In a 10 L vessel, HI was heated to attain equilibrium. At equilibrium 327.68 g HI, 406.4 g `I_(2)` 15.6g `H_(2)` were present in the mixture. Calculate `K_(c)` if the mixture is transferred to 5L vessel.

Options

A. `0.029`
B. `0.059`
C. `0.019`
D. `0.91`

Answer

C

Explanation

`2HI hArr H_(2) + I_(2), " Now " K_(c )= ([H_(2)][I_(2)])/([HI]^(2))`
Give that `[H_(2)] = (w_(H_(2)))/(m_(H_(2)) xx V_(L)) = (15.6)/(2 xx 10) = 0.78 " "[I_(2)] = (w_(I_(2)))/(m_(I_(2)) xx V_(L)) = (406.4)/(254 xx 10) = 0.16`
`[HI] = (w_(HI))/(m_(HI) xx V_(L)) = (3276.8)/(128 xx 10) = 2.56` or `K_(c) = (0.78 xx 0.16)/((2.56)^(2)) = 0.019`
`K_(c )` is constant at a given temperature.