Question

Two moles of HI were heated in a sealed tube at `440^@ C` till the equilibrium was reached. HI was found to be 22% decomposed.The equilibrium constant for disssociation is :

Answer 1

Correct Answer - `K_(c) =0.0198`
The equation for the decomposition of HI (g) is :
`2HI(g) hArr H_(2) +I_(2) (g)`
Number of moles of HI initially present =2 mol
Percentage of HI dissociated =22%
`:.` Number of moles of HI dissociated `=(2xx22)/(100) =0.44` mol
According to the equation.
Number to moles of`H_(2) " and "I_(2)` formed `=(0.44)/(2) =0.22`mol
The molar conc. of the different species before the reaction and at the equilibrium point is :
`{:(,2HI(g) ,hArr,H_(2)(g),+,I_(2)(g)),("Initial molar conc.",2,,0,,0),("Equilibrium molar conc.",2-0.44,,0.22,,0.22),(,=1.56,,,,):}`
Applying Law of chemical equilibrium.
`K_(C) =[[H_(2)(g)][I_(2)(g)]]/[[HI(g)]^(2)] =(0.22 xx 0.22)/(1.56xx1.56)=0.0198`