The value of `K_c` for the reaction `3O_(2(g)) hArr 2O_(3(g))` is `2.0 xx10^(-50)` at `25^@C`. If the equilibrium concentration of `O_2` in air at `25^@C` is `1.6 xx 10^(-2)`, what is the concentration of `O_3` ?

Explanation

`{:("Equilibrium reaction :",3O_(2(g)) hArr, 2O_(3(g))),("Constant of equilibrium :", 1.6xx10^(-2) , x):}`
According to chemical equilibrium : `K_c=[O_3]^2/[O_2]^3`
where, `K_c=2.0xx10^(-50)`
`therefore 2.0xx10^(-50) = [O_3]^2/[1.6xx10^(-2)]^3`
`[O_2]=1.6xx10^(-2)`
`therefore [O_3]^2=2.0xx10^(-50) (1.6xx10^(-2))^3=8.192xx10^(-56)`
`therefore [O_3]=(8.192xx10^(-56))^(1/2)`
`=2.862xx10^(-28) "mol L"^(-1)`