Question

For the equilibrium
`2NOCl(g) hArr 2NO(g)+Cl_(2)(g)`
the value of the equilibrium constant, `K_(c)` is `3.75 xx 10^(-6)` at `1069 K`. Calcualate the `K_(p)` for the reaction at this temperature?

Answer 1

We know that,
`K_(p) = K_(c)(RT)^(Deltan)`
For the above reaction,
`Deltan = (2+1)-2=1`
`K_(p) = 3.75 xx 10^(-6)(0.0831 xx 1069)`
`K_(P) = 0.033`