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 `:`

Options

A. 0.5 mol
B. 1.5mol
C. 0.25 mol
D. 3.0 mol

Answer

B

Explanation

`K_(p)=K_(c)(RT)^(Deltan)`
Here, `Deltan=0`
`K_(c)=K_(p)=9`
`CO(g)+H_(2)O(g)hArr CO_(2)(g)+H_(2)(g)`
`{:(,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=3-3x`
`4x=2`
`x=(1)/(2)=0.5`
`:. `No. of moles of `CO_(2)` at eqm. `=1+x`
`=1+0.5=1.5`