Find out the value of `K_(c)` for each of the following equilibrium from the value of `K_(p)` :
`2NOCl(g) hArr 2NO (g) + Cl_(2)(g), K_(p) = 1.8 xx 10^(-2)` at 500 K.

Explanation

We are given that
`K_(p) = 1.8 xx 10^(-2) " "T = 500 k`
We know `Kp (RT)^(-Delta n_(g))`
For this reaction, `Delta n_(g) = (2+1)-(2) = +1`
`K_(c) = K_(p)(RT)^(-1)`
`= 1.8 xx 10^(-2) xx (0.082 xx 500)^(-1)`
`= 4.4 xx 10^(-4)`