In the dissociation of HI, `20%` of HI is dissociated at equilibrium. Calculate `K_(p)` for
`HI(g) hArr 1//2 H_(2)(g)+1//2 I_(2)(g)`

Options

A. 1.25
B. 0.125
C. 12.5
D. 0.0125

Answer

B

Explanation

`HI Leftrightarrow 1/2 H_(2) +1/2 I_(2)`
`[{:(,"Initial",1,0,0),(,"Mole at equilibrium",(1-alpha),alpha/2,alpha/2):}]`
Where `alpha` is degree of dissociation and volume of container is V litre
`K_(p)=K_(c)=((alpha)/(2V))^(1//2) ((alpha)/(2V))^(1//2)/((1-alpha)/(V))=(alpha)/(2(1-alpha))`
`K_(p)=K_(c)=(0.2)/(2(1-0.2))" "(therefore alpha=0.2)`
`K_(p)=K_(c)=0.125`