# Find out the value of Kc for each of the following equilibria from the value of Kp:

Question:

Find out the value of Kc for each of the following equilibria from the value of Kp:

(i) $\quad 2 \mathrm{NOCl}(\mathrm{g}) \longleftrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \quad K_{p}=1.8 \times 10^{-2}$ at $500 \mathrm{~K}$

(ii) $\mathrm{CaCO}_{3}$ (s) $\longleftrightarrow \mathrm{CaO}$ (s) $+\mathrm{CO}_{2}$ (g); $\quad \mathrm{K}_{p}=167$ at $1073 \mathrm{~K}$

Solution:

The relation between $K_{p}$ and $K_{c}$ is given as:

$K_{p}=K_{c}(\mathrm{RT})^{\Delta n}$

(a) Here,

$\Delta n=3-2=1$

$R=0.0831$ barLmol $^{-1} \mathrm{~K}^{-1}$

$T=500 \mathrm{~K}$

$K_{\rho}=1.8 \times 10^{-2}$

Now,

$K_{p}=K_{c}(R T)^{\Delta n}$

$\Rightarrow 1.8 \times 10^{-2}=K_{c}(0.0831 \times 500)^{1}$

$\Rightarrow K_{c}=\frac{1.8 \times 10^{-2}}{0.0831 \times 500}$

$=4.33 \times 10^{-4}($ approximately $)$

(b) Here,

$\Delta n=2-1=1$

$R=0.0831$ bar $\mathrm{Lmol}^{-1} \mathrm{~K}^{-1}$

$T=1073 \mathrm{~K}$

$K_{p}=167$

Now,

$K_{p}=K_{c}(R T)^{\Delta n}$

$\Rightarrow 167=K_{c}(0.0831 \times 1073)^{\Delta n}$

$\Rightarrow K_{c}=\frac{167}{0.0831 \times 1073}$

$=1.87$ (approximately)