At equilibrium, the concentrations of `N_(2)=3.0xx10^(-3)M, O_(2)=4.2xx10^(-3) M,` and `NO=2.8xx10^(-3) M` in a sealed vessel at `800K`. What will be `K_(c)` for the reaction
`N_(2)(g)+O_(2)(g)N_(2)(g)+O_(2)(g)hArr2NO(g)2NO(g)`

Explanation

For the reaction equilibrium constant.
`K_(c)` can be written as,
`K_(c) = ([NO]^(2))/([N_(2)][O_(2)])`
`((2.8 xx 10^(-3)M)^(2))/((3.0 xx 10^(-3)M)(4.2 xx 10^(-3) M))`
`= 0.622`