What is `K_(c )` for the following equilibrium concentration of each substance is:
`[SO_(2)]=0.60 M, [O_(2)]=0.82 M` and `[SO_(3)]=1.90 M`?

`2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`

Answer

`12.229`

Explanation

The equilibrium constant `(K_(c))` for the give reaction is:
`K_(c) = ([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])`
`= ((1.90)^(2)M^(2))/((0.60)^(2)(0.821)M^(3))`
`= 12.239 M^(-1)` (approximately)
Hence, `K_(c)` for the equilibrium is `12.239 M^(-1)`