`{:(H_(2),+,I_(2),hArr,2HI,),(1 mol,,1mol,,0, ("प्रारम्भ में")),((1-x)mol,,(1-x)mol,,2xmol, ("साम्यावस्था में") ):}`
`K = ([HI]^(2))/([H_(2)] xx [I_(2)])`
या `49.0 = ((2x)^(2))/((1-x)(1-x)) = ((2x)/(1-x))^(2)`
`:. (2x)/(1-x) = sqrt(49.0) = 7.0`
अतः, साम्यावस्था में `[HI] = 2x = (2 xx 0.77) = - 1.54 mol L^(-1)`
`" [H_(2)] = 1 - x = (1-0.77) = 0.23mol L^(-1)`
`" [I_(2)] = 1-x = 1-0.77 = 0.23 mol L^(-1)` |