At T (K) the equilibrium constant of `H_(2)(g)+I_(2)(g) rarr 2HI(g) ` is 49 . If `[H_(2)],[I_(2)]` at equilibrium at the same temperature are `2.0xx 10^(-2) ` M and `8.0xx 10^(-2)` M respectively the [HI] at equilibrium in mol`L^(-I)` is

Options

A. 2.8
B. 0.28
C. 0.14
D. 1.4

Answer

B

Explanation

Given `K_(C) = 49`
`[H_(2)]=2.0xx10^(-2)M`
`[I_(2)]=8.0xx10^(-2)`M
For the reaction
`H_(2)(g)+I_(2) (g) rarr 2HI(g)`
`K_(C)= ([HI]^(2))/([H_(2)][I_(2)])or [HI]^(2)=K_(c)[H_(2)[I_(2)]`
On putting the values in the above equation we have `[HI]^(2)= 49xx16xx10^(-4)`
`HI=sqrt(49xx16+10^(-4))=0.28`mol