The equilibrium constant for the reaction, `H_(2)(g) + I_(2)(g) hArr 2HI(g)` is 64 at a certain temperature. The equilibrium concentrations of `H_(2)` and HI are `2 mol//L and 16 mol//L` respectively. What is the equilibrium concentration (`"in "mol//L`) of `I_(2)` ?

Options

A. 16
B. 4
C. 8
D. 2

Answer

D

Explanation

Given that, `H_(2) + I_(2) hArr 2HI`
`K_(c) = 64`
`[H_(2)] = 2 mol/L`
`[HI] = 16 mol/L`
`:. K_(c) = [HI]_(2)/([H_(2)][I_(2)])`
`64 = (16)^(2)/2 xx [I_(2)]`
`implies [I_(2)] = (16)^(2)/2xx64`
`[I_(2)] = `2 mol/L