Question

The equilibrium of formation of phosgene is represented as :
`CO(g)+Cl_(2)(g)hArrCOCl_(2)(g)`
The reaction is carried out in a `500 mL` flask. At equilibrium, `0.3` mol of phosgene, `0.1 mol` of `CO`, and `0.1` mol of `Cl_(2)` are present.
The equilibrium constant of the reaction is
A. `30`
B. `15`
C. `5`
D. `25`

Answer 1

Correct Answer - B
For the reaction
`CO(g)+Cl_(2)(g)hArrCOCl_(2)(g)`
The equation constant, `K_(c )=([COCl_(2)])/([CO][Cl_(2)])…(i)`
The concentration of `[COCl_(2)]=("mol")/(V(L))=0.3/0.5 "mol" L^(-1)`
`[CO]=0.1/0.5 "mol" L^(-1)`
`[Cl_(2)]=0.1/0.5 "mol" L^(-1)`
Therefore, on substituting all the value in expression (1), we get
`(0.3/0.5)/((0.1/0.5)(0.1/0.5))=0.3/0.5xx0.5/0.1xx0.5/0.1`
`=0.15/0.01=15`