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)`?

Answer

`2.86 xx 10^(-28) M`

Explanation

The given reaction is:
`3O_(2(g)) harr 2O_(3(g))`
Then, `K_(c) = ([O_(3(g))]^(2))/([O_(2(g))]^(3))`
It is given that `K_(c) = 2.0 xx 10^(-50)` and `[O_(2(g))] = 1.6 xx 10^(-2)`.
Then, we have
`2.0 xx 10^(-50) = ([O_(3(g))]^(2))/([1.6 xx 10^(-2)]^(3))`
`rArr [O_(3(g))]^(2) = 2.0 xx 10^(-50) xx (1.6 xx 10^(-2))^(3)`
`rArr [O_(3(g))]^(2) = 8.192 xx 10^(-56)`
`rArr [O_(3(g))] = 2.86 xx 10^(-28) M`
Hence, the concentration of `O_(3)`is `2.86 xx 10^(-28) M`.