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 isA. `30`B. `15`C. `5`D. `25`

The equilibrium of formation of phosgene is represented as : `CO(g)+Cl

1.	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 isA. `30`B. `15`C. `5`D. `25`
Answer» 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`

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