At a particular temperature, CO(g) reacts with `Cl_(2)(g)` in a closed container to produce `COCl_(2)(g)`. In the equilibrium mixture, partial pressures of CO(g), `Cl_(2)(g)` and `COCl_(2)(g)` are found to be 0.12, 1.2 and 0.58 atm respectively. Find the value of `K_(p)` of the reaction, `CO(g)+Cl_(2)(g)hArr COCl_(2)(g)`.

Explanation

For the given equilibrium, `K_p=(P_(COCl_(2)))/(p_(CO)xxp_(Cl_(2)))`
As given, `p_(CO)=0.12atm,p_(Cl_(2))=1.2` atm and
`p_(COCl_(2))=0.58` atm at equilibrium
`therefore K_(p)=(0.58)/(0.12xx1.2)=4.03`