1 mole each of `CO(g), H_(2)O(g), H_(2)(g)` and `CO_(2)(g)` are placed in one placed in one litre flask at `25^(@)C`. When following equilibrium is set -up `CO(g)+H_(2)O(g)hArr CO_(2)(g)+H_(2)(g)`
`K_(p)=0`. Hence `CO_(2)` present at equilibrium is `:`

Answer

`1.5`

Explanation

`K_(p) = K_(c) (RT)^(Delta n)`
Here `Delta n = 0`
`:. K_(c) = K_(p) = 9`
`{:(,CO(g) + H_(2) O (g) hArr CO_(2) (g) + H_(2) (g)),("Initial"," 1 1 1 1"),("At eqm.","1 – x 1 – x 1 + x 1 + x"):}`
Let volume of container = V
`K_(c) = ([CO_(2)][H_(2)])/([CO][H_(2)O])`
` 9 = ((1 + x)^(2) // V^(2))/( (1 - x)^(2)// V^(2))`
`(1 + x)/( 1 - x) = 3`
1 + x = 3 - 3 x
4 x = 2
`x = (1)/(2) = 0.5`
`:.` No. of moles of `CO_(2)` at eqm.= 1+ x
= 1 + 0.5 = 1.5 .