The value of `K_(p)` for dissociation of `2HI hArr H_(2)+I_(2)` is `1.84xx10^(-2)`. If the equilibrium concentration of `H_(2)` is `0.4789` mol `L^(-1)`, calculate the concentration of `HI` at equilibrium.

Answer

A, C

Explanation

Given, `K_(p)=1.84xx10^(-2)`
`:' Deltan=0 :. K_(p)=K_(c)=1.84xx10^(-2)`
`{:(,2HI,hArr,H_(2),+,I_(2)),("Conc at Eq",a,,0.4789,,0.4789 "mol" L^(-1)):}`
Since given `[H_(2)]=0.4789`
`:. [I_(2)]=0.4789 "mol" L^(-1)`
Concentration of HI at equilibrium is a "mole" `L^(-1)`
`:. K_(C)=([H_(2)][I_(2)])/([HI]^(2))=(0.4789xx0.4789)/a^(2)`
`:. a^(2)=(0.4789xx0.4789)/(1.84xx10^(-2)) ( :' K_(C)=1.84xx10^(-2))`
`:. a=3.53`
or [HI] at equilibrium `=3.53 "mol" L^(-1)`