`1` mol of `H_(2), 2` mol of `I_(2)` and `3` mol of HI were taken in a `1-L` flask. If the value of `K_(c)` for the equation `H_(2)(g)+I_(2)(g) hArr 2HI(g)` is `50` at `440^(@)C`, what will be the concentration of each specie at equilibrium?

Answer

A, B, C, D

Explanation

Suppose that `x` mol each of `H_(2)` and `I_(2)` is converted to `HI` at equilibrium.
`{:(1,,2,,3,"Initial mole"),(H_(2),+,I_(2),hArr,2HI,),((1-x),,(2-x),,(3+2x),"mole at Eq"),(((1-x))/(1),,((2-x))/(1),,((3+2x))/(1),"molar conc at Eq"):}`
`K_(c )=([HI]^(2))/([H_(2)][I+_(2)])=((3+2x)^(2))/((1-x)(2-x))=50`
`:. x=0.7 ("approx")`
`:. [H_(2)]=(1-x)=1-0.7=0.3`
`[I_(2)]=2-x=2-0.7=1.3`
`[HI]=3+2x=3+1.4=4.4`