Question

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)`

Answer 1

For the reaction equilibrium constant `K_(c )` can be written as
`K_(c )=([NO]^(2))/([N_(2)][O_(2)])=((2.8xx10^(-3)M)^(2))/((3.0xx10^(-3)M)(4.2xx10^(-3)M))=0.622`