Find out the value of `K_(c )` for each of the following equilibrium from the value of `K_(p)`:
a. `2NOCl(g) hArr 2NO(g)+Cl_(2)(g), K_(p)=1.8xx10^(-2)` at `500 K`
b. `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g), K_(p)=167` at `1073 K`

Answer

(i) `4.33 xx 10^(-4)` (ii) `1.90`

Explanation

The relation between `K_(p)` and `K_(c)` is given as :
`K_(p) = K_(c) (RT)^(Deltan)`
(a) Here,
`Deltan = 3-2 = 1`
`R = 0.0831 "bar L mol"^(-1) K^(-1)`
`T = 500 K`
`K_(p) = 1.8 xx 10^(-2)`
Now,
`K_(p) = K_(c)(RT)^(Deltan)`
`rArr 1.8 xx 10^(-2) = K_(c) (0.0831 xx 500)^(1)`
`rArr K_(c) = (1.8 xx 10^(-2))/(0.0831 xx 500)`
`= 4.33 xx 10^(-4)` (approximately)
(b) Here,
`Deltan = 2-1 = 1`
`R = 0.0831 "bar L mol"^(-1) K^(-1)`
`T = 1073 K`
`K_(p) = 167`
Now, `K_(p) = K_(c) = (RT)^(Deltan)`
`rArr 167 = K_(c) (0.0831 xx 1073)^(Deltan)`
`K_(c) = (167)/(0.0831 xx 1073)`
`= 1.87` (approximately)