HI were heated in a sealed tube...

2 mol of HI were heated in a sealed tube at `440^(@)C` until the equilibrium was reched. HI was found to be 22% dissociated. Calculate the equilibrium constant for the reaction `2HI(g)hArr H_(2)(g)+I_(2)(g)`.

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For the given reaction, `K_(c)=([H_(2)]xx[I_(2)])/([HI]^(2))`
As given in the question, HI(g) undergoes 22% dissociation, hence, out of 2 mol `HI(g),2xx0.022=0.44` mol HI(g) dissociates.
As obtained from the equation, 2 mol HI(g) dissociates to produce 1 mol `H_(2)(g) and 1 " mol "I_(2)(g)`. therefore, 0.44 mol HI(g) dissociates to form 0.22 mol of each of `H_(2)(g) and I_(2)(g)`. If the volume of the container be V L, the equilibrium molar concentrations of different constituents will be as follows:
`2HI(g)hArrH_(2)(g)+I_(2)(g)`
Equilibriu concn. `(mol*L^(-1)):(2-0.44)//V" "(0.22)//V" "(0.22)//V`
`=1.56//V`
`therefore K_(c)=([H_(2)]xx[I_(2)])/([HI]^(2))=((0.22)/(V)xx(0.22)/(V))/(((1.56)/(V))^(2))=0.0198`
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