The value of `K_(c)` for the reaction `3O_(2)(g)hArr 2O_(3)(g)` is `2.0xx10^(-50)` at `25^(@)C`. If the equilibrium concentration of `O_(2)` in air at `25^(@)C` is `1.6xx10^(-2)`, what is the concentration of `O_(3)` ?

Explanation

`K_(c)=([O_(3)])/([O_(2)]^(3))=2xx10^(-50)`
Given: `[O_(2)]=1.6xx10^(-2)`
`therefore [O_(3)]^(2)=K_(c)xx[O_(2)]^(3)=2xx10^(-50)xx(1.6xx10^(-2))^(3)`
or, `[O_(3)]=2.86xx10^(-28)mol*L^(-1)`