Equilivalent amounts of `H_(2)` and `I_(2)` are heated in a closed vessel till equilibrium is obtained. If `80%` of the hydrogen is converted to `HI`, the `K_(c)` at this temperature is

Options

A. 64
B. 16
C. 0.25
D. 14

Answer

A

Explanation

`{:(,H_(2),+,I_(2),hArr,2HI),("Initial",a" mol",,a " mol",,),("At eqm.",a-0.8a=2a,,0-0.8=0.2a,,2xx0.8a=1.6a),("Molar conc.",(0.2a)/(V),,(0.2a)/(V),,(1.6a)/(V)):}`
`K_(c)=([HI]^(2))/([H_(2)][I_(2)])=((1.6a//V)^(2))/((0.2a//V)(0.20//V))`
`=((16)^(2))/((0.2)^(2))=(2.56)/(0.04)=64`